Matrices — Exercise 9(C)

If A is a matrix of order m × 3, B is a matrix of order 3 × 2 and R is a matrix of order 5 × n such that AB = R, the values of m and n are :

1. m = -5 and n = -2

2. m = 5 and n = 2

3. m = 5 and n = -2

4. m = 2 and n = 5

Answer

Given,

AB = R

A_{m × 3}B_{3 × 2} = R_{5 × n}

We know that,

For matrix multiplication :

No. of columns in A must be equal to the number of rows in B.

Resultant matrix order = No. of rows in A × No. of columns in B.

∴ m = 5 and n = 2.

Hence, Option 2 is the correct option.

If A = $\begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*}, B = \begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*}$ and A = B², the value of x is :

1. 38

2. -6

3. -36

4. 36

Answer

Substituting value of A and B in A = B², we get :

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

∴ x = 36.

Hence, Option 4 is the correct option.

A, B and C are three square matrices each of order 3; the order of matrix CA + B² is :

1. 3 × 1

2. 3 × 3

3. 1 × 3

4. 2 × 3

Answer

Let resultant matrix of CA be R.

C_{3 × 3} × A_{3 × 3} = R_{m × n}

We know that,

For matrix multiplication :

No. of columns in C must be equal to the number of rows in A.

Resultant matrix order = No. of rows in C × No. of columns in A.

∴ m = 3 and n = 3.

∴ Order of matrix R will be 3 × 3.

Let resultant matrix of B² be S.

B_{3 × 3} × B_{3 × 3} = S_{g × h}

Resultant matrix order = No. of rows in B × No. of columns in B.

∴ g = 3 and h = 3.

∴ Order of matrix S will be 3 × 3.

⇒ CA + B²

⇒ R + S

Addition of matrix possible between matrix of same order and same is the order of resultant matrix.

Hence, resultant matrix's order = 3 × 3.

Hence, Option 2 is the correct option.

If A = $\begin{bmatrix*}[r] 5 & -2 \\ 7 & 0 \end{bmatrix*} \text{ and B} = \begin{bmatrix*}[r] 8 \\ 3 \end{bmatrix*}$, then which of the following is not possible ?

1. A²

2. AB

3. BA

4. 15A

Answer

BA is not possible because no. of columns in B (1) is not equal to the no. of rows in A (2).

Hence, Option 3 is the correct option.

If A = $\begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*}, B = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}\text{ and C} = \begin{bmatrix*}[r] 1 & 1 \\ 0 & 0 \end{bmatrix*}$, the matrix A² + 2B - 3C is :

1. $\begin{bmatrix*}[r] -2 & -1 \\ 4 & 1 \end{bmatrix*}$

2. $\begin{bmatrix*}[r] 2 & -1 \\ 4 & 1 \end{bmatrix*}$

3. $\begin{bmatrix*}[r] 2 & 1 \\ 4 & 1 \end{bmatrix*}$

4. $\begin{bmatrix*}[r] 2 & 1 \\ -4 & -1 \end{bmatrix*}$

Answer

Substituting values of A, B and C in A² + 2B - 3C, we get :

$\Rightarrow A^2 + 2B - 3C = \begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*} + 2\begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*} - 3\begin{bmatrix*}[r] 1 & 1 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 1 & 1 \times 0 + 0 \times 1 \\ 1 \times 1 + 1 \times 1 & 1 \times 0 + 1 \times 1 \end{bmatrix*} + \begin{bmatrix*}[r] 0 & 2 \\ 2 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 3 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 0 & 0 + 0 \\ 1 + 1 & 0 + 1 \end{bmatrix*} + \begin{bmatrix*}[r] 0 & 2 \\ 2 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 3 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 0 & 2 \\ 2 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 3 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 0 - 3 & 0 + 2 - 3 \\ 2 + 2 - 0 & 1 + 0 - 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 & -1 \\ 4 & 1 \end{bmatrix*}$

Hence, Option 1 is the correct option.

Evaluate : if possible :

$\begin{bmatrix*}[r] 3 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 0 \end{bmatrix*}$

Answer

$\begin{bmatrix*}[r] 3 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 0 \end{bmatrix*}$

For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.

Hence, matrix multiplication is possible.

$\Rightarrow \begin{bmatrix*}[r] 3 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 3 \times 2 + 2 \times 0 \\ \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 \\ \end{bmatrix*}.$

Hence, $\begin{bmatrix*}[r] 3 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 6 \end{bmatrix*}.$

Evaluate : if possible :

$\begin{bmatrix*}[r] 1 & -2 \\ \end{bmatrix*}\begin{bmatrix*}[r] -2 & 3 \\ -1 & 4 \end{bmatrix*}$

Answer

$\begin{bmatrix*}[r] 1 & -2 \\ \end{bmatrix*}\begin{bmatrix*}[r] -2 & 3 \\ -1 & 4 \end{bmatrix*}$

For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.

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

Hence, $\begin{bmatrix*}[r] 1 & -2 \\ \end{bmatrix*}\begin{bmatrix*}[r] -2 & 3 \\ -1 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -5 \\ \end{bmatrix*}.$

Evaluate : if possible :

$\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 3 \end{bmatrix*}$

Answer

$\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 3 \end{bmatrix*}$

For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.

$\Rightarrow \begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 6 \times (-1) + 4 \times 3 \\ 3 \times (-1) + (-1) \times 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + 12 \\ -3 + (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 \\ -6 \end{bmatrix*}.$

Hence, $\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 6 \\ -6 \end{bmatrix*}.$

Evaluate : if possible :

$\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 & 3 \\ \end{bmatrix*}$

Answer

$\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 & 3 \\ \end{bmatrix*}$

For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.

The above matrix multiplication is not possible as the no. of columns in first matrix is not equal to no. of rows in the second matrix.

If A = $\begin{bmatrix*}[r] 0 & 2 \\ 5 & -2 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & -1 \\ 3 & 2 \end{bmatrix*}$ and I is a unit matrix of order 2 × 2, find :

(i) AB

(ii) BA

(iii) AI

Answer

(i) Substituting value in AB,

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

Hence, AB = $\begin{bmatrix*}[r] 6 & 4 \\ -1 & -9 \end{bmatrix*}.$

(ii) Substituting value in BA,

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

Hence, BA = $\begin{bmatrix*}[r] -5 & 4 \\ 10 & 2 \end{bmatrix*}.$

(iii) Substituting value in AI,

$\Rightarrow AI = \begin{bmatrix*}[r] 0 & 2 \\ 5 & -2 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 \times 1 + 2 \times 0 & 0 \times 0 + 2 \times 1 \\ 5 \times 1 + (-2) \times 0 & 5 \times 0 + (-2) \times 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 2 \\ 5 & -2 \end{bmatrix*} = A$

Hence, AI = matrix A.

If A = $\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} \text{ and B }= \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*}$, find x and y when A² = B.

Answer

Given,

$\Rightarrow A^2 = B \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times 3 + x \times 0 & 3 \times x + x \times 1 \\ 0 \times 3 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 + 0 & 3x + x \\ 0 + 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 4x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*}$

By definition of equality of matrices we get,

4x = 16
⇒ x = 4.

1 = -y
⇒ y = -1.

Hence, x = 4 and y = -1.

Find x and y, if :

$\begin{bmatrix*}[r] x & 0 \\ -3 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 1 \\ 0 & y \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 2 \\ -3 & -2 \end{bmatrix*}$

Answer

Given,

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

By definition of equality of matrices we get,

x = 2

-3 + y = -2
⇒ y = -2 + 3
⇒ y = 1.

Hence, x = 2 and y = 1.

If A = $\begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & 2 \\ 4 & 3 \end{bmatrix*} \text{ and } C = \begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*}$, find :

(i) (AB)C

(ii) A(BC)

Is A(BC) = (AB)C ?

Answer

(i) Substituting value of A, B and C in (AB)C we get,

$\Rightarrow \Big(\begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 2 \\ 4 & 3 \end{bmatrix*}\Big)\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*} \\[1em] = \Big(\begin{bmatrix*}[r] 1 \times 1 + 3 \times 4 & 1 \times 2 + 3 \times 3 \\ 2 \times 1 + 4 \times 4 & 2 \times 2 + 4 \times 3 \end{bmatrix*}\Big)\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*} \\[1em] = \Big(\begin{bmatrix*}[r] 1 + 12 & 2 + 9 \\ 2 + 16 & 4 + 12 \end{bmatrix*}\Big)\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 13 & 11 \\ 18 & 16 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 13 \times 4 + 11 \times 1 & 13 \times 3 + 11 \times 2 \\ 18 \times 4 + 16 \times 1 & 18 \times 3 + 16 \times 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 52 + 11 & 39 + 22 \\ 72 + 16 & 54 + 32 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 63 & 61 \\ 88 & 86 \end{bmatrix*}.$

Hence, (AB)C = $\begin{bmatrix*}[r] 63 & 61 \\ 88 & 86 \end{bmatrix*}.$

(ii) Substituting value of A, B and C in A(BC) we get,

$\Rightarrow \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\Big(\begin{bmatrix*}[r] 1 & 2 \\ 4 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*}\Big) \\[1em] = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\Big(\begin{bmatrix*}[r] 1 \times 4 + 2\times 1 & 1 \times 3 + 2 \times 2 \\ 4 \times 4 + 3 \times 1 & 4 \times 3 + 3 \times 2 \end{bmatrix*}\Big) \\[1em] = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 4 + 2 & 3 + 4 \\ 16 + 3 & 12 + 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 6 & 7 \\ 19 & 18 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 6 + 3 \times 19 & 1 \times 7 + 3 \times 18 \\ 2 \times 6 + 4 \times 19 & 2 \times 7 + 4 \times 18 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 + 57 & 7 + 54 \\ 12 + 76 & 14 + 72 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 63 & 61 \\ 88 & 86 \end{bmatrix*}$

Hence, A(BC) = $\begin{bmatrix*}[r] 63 & 61 \\ 88 & 86 \end{bmatrix*}$.

Hence, A(BC) = (AB)C.

Let A = $\begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*}, B = \begin{bmatrix*}[r] 4 & 1 \\ -3 & -2 \end{bmatrix*} \text{ and C }= \begin{bmatrix*}[r] -3 & 2 \\ -1 & 4 \end{bmatrix*}$. Find A² + AC - 5B.

Answer

Substituting values of A, B and C in A² + AC - 5B we get,

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

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

If M = $\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*}$ and I is a unit matrix of the same order as that of M; show that :

M² = 2M + 3I

Answer

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

M = $\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*}$

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

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

Hence, proved that M² = 2M + 3I.

If A = $\begin{bmatrix*}[r] a & 0 \\ 0 & 2 \end{bmatrix*}, B = \begin{bmatrix*}[r] 0 & -b \\ 1 & 0 \end{bmatrix*}, M = \begin{bmatrix*}[r] 1 & -1 \\ 1 & 1 \end{bmatrix*}$ and BA = M², find the values of a and b.

Answer

Given,

$\Rightarrow BA = M^2 \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -b \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] a & 0 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & -1 \\ 1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -1 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 \times a + (-b) \times 0 & 0 \times 0 + (-b) \times 2 \\ 1 \times a + 0 \times 0 & 1 \times 0 + 0 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 \times 1 + (-1) \times 1 & 1 \times (-1) + (-1) \times 1 \\ 1 \times 1 + 1 \times 1 & 1 \times (-1) + 1 \times 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & 0 - 2b \\ a + 0 & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 1 - 1 & -1 - 1 \\ 1 + 1 & -1 + 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -2b \\ a & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -2 \\ 2 & 0 \end{bmatrix*}$

By definition of equality of matrices we get,

-2b = -2
⇒ b = 1.

a = 2.

Hence, a = 2 and b = 1.

Find the matrix A, if B = $\begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} \text{ and } B^2 = B + \dfrac{1}{2}A$.

Answer

Given,

$\Rightarrow B^2 = B + \dfrac{1}{2}A$

Substituting value of B in above equation we get,

$\Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} + \dfrac{1}{2}A \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + 1 \times 0 & 2 \times 1 + 1 \times 1\ 0 \times 2 + 1 \times 0 & 0 \times 1 + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} + \dfrac{1}{2}A \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 0 & 2 + 1 \\ 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} + \dfrac{1}{2}A \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 3 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} + \dfrac{1}{2}A \\[1em] \Rightarrow \dfrac{1}{2}A = \begin{bmatrix*}[r] 4 & 3 \\ 0 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \dfrac{1}{2}A = \begin{bmatrix*}[r] 4 - 2 & 3 - 1 \\ 0 - 0 & 1 - 1 \end{bmatrix*} \\[1em] \Rightarrow \dfrac{1}{2}A = \begin{bmatrix*}[r] 2 & 2 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow A = 2\begin{bmatrix*}[r] 2 & 2 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 4 & 4 \\ 0 & 0 \end{bmatrix*}.$

Hence, A = $\begin{bmatrix*}[r] 4 & 4 \\ 0 & 0 \end{bmatrix*}.$

If A = $\begin{bmatrix*}[r] -1 & 1 \\ a & b \end{bmatrix*}$ and A² = I, find a and b.

Answer

Given, A² = I

$\therefore \begin{bmatrix*}[r] -1 & 1 \\ a & b \end{bmatrix*}\begin{bmatrix*}[r] -1 & 1 \\ a & b \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -1 \times -1 + 1 \times a & -1 \times 1 + 1 \times b \\ a \times (-1) + b \times a & a \times 1 + b \times b \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 + a & -1 + b \\ -a + ab & a + b^2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$

By definition of equality of matrices we get,

1 + a = 1
⇒ a = 0.

-1 + b = 0
⇒ b = 1.

Hence, a = 0 and b = 1

Solve for x and y :

$\begin{bmatrix*}[r] 2 & 5 \\ 5 & 2 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -7 \\ 14 \end{bmatrix*}$

Answer

Given,

$\Rightarrow \begin{bmatrix*}[r] 2 & 5 \\ 5 & 2 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -7 \\ 14 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times x + 5 \times y \\ 5 \times x + 2 \times y \end{bmatrix*} = \begin{bmatrix*}[r] -7 \\ 14 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 5y \\ 5x + 2y \end{bmatrix*} = \begin{bmatrix*}[r] -7 \\ 14 \end{bmatrix*}$

By definition of equality of matrices we get,

2x + 5y = -7

⇒ 2x = -(7 + 5y)

⇒ x = $\dfrac{-(7 + 5y)}{2}$ ......(i)

5x + 2y = 14

Substituting value of x from (i) in above equation we get,

$\Rightarrow 5 \times \dfrac{-(7 + 5y)}{2} + 2y = 14 \\[1em] \Rightarrow \dfrac{-35 - 25y}{2} + 2y = 14 \\[1em] \Rightarrow \dfrac{-35 - 25y + 4y}{2} = 14 \\[1em] \Rightarrow -35 - 21y = 28 \\[1em] \Rightarrow -21y = 28 + 35 \\[1em] \Rightarrow -21y = 63 \\[1em] \Rightarrow y = -3.$

Substituting y = -3 in (i) we get,

$x = \dfrac{-(7 + 5(-3))}{2} = \dfrac{-(7 - 15)}{2} = \dfrac{8}{2} = 4$.

Hence, x = 4 and y = -3.

Solve for x and y :

$\begin{bmatrix*}[r] x + y & x - 4 \end{bmatrix*}\begin{bmatrix*}[r] -1 & -2 \\ 2 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*}$

Answer

Given,

$\Rightarrow \begin{bmatrix*}[r] x + y & x - 4 \end{bmatrix*}\begin{bmatrix*}[r] -1 & -2 \\ 2 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] (x + y) \times -1 + (x - 4) \times 2 & (x + y) \times -2 + (x - 4) \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x - y + 2x - 8 & -2x - 2y + 2x - 8 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x - y - 8 & - 2y - 8 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*}$

By definition of equality of matrices we get,

-2y - 8 = -11

⇒ -2y = -3

⇒ y = $\dfrac{3}{2}$.

x - y - 8 = -7

⇒ x - y = 1

⇒ x = 1 + y

⇒ x = $1 + \dfrac{3}{2} = \dfrac{5}{2}$.

Hence, $x = \dfrac{5}{2}, y = \dfrac{3}{2}.$

Solve for x and y :

$\begin{bmatrix*}[r] -2 & 0 \\ 3 & 1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 2x \end{bmatrix*} + 3\begin{bmatrix*}[r] -2 \\ 1 \end{bmatrix*} = 2\begin{bmatrix*}[r] y \\ 3 \end{bmatrix*}$.

Answer

Given,

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

By definition of equality of matrices we get,

2y = -4
⇒ y = -2.

2x = 6
⇒ x = 3.

Hence, x = 3 and y = -2.

In each case given below, find :

(a) the order of matrix M.

(b) the matrix M.

(i) $M \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} =\begin{bmatrix*}[r] 1 & 2 \end{bmatrix*}$

(ii) $\begin{bmatrix*}[r] 1 & 4 \\ 2 & 1 \end{bmatrix*} \times M = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*}$

Answer

(i) Let order of matrix M be a × b.

$M_{a \times b} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*}_{2 \times 2} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*}_{1 \times 2}$

Since, the product of matrices is possible, only when the number of columns in the first matrix is equal to the number of rows in the second.

∴ b = 2

Also, the no. of rows of product (resulting) matrix is equal to no. of rows of first matrix.

∴ a = 1

Order of matrix M = a × b = 1 × 2.

Let M = $\begin{bmatrix*}[r] x & y \end{bmatrix*}$.

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

By definition of equality of matrices we get,

x = 1

x + 2y = 2
⇒ 1 + 2y = 2
⇒ 2y = 1
⇒ y = $\dfrac{1}{2}$.

∴ M = $\begin{bmatrix*}[r] x & y \end{bmatrix*} = \begin{bmatrix*}[r] 1 & \dfrac{1}{2} \end{bmatrix*}$.

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

(ii) Let order of matrix M be a × b.

i.e. $\begin{bmatrix*}[r] 1 & 4 \\ 2 & 1 \end{bmatrix*}_{2 \times 2} \times M_{a \times b} = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*}_{2 \times 1}$

Since product of matrix is possible, only when the number of columns in the first matrix is equal to no. of rows in second.

∴ a = 2.

Also the no. of columns of product (resulting matrix) is equal to no. of columns of second matrix.

∴ b = 1.

Hence, order of matrix = 2 × 1.

Let M = $\begin{bmatrix*}[r] x \\ y \end{bmatrix*}$

$\Rightarrow \begin{bmatrix*}[r] 1 & 4 \\ 2 & 1 \end{bmatrix*} \times \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 \times x + 4 \times y \\ 2 \times x + 1 \times y \end{bmatrix*} = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x + 4y \\ 2x + y \end{bmatrix*} = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*}$

By definition of equality of matrices we get,

x + 4y = 13
⇒ x = 13 - 4y ......(i)

2x + y = 5
⇒ 2(13 - 4y) + y = 5
⇒ 26 - 8y + y = 5
⇒ -7y = -21
⇒ y = 3.

⇒ x = 13 - 4y = 13 - 4(3) = 13 - 12 = 1.

$\therefore M = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 1 \\ 3 \end{bmatrix*}.$

Hence, $M = \begin{bmatrix*}[r] 1 \\ 3 \end{bmatrix*}.$

If A = $\begin{bmatrix*}[r] 2 & x \\ 0 & 1 \end{bmatrix*} \text{and } B = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*};$ find the value of x, given that : A² = B.

Answer

Given,

⇒ A² = B

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

By definition of equality of matrices we get,

3x = 36
⇒ x = 12.

Hence, x = 12.

If A and B are any two 2 × 2 matrices such that AB = BA = B and B is not a zero matrix, what can you say about the matrix A?

Answer

Since, AB = BA = B and B is not a zero matrix.

It means that A is a unit or identity matrix.

Given A = $\begin{bmatrix*}[r] 3 & 0 \\ 0 & 4 \end{bmatrix*}, B = \begin{bmatrix*}[r] a & b \\ 0 & c \end{bmatrix*}$ and that AB = A + B; find the values of a, b and c.

Answer

Given,

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