Exercise 2.4 [Pages 59 - 60] Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 2 Matrices

MATHEMATICS

Balbharati Solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board

Chapter 2: Matrices - Exercise 2.4 [Pages 59 - 60]

Exercise 2.4 | Q 1.1 | Page 59

QUESTION

Find $A^T$, if $A = \begin{bmatrix} 1 & 3 \\ -4 & 5 \end{bmatrix}$

SOLUTION

$A = \begin{bmatrix} 1 & 3 \\ -4 & 5 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 1 & -4 \\ 3 & 5 \end{bmatrix}$.

Exercise 2.4 | Q 1.2 | Page 59

QUESTION

Find $A^T$, if $A = \begin{bmatrix} 2 & -6 & 1 \\ -4 & 0 & 5 \end{bmatrix}$

SOLUTION

$A = \begin{bmatrix} 2 & -6 & 1 \\ -4 & 0 & 5 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 2 & -4 \\ -6 & 0 \\ 1 & 5 \end{bmatrix}$.

Exercise 2.4 | Q 2 | Page 59

QUESTION

If $A = [a_{ij}]_{3 \times 3}$, where $a_{ij} = 2(i - j)$, find A and $A^T$. State whether A and $A^T$ both are symmetric or skew-symmetric matrices?

SOLUTION

$A = [a_{ij}]_{3 \times 3} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$

Given, $a_{ij} = 2(i - j)$
$\therefore a_{11} = 2(1 - 1) = 0, \quad a_{12} = 2(1 - 2) = -2$
$a_{13} = 2(1 - 3) = -4, \quad a_{21} = 2(2 - 1) = 2,$
$a_{22} = 2(2 - 2) = 0, \quad a_{23} = 2(2 - 3) = -2,$
$a_{31} = 2(3 - 1) = 4, \quad a_{32} = 2(3 - 2) = 2,$
$a_{33} = 2(3 - 3) = 0$

$\therefore A = \begin{bmatrix} 0 & -2 & -4 \\ 2 & 0 & -2 \\ 4 & 2 & 0 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 0 & 2 & 4 \\ -2 & 0 & 2 \\ -4 & -2 & 0 \end{bmatrix}$

$= -\begin{bmatrix} 0 & -2 & -4 \\ 2 & 0 & -2 \\ 4 & 2 & 0 \end{bmatrix} = -A$

$\therefore A^T = -A$ and $A = -A^T$
$\therefore A$ and $A^T$ both are skew-symmetric matrices.

Exercise 2.4 | Q 3 | Page 59

QUESTION

If $A = \begin{bmatrix} 5 & -3 \\ 4 & -3 \\ -2 & 1 \end{bmatrix}$, prove that $(A^T)^T = A$.

SOLUTION

$A = \begin{bmatrix} 5 & -3 \\ 4 & -3 \\ -2 & 1 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 5 & 4 & -2 \\ -3 & -3 & 1 \end{bmatrix}$

$\therefore (A^T)^T = \begin{bmatrix} 5 & -3 \\ 4 & -3 \\ -2 & 1 \end{bmatrix}$
$= A$.

Exercise 2.4 | Q 4 | Page 59

QUESTION

If $A = \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix}$, prove that $A^T = A$.

SOLUTION

$A = \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9 \end{bmatrix}$
$=A$.

Exercise 2.4 | Q 5.1 | Page 59

QUESTION

If $A = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 1 \\ 4 & -1 \\ -3 & 3 \end{bmatrix}$, $C = \begin{bmatrix} 1 & 2 \\ -1 & 4 \\ -2 & 3 \end{bmatrix}$, then show that $(A + B)^T = A^T + B^T$.

SOLUTION

$A+B = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} + \begin{bmatrix} 2 & 1 \\ 4 & -1 \\ -3 & 3 \end{bmatrix}$

$= \begin{bmatrix} 2+2 & -3+1 \\ 5+4 & -4-1 \\ -6-3 & 1+3 \end{bmatrix}$

$= \begin{bmatrix} 4 & -2 \\ 9 & -5 \\ -9 & 4 \end{bmatrix}$

$\therefore (A + B)^T = \begin{bmatrix} 4 & 9 & -9 \\ -2 & -5 & 4 \end{bmatrix} \quad \dots(i)$

Now, $A^T = \begin{bmatrix} 2 & 5 & -6 \\ -3 & -4 & 1 \end{bmatrix}$ and $B^T = \begin{bmatrix} 2 & 4 & -3 \\ 1 & -1 & 3 \end{bmatrix}$

$\therefore A^T + B^T = \begin{bmatrix} 2 & 5 & -6 \\ -3 & -4 & 1 \end{bmatrix} + \begin{bmatrix} 2 & 4 & -3 \\ 1 & -1 & 3 \end{bmatrix}$

$= \begin{bmatrix} 2+2 & 5+4 & -6-3 \\ -3+1 & -4-1 & 1+3 \end{bmatrix}$

$= \begin{bmatrix} 4 & 9 & -9 \\ -2 & -5 & 4 \end{bmatrix} \quad \dots(ii)$
From (i) and (ii), we get
$(A + B)^T = A^T + B^T$.

Exercise 2.4 | Q 5.2 | Page 59

QUESTION

If $A = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 1 \\ 4 & -1 \\ -3 & 3 \end{bmatrix}$, $C = \begin{bmatrix} 1 & 2 \\ -1 & 4 \\ -2 & 3 \end{bmatrix}$, then show that $(A - C)^T = A^T - C^T$.

SOLUTION

$A - C = \begin{bmatrix} 2 & -3 \\ 5 & -4 \\ -6 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ -1 & 4 \\ -2 & 3 \end{bmatrix}$

$= \begin{bmatrix} 2-1 & -3-2 \\ 5-(-1) & -4-4 \\ -6-(-2) & 1-3 \end{bmatrix} = \begin{bmatrix} 2-1 & -3-2 \\ 5+1 & -4-4 \\ -6+2 & 1-3 \end{bmatrix}$

$= \begin{bmatrix} 1 & -5 \\ 6 & -8 \\ -4 & -2 \end{bmatrix}$

$\therefore (A - C)^T = \begin{bmatrix} 1 & 6 & -4 \\ -5 & -8 & -2 \end{bmatrix} \quad \dots(i)$

Now, $A^T = \begin{bmatrix} 2 & 5 & -6 \\ -3 & -4 & 1 \end{bmatrix}$ and $C^T = \begin{bmatrix} 1 & -1 & -2 \\ 2 & 4 & 3 \end{bmatrix}$

$\therefore A^T - C^T = \begin{bmatrix} 2 & 5 & -6 \\ -3 & -4 & 1 \end{bmatrix} - \begin{bmatrix} 1 & -1 & -2 \\ 2 & 4 & 3 \end{bmatrix}$

$= \begin{bmatrix} 2-1 & 5-(-1) & -6-(-2) \\ -3-2 & -4-4 & 1-3 \end{bmatrix} = \begin{bmatrix} 2-1 & 5+1 & -6+2 \\ -3-2 & -4-4 & 1-3 \end{bmatrix}$

$= \begin{bmatrix} 1 & 6 & -4 \\ -5 & -8 & -2 \end{bmatrix} \quad \dots(ii)$

From (i) and (ii), we get
$(A - C)^T = A^T - C^T$.

Exercise 2.4 | Q 6 | Page 59

QUESTION

If $A = \begin{bmatrix} 5 & 4 \\ -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 3 \\ 4 & -1 \end{bmatrix}$, then find $C^T$, such that $3A - 2B + C = I$, where I is the unit matrix of order 2.

SOLUTION

$3A - 2B + C = I$
$\therefore C = I + 2B - 3A$

$\therefore C = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + 2\begin{bmatrix} -1 & 3 \\ 4 & -1 \end{bmatrix} - 3\begin{bmatrix} 5 & 4 \\ -2 & 3 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} -2 & 6 \\ 8 & -2 \end{bmatrix} - \begin{bmatrix} 15 & 12 \\ -6 & 9 \end{bmatrix}$

$= \begin{bmatrix} 1-2-15 & 0+6-12 \\ 0+8-(-6) & 1-2-9 \end{bmatrix} = \begin{bmatrix} 1-2-15 & 0+6-12 \\ 0+8+6 & 1-2-9 \end{bmatrix}$

$\therefore C = \begin{bmatrix} -16 & -6 \\ 14 & -10 \end{bmatrix}$

$\therefore C^T = \begin{bmatrix} -16 & 14 \\ -6 & -10 \end{bmatrix}$.

Exercise 2.4 | Q 7.1 | Page 59

QUESTION

If $A = \begin{bmatrix} 7 & 3 & 0 \\ 0 & 4 & -2 \end{bmatrix}$, $B = \begin{bmatrix} 0 & -2 & 3 \\ 2 & 1 & -4 \end{bmatrix}$, then find $A^T + 4B^T$.

SOLUTION

$A = \begin{bmatrix} 7 & 3 & 0 \\ 0 & 4 & -2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -2 & 3 \\ 2 & 1 & -4 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix}$ and $B^T = \begin{bmatrix} 0 & 2 \\ -2 & 1 \\ 3 & -4 \end{bmatrix}$

$A^T + 4B^T = \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix} + 4\begin{bmatrix} 0 & 2 \\ -2 & 1 \\ 3 & -4 \end{bmatrix}$

$= \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix} + \begin{bmatrix} 0 & 8 \\ -8 & 4 \\ 12 & -16 \end{bmatrix}$

$= \begin{bmatrix} 7+0 & 0+8 \\ 3-8 & 4+4 \\ 0+12 & -2-16 \end{bmatrix}$

$= \begin{bmatrix} 7 & 8 \\ -5 & 8 \\ 12 & -18 \end{bmatrix}$.

Exercise 2.4 | Q 7.2 | Page 59

QUESTION

If $A = \begin{bmatrix} 7 & 3 & 0 \\ 0 & 4 & -2 \end{bmatrix}$, $B = \begin{bmatrix} 0 & -2 & 3 \\ 2 & 1 & -4 \end{bmatrix}$, then find $5A^T - 5B^T$.

SOLUTION

$A = \begin{bmatrix} 7 & 3 & 0 \\ 0 & 4 & -2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -2 & 3 \\ 2 & 1 & -4 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix}$ and $B^T = \begin{bmatrix} 0 & 2 \\ -2 & 1 \\ 3 & -4 \end{bmatrix}$

$5A^T - 5B^T = 5(A^T - B^T)$

$= 5\left( \begin{bmatrix} 7 & 0 \\ 3 & 4 \\ 0 & -2 \end{bmatrix} - \begin{bmatrix} 0 & 2 \\ -2 & 1 \\ 3 & -4 \end{bmatrix} \right)$

$= 5\begin{bmatrix} 7-0 & 0-2 \\ 3-(-2) & 4-1 \\ 0-3 & -2-(-4) \end{bmatrix} = 5\begin{bmatrix} 7 & -2 \\ 3+2 & 3 \\ -3 & -2+4 \end{bmatrix}$

$= 5\begin{bmatrix} 7 & -2 \\ 5 & 3 \\ -3 & 2 \end{bmatrix}$

$= \begin{bmatrix} 35 & -10 \\ 25 & 15 \\ -15 & 10 \end{bmatrix}$.

Exercise 2.4 | Q 8 | Page 59

QUESTION

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 3 & 1 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 2 & 1 & -4 \\ 3 & 5 & -2 \end{bmatrix}$ and $C = \begin{bmatrix} 0 & 2 & 3 \\ -1 & -1 & 0 \end{bmatrix}$, verify that $(A + 2B + 3C)^T = A^T + 2B^T + 3C^T$.

SOLUTION

$A + 2B + 3C$

$= \begin{bmatrix} 1 & 0 & 1 \\ 3 & 1 & 2 \end{bmatrix} + 2\begin{bmatrix} 2 & 1 & -4 \\ 3 & 5 & -2 \end{bmatrix} + 3\begin{bmatrix} 0 & 2 & 3 \\ -1 & -1 & 0 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 & 1 \\ 3 & 1 & 2 \end{bmatrix} + \begin{bmatrix} 4 & 2 & -8 \\ 6 & 10 & -4 \end{bmatrix} + \begin{bmatrix} 0 & 6 & 9 \\ -3 & -3 & 0 \end{bmatrix}$

$= \begin{bmatrix} 1+4+0 & 0+2+6 & 1-8+9 \\ 3+6-3 & 1+10-3 & 2-4+0 \end{bmatrix}$

$\therefore A + 2B + 3C = \begin{bmatrix} 5 & 8 & 2 \\ 6 & 8 & -2 \end{bmatrix}$

$\therefore (A + 2B + 3C)^T = \begin{bmatrix} 5 & 6 \\ 8 & 8 \\ 2 & -2 \end{bmatrix} \quad \dots(i)$

Now, $A^T = \begin{bmatrix} 1 & 3 \\ 0 & 1 \\ 1 & 2 \end{bmatrix}$, $B^T = \begin{bmatrix} 2 & 3 \\ 1 & 5 \\ -4 & -2 \end{bmatrix}$ and $C^T = \begin{bmatrix} 0 & -1 \\ 2 & -1 \\ 3 & 0 \end{bmatrix}$

$\therefore A^T + 2B^T + 3C^T$

$= \begin{bmatrix} 1 & 3 \\ 0 & 1 \\ 1 & 2 \end{bmatrix} + 2\begin{bmatrix} 2 & 3 \\ 1 & 5 \\ -4 & -2 \end{bmatrix} + 3\begin{bmatrix} 0 & -1 \\ 2 & -1 \\ 3 & 0 \end{bmatrix}$

$= \begin{bmatrix} 1 & 3 \\ 0 & 1 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} 4 & 6 \\ 2 & 10 \\ -8 & -4 \end{bmatrix} + \begin{bmatrix} 0 & -3 \\ 6 & -3 \\ 9 & 0 \end{bmatrix}$

$= \begin{bmatrix} 1+4+0 & 3+6-3 \\ 0+2+6 & 1+10-3 \\ 1-8+9 & 2-4+0 \end{bmatrix}$

$\therefore A^T + 2B^T + 3C^T = \begin{bmatrix} 5 & 6 \\ 8 & 8 \\ 2 & -2 \end{bmatrix} \quad \dots(ii)$

From (i) and (ii), we get
$(A + 2B + 3C)^T = A^T + 2B^T + 3C^T$.

Exercise 2.4 | Q 9 | Page 59

QUESTION

If $A = \begin{bmatrix} -1 & 2 & 1 \\ -3 & 2 & -3 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ -3 & 2 \\ -1 & 3 \end{bmatrix}$, prove that $(A + B^T)^T = A^T + B$.

SOLUTION

$A = \begin{bmatrix} -1 & 2 & 1 \\ -3 & 2 & -3 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ -3 & 2 \\ -1 & 3 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} -1 & -3 \\ 2 & 2 \\ 1 & -3 \end{bmatrix}$ and $B^T = \begin{bmatrix} 2 & -3 & -1 \\ 1 & 2 & 3 \end{bmatrix}$

$\therefore A + B^T = \begin{bmatrix} -1 & 2 & 1 \\ -3 & 2 & -3 \end{bmatrix} + \begin{bmatrix} 2 & -3 & -1 \\ 1 & 2 & 3 \end{bmatrix}$

$= \begin{bmatrix} -1+2 & 2-3 & 1-1 \\ -3+1 & 2+2 & -3+3 \end{bmatrix}$

$= \begin{bmatrix} 1 & -1 & 0 \\ -2 & 4 & 0 \end{bmatrix}$

$\therefore (A + B^T)^T = \begin{bmatrix} 1 & -2 \\ -1 & 4 \\ 0 & 0 \end{bmatrix} \quad \dots(i)$

Now, $A^T + B = \begin{bmatrix} -1 & -3 \\ 2 & 2 \\ 1 & -3 \end{bmatrix} + \begin{bmatrix} 2 & 1 \\ -3 & 2 \\ -1 & 3 \end{bmatrix}$

$= \begin{bmatrix} -1+2 & -3+1 \\ 2-3 & 2+2 \\ 1-1 & -3+3 \end{bmatrix}$

$= \begin{bmatrix} 1 & -2 \\ -1 & 4 \\ 0 & 0 \end{bmatrix} \quad \dots(ii)$
From (i) and (ii), we get
$(A + B^T)^T = A^T + B$.

Exercise 2.4 | Q 10.1 | Page 59

QUESTION

Prove that $A + A^T$ is a symmetric and $A - A^T$ is a skew symmetric matrix, where $A = \begin{bmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ -2 & -3 & 2 \end{bmatrix}$

SOLUTION

$A = \begin{bmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ -2 & -3 & 2 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 1 & 3 & -2 \\ 2 & 2 & -3 \\ 4 & 1 & 2 \end{bmatrix}$

$\therefore A + A^T = \begin{bmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ -2 & -3 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3 & -2 \\ 2 & 2 & -3 \\ 4 & 1 & 2 \end{bmatrix}$

$= \begin{bmatrix} 1+1 & 2+3 & 4-2 \\ 3+2 & 2+2 & 1-3 \\ -2+4 & -3+1 & 2+2 \end{bmatrix}$

$\therefore A + A^T = \begin{bmatrix} 2 & 5 & 2 \\ 5 & 4 & -2 \\ 2 & -2 & 4 \end{bmatrix}$

Let $P = A + A^T$. Then $P^T = (A + A^T)^T = \begin{bmatrix} 2 & 5 & 2 \\ 5 & 4 & -2 \\ 2 & -2 & 4 \end{bmatrix}^T = \begin{bmatrix} 2 & 5 & 2 \\ 5 & 4 & -2 \\ 2 & -2 & 4 \end{bmatrix} = P$.
Since $P^T = P$, $A + A^T$ is a symmetric matrix.

$A - A^T = \begin{bmatrix} 1 & 2 & 4 \\ 3 & 2 & 1 \\ -2 & -3 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 3 & -2 \\ 2 & 2 & -3 \\ 4 & 1 & 2 \end{bmatrix}$

$= \begin{bmatrix} 1-1 & 2-3 & 4-(-2) \\ 3-2 & 2-2 & 1-(-3) \\ -2-4 & -3-1 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 6 \\ 1 & 0 & 4 \\ -6 & -4 & 0 \end{bmatrix}$

Let $Q = A - A^T$. Then $Q^T = (A - A^T)^T = \begin{bmatrix} 0 & -1 & 6 \\ 1 & 0 & 4 \\ -6 & -4 & 0 \end{bmatrix}^T = \begin{bmatrix} 0 & 1 & -6 \\ -1 & 0 & -4 \\ 6 & 4 & 0 \end{bmatrix}$
$= -\begin{bmatrix} 0 & -1 & 6 \\ 1 & 0 & 4 \\ -6 & -4 & 0 \end{bmatrix} = -Q$.
Since $Q^T = -Q$, $A - A^T$ is a skew symmetric matrix.

Exercise 2.4 | Q 10.2 | Page 59

QUESTION

Prove that $A + A^T$ is a symmetric and $A - A^T$ is a skew symmetric matrix, where $A = \begin{bmatrix} 5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3 \end{bmatrix}$

SOLUTION

$A = \begin{bmatrix} 5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 5 & 3 & 4 \\ 2 & -7 & -5 \\ -4 & 2 & -3 \end{bmatrix}$

$\therefore A + A^T = \begin{bmatrix} 5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3 \end{bmatrix} + \begin{bmatrix} 5 & 3 & 4 \\ 2 & -7 & -5 \\ -4 & 2 & -3 \end{bmatrix}$

$= \begin{bmatrix} 5+5 & 2+3 & -4+4 \\ 3+2 & -7-7 & 2-5 \\ 4-4 & -5+2 & -3-3 \end{bmatrix}$

$= \begin{bmatrix} 10 & 5 & 0 \\ 5 & -14 & -3 \\ 0 & -3 & -6 \end{bmatrix}$

Let $P = A+A^T$. Then $P^T = (A+A^T)^T = \begin{bmatrix} 10 & 5 & 0 \\ 5 & -14 & -3 \\ 0 & -3 & -6 \end{bmatrix}^T = \begin{bmatrix} 10 & 5 & 0 \\ 5 & -14 & -3 \\ 0 & -3 & -6 \end{bmatrix} = P$.
Since $P^T=P$, $A + A^T$ is a symmetric matrix.

$A - A^T = \begin{bmatrix} 5 & 2 & -4 \\ 3 & -7 & 2 \\ 4 & -5 & -3 \end{bmatrix} - \begin{bmatrix} 5 & 3 & 4 \\ 2 & -7 & -5 \\ -4 & 2 & -3 \end{bmatrix}$

$= \begin{bmatrix} 5-5 & 2-3 & -4-4 \\ 3-2 & -7-(-7) & 2-(-5) \\ 4-(-4) & -5-2 & -3-(-3) \end{bmatrix} = \begin{bmatrix} 0 & -1 & -8 \\ 1 & 0 & 7 \\ 8 & -7 & 0 \end{bmatrix}$

Let $Q = A-A^T$. Then $Q^T = (A-A^T)^T = \begin{bmatrix} 0 & -1 & -8 \\ 1 & 0 & 7 \\ 8 & -7 & 0 \end{bmatrix}^T = \begin{bmatrix} 0 & 1 & 8 \\ -1 & 0 & -7 \\ -8 & 7 & 0 \end{bmatrix}$
$= -\begin{bmatrix} 0 & -1 & -8 \\ 1 & 0 & 7 \\ 8 & -7 & 0 \end{bmatrix} = -Q$.
Since $Q^T = -Q$, $A - A^T$ is a skew symmetric matrix.

Exercise 2.4 | Q 11.1 | Page 59

QUESTION

Express each of the following matrix as the sum of a symmetric and a skew symmetric matrix: $\begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix}$.

SOLUTION

A square matrix A can be expressed as the sum of a symmetric and a skew-symmetric matrix as $A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$

Let $A = \begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 4 & 3 \\ -2 & -5 \end{bmatrix}$

$\therefore A + A^T = \begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix} + \begin{bmatrix} 4 & 3 \\ -2 & -5 \end{bmatrix}$ $= \begin{bmatrix} 4+4 & -2+3 \\ 3-2 & -5-5 \end{bmatrix} = \begin{bmatrix} 8 & 1 \\ 1 & -10 \end{bmatrix}$

Also, $A - A^T = \begin{bmatrix} 4 & -2 \\ 3 & -5 \end{bmatrix} - \begin{bmatrix} 4 & 3 \\ -2 & -5 \end{bmatrix}$ $= \begin{bmatrix} 4-4 & -2-3 \\ 3-(-2) & -5-(-5) \end{bmatrix} = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix}$

Let $P = \frac{1}{2}(A + A^T) = \frac{1}{2}\begin{bmatrix} 8 & 1 \\ 1 & -10 \end{bmatrix} = \begin{bmatrix} 4 & 1/2 \\ 1/2 & -5 \end{bmatrix}$.
$P^T = \begin{bmatrix} 4 & 1/2 \\ 1/2 & -5 \end{bmatrix} = P$. So, P is symmetric.

Let $Q = \frac{1}{2}(A - A^T) = \frac{1}{2}\begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -5/2 \\ 5/2 & 0 \end{bmatrix}$.
$Q^T = \begin{bmatrix} 0 & 5/2 \\ -5/2 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & -5/2 \\ 5/2 & 0 \end{bmatrix} = -Q$. So, Q is skew-symmetric.

$\therefore A = P + Q = \begin{bmatrix} 4 & 1/2 \\ 1/2 & -5 \end{bmatrix} + \begin{bmatrix} 0 & -5/2 \\ 5/2 & 0 \end{bmatrix}$.

Exercise 2.4 | Q 11.2 | Page 59

QUESTION

Express each of the following matrix as the sum of a symmetric and a skew symmetric matrix: $\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}$.

SOLUTION

A square matrix A can be expressed as the sum of a symmetric and a skew-symmetric matrix as $A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$

Let $A = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}$

$\therefore A + A^T = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} + \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}$ $= \begin{bmatrix} 3+3 & 3-2 & -1-4 \\ -2+3 & -2-2 & 1-5 \\ -4-1 & -5+1 & 2+2 \end{bmatrix} = \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix}$

Also, $A - A^T = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} - \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}$ $= \begin{bmatrix} 3-3 & 3-(-2) & -1-(-4) \\ -2-3 & -2-(-2) & 1-(-5) \\ -4-(-1) & -5-1 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix}$

Let $P = \frac{1}{2}(A + A^T) = \frac{1}{2}\begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix}$.
$P^T = P$. So, P is symmetric.

Let $Q = \frac{1}{2}(A - A^T) = \frac{1}{2}\begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{bmatrix}$.
$Q^T = -Q$. So, Q is skew-symmetric.

$\therefore A = P + Q = \begin{bmatrix} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 5/2 & 3/2 \\ -5/2 & 0 & 3 \\ -3/2 & -3 & 0 \end{bmatrix}$.

Exercise 2.4 | Q 12.1 | Page 60

QUESTION

If $A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \\ 4 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 3 & -4 \\ 2 & -1 & 1 \end{bmatrix}$, verify that $(AB)^T = B^T A^T$.

SOLUTION

$A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \\ 4 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 3 & -4 \\ 2 & -1 & 1 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 2 & 3 & 4 \\ -1 & -2 & 1 \end{bmatrix}$ and $B^T = \begin{bmatrix} 0 & 2 \\ 3 & -1 \\ -4 & 1 \end{bmatrix}$

$AB = \begin{bmatrix} 2 & -1 \\ 3 & -2 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 0 & 3 & -4 \\ 2 & -1 & 1 \end{bmatrix}$

$= \begin{bmatrix} (2)(0)+(-1)(2) & (2)(3)+(-1)(-1) & (2)(-4)+(-1)(1) \\ (3)(0)+(-2)(2) & (3)(3)+(-2)(-1) & (3)(-4)+(-2)(1) \\ (4)(0)+(1)(2) & (4)(3)+(1)(-1) & (4)(-4)+(1)(1) \end{bmatrix}$

$= \begin{bmatrix} 0-2 & 6+1 & -8-1 \\ 0-4 & 9+2 & -12-2 \\ 0+2 & 12-1 & -16+1 \end{bmatrix} = \begin{bmatrix} -2 & 7 & -9 \\ -4 & 11 & -14 \\ 2 & 11 & -15 \end{bmatrix}$

$\therefore (AB)^T = \begin{bmatrix} -2 & -4 & 2 \\ 7 & 11 & 11 \\ -9 & -14 & -15 \end{bmatrix} \quad \dots(i)$

$B^T A^T = \begin{bmatrix} 0 & 2 \\ 3 & -1 \\ -4 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \\ -1 & -2 & 1 \end{bmatrix}$

$= \begin{bmatrix} (0)(2)+(2)(-1) & (0)(3)+(2)(-2) & (0)(4)+(2)(1) \\ (3)(2)+(-1)(-1) & (3)(3)+(-1)(-2) & (3)(4)+(-1)(1) \\ (-4)(2)+(1)(-1) & (-4)(3)+(1)(-2) & (-4)(4)+(1)(1) \end{bmatrix}$

$= \begin{bmatrix} 0-2 & 0-4 & 0+2 \\ 6+1 & 9+2 & 12-1 \\ -8-1 & -12-2 & -16+1 \end{bmatrix} = \begin{bmatrix} -2 & -4 & 2 \\ 7 & 11 & 11 \\ -9 & -14 & -15 \end{bmatrix} \quad \dots(ii)$
From (i) and (ii), we get
$(AB)^T = B^T A^T$.

Exercise 2.4 | Q 12.2 | Page 60

QUESTION

If $A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \\ 4 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 3 & -4 \\ 2 & -1 & 1 \end{bmatrix}$, verify that $(BA)^T = A^T B^T$.

SOLUTION

$A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \\ 4 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 3 & -4 \\ 2 & -1 & 1 \end{bmatrix}$

$\therefore A^T = \begin{bmatrix} 2 & 3 & 4 \\ -1 & -2 & 1 \end{bmatrix}$ and $B^T = \begin{bmatrix} 0 & 2 \\ 3 & -1 \\ -4 & 1 \end{bmatrix}$

$BA = \begin{bmatrix} 0 & 3 & -4 \\ 2 & -1 & 1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 3 & -2 \\ 4 & 1 \end{bmatrix}$

$= \begin{bmatrix} (0)(2)+(3)(3)+(-4)(4) & (0)(-1)+(3)(-2)+(-4)(1) \\ (2)(2)+(-1)(3)+(1)(4) & (2)(-1)+(-1)(-2)+(1)(1) \end{bmatrix}$

$= \begin{bmatrix} 0+9-16 & 0-6-4 \\ 4-3+4 & -2+2+1 \end{bmatrix} = \begin{bmatrix} -7 & -10 \\ 5 & 1 \end{bmatrix}$

$\therefore (BA)^T = \begin{bmatrix} -7 & 5 \\ -10 & 1 \end{bmatrix} \quad \dots(i)$

$A^T B^T = \begin{bmatrix} 2 & 3 & 4 \\ -1 & -2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 3 & -1 \\ -4 & 1 \end{bmatrix}$

$= \begin{bmatrix} (2)(0)+(3)(3)+(4)(-4) & (2)(2)+(3)(-1)+(4)(1) \\ (-1)(0)+(-2)(3)+(1)(-4) & (-1)(2)+(-2)(-1)+(1)(1) \end{bmatrix}$

$= \begin{bmatrix} 0+9-16 & 4-3+4 \\ 0-6-4 & -2+2+1 \end{bmatrix} = \begin{bmatrix} -7 & 5 \\ -10 & 1 \end{bmatrix} \quad \dots(ii)$
From (i) and (ii), we get
$(BA)^T = A^T B^T$.