Matrices

(i) It is a square matrix of order 2.

(ii) It is a row matrix of order 1 x 3.

(iii) It is a column matrix of order 3 x 1.

(iv) It is a 3 x 2 matrix.

(v) It is a 2 x 3 matrix.

(vi) It is a zero matrix of order 2 x 3.

If a matrix has 4 elements the possible orders are,

1 x 4, 2 x 2, 4 x 1.

If a matrix has 8 elements the possible orders are,

1 x 8, 4 x 2, 2 x 4, 8 x 1.

(i) Given a_ij = 2i - j,

∴ a₁₁ = 2(1) - 1 = 1, a₁₂ = 2(1) - 2 = 0,
    a₂₁ = 2(2) - 1 = 3, a₂₂ = 2(2) - 2 = 2.

Hence, the required matrix =$$\begin{bmatrix} 1 & 0 \\ 3 & 2 \end{bmatrix}$$.

(ii) Given a_ij = i.j,

∴ a₁₁ = 1.1 = 1, a₁₂ = 1.2 = 2,
    a₂₁ = 2.1 = 2, a₂₂ = 2.2 = 4.

Hence, the required matrix =$$\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$$.

Given $$\begin{bmatrix*}[r] 2x + y \\ 3x - 2y \end{bmatrix*} = \begin{bmatrix} 5 \\ 4 \end{bmatrix}$$.

By definition of equality of matrices, we get

2x + y = 5    [....Eq 1]
3x - 2y = 4    [....Eq 2]

2x + y = 5
⇒ y = 5 - 2x

Putting value of y in Eq 2,

3x - 2(5 - 2x) = 4
⇒ 3x - 10 + 4x = 4
⇒ 7x = 14
⇒ x = 2.

∴ x = 2, y = 5 - 2x = 5 - 4 = 1.

Hence, the value of x = 2 and y = 1.

Given $$\begin{bmatrix*}[r] 3x + y & -y \\ 2y - x & 3 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \\ -5 & 3 \end{bmatrix*}$$

By definition of equality of matrices, we get

-y = 2 or y = -2 and
3x + y = 1

Putting value of y = -2 in above equation

⇒ 3x - 2 = 1
⇒ 3x = 3
⇒ x = 1

Hence, the value of x = 1.

Given $$\begin{bmatrix*}[r] x + 3 & 4 \\ y - 4 & x + y \end{bmatrix*} = \begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix}$$.

By definition of equality of matrices, we get

    x + 3 = 5, y - 4 = 3 and x + y = 9
⇒ x = 2, y = 7 and x + y = 9.

Note that x = 2 and y = 7 satisfies the equation x + y = 9.

Hence, the value of x = 2 and y = 7.

Given $$\begin{bmatrix*}[r] x + 2 & 6 \\ 3 & 5z \end{bmatrix*} = \begin{bmatrix*}[r] -5 & y^2 + y \\ 3 & -20 \end{bmatrix*} .$$

By definition of equality of matrices, we get

x + 2 = -5 or x = -7,
y² + y = 6       [....Eq 1],
5z = -20 or z = -4.

From Eq 1 we get,

$$\begin{aligned}\Rightarrow y^2 + y - 6 = 0 \\[0.5em] \Rightarrow y^2 + 3y - 2y - 6 = 0 \\[0.5em] \Rightarrow y(y + 3) - 2(y + 3) = 0 \\[0.5em] \Rightarrow (y - 2)(y + 3) = 0 \\[0.5em] \Rightarrow y = 2 \text{ or } y = -3. \\[0.5em]\end{aligned}$$

∴ x = -7, y = 2 or -3 and z = -4.

Hence, the values are:
x = -7, y = 2 and z = -4
OR
x = -7, y = -3 and z = -4.

Given, $$\begin{bmatrix*}[r] x - 2 & y \\ a + 2b & 3a - b \end{bmatrix*} = \begin{bmatrix*}[r] 3 & 1 \\ 5 & 1 \end{bmatrix*}$$.

By definition of equality of matrices, we get

x - 2 = 3 or x = 5,
y = 1,
a + 2b = 5       [...Eq 1],
3a - b = 1        [...Eq 2].

From Eq 1 we get,

⇒ a + 2b = 5
⇒ a = 5 - 2b

Putting this value of a in Eq 2,

⇒ 3(5 - 2b) - b = 1
⇒ 15 - 6b - b = 1
⇒ 7b = 14
⇒ b = 2.

Using value of b to find value of a,

⇒ a = 5 - 2b
⇒ a = 5 - 2(2)
⇒ a = 5 - 4
⇒ a = 1.

∴ x = 5, y = 1, a = 1 and b = 2.

Hence, the values are:
x = 5, y = 1, a = 1 and b = 2.

Given $$\begin{bmatrix*}[r] a + b & 3 \\ 5 + c & ab \end{bmatrix*} = \begin{bmatrix*}[r] 6 & d \\ -1 & 8 \end{bmatrix*} .$$

By definition of equality of matrices, we get

d = 3,
5 + c = -1 or c = -6,
a + b = 6       [...Eq 1],
ab = 8            [...Eq 2].

Putting value of b from Eq 1 in Eq 2,

$$\begin{aligned}a + b = 6 \text{ or } b = 6 - a \\[1em] \Rightarrow a(6 - a) = 8 \\[0.5em] \Rightarrow 6a - a^2 = 8 \\[0.5em] \Rightarrow a^2 - 6a + 8 = 0 \\[0.5em] \Rightarrow a^2 - 4a - 2a + 8 = 0 \\[0.5em] \Rightarrow a(a - 4) - 2(a - 4) = 0 \\[0.5em] \Rightarrow (a - 2)(a - 4) = 0 \\[0.5em] \Rightarrow a = 2 \text{ or } a = 4.\end{aligned}$$

Now, finding value of b = 6 - a,

if,  a = 2, b = 6 - 2 = 4.

or, a = 4, b = 6 - 4 = 2.

∴ a = 2 or 4, b = 4 or 2, c = -6 and d = 3.

Hence, the values are
a = 2, b = 4, c = -6, d = 3
OR
a = 4, b = 2, c = -6, d = 3.

$$\text{M + 2N }= \begin{bmatrix*}[r] 2 & 0 \\ 1 & 2 \end{bmatrix*} + 2\begin{bmatrix*}[r] 2 & 0 \\ -1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 0 \\ 1 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 0 \\ -2 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + 4 & 0 + 0 \\ 1 - 2 & 2 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 & 0 \\ -1 & 6 \end{bmatrix*} \\[1em]$$

Hence, the matrix M + 2N = $$\begin{bmatrix*}[r] 6 & 0 \\ -1 & 6 \end{bmatrix*}$$.

$$\text{2A - 3B }= 2\begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} - 3\begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 0 \\ -6 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & 3 \\ -6 & 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 0 & 0 - 3 \\ -6 - (-6) & 2 - 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & -3 \\ 0 & -7 \end{bmatrix*} \\[1em]$$

Hence, the matrix 2A - 3B = $$\begin{bmatrix*}[r] 4 & -3 \\ 0 & -7 \end{bmatrix*}.$$

Given,

$$\Rightarrow \text{sin A}\begin{bmatrix*}[r] \text{sin A} & -\text{cos A} \\ \text{cos A} & \text{sin A} \end{bmatrix*} + \text{cos A}\begin{bmatrix*}[r] \text{cos A} & \text{sin A} \\ -\text{sin A} & \text{cos A} \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \text{sin}^2\text{ A} & -\text{sin A}.\text{cos A} \\ \text{sin A}.\text{cos A} & \text{sin}^2\text{ A} \end{bmatrix*} + \begin{bmatrix*}[r] \text{cos}^2\text{ A} & \text{cos A}.\text{sin A} \\ -\text{cos A}.\text{sin A} & \text{cos}^2\text{ A} \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] \text{sin}^2\text{ A} + \text{cos}^2\text{ A} & -\text{sin A}.\text{cos A} + \text{cos A}.\text{sin A} \\ \text{sin A}.\text{cos A} - \text{cos A}.\text{sin A} & \text{sin}^2\text{ A} + \text{cos}^2\text{ A} \end{bmatrix*} \\[1em] \because \text{sin}^2\text{ A} + \text{cos}^2\text{ A} = 1 \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em]$$

Hence, on simplifying, the resultant matrix = $$\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}.$$

$$\text{A + 2B - 3C }= \begin{bmatrix*}[r] 1 & 2 \\ -2 & 3 \end{bmatrix*} + 2\begin{bmatrix*}[r] -2 & -1 \\ 1 & 2 \end{bmatrix*} - 3\begin{bmatrix*}[r] 0 & 3 \\ 2 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 2 \\ -2 & 3 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & -2 \\ 2 & 4 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & 9 \\ 6 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + (-4) - 0 & 2 + (-2) - 9 \\ -2 + 2 - 6 & 3 + 4 - (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 & -9 \\ -6 & 10 \end{bmatrix*}$$

Hence, the matrix A + 2B - 3C = $$\begin{bmatrix*}[r] -3 & -9 \\ -6 & 10 \end{bmatrix*}.$$

(i) 3A + X = B

⇒ X = B - 3A

$$\Rightarrow X = \begin{bmatrix*}[r] 1 & 2 \\ -1 & 1 \end{bmatrix*} - 3\begin{bmatrix*}[r] 0 & -1 \\ 1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 2 \\ -1 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & -3 \\ 3 & 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 - 0 & 2 - (-3) \\ -1 - 3 & 1 - 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 5 \\ -4 & -5 \end{bmatrix*}$$

Hence, matrix X = $$\begin{bmatrix*}[r] 1 & 5 \\ -4 & -5 \end{bmatrix*}.$$

(ii) X - 3B = 2A

⇒ X = 2A + 3B

$$X = 2\begin{bmatrix*}[r] 0 & -1 \\ 1 & 2 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 2 \\ -1 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & -2 \\ 2 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 6 \\ -3 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 + 3 & -2 + 6 \\ 2 + (-3) & 4 + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3 & 4 \\ -1 & 7 \end{bmatrix*}$$

Hence, matrix X = $$\begin{bmatrix*}[r] 3 & 4 \\ -1 & 7 \end{bmatrix*}.$$

Given,

$$\Rightarrow \begin{bmatrix} 2 & 1 \\ 5 & 0 \end{bmatrix} - 3X = \begin{bmatrix*}[r] -7 & 4 \\ 2 & 6 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix} 2 & 1 \\ 5 & 0 \end{bmatrix} - \begin{bmatrix*}[r] -7 & 4 \\ 2 & 6 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 2 - (-7) & 1 - 4 \\ 5 - 2 & 0 - 6 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 9 & -3 \\ 3 & -6 \end{bmatrix*} \\[1em] \Rightarrow X = \frac{1}{3}\begin{bmatrix*}[r] 9 & -3 \\ 3 & -6 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 3 & -1 \\ 1 & -2 \end{bmatrix*}$$

Hence, matrix X = $$\begin{bmatrix*}[r] 3 & -1 \\ 1 & -2 \end{bmatrix*} .$$

Given,

$$\Rightarrow \begin{bmatrix*}[r] 1 & 4 \\ -2 & 3 \end{bmatrix*} + 2\text{M} = 3\begin{bmatrix*}[r] 3 & 2 \\ 0 & -3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 & 4 \\ -2 & 3 \end{bmatrix*} + 2\text{M} = \begin{bmatrix*}[r] 9 & 6 \\ 0 & -9 \end{bmatrix*} \\[1em] \Rightarrow 2\text{M} = \begin{bmatrix*}[r] 9 & 6 \\ 0 & -9 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 4 \\ -2 & 3 \end{bmatrix*} \\[1em] \Rightarrow 2\text{M} = \begin{bmatrix*}[r] 9 - 1 & 6 - 4 \\ 0 - (-2) & -9 - 3 \end{bmatrix*} \\[1em] \Rightarrow \text{M} = \frac{1}{2}\begin{bmatrix*}[r] 8 & 2 \\ 2 & -12 \end{bmatrix*} \\[1em] \Rightarrow \text{M} = \begin{bmatrix*}[r] 4 & 1 \\ 1 & -6 \end{bmatrix*} \\[1em]$$

Hence, matrix M = $$\begin{bmatrix*}[r] 4 & 1 \\ 1 & -6 \end{bmatrix*}.$$

Putting values of A, B and C in A + 2X = 2B + C,

$$\Rightarrow \begin{bmatrix*}[r] 2 & -6 \\ 2 & 0 \end{bmatrix*} + 2X = 2\begin{bmatrix*}[r] -3 & 2 \\ 4 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 0 \\ 0 & 2 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] -6 & 4 \\ 8 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 0 \\ 0 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & -6 \\ 2 & 0 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] -6 + 4 - 2 & 4 + 0 - (-6) \\ 8 + 0 - 2 & 0 + 2 - 0 \end{bmatrix*} \\[1em] \Rightarrow X = \frac{1}{2}\begin{bmatrix*}[r] -4 & 10 \\ 6 & 2 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -2 & 5 \\ 3 & 1 \end{bmatrix*}$$

Hence, matrix X = $$\begin{bmatrix*}[r] -2 & 5 \\ 3 & 1 \end{bmatrix*}$$.

Given,

X + Y = $$\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}$$        (...Eq1)

X - Y = $$\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$$         (...Eq2)

Adding both the equations,

$$\Rightarrow X + Y + X - Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix} 7 + 3 & 0 + 0 \\ 2 + 0 & 5 + 3 \end{bmatrix} \\[1em] \Rightarrow X = \frac{1}{2}\begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \\[1em]$$

From Eq 1 we get,

$$\Rightarrow Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} - X \\[1em] = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 - 5 & 0 - 0 \\ 2 - 1 & 5 - 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \\[1.5em] \therefore X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} , Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}.$$

Hence, $$X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \text{ and } Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}.$$

Given,

$$\Rightarrow 2\begin{bmatrix*}[r] x & 7 \\ 9 & y - 5 \end{bmatrix*} + \begin{bmatrix*}[r] 6 & -7 \\ 4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 7 \\ 22 & 15 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x & 14 \\ 18 & 2y - 10 \end{bmatrix*} + \begin{bmatrix*}[r] 6 & -7 \\ 4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 7 \\ 22 & 15 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 6 & 14 + (-7) \\ 18 + 4 & 2y - 10 + 5 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 7 \\ 22 & 15 \end{bmatrix*} \\[1em] \Rightarrow 2x + 6 = 10 \text{ and } 2y - 5 = 15 \\[1em] \Rightarrow 2x = 4 \text{ and } 2y = 20 \\[1em]$$

∴ x = 2 and y = 10.

Hence, the value of x = 2 and y = 10.

Given,

$$\Rightarrow 2\begin{bmatrix*}[r] 3 & 4 \\ 5 & x \end{bmatrix*} + \begin{bmatrix*}[r] 1 & y \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 8 \\ 10 & 2x \end{bmatrix*} + \begin{bmatrix*}[r] 1 & y \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 1 & 8 + y \\ 10 + 0 & 2x + 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 7 & 8 + y \\ 10 & 2x + 1 \end{bmatrix*} = \begin{bmatrix*}[r] z & 0 \\ 10 & 5 \end{bmatrix*} \\[1em] \Rightarrow 7 = z, 8 + y = 0 \text{ and } 2x + 1 = 5 \\[0.5em] \Rightarrow z = 7, y = -8 \text{ and } 2x = 4 \\[0.5em] \Rightarrow z = 7, y = -8 \text{ and } x = 2.$$

∴ x = 2, y = -8 and z = 7.

Hence, the values are x = 2, y = -8 and z = 7.

Given,

$$\Rightarrow \begin{bmatrix*}[r] 5 & 2 \\ -1 & y + 1 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 2x - 1 \\ 3 & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 5 & 2 \\ -1 & y + 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4x - 2 \\ 6 & -4 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 5 - 2 & 2 - 4x + 2 \\ -1 - 6 & y + 1 - (-4) \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & 4 - 4x \\ -7 & y + 5 \end{bmatrix*} = \begin{bmatrix*}[r] 3 & -8 \\ -7 & 2 \end{bmatrix*} \\[1em] \Rightarrow 4 - 4x = -8 \text{ and } y + 5 = 2 \\[0.5em] \Rightarrow 4x = 8 + 4 \text{ and } y = 2 - 5 \\[0.5em] \Rightarrow 4x = 12 \text{ and } y = -3 \\[0.5em]$$

∴ x = 3 and y = -3.

Hence, the value of x = 3 and y = -3.

Given,

$$\Rightarrow \begin{bmatrix*}[r] a & 3 \\ 4 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 2 & b \\ 1 & -2 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 1 \\ -2 & c \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 0 \\ 7 & 3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a + 2 - 1 & 3 + b - 1 \\ 4 + 1 -(-2) & 2 + (-2) - c \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 0 \\ 7 & 3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a + 1 & b + 2 \\ 7 & -c \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 0 \\ 7 & 3 \end{bmatrix*} \\[1em] \Rightarrow a + 1 = 5, b + 2 = 0 \text{ and } -c = 3 \\[1em]$$

∴ a = 4, b = -2 and c = -3.

Hence, the values are a = 4, b = -2 and c = -3.

Given, 5A + 2B = C

$$\Rightarrow 5\begin{bmatrix*}[r] 2 & a \\ -3 & 5 \end{bmatrix*} + 2\begin{bmatrix*}[r] -2 & 3 \\ 7 & b \end{bmatrix*} = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 10 & 5a \\ -15 & 25 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & 6 \\ 14 & 2b \end{bmatrix*} = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 10 + (-4) & 5a + 6 \\ -15 + 14 & 25 + 2b \end{bmatrix*} = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 5a + 6 \\ -1 & 25 + 2b \end{bmatrix*} = \begin{bmatrix*}[r] c & 9 \\ -1 & -11 \end{bmatrix*} \\[1em] \Rightarrow 6 = c, 5a + 6 = 9 \text{ and } 25 + 2b = -11 \\[0.5em] \Rightarrow c = 6, 5a = 3 \text{ and } 2b = -11 - 25 \\[0.5em] \Rightarrow c = 6, a = \frac{3}{5} \text{ and } 2b = -36 \\[0.5em] \Rightarrow c = 6, a = \frac{3}{5} \text{ and } b = -18 \\[0.5em] \therefore a = \frac{3}{5}, b = -18 \text{ and } c = 6.$$

Hence, the values are a = $\bold{\frac{3}{5}},$ b = -18 and c = 6.

The product is possible because number of rows in A = number of columns in B =

$$AB = \begin{bmatrix*}[r] 3 & 5 \\ 4 & -2 \end{bmatrix*} \begin{bmatrix*}[r] 2 \\ 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3.2 + 5.4 \\ 4.2 + (-2).4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 26 \\ 0 \end{bmatrix*}$$

Hence, the matrix AB = $$\begin{bmatrix*}[r] 26 \\ 0 \end{bmatrix*}.$$

$$\text{ AB } = \begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 1 & -1 \\ -3 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2.1 + 5.(-3) & 2.(-1) + 5.2 \\ 1.1 + 3.(-3) & 1.(-1) + 3.2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -13 & 8 \\ -8 & 5 \end{bmatrix*} \\[1em] \text{ BA } = \begin{bmatrix*}[r] 1 & -1 \\ -3 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 2 + (-1) \times 1 & 1 \times 5 + (-1) \times 3 \\ -3 \times 2 + 2 \times 1 & -3 \times 5 + 2 \times 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 - 1 & 5 - 3 \\ -6 + 2 & -15 + 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 2 \\ -4 & -9 \end{bmatrix*}$$

The matrix AB = $$\begin{bmatrix*}[r] -13 & 8 \\ -8 & 5 \end{bmatrix*} \text{ and BA } = \begin{bmatrix*}[r] 1 & 2 \\ -4 & -9 \end{bmatrix*}.$$ AB ≠ BA.

$$AB - 5C = \begin{bmatrix*}[r] 3 & 7 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 0 & 2 \\ 5 & 3 \end{bmatrix*} - 5\begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3 \times 0 + 7 \times 5 & 3 \times 2 + 7 \times 3 \\ 2 \times 0 + 4 \times 5 & 2 \times 2 + 4 \times 3 \end{bmatrix*} - \begin{bmatrix*}[r] 5 & -25 \\ -20 & 30 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 + 35 & 6 + 21 \\ 0 + 20 & 4 + 12 \end{bmatrix*} - \begin{bmatrix*}[r] 5 & -25 \\ -20 & 30 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 35 - 5 & 27 - (-25) \\ 20 - (-20) & 16 - 30 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 30 & 52 \\ 40 & -14 \end{bmatrix*}$$

Hence, the matrix AB - 5C = $$\begin{bmatrix*}[r] 30 & 52 \\ 40 & -14 \end{bmatrix*}.$$

$$\text{BA } = \begin{bmatrix*}[r] 2 & 1 \\ 1 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 1 + 1 \times 2 & 2 \times 2 + 1 \times 1 \\ 1 \times 1 + 2 \times 2 & 1 \times 2 + 2 \times 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + 2 & 4 + 1 \\ 1 + 4 & 2 + 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} \\[1.5em] \therefore \text{A(BA) } = \text{A} \times \text{BA} \\[1em] = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 4 + 2 \times 5 & 1 \times 5 + 2 \times 4 \\ 2 \times 4 + 1 \times 5 & 2 \times 5 + 1 \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 10 & 5 + 8 \\ 8 + 5 & 10 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 & 13 \\ 13 & 14 \end{bmatrix*}$$

Hence, the matrix A(BA) = $$\begin{bmatrix*}[r] 14 & 13 \\ 13 & 14 \end{bmatrix*}.$$

(i)

$$\text{ABC } = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*}\begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 3 + 1 \times (-1) & 2 \times 4 + 1 \times (-2) \\ 4 \times 3 + 2 \times (-1) & 4 \times 4 + 2 \times (-2) \end{bmatrix*} \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 - 1 & 8 - 2 \\ 12 - 2 & 16 - 4 \end{bmatrix*} \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix} 5 & 6 \\ 10 & 12 \end{bmatrix} \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 \times (-3) + 6 \times 0 & 5 \times 1 + 6 \times (-2) \\ 10 \times (-3) + 12 \times 0 & 10 \times 1 + 12 \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix} -15 + 0 & 5 - 12 \\ -30 + 0 & 10 + (-24) \end{bmatrix} \\[1em] = \begin{bmatrix} -15 & -7 \\ -30 & -14 \end{bmatrix} \\[1em]$$

Hence, the matrix ABC = $$\begin{bmatrix} -15 & -7 \\ -30 & -14 \end{bmatrix} .$$

(ii)

$$\text{ACB } = \begin{bmatrix*}[r] 2 & 1 \\ 4 & 2 \end{bmatrix*} \begin{bmatrix*}[r] -3 & 1 \\ 0 & -2 \end{bmatrix*} \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times (-3) + 1 \times 0 & 2 \times 1 + 1 \times (-2) \\ 4 \times (-3) + 2 \times 0 & 4 \times 1 + 2 \times (-2) \end{bmatrix*} \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + 0 & 2 - 2 \\ -12 + 0 & 4 - 4 \end{bmatrix*} \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 & 0 \\ -12 & 0 \end{bmatrix*} \begin{bmatrix*}[r] 3 & 4 \\ -1 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 \times 3 + 0 \times (-1) & -6 \times 4 + 0 \times (-2) \\ -12 \times 3 + 0 \times (-1) & -12 \times 4 + 0 \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -18 + 0 & -24 + 0 \\ -36 + 0 & -48 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -18 & -24 \\ -36 & -48 \end{bmatrix*} \\[1em]$$

Hence, the matrix ACB = $$\begin{bmatrix} -18 & -24 \\ -36 & -48 \end{bmatrix} ,$$ and matrix ABC ≠ ACB.

$$\begin{bmatrix} \text{4 sin 30° } & \text {2 cos 60°} \\ \text{ sin 90° } & \text{ 2 cos 0°} \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \\[1em] \text{ Since, sin 30° = cos 60° } = \frac{1}{2}, \text { sin 90° = cos 0° = 1.} \\[1em] \Rightarrow \begin{bmatrix} 4 \times \frac{1}{2} & 2 \times \frac{1}{2} \\ 1 & 2 \times 1 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 \times 4 + 1 \times 5 & 2 \times 5 + 1 \times 4 \\ 1 \times 4 + 2 \times 5 & 1 \times 5 + 2 \times 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 8 + 5 & 10 + 4 \\ 4 + 10 & 5 + 8 \end{bmatrix} \\[1em] = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix} .$$

Hence, the resultant matrix is $$\begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}.$$

$$\text{ AB } = \begin{bmatrix*}[r] -1 & 3 \\ 2 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 2 & -3 \\ -4 & -6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 \times 2 + 3 \times -4 & -1 \times (-3) + 3 \times (-6) \\ 2 \times 2 + 4 \times (-4) & 2 \times (-3) + 4 \times (-6) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 - 12 & 3 - 18 \\ 4 - 16 & -6 - 24 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -14 & -15 \\ -12 & -30 \end{bmatrix*} \\[1em] \text{BA } = \begin{bmatrix*}[r] 2 & -3 \\ -4 & -6 \end{bmatrix*} \begin{bmatrix*}[r] -1 & 3 \\ 2 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times (-1) + (-3) \times 2 & 2 \times 3 + (-3) \times 4 \\ (-4) \times (-1) + (-6) \times 2 & (-4) \times 3 + (-6) \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 - 6 & 6 - 12 \\ 4 - 12 & -12 - 24 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -8 & -6 \\ -8 & -36 \end{bmatrix*} \\[1em] \text{Given, AB + BA } = \begin{bmatrix*}[r] -14 & -15 \\ -12 & -30 \end{bmatrix*} + \begin{bmatrix*}[r] -8 & -6 \\ -8 & -36 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -14 + (-8) & -15 + (-6) \\ -12 + (-8) & -30 + (-36) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -22 & -21 \\ -20 & -66 \end{bmatrix*}.$$

Hence, the matrix AB + BA = $$\begin{bmatrix*}[r] -22 & -21 \\ -20 & -66 \end{bmatrix*}.$$

$$2B = 2\begin{bmatrix*}[r] 3 & 2 \\ -2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 & 4 \\ -4 & 2 \end{bmatrix*} \\[1em] A^2 = \begin{bmatrix*}[r] 1 & -2 \\ 2 & -1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & -2 \\ 2 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + (-2) \times 2 & 1 \times (-2) + (-2) \times (-1) \\ 2 \times 1 + (-1) \times 2 & 2 \times (-2) + (-1)\times (-1) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 - 4 & -2 + 2 \\ 2 - 2 & -4 + 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 & 0 \\ 0 & -3 \end{bmatrix*} \\[1em] \therefore 2B - A^2 = \begin{bmatrix*}[r] 6 & 4 \\ -4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] -3 & 0 \\ 0 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 - (-3) & 4 - 0 \\ -4 - 0 & 2 - (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ -4 & 5 \end{bmatrix*}$$

Hence, the matrix 2B - A² = $$\begin{bmatrix*}[r] 9 & 4 \\ -4 & 5 \end{bmatrix*}.$$

(i) A(B + C)

$$\text{B + C } = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 5 & 1 \\ 7 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 + 5 & 1 + 1 \\ 4 + 7 & 2 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \\[1em] \therefore \text{ A(B + C) } = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 \times 7 + 2 \times 11 & 1 \times 2 + 2 \times 6 \\ 3 \times 7 + 4 \times 11 & 3 \times 2 + 4 \times 6 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 + 22 & 2 + 12 \\ 21 + 44 & 6 + 24 \end{bmatrix} \\[1em] = \begin{bmatrix} 29 & 14 \\ 65 & 30 \end{bmatrix}. \\[1em]$$

Hence, the matrix A(B + C) = $$\begin{bmatrix} 29 & 14 \\ 65 & 30 \end{bmatrix}.$$

(ii) (B + C)A

$$\text{B + C } = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 5 & 1 \\ 7 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 + 5 & 1 + 1 \\ 4 + 7 & 2 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \\[1em] \therefore \text{(B + C)A } = \begin{bmatrix} 7 & 2 \\ 11 & 6 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 \times 1 + 2 \times 3 & 7 \times 2 + 2 \times 4 \\ 11 \times 1 + 6 \times 3 & 11 \times 2 + 6 \times 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 + 6 & 14 + 8 \\ 11 + 18 & 22 + 24 \end{bmatrix} \\[1em] = \begin{bmatrix} 13 & 22 \\ 29 & 46 \end{bmatrix}.$$

Hence, the matrix (B + C)A = $$\begin{bmatrix} 13 & 22 \\ 29 & 46 \end{bmatrix}.$$

$$B - A = \begin{bmatrix*}[r] 2 & 1 \\ 3 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 2 \\ 2 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 - 1 & 1 - 2 \\ 3 - 2 & 2 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & -1 \\ 1 & -1 \end{bmatrix*} \\[1.5em] \therefore C(B - A) = \begin{bmatrix*}[r] 1 & 3 \\ 3 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & -1 \\ 1 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 3 \times 1 & 1 \times (-1) + 3 \times (-1) \\ 3 \times 1 + 1 \times 1 & 3 \times (-1) + 1 \times (-1) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 3 & -1 - 3 \\ 3 + 1 & -3 - 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & -4 \\ 4 & -4 \end{bmatrix*}.$$

Hence, the matrix C(B - A) = $$\begin{bmatrix} 4 & -4 \\ 4 & -4 \end{bmatrix}$$.

$$A^2 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 \times 1 + 0 \times 2 & 1 \times 0 + 0 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 0 + 1 \times 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 + 0 & 0 + 0 \\ 2 + 2 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \\[1em] AB = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix} 1 \times 2 + 0 \times (-1) & 1 \times 3 + 0 \times 0 \\ 2 \times 2 + 1 \times (-1) & 2 \times 3 + 1 \times 0 \end{bmatrix} \\[1em] = \begin{bmatrix*}[r] 2 + 0 & 3 + 0 \\ 4 - 1 & 6 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 3 \\ 3 & 6 \end{bmatrix*} \\[1em] B^2 = \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix} 2 \times 2 + 3 \times (-1) & 2 \times 3 + 3 \times 0 \\ (-1) \times 2 + 0 \times (-1) & (-1) \times 3 + 0 \times 0 \end{bmatrix} \\[1em] = \begin{bmatrix*}[r] 4 - 3 & 6 + 0 \\ -2 + 0 & -3 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*}. \\[1em] \therefore A^2 + AB + B^2 = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} + \begin{bmatrix*}[r] 2 & 3 \\ 3 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 2 + 1 & 0 + 3 + 6 \\ 4 + 3 + (-2) & 1 + 6 + (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix}.$$

Hence, the matrix A² + AB + B² = $$\begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix}$$.

Given, A = $$\begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*}$$

$$\Rightarrow A^2 =\begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*} \times \begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 3 \times 3 + 0 \times 5 & 3 \times 0 + 0 \times 1 \\ 5 \times 3 + 1 \times 5 & 5 \times 0 + 1 \times 1 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 9 + 0 & 0 + 0 \\ 15 + 5 & 0 + 1 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 9 & 0 \\ 20 & 1 \end{bmatrix*}\\[1em] \Rightarrow \text{2AB} = 2 \times \begin{bmatrix*}[r] 3 & 0 \\ 5 & 1 \end{bmatrix*} \times \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 6 & 0 \\ 10 & 2 \end{bmatrix*} \times \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 6 \times (-4) + 0 \times 1 & 6 \times 2 + 0 \times 0 \\ 10 \times (-4) + 2 \times 1 & 10 \times 2 + 2 \times 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] -24 + 0 & 12 + 0 \\ -40 + 2 & 20 + 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] -24 & 12 \\ -38 & 20 \end{bmatrix*}\\[1em] \Rightarrow \text{B}^2 = \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*} \times \begin{bmatrix*}[r] -4 & 2 \\ 1 & 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] -4 \times (-4) + 2 \times 1 & -4 \times 2 + 2 \times 0 \\ 1 \times (-4) + 0 \times 1 & 1 \times 2 + 0 \times 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 16 + 2 & -8 + 0 \\ -4 + 0 & 2 + 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 18 & -8 \\ -4 & 2 \end{bmatrix*}$$

Substituting values in A² - 2AB + B², we get :

$$A^2 - 2AB + B^2 = \begin{bmatrix*}[r] 9 & 0 \\ 20 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] -24 & 12 \\ -38 & 20 \end{bmatrix*} + \begin{bmatrix*}[r] 18 & -8 \\ -4 & 2 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 9 - (-24) + 18 & 0 - 12 + (-8) \\ 20 - (-38) + (-4) & 1 - 20 + 2 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 51 & -20 \\ 54 & -17 \end{bmatrix*}$$

Hence, the value of A² - 2AB + B² = $$\begin{bmatrix*}[r] 51 & -20 \\ 54 & -17 \end{bmatrix*}$$

$$A^2 = \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 2 + 1 \times 0 & 2 \times 1 + 1 \times (-2) \\ 0 \times 2 + (-2) \times 0 & 0 \times 1 + (-2) \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 0 & 2 - 2 \\ 0 + 0 & 0 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*}. \\[1.5em] AC = \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*} \begin{bmatrix*}[r] -3 & 2 \\ -1 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times (-3) + 1 \times (-1) & 2 \times 2 + 1 \times 4 \\ 0 \times (-3) + (-2) \times (-1) & 0 \times 2 + (-2) \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + (-1) & 4 + 4 \\ 0 + 2 & 0 + (-8) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -7 & 8 \\ 2 & -8 \end{bmatrix*}. \\[1.5em] 5B = 5 \begin{bmatrix*}[r] 4 & 1 \\ -3 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 20 & 5 \\ -15 & -10 \end{bmatrix*} \\[1.5em] \therefore A^2 + AC - 5B = \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] -7 & 8 \\ 2 & -8 \end{bmatrix*} - \begin{bmatrix*}[r] 20 & 5 \\ -15 & -10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + (-7) - 20 & 0 + 8 - 5 \\ 0 + 2 - (-15) & 4 + (-8) - (-10) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -23 & 3 \\ 17 & 6 \end{bmatrix*} .$$

Hence, the matrix A² + AC - 5B = $$\begin{bmatrix*}[r] -23 & 3 \\ 17 & 6 \end{bmatrix*} .$$

$$AC = \begin{bmatrix*}[r] 2 & 3 \\ 5 & 7 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ -1 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 1 + 3 \times (-1) & 2 \times 0 + 3 \times 4 \\ 5 \times 1 + 7 \times (-1) & 5 \times 0 + 7 \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 - 3 & 0 + 12 \\ 5 - 7 & 0 + 28 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 & 12 \\ -2 & 28 \end{bmatrix*}.$$

$$B^2 = \begin{bmatrix*}[r] 0 & 4 \\ -1 & 7 \end{bmatrix*} \begin{bmatrix*}[r] 0 & 4 \\ -1 & 7 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 \times 0 + 4 \times (-1) & 0 \times 4 + 4 \times 7 \\ (-1) \times 0 + 7 \times (-1) & (-1) \times 4 + 7 \times 7 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 - 4 & 0 + 28 \\ 0 - 7 & -4 + 49 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -4 & 28 \\ -7 & 45 \end{bmatrix*}.$$

$$10C = 10 \begin{bmatrix*}[r] 1 & 0 \\ -1 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 10 & 0 \\ -10 & 40 \end{bmatrix*}. \\[1em] \therefore AC + B^2 - 10C = \begin{bmatrix*}[r] -1 & 12 \\ -2 & 28 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & 28 \\ -7 & 45 \end{bmatrix*} - \begin{bmatrix*}[r] 10 & 0 \\ -10 & 40 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 + (-4) - 10 & 12 + 28 - 0 \\ -2 + (-7) - (-10) & 28 + 45 - 40 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -15 & 40 \\ 1 & 33 \end{bmatrix*}$$

Hence, the matrix AC + B² - 10C = $$\begin{bmatrix*}[r] -15 & 40 \\ 1 & 33 \end{bmatrix*}$$.

$$A^2 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 0 & 1 \times 0 + 0 \times (-1) \\ 0 \times 1 + (-1) \times 0 & 0 \times 0 + (-1) \times (-1) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$$

$$A^3 = A^2 \times A \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 0 & 1 \times 0 + 0 \times (-1) \\ 0 \times 1 + 1 \times 0 & 0 \times 0 + 1 \times (-1) \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] 1 + 0 & 0 + 0 \\ 0 + 0 & 0 -1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*}$$

Hence, the matrix $$A^2 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \text{ and } A^3 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*}.$$
Thus, A³ = A.

We have to prove 6X - X² = 9I,

$$\text{L.H.S. } = 6X - X^2 \\[1em] 6X - X^2 = 6\begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 & 6 \\ -6 & 12 \end{bmatrix*} - \begin{bmatrix*}[r] 4 \times 4 + 1 \times (-1) & 4 \times 1 + 1 \times 2 \\ (-1) \times 4 + 2 \times (-1) & (-1) \times 1 + 2 \times 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 & 6 \\ -6 & 12 \end{bmatrix*} - \begin{bmatrix*}[r] 16 - 1 & 4 + 2 \\ -4 - 2 & -1 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 & 6 \\ -6 & 12 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & 6 \\ -6 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 - 15 & 6 - 6 \\ -6 + 6 & 12 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 0 \\ 0 & 9 \end{bmatrix*}$$

$$\text{R.H.S. } = 9I \\[1em] 9I = 9\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 0 \\ 0 & 9 \end{bmatrix*}.$$

Since, L.H.S. = $$\begin{bmatrix*}[r] 9 & 0 \\ 0 & 9 \end{bmatrix*}$$ = R.H.S.
Hence, proved that 6X - X² = 9I.

$$I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}, X = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \\[1em]$$

Given,

X² - 2X - 3I = 0

Putting value of X and I in above equation we get,

$$\text{L.H.S. } = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} - 2 \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} - 3 \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 2 \times 2 & 1 \times 2 + 2 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 2 + 1 \times 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4 \\ 4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] 1 + 4 & 2 + 2 \\ 2 + 2 & 4 + 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4 \\ 4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] 5 & 4 \\ 4 & 5 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4 \\ 4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 - 2 - 3 & 4 - 4 - 0 \\ 4 - 4 - 0 & 5 - 2 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}$$

Since, L.H.S. = $$\begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}$$ = R.H.S.
Hence, proved that X² - 2X -3I = 0.

∴$$\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*}$$ is a solution of the matrix equation X² - 2X - 3I = 0

Given,

$$\begin{bmatrix*}[r] 3 & 7 \\ 2 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 0 & 2 \\ 5 & 3 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*}, \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times 0 + 7 \times 5 & 3 \times 2 + 7 \times 3 \\ 2 \times 0 + 4 \times 5 & 2 \times 2 + 4 \times 3 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 + 35 & 6 + 21 \\ 0 + 20 & 4 + 12 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 35 & 27 \\ 20 & 16 \end{bmatrix*} + 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] 1 & -5 \\ -4 & 6 \end{bmatrix*} - \begin{bmatrix*}[r] 35 & 27 \\ 20 & 16 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] 1 - 35 & -5 - 27 \\ -4 - 20 & 6 - 16 \end{bmatrix*} \\[1em] \Rightarrow 2X = \begin{bmatrix*}[r] -34 & -32 \\ -24 & -10 \end{bmatrix*} \\[1em] \Rightarrow X = \frac{1}{2} \begin{bmatrix*}[r] -34 & -32 \\ -24 & -10 \end{bmatrix*} \\[1em] X = \begin{bmatrix*}[r] -17 & -16 \\ -12 & -5 \end{bmatrix*}.$$

Hence, the matrix X = $$\begin{bmatrix*}[r] -17 & -16 \\ -12 & -5 \end{bmatrix*}$$.

Given, A² = O.

$$\text{or}, \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*} \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \text{L.H.S.} = \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*} \begin{bmatrix*}[r] 1 & 1 \\ x & x \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 1 \times x & 1 \times 1 + 1 \times x \\ x \times 1 + x \times x & x \times 1 + x \times x \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + x & 1 + x \\ x + x^2 & x + x^2 \end{bmatrix*}$$

Comparing with R.H.S. we get,

⇒ 1 + x = 0 or x = -1      (...Eq 1)

⇒ x + x² = 0                    (...Eq 2)

Putting the value x = -1 in equation 2,

⇒ -1 + (-1)² = -1 + 1 = 0.

Since, x = -1 satisfies the equation,

Hence, the required value of x = -1.

Given,

$$\begin{bmatrix*}[r] 2x & x \\ y & 3y \end{bmatrix*} \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} = \begin{bmatrix*}[r] 16 \\ 9 \end{bmatrix*} \\[1em] \text{L.H.S.} = \begin{bmatrix*}[r] 2x & x \\ y & 3y \end{bmatrix*} \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2x \times 3 + x \times 2 \\ y \times 3 + 3y \times 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6x + 2x \\ 3y + 6y \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 8x \\ 9y \end{bmatrix*}$$

Comparing L.H.S. with R.H.S. we get,

$$\Rightarrow \begin{bmatrix*}[r] 8x \\ 9y \end{bmatrix*} = \begin{bmatrix*}[r] 16 \\ 9 \end{bmatrix*} \\[1em] \Rightarrow 8x = 16 \text{ and } 9y = 9 \\[1em] \therefore x = 2 \text{ and } y = 1.$$

Hence, the value of x = 2 and y = 1.

Given,

$$\begin{bmatrix*}[r] x + y & y \\ 2x & x - y \end{bmatrix*} \begin{bmatrix*}[r] 2 \\ -1 \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] (x + y) \times 2 + y \times (-1) \\ 2x \times 2 + (x - y) \times (-1) \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 2y - y \\ 4x - x + y \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + y \\ 3x + y \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ 2 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we have,

⇒ 2x + y = 3 or y = 3 - 2x     (...Eq 1)
⇒ 3x + y = 2                         (...Eq 2)

Putting value of y from equation 1 in equation 2,

⇒ 3x + 3 - 2x = 2
⇒ x + 3 = 2
⇒ x = -1.

Now finding value of y,

⇒ y = 3 - 2x
⇒ y = 3 - 2(-1)
⇒ y = 3 + 2
⇒ y = 5.

Hence, the value of x = -1 and y = 5.

Given,

$$\begin{bmatrix*}[r] 1 & 2 \\ 3 & 3 \end{bmatrix*} \begin{bmatrix*}[r] x & 0 \\ 0 & y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 \times x + 2 \times 0 & 1 \times 0 + 2 \times y \\ 3 \times x + 3 \times 0 & 3 \times 0 + 3 \times y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x + 0 & 0 + 2y \\ 3x + 0 & 0 + 3y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & 2y \\ 3x & 3y \end{bmatrix*} = \begin{bmatrix*}[r] x & 0 \\ 9 & 0 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices,

⇒ 2y = 0, 3x = 9 and 3y = 0
⇒ y = 0, x = 3 and y = 0.

Hence, the values are x = 3 and y = 0.

Given,

$$\begin{bmatrix*}[r] 3 & 4 \\ 2 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & 4 \\ 2 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] a \times 1 + b \times 0 & a \times 0 + b \times 1 \\ c \times 1 + d \times 0 & c \times 0 + d \times 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & 4 \\ 2 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*}$$

By definition of equality of matrices,

⇒ a = 3, b = 4, c = 2 and d = 5

Hence, the value of a = 3, b = 4, c = 2 and d = 5.

$$\text{Given } A^2 = B, \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + 12 \times 0 & 2 \times 12 + 12 \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times 12 + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 0 & 24 + 12 \\ 0 + 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices,

⇒ x = 36.

Hence, the value of x = 36.

Given,

A² = B

$$\Rightarrow \begin{bmatrix*}[r] 2 & x \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 2 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + x \times 0 & 2 \times x + x \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 0 & 2x + x \\ 0 + 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 3x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

⇒ 3x = 36
∴ x = 12.

Hence, the value of x = 12.

Given,

A² = B

$$\Rightarrow \begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times 3 + x \times 0 & 3 \times x + x \times 1 \\ 0 \times 3 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 + 0 & 3x + x \\ 0 + 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 4x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

⇒ 4x = 16 and -y = 1
∴  x = 4 and y = -1.

Hence, the values are x = 4 and y = -1.

Given,

$$\begin{bmatrix*}[r] -2 & 0 \\ 3 & 1 \end{bmatrix*} \begin{bmatrix*}[r] -1 \\ 2x \end{bmatrix*} + 3\begin{bmatrix*}[r] -2 \\ 1 \end{bmatrix*} = 2\begin{bmatrix*}[r] y \\ 3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] (-2) \times (-1) + 0 \times 2x \\ 3 \times (-1) + 1 \times 2x \end{bmatrix*} + \begin{bmatrix*}[r] -6 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \\ -3 + 2x \end{bmatrix*} + \begin{bmatrix*}[r] -6 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 + (-6) \\ -3 + 2x + 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -4 \\ 2x \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

2y = -4 and 2x = 6
∴  y = -2 and x = 3.

Hence, the values are x = 3 and y = -2.

Given,

$$\begin{bmatrix*}[r] a & 1 \\ 1 & 0 \end{bmatrix*} \begin{bmatrix*}[r] 4 & 3 \\ -3 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a \times 4 + 1 \times (-3) & a \times 3 + 1 \times 2 \\ 1 \times 4 + 0 \times (-3) & 1 \times 3 + 0 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4a - 3 & 3a + 2 \\ 4 + 0 & 3 + 0 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4a - 3 & 3a + 2 \\ 4 & 3 \end{bmatrix*} = \begin{bmatrix*}[r] b & 11 \\ 4 & c \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

4a - 3 = b        (...Eq 1)
3a + 2 = 11     (...Eq 2)
c = 3.

Solving (Eq 2) first,

⇒ 3a + 2 = 11
⇒ 3a = 9
⇒ a = 3.

Putting value of a in Eq 1,

⇒ 4a - 3 = b
⇒ 4(3) - 3 = b
⇒ 12 - 3 = b
⇒ b = 9

∴ a = 3, b = 9 and c = 3.

Hence, the value of a = 3, b = 9 and c = 3.

Given,

$$\text{A } = \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*}, \text{ I } = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \text{A}^2 = x\text{A} + y\text{I } \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*} = x \begin{bmatrix*}[r] 2 & 3 \\ 1 & 2 \end{bmatrix*} + y \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + 3 \times 1 & 2 \times 3 + 3 \times 2 \\ 1 \times 2 + 2 \times 1 & 1 \times 3 + 2 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 2x & 3x \\ x & 2x \end{bmatrix*} + \begin{bmatrix*}[r] y & 0 \\ 0 & y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4 \end{bmatrix*} = \begin{bmatrix*}[r] 2x + y & 3x + 0 \\ x + 0 & 2x + y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 7 & 12 \\ 4 & 7 \end{bmatrix*} = \begin{bmatrix*}[r] 2x + y & 3x + 0 \\ x + 0 & 2x + y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 7 & 12 \\ 4 & 7 \end{bmatrix*} = \begin{bmatrix*}[r] 2x + y & 3x \\ x & 2x + y \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

3x = 12 or x = 4     (...Eq 1)

2x + y = 7              (...Eq 2)

Putting value of x from Eq1 in Eq 2,

⇒ 2(4) + y = 7
⇒ 8 + y = 7
⇒ y = 7 - 8
⇒ y = -1.

Hence, the value of x = 4 and y = -1.

Given,

PQ = 0

$$\Rightarrow \begin{bmatrix*}[r] 2 & 6 \\ 3 & 9 \end{bmatrix*} \begin{bmatrix*}[r] 3 & x \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 3 + 6 \times y & 2 \times x + 6 \times 2 \\ 3 \times 3 + 9 \times y & 3 \times x + 9 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 6y & 2x + 12 \\ 9 + 9y & 3x + 18 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

⇒ 6 + 6y = 0 and 2x + 12 = 0
⇒ 6y = -6 and 2x = -12
⇒ y = -1 and x = -6.

Checking whether x = -6 and y = -1, satisfies other equations 9 + 9y = 0 and 3x + 18 = 0,

⇒ 9 + 9y = 0
⇒ 9 + 9(-1) = 0
⇒ 9 - 9 = 0 (L.H.S. = R.H.S.)

⇒ 3x + 18 = 0
⇒ 3(-6) + 18 = 0
⇒ -18 + 18 = 0 (L.H.S. = R.H.S.)

∴ x = -6 and y = -1.

Hence, the value of x = -6 and y = -1.

(i)

Since,

$$\text{M} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \text{M} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} \text{is a } 1 \times 2 \text{ matrix, but} \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \Rightarrow \text{M is a } 1 \times 2 \text{ matrix}.$$

The order of matrix M is 1 × 2.

(ii)

$$\text{Let M =} \begin{bmatrix*}[r] x & y \end{bmatrix*} \\[1em] \text{Given, } \text{M} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & y \end{bmatrix*} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 1 + y \times 0 & x \times 1 + y \times 2 \\ \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & x + 2y \\ \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

x = 1                (...Eq 1)
x + 2y = 2        (...Eq 2)

Putting value of x from Eq 1 in Eq 2,

⇒ x + 2y = 2
⇒ 1 + 2y = 2
⇒ 2y = 2 - 1
⇒ 2y = 1
⇒ y = $\frac{1}{2}$

$$\text{Since, M }= \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \\[1em] \therefore \text{M} = \begin{bmatrix*}[r] 1 & \frac{1}{2} \\ \end{bmatrix*}.$$

Hence, the matrix M = $$\begin{bmatrix*}[r] 1 & \frac{1}{2} \end{bmatrix*}$$.

(i) Since,

$$\begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*}X = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*}X \text{ is a } 2 \times 1 \text{ matrix, but} \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \Rightarrow \text{X is a } 2 \times 1 \text{ matrix}.$$

The order of the matrix is 2 × 1.

(ii) Let $$\text{X} = \begin{bmatrix*}[r] x \\ y \end{bmatrix*}$$

Given,

$$\begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*}X = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ -3 & 4 \end{bmatrix*} \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times x + 1 \times y \\ -3 \times x + 4 \times y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + y \\ -3x + 4y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ 6 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

2x + y = 7 or y = 7 - 2x     (...Eq 1)

-3x + 4y = 6                      (...Eq 2)

Putting value of y from Eq 1 in Eq 2

⇒ -3x + 4y = 6
⇒ -3x + 4(7 - 2x) = 6
⇒ -3x + 28 - 8x = 6
⇒ -11x = 6 - 28
⇒ -11x = -22
⇒ x = 2.

∴ x = 2 and y = 7 - 2x = 7 - 2(2) = 7 - 4 = 3.

Since, $$\text{X} = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} \\[1em] \therefore \text{X} = \begin{bmatrix*}[r] 2 \\ 3 \end{bmatrix*}$$

Hence, the matrix $$\text{X} = \begin{bmatrix*}[r] 2 \\ 3 \end{bmatrix*} .$$

Since,

$$\begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} X = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} X \text{ is a } 2 \times 2 \text{ matrix, but} \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} \text{ is a } 2 \times 1 \text{ matrix}. \\[1em] \Rightarrow \text{X is a } 1 \times 2 \text{ matrix}.$$

We know that X matrix will be of order 1 × 2. So, let matrix X be $$\begin{bmatrix*}[r] x & y \end{bmatrix*}.$$

Given,

$$\begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} X = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*} \begin{bmatrix*}[r] x & y \end{bmatrix*} = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4x & 4y \\ x & y \end{bmatrix*} = \begin{bmatrix*}[r] -4 & 8 \\ -1 & 2 \end{bmatrix*} \\[1em]$$

From definition of equality of matrices we get,

⇒ x = -1 and y = 2.

$$\text{Since, X }= \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \\[1em] \therefore \text{X } = \begin{bmatrix*}[r] -1 & 2 \\ \end{bmatrix*}$$

Hence, the matrix $$\text{X} = \begin{bmatrix*}[r] -1 & 2 \end{bmatrix*}.$$

Given,

AC = B

$$\Rightarrow \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} C = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} C \text{ is a } 2 \times 1 \text{ matrix, but} \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[0.5em] \therefore \text{C is a } 1 \times 2 \text{ matrix}. \\[0.5em] \text{Let matrix C } = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 \times x + (-1) \times y \\ -4 \times x + 5 \times y \end{bmatrix*} = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2x - y \\ -4x + 5y \end{bmatrix*} = \begin{bmatrix*}[r] -3 \\ 2 \end{bmatrix*} \\[0.5em]$$

By definition of equality of matrices we get,

⇒ 2x - y = -3 or y = 2x + 3     (...Eq 1)

⇒ -4x + 5y = 2                        (...Eq 2)

Putting value of y from Eq 1 in Eq 2,

⇒ -4x + 5y = 2
⇒ -4x + 5(2x + 3) = 2
⇒ -4x + 10x + 15 = 2
⇒ 6x = 2 - 15
⇒ 6x = -13
⇒ x = $-\frac{13}{6}$

Now finding value of y,

$$\begin{aligned}y = 2x + 3 \\[1em] = 2\Big(-\frac{13}{6}\Big) + 3 \\[1em] = -\frac{26}{6} + 3 \\[1em] = \frac{-26 + 18}{6} \\[1em] = -\frac{8}{6} \\[1em] = -\frac{4}{3}.\end{aligned}$$

∴ x = $-\frac{13}{6} \text{ and y } = -\frac{4}{3}.$

Since,

$$\text{C }= \begin{bmatrix*}[r] x \\ y \end{bmatrix*} \\[1em] \therefore C = \begin{bmatrix*}[r] -\frac{13}{6} \\ -\frac{4}{3} \end{bmatrix*}$$

Hence, the matrix C = $$\begin{bmatrix*}[r] -\frac{13}{6} \\ -\frac{4}{3} \end{bmatrix*}.$$

Given,

CA = B.

$$C\begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*} \\[1em] \Rightarrow C\begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{ is a } 1 \times 2 \text{ matrix, but} \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{C is a } 1 \times 2 \text{ matrix}. \\[1em] \text{Let matrix C } = \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \begin{bmatrix*}[r] 2 & -1 \\ -4 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 2 + y \times (-4) & x \times (-1) + y \times 5 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x - 4y & -x + 5y \\ \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ \end{bmatrix*}$$

By definition of equality of matrices we get,

2x - 4y = 0                             (...Eq 1)

-x + 5y = -3 or x = 5y + 3     (...Eq 2)

Putting value of x from Eq 2 in Eq 1,

⇒ 2x - 4y = 0
⇒ 2(5y + 3) - 4y = 0
⇒ 10y + 6 - 4y = 0
⇒ 6y + 6 = 0
⇒ 6y = -6
⇒ y = -1.

Now finding value of x,

⇒ x = 5y + 3 = 5(-1) + 3 = -5 + 3 = -2.

∴ x = -2, y = -1.

$$\text{Since, C }= \begin{bmatrix*}[r] x & y \\ \end{bmatrix*} \\[0.5em] \therefore C = \begin{bmatrix*}[r] -2 & -1 \\ \end{bmatrix*}$$

Hence, the matrix C = $$\begin{bmatrix*}[r] -2 & -1 \\ \end{bmatrix*}$$.

Given,

BA = I

$$\text{B}\begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \text{B}\begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix, and} \begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{B is a } 2 \times 2 \text{ matrix}. \\[1em] \text{I } = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$$

We know that B will be of order 2 × 2. So, let

$$\text{B} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \begin{bmatrix*}[r] 3 & -4 \\ -1 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a \times 3 + b \times (-1) & a \times (-4) + b \times 2 \\ c \times 3 + d \times (-1) & c \times (-4) + d \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3a - b & -4a + 2b \\ 3c - d & -4c + 2d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

3a - b = 1          (...Eq 1)

-4a + 2b = 0
⇒ 4a = 2b
⇒ b = 2a           (...Eq 2)

3c - d = 0
⇒ d = 3c           (...Eq 3)

-4c + 2d = 1     (...Eq 4)

Putting value of b from Eq 2 in Eq 1

⇒ 3a - b = 1
⇒ 3a - 2a = 1
⇒ a = 1

∴ a = 1, b = 2a = 2.

Putting value of d from Eq 3 in Eq 4

⇒ -4c + 2d = 1
⇒ -4c + 2(3c) = 1
⇒ -4c + 6c = 1
⇒ 2c = 1
⇒ c = $\frac{1}{2}$

∴ c = $\frac{1}{2}$, d = 3c = $\frac{3}{2}$.

$$\text{Since, B }= \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \therefore \text{B} = \begin{bmatrix*}[r] 1 & 2 \\ \frac{1}{2} & \frac{3}{2} \end{bmatrix*}$$

Hence, the matrix B = $$\begin{bmatrix*}[r] 1 & 2 \\ \frac{1}{2} & \frac{3}{2} \end{bmatrix*}$$.

(i) Given,

$$\begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} M = 6I \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} M \text{ is a } 2 \times 2 \text{ matrix, and} \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{M is a } 2 \times 2 \text{ matrix}. \\[1em]$$

Hence, the matrix M is of order 2 × 2.

(ii) Let matrix M be $$\begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*}.$$

Given,

$$\begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} M = 6I \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 2 \\ -1 & 1 \end{bmatrix*} \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = 6\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \times a + 2 \times c & 4 \times b + 2 \times d \\ (-1) \times a + 1 \times c & (-1) \times b + 1 \times d \end{bmatrix*} = \begin{bmatrix*}[r] 6 & 0 \\ 0 & 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4a + 2c & 4b + 2d \\ -a + c & -b + d \end{bmatrix*} = \begin{bmatrix*}[r] 6 & 0 \\ 0 & 6 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

4a + 2c = 6     (...Eq 1)

4b + 2d = 0
⇒ d = -2b       (...Eq 2)

-a + c = 0
⇒ a = c           (...Eq 3)

-b + d = 6       (...Eq 4)

Putting value of a from Eq 3 in Eq 1

⇒ 4a + 2c = 6
⇒ 4c + 2c = 6
⇒ 6c = 6
⇒ c = 1.

∴ c = 1 and a = c = 1.

Putting value of d from Eq 2 in Eq 4

⇒ -b + d = 6
⇒ -b + (-2b) = 6
⇒ -3b = 6
⇒ b = -2.

∴ b = -2 and d = -2b = 4.

Since,

$$\text{M} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \therefore \text{M} = \begin{bmatrix*}[r] 1 & -2 \\ 1 & 4 \end{bmatrix*}$$

Hence, the matrix $$\text{M} = \begin{bmatrix*}[r] 1 & -2 \\ 1 & 4 \end{bmatrix*}$$.

Given,

AB = C

$$\Rightarrow A\begin{bmatrix*}[r] -4 & 2 \\ 5 & -1 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix, and} \begin{bmatrix*}[r] 4 & 2 \\ 5 & -1 \end{bmatrix*} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{A is a } 2 \times 2 \text{ matrix}.$$

We know that matrix A will be of order 2 × 2. Let matrix be

$$\text{A} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \begin{bmatrix*}[r] -4 & 2 \\ 5 & -1 \end{bmatrix*} = \begin{bmatrix*}[r] 17 & -1 \\ 47 & -13 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] a \times (-4) + b \times 5 & a \times 2 + b \times (-1) \\ c \times (-4) + d \times 5 & c \times 2 + d \times (-1) \end{bmatrix*} = \begin{bmatrix*}[r] 17 & -1 \\ 47 & -13 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -4a + 5b & 2a - b \\ -4c + 5d & 2c - d \end{bmatrix*} = \begin{bmatrix*}[r] 17 & -1 \\ 47 & -13 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

-4a + 5b = 17    (...Eq 1)

2a - b = -1
⇒ b = 2a + 1     (...Eq 2)

-4c + 5d = 47    (...Eq 3)

2c - d = -13
⇒ d = 2c + 13    (...Eq 4)

Putting value of b from Eq 2 in Eq 1

⇒ -4a + 5b = 17
⇒ -4a + 5(2a + 1) = 17
⇒ -4a + 10a + 5 = 17
⇒ 6a + 5 = 17
⇒ 6a = 12
⇒ a = 2.

∴ a = 2, b = 2a + 1 = 2(2) + 1 = 5.

Putting value of d from Eq 4 in Eq 3

⇒ -4c + 5d = 47
⇒ -4c + 5(2c + 13) = 47
⇒ -4c + 10c + 65 = 47
⇒ 6c + 65 = 47
⇒ 6c = -18
⇒ c = -3.

∴ c = -3, d = 2c + 13 = 2(-3) + 13 = 7.

Since,

$$\text{A} = \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} \\[1em] \therefore \text{A} = \begin{bmatrix*}[r] 2 & 5 \\ -3 & 7 \end{bmatrix*}$$

Hence, the matrix A = $$\begin{bmatrix*}[r] 2 & 5 \\ -3 & 7 \end{bmatrix*}$$.

Given,

⇒ A² = pA

$$\Rightarrow \begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \times 4 + (-4) \times (-4) & 4 \times (-4) + (-4) \times 4 \\ -4 \times 4 + 4 \times (-4) & (-4) \times (-4) + 4 \times 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 16 + 16 & -16 - 16 \\ -16 - 16 & 16 + 16 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 32 & -32 \\ -32 & 32 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow 32\begin{bmatrix*}[r] 1 & -1 \\ -1 & 1 \end{bmatrix*} = 4p\begin{bmatrix*}[r] 1 & -1 \\ -1 & 1 \end{bmatrix*} \\[1em] \Rightarrow 32 = 4p \\[1em] \Rightarrow p = \frac{32}{4} = 8.$$

Hence, p = 8.

Given,

$$\begin{bmatrix*}[r] x + 3 & 4 \\ y - 4 & x + y \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 4 \\ 3 & 9 \end{bmatrix*} \\[0.5em]$$

By definition of equality of matrices we get,

⇒ x + 3 = 5 or x = 5 - 3 = 2.

⇒ y - 4 = 3 or y = 3 + 4 = 7.

⇒ x + y = 9.

Since, x = 2 and y = 7 satisfies the equation x + y = 9, hence, value of x = 2 and y = 7.

∴ Option 1 is the correct option.

Given,

$$\begin{bmatrix*}[r] x + 2y & -y \\ 3x & 4 \end{bmatrix*} = \begin{bmatrix*}[r] -4 & 3 \\ 6 & 4 \end{bmatrix*} \\[0.5em]$$

By definition of equality of matrices we get,

⇒ x + 2y = -4     (...Eq 1)

⇒ -y = 3 or y = -3

⇒ 3x = 6 or x = 2

Putting, x = 2 and y = -3 in Eq 1,

⇒ x + 2y = -4
⇒ L.H.S. = 2 + 2(-3) = 2 - 6 = -4 = R.H.S.

Since, x = 2 and y = -3 satisfies the equation x + 2y = -4,

∴ x = 2 and y = -3.

∴ Option 2 is the correct option.

Given,

$$\begin{bmatrix*}[r] x - 2y & 5 \\ 3 & y \end{bmatrix*} = \begin{bmatrix*}[r] 6 & 5 \\ 3 & -2 \end{bmatrix*} \\[0.5em]$$

By definition of equality of matrices,

⇒ x - 2y = 6     (...Eq 1)

⇒ y = -2.

Putting value of y in Eq 1 we get,

⇒ x - 2y = 6
⇒ x - 2(-2) = 6
⇒ x + 4 = 6
⇒ x = 6 - 4
⇒ x = 2.

∴ x = 2.

∴ Option 4 is the correct option .

Given,

$$\begin{bmatrix*}[r] x + 2y & 3y \\ 4x & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -3 \\ 8 & 2 \end{bmatrix*} \\[0.5em]$$

By definition of equality of matrices we get,

⇒ x + 2y = 0     (...Eq 1)

⇒ 3y = -3 or y = -1

⇒ 4x = 8 or x = 2

Putting the value of x = 2 and y = -1 in Eq 1

⇒ x + 2y = 0
⇒ L.H.S. = 2 + 2(-1) = 2 - 2 = 0 = R.H.S.

Since, x = 2 and y = -1 satisfies Eq 1

∴ x = 2, y = -1 and x - y = 2 - (-1) = 2 + 1 = 3.

∴ Option 3 is the correct option.

Given,

$$x\begin{bmatrix*}[r] 2 \\ 3 \end{bmatrix*} + y\begin{bmatrix*}[r] -1 \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 10 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x \\ 3x \end{bmatrix*} + \begin{bmatrix*}[r] -y \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 10 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x - y \\ 3x \end{bmatrix*} = \begin{bmatrix*}[r] 10 \\ 6 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

⇒ 2x - y = 10          (...Eq 1)

⇒ 3x = 6 or x = 2.

Putting value of x in Eq 1

⇒ 2x - y = 10
⇒ 2(2) - y = 10
⇒ 4 - y = 10
⇒ y = 4 - 10
⇒ y = -6

∴ x = 2 and y = -6.

∴ Option 2 is the correct option.

Given,

$$\text{A} = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*} \\[1em] \Rightarrow \text{A}^2 = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 \times 0 + 1 \times 1 & 0 \times 1 + 1 \times 0 \\ 1 \times 0 + 0 \times 1 & 1 \times 1 + 0 \times 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \therefore \text{A}^2 = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}.$$

∴ Option 4 is the correct option.

Given,

$$\text{A} = \begin{bmatrix*}[r] 0 & 0 \\ 1 & 0 \end{bmatrix*} \\[1em] \Rightarrow \text{A}^2 = \begin{bmatrix*}[r] 0 & 0 \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] 0 & 0 \\ 1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 \times 0 + 0 \times 1 & 0 \times 0 + 0 \times 0 \\ 1 \times 0 + 0 \times 1 & 1 \times 0 + 0 \times 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \therefore \text{A}^2 = O.$$

∴ Option 2 is the correct option.

Given,

$$\text{A} = \begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*} \\[0.5em] \Rightarrow \text{A}^2 = \begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 1 & 1 \times 0 + 0 \times 1 \\ 1 \times 1 + 1 \times 1 & 1 \times 0 + 1 \times 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*} \\[0.5em] \therefore \text{A}^2 = \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*}.$$

∴ Option 3 is the correct option.

Given,

$$\text{A} = \begin{bmatrix*}[r] 3 & 1 \\ -1 & 2 \end{bmatrix*} \\[0.5em] \Rightarrow \text{A}^2 = \begin{bmatrix*}[r] 3 & 1 \\ -1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 3 & 1 \\ -1 & 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 3 \times 3 + 1 \times (-1) & 3 \times 1 + 1 \times 2 \\ (-1) \times 3 + 2 \times (-1) & (-1) \times 1 + 2 \times 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 8 & 5 \\ -5 & 3 \end{bmatrix*} \\[0.5em] \therefore \text{A}^2 = \begin{bmatrix*}[r] 8 & 5 \\ -5 & 3 \end{bmatrix*}.$$

∴ Option 1 is the correct option.

$$\phantom{\Rightarrow} A^2 = \begin{bmatrix*}[r] 2 & 2 \\ 0 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 2 \\ 0 & 2 \end{bmatrix*} \\[1em] \Rightarrow A^2 = \begin{bmatrix*}[r] 2 \times 2 + 2\times 0 & 2 \times 2 + 2 \times 2 \\ 0 \times 2 + 2 \times 0 & 0 \times 2 + 2 \times 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 + 0 & 4 + 4 \\ 0 + 0 & 0 + 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 8 \\ 0 & 4 \end{bmatrix*} \\[1em] \Rightarrow x = 8.$$

Hence, Option 3 is the correct option.

Given,

$$\text{A}^2 = pA \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} = p\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + (-2) \times (-2) & 2 \times (-2) + (-2) \times 2 \\ -2 \times 2 + 2 \times (-2) & (-2) \times (-2) + 2 \times 2 \end{bmatrix*} = p\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 4 + 4 & -4 - 4 \\ -4 - 4 & 4 + 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 8 & -8 \\ -8 & 8 \end{bmatrix*} = \begin{bmatrix*}[r] 2p & -2p \\ -2p & 2p \end{bmatrix*} \\[0.5em]$$

By definition of equality of matrices we get,

⇒ 2p = 8

∴ p = 4.

∴ Option 2 is the correct option.

Matrix multiplication is possible if the number of columns of the first matrix is the same as the number of rows as the second matrix.

So, AB is possible if the number of columns of A is equal to the number of rows of B.

Hence, option 3 is the correct option.

Order of A : 1 × 2 and order of B : 2 × 2

Since, no. of columns in A is same as the no. of rows in B.

∴ AB is possible.

Hence, Option 3 is the correct option.

Given, A, B and C are square matrices of order 2 such that AB = C.

Matrix multiplication is not always commutative.

So, reason (R) is true.

⇒ AB ≠ BA

So, BA is not necessarily equal to C.

So, assertion (A) is false.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Given, A, B and C are square matrices of order 2 such that AB = C.

Matrix multiplication is not always commutative.

∴ AB is not necessarily equal to BA.

So, BA is not necessarily equal to C.

So, assertion (A) is true.

Matrix multiplication is associative.

So, reason (R) is false.

Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

Given, A = $$\begin{bmatrix*}[r] 3 & -2 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] -1 & 4 \\ 2 & 0 \end{bmatrix*}$$

To multiply two matrices, the number of columns of the first matrix must equal the number of rows of the second matrix.

Matrix A has 1 row and 2 columns.

Matrix B has 2 rows and 2 columns.

So, reason (R) is true.

The number of columns in A is 2 and number of rows in B is also 2.

So, product AB is defined.

So, both A and R are true and R is the correct explanation of assertion A.

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

By definition of equality of matrices we get,

⇒ a + 3 = 2a + 1 or a = 2

⇒ b² + 2 = 3b     (...Eq 1)

⇒ b² - 5b = -6     (...Eq 2)

Solving Eq 1 for b,

⇒ b² + 2 = 3b
⇒ b² - 3b + 2 = 0
⇒ b² - 2b - b + 2 = 0
⇒ b(b - 2) - 1(b - 2) = 0
⇒ (b - 1)(b - 2) = 0
⇒ b = 1 or b = 2.

Checking whether the value of b = 1 satisfies Eq 2

⇒ b² - 5b = -6

L.H.S. = b² - 5b = (1)² - 5(1) = -4.

L.H.S. $\neq$ R.H.S., so b = 1 is not the solution.

Checking whether the value of b = 2 satisfies Eq 2

⇒ b² - 5b = -6

L.H.S. = b² - 5b
= (2)² - 5(2)
= 4 - 10
= -6 = R.H.S..

∴ a = 2 and b = 2.

Hence, the values are a = 2 and b = 2.

Given,

$$3\begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 4 & a + b \\ c + d & 3 \end{bmatrix*} + \begin{bmatrix*}[r] a & 6 \\ -1 & 2d \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 3a & 3b \\ 3c & 3d \end{bmatrix*} = \begin{bmatrix*}[r] 4 + a & a + b + 6 \\ c + d - 1 & 3 + 2d \end{bmatrix*} \\[0.5em]$$

By definition of equality of matrices

⇒ 3a = 4 + a     (...Eq 1)

⇒ 3b = a + b + 6     (...Eq 2)

⇒ 3c = c + d - 1     (...Eq 3)

⇒ 3d = 3 + 2d     (...Eq 4)

Solving Eq 1 we get,

⇒ 3a = 4 + a
⇒ 3a - a = 4
⇒ 2a = 4
⇒ a = 2.

Solving Eq 2 by putting value of a from Eq 1 we get,

⇒ 3b = a + b + 6
⇒ 3b = 2 + b + 6
⇒ 3b = b + 8
⇒ 3b - b = 8
⇒ 2b = 8
⇒ b = 4.

Solving Eq 4 we get,

⇒ 3d = 3 + 2d
⇒ 3d - 2d = 3
⇒ d = 3.

Solving Eq 3 by putting value of d from Eq 4 we get,

⇒ 3c = c + d - 1
⇒ 3c = c + 3 - 1
⇒ 3c - c = 2
⇒ 2c = 2
⇒ c = 1.

∴ a = 2, b = 4, c = 1 and d = 3.

Hence, the value of a = 2, b = 4, c = 1 and d = 3.

Given,

$$\text{A + 2B} = \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} \qquad \text{....[Eq 1] } \\[1em] \text{2A - B} = \begin{bmatrix*}[r] 2 & -1 \\ 2 & -1 \end{bmatrix*} \qquad \text{....[Eq 2] } \\[1em]$$

Multiplying Eq 1 by 2,

$$\Rightarrow 2A + 4B = \begin{bmatrix*}[r] 2 & 4 \\ 12 & -6 \end{bmatrix*} \\[1em]$$

Subtracting Eq 2 from above equation we get,

$$\Rightarrow 2A + 4B - (2A - B) = \begin{bmatrix*}[r] 2 & 4 \\ 12 & -6 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & -1 \\ 2 & -1 \end{bmatrix*} \\[1em] \Rightarrow 2A - 2A + 4B - (-B) = \begin{bmatrix*}[r] 2 - 2 & 4 - (-1) \\ 12 - 2 & -6 - (-1) \end{bmatrix*} \\[1em] \Rightarrow 5B = \begin{bmatrix*}[r] 0 & 5 \\ 10 & -5 \end{bmatrix*} \\[1em] \Rightarrow B = \frac{1}{5}\begin{bmatrix*}[r] 0 & 5 \\ 10 & -5 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*}. \\[1em]$$

Putting value of matrix B in Eq 1,

$$\Rightarrow A + 2 \begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} \\[1em] \Rightarrow A + \begin{bmatrix*}[r] 0 & 2 \\ 4 & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 1 & 2 \\ 6 & -3 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & 2 \\ 4 & -2 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 1 - 0 & 2 - 2 \\ 6 - 4 & -3 - (-2) \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 1 & 0 \\ 2 & -1 \end{bmatrix*}. \\[1em] \therefore A = \begin{bmatrix*}[r] 1 & 0 \\ 2 & -1 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*}.$$

Hence, the value of $$\text{A} = \begin{bmatrix*}[r] 1 & 0 \\ 2 & -1 \end{bmatrix*} \text{and B} = \begin{bmatrix*}[r] 0 & 1 \\ 2 & -1 \end{bmatrix*}.$$

Given, A² = A + 2B.

$$\Rightarrow \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} + 2B \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 4 \times 4 + 1 \times 2 & 4 \times 1 + 1 \times 3 \\ 2 \times 4 + 3 \times 2 & 2 \times 1 + 3 \times 3 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} + 2B \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 16 + 2 & 4 + 3 \\ 8 + 6 & 2 + 9 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} + 2B \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 18 & 7 \\ 14 & 11 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} + 2B \\[0.5em] \Rightarrow 2B = \begin{bmatrix*}[r] 18 & 7 \\ 14 & 11 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & 1 \\ 2 & 3 \end{bmatrix*} \\[0.5em] \Rightarrow 2B = \begin{bmatrix*}[r] 18 - 4 & 7 - 1 \\ 14 - 2 & 11 - 3 \end{bmatrix*} \\[0.5em] \Rightarrow 2B = \begin{bmatrix*}[r] 14 & 6 \\ 12 & 8 \end{bmatrix*} \\[0.5em] \therefore B = \begin{bmatrix*}[r] 7 & 3 \\ 6 & 4 \end{bmatrix*}.$$

Hence, the matrix B = $$\begin{bmatrix*}[r] 7 & 3 \\ 6 & 4 \end{bmatrix*}.$$

We need to find the value of A(4B - 3C).

$$\text{A(4B - 3C)} = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*} \Big(4\begin{bmatrix*}[r] 0 & 1 \\ -2 & 5 \end{bmatrix*} - 3\begin{bmatrix*}[r] -2 & 0 \\ -1 & 1 \end{bmatrix*}\Big) \\[0.5em] = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*} \Big(\begin{bmatrix*}[r] 0 & 4 \\ -8 & 20 \end{bmatrix*} - \begin{bmatrix*}[r] -6 & 0 \\ -3 & 3 \end{bmatrix*}\Big) \\[0.5em] = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 0 - (-6) & 4 - 0 \\ -8 - (-3) & 20 - 3 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 1 & 2 \\ -3 & 4 \end{bmatrix*} \begin{bmatrix*}[r] 6 & 4 \\ -5 & 17 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 1 \times 6 + 2 \times (-5) & 1 \times 4 + 2 \times 17 \\ -3 \times 6 + 4 \times (-5) & -3 \times 4 + 4 \times 17 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 6 - 10 & 4 + 34 \\ -18 - 20 & -12 + 68 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] -4 & 38 \\ -38 & 56 \end{bmatrix*}.$$

Hence, the value of A(4B - 3C) = $$\begin{bmatrix*}[r] -4 & 38 \\ -38 & 56 \end{bmatrix*}.$$

B + C = $$\begin{bmatrix*}[r] 1 & 2 \\ 2 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 1 \\ 1 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*}$$

Substituting values we get :

$$A(B + C) - 14I = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*} - 14\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 5 + 3 \times 3 & 1 \times 3 + 3 \times 9 \\ 2 \times 5 + 4 \times 3 & 2 \times 3 + 4 \times 9 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 + 9 & 3 + 27 \\ 10 + 12 & 6 + 36 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 & 30 \\ 22 & 42 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 - 14 & 30 - 0 \\ 22 - 0 & 42 - 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}.$$

Hence, A(B + C) - 14I = $$\begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}.$$

(i) We need to find the value of (A + B)(A - B)

$$(A + B)(A - B) = \Big(\begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*}\Big)\Big(\begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*}\Big) \\[0.5em] = \begin{bmatrix*}[r] 3 + 1 & 2 + 0 \\ 0 + 1 & 5 + 2 \end{bmatrix*} \begin{bmatrix*}[r] 3 - 1 & 2 - 0 \\ 0 - 1 & 5 - 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 4 & 2 \\ 1 & 7 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 2 \\ -1 & 3 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 4 \times 2 + 2 \times (-1) & 4 \times 2 + 2 \times 3 \\ 1 \times 2 + 7 \times (-1) & 1 \times 2 + 7 \times 3 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 8 - 2 & 8 + 6 \\ 2 - 7 & 2 + 21 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 6 & 14 \\ -5 & 23 \end{bmatrix*}.$$

Hence, the value of (A + B)(A - B) = $$\begin{bmatrix*}[r] 6 & 14 \\ -5 & 23 \end{bmatrix*}$$.

(ii) We need to find the value of A² - B²

$$A^2 - B^2 = \begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} \begin{bmatrix*}[r] 3 & 2 \\ 0 & 5 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 1 & 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 3 \times 3 + 2 \times 0 & 3 \times 2 + 2 \times 5 \\ 0 \times 3 + 5 \times 0 & 0 \times 2 + 5 \times 5 \end{bmatrix*} - \begin{bmatrix*}[r] 1 \times 1 + 0 \times 1 & 1 \times 0 + 0 \times 2 \\ 1 \times 1 + 2 \times 1 & 1 \times 0 + 2 \times 2 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 9 & 16 \\ 0 & 25 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 0 \\ 3 & 4 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 9 - 1 & 16 - 0 \\ 0 - 3 & 25 - 4 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 8 & 16 \\ -3 & 21 \end{bmatrix*} .$$

Hence, the value of $$A^2 - B^2 = \begin{bmatrix*}[r] 8 & 16 \\ -3 & 21 \end{bmatrix*} \text{ and (A + B)(A - B) } \neq A^2 - B^2.$$

We need to find the value of A² - 5A - 14I.

$$A^2 - 5A -14I = \begin{bmatrix*}[r] 3 & -5 \\ -4 & 2 \end{bmatrix*} \begin{bmatrix*}[r] 3 & -5 \\ -4 & 2 \end{bmatrix*} - 5\begin{bmatrix*}[r] 3 & -5 \\ -4 & 2 \end{bmatrix*} - 14\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 3 \times 3 + (-5) \times (-4) & 3 \times (-5) + (-5) \times 2 \\ (-4) \times 3 + 2 \times (-4) & (-4) \times (-5) + 2 \times 2 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & -25 \\ -20 & 10 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 9 + 20 & -15 - 10 \\ -12 - 8 & 20 + 4 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & -25 \\ -20 & 10 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 29 & -25 \\ -20 & 24 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & -25 \\ -20 & 10 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 29 - 15 - 14 & -25 - (-25) - 0 \\ -20 - (-20) - 0 & 24 - 10 - 14 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 29 - 29 & -25 + 25 \\ -20 + 20 & 24 - 24 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[0.5em]$$

Hence, the value of $$A^2 - 5A - 14I = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}.$$

Given, A² = O.

$$\Rightarrow \begin{bmatrix*}[r] 3 & 3 \\ p & q \end{bmatrix*} \begin{bmatrix*}[r] 3 & 3 \\ p & q \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 3 \times 3 + 3 \times p & 3 \times 3 + 3 \times q \\ p \times 3 + q \times p & p \times 3 + q \times q \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 9 + 3p & 9 + 3q \\ 3p + qp & 3p + q^2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[0.5em]$$

By definition of equality of matrices we get,

⇒ 9 + 3p = 0 or 3p = -9 or p = -3

⇒ 9 + 3q = 0 or 3q = -9 or q = -3

⇒ 3p + qp = 0     (Eq 1)

⇒ 3p + q² = 0     (Eq 2)

Checking whether p = -3 and q = -3 satisfy Eq 1,

⇒ 3p + qp = 0

L.H.S. = 3(-3) + (-3)(-3) = -9 + 9 = 0 = R.H.S.

Checking whether p = -3 and q = -3 satisfy Eq 2,

⇒ 3p + q² = 0

L.H.S. = 3(-3) + (-3)² = -9 + 9 = 0 = R.H.S.

Since, p = -3 and q = -3 satisfies Eq 1 and Eq 2,

∴ p = -3 and q = -3.

Hence, the values are p = -3 and q = -3.

Given,

$$\begin{bmatrix*}[r] -1 & 0 \\ 0 & 1 \end{bmatrix*} \begin{bmatrix*}[r] a & b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] -1 \times a + 0 \times c & -1 \times b + 0 \times d \\ 0 \times a + 1 \times c & 0 \times b + 1 \times d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] -a & -b \\ c & d \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & -1 \end{bmatrix*} \\[0.5em]$$

By definition of equality of matrices we get,

⇒ -a = 1 or a = -1

⇒ -b = 0 or b = 0

⇒ c = 0

⇒ d = -1.

Hence, the value of a = -1, b = 0, c = 0 and d = -1.

Given,

$$\begin{bmatrix*}[r] a - b & b - 4 \\ b + 4 & a - 2 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 0 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & -2 \\ 14 & 0 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] (a - b) \times 2 + (b - 4) \times 0 & (a - b) \times 0 + (b - 4) \times 2 \\ (b + 4) \times 2 + (a - 2) \times 0 & (b + 4) \times 0 + (a - 2) \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & -2 \\ 14 & 0 \end{bmatrix*} \\[0.5em] \Rightarrow \begin{bmatrix*}[r] 2a - 2b & 2b - 8 \\ 2b + 8 & 2a - 4 \end{bmatrix*} = \begin{bmatrix*}[r] -2 & -2 \\ 14 & 0 \end{bmatrix*} \\[0.5em]$$

By definition of equality of matrices we get,

⇒ 2a - 4 = 0 or 2a = 4 or a = 2

⇒ 2b - 8 = -2 or 2b = -2 + 8 = 6 or b = 3

⇒ 2a - 2b = -2     (Eq 1)

Checking whether a = 2 and b = 3 satisfies Eq 1,

⇒ 2a - 2b = -2

L.H.S. = 2a - 2b = 2(2) - 2(3) = 4 - 6 = -2 = R.H.S.

∴ a = 2 and b = 3.

Hence, the values are a = 2 and b = 3.

(i) Given,

$$\text{A} = \begin{bmatrix*}[r] \text{sec 60°} & \text{cos 90°} \\ \text{-3 tan 45°} & \text{sin 90°} \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \\[1em] \text{B} = \begin{bmatrix*}[r] 0 & \text{cot 45°} \\ -2 & \text{3 sin 90°} \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*} \\[1em] \text{2A - 3B} = 2\begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} - 3 \begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 0 \\ -6 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 0 & 3 \\ -6 & 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 - 0 & 0 - 3 \\ -6 - (-6) & 2 - 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & -3 \\ 0 & -7 \end{bmatrix*}.$$

Hence, the value of 2A - 3B $$= \begin{bmatrix*}[r] 4 & -3 \\ 0 & -7 \end{bmatrix*}.$$

(ii) Given,

$$\text{A}^2 = \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 2 \times 2 + 0 \times (-3) & 2 \times 0 + 0 \times 1 \\ (-3) \times 2 + 1 \times (-3) & (-3) \times 0 + 1 \times 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 4 + 0 & 0 + 0 \\ -6 - 3 & 0 + 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 4 & 0 \\ -9 & 1 \end{bmatrix*}.$$

Hence, the value of $$\text{A}^2 = \begin{bmatrix*}[r] 4 & 0 \\ -9 & 1 \end{bmatrix*}.$$

(iii)

$$\text{BA } = \begin{bmatrix*}[r] 0 & 1 \\ -2 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 2 & 0 \\ -3 & 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] 0 \times 2 + 1 \times (-3) & 0 \times 0 + 1 \times 1 \\ -2 \times 2 + 3 \times (-3) & -2 \times 0 + 3 \times 1 \end{bmatrix*} \\[0.5em] = \begin{bmatrix*}[r] -3 & 1 \\ -13 & 3 \end{bmatrix*}.$$

Hence, the value of matrix BA = $$\begin{bmatrix*}[r] -3 & 1 \\ -13 & 3 \end{bmatrix*}.$$

Given,

X² = 4X + 5I

Solving for L.H.S.,

$$X^2 = \begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 1 \times 8 & 1 \times 1 + 1\times 3 \\ 8 \times 1 + 3 \times 8 & 8 \times 1 + 3 \times 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 8 & 1 + 3 \\ 8 + 24 & 8 + 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ 32 & 17 \end{bmatrix*}.$$

Solving for R.H.S.,

$$4X + 5I = 4\begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} + 5\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 4 \\ 32 & 12 \end{bmatrix*} + \begin{bmatrix*}[r] 5 & 0 \\ 0 & 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 5 & 4 + 0 \\ 32 + 0 & 12 + 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ 32 & 17 \end{bmatrix*}.$$

Since, L.H.S. = R.H.S.

Hence, proved that X² = 4X + 5I.