Exercise 1.3

Solution:

Solving D

D = $\left[ \begin{matrix} 2+1 \\ 3 \\ \end{matrix} \right]$

D = $\left[ \begin{matrix} 3 \\ 3 \\ \end{matrix} \right]$

 

Solving F

F = $\left[ \begin{matrix} 3 & 2 \\ 1+1 & -4 \\ 3+2 & 2+1 \\ \end{matrix} \right]$

F = $\left[ \begin{matrix} 3 & 2 \\ 2 & -4 \\ 5 & 3 \\ \end{matrix} \right]$

 

Matrices that are of same order are conformable for addition. So, according to this definition:

 

(i) Matrices A and E are conformable for addition (because both have order 2-by-2).

(ii) Matrices B and D are conformable for addition (because both have order 1-by-1).

(iii) Matrices C and F are conformable for addition (because both have order 3-by-2).

Solution:

The additive inverse of a matrix is obtained by changing the sign of each entry. So, according to the definition:

 

(i) Additive inverse of A = $-$A = $\left[ \begin{matrix} -2 & -4 \\ 2 & -1 \\ \end{matrix} \right]$

(ii) Additive inverse of B = $-$B = $\left[ \begin{matrix} -1 & 0 & 1 \\ -2 & 1 & -3 \\ -3 & 2 & -1 \\ \end{matrix} \right]$

(iii) Additive inverse of C = $-$C = $\left[ \begin{matrix} -4 \\ 2 \\ \end{matrix} \right]$

(iv) Additive inverse of D = $-$D = $\left[ \begin{matrix} -1 & 0 \\ 3 & 2 \\ -2 & -1 \\ \end{matrix} \right]$

(v) Additive inverse of E = $-$E = $\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} \right]$

(vi) Additive inverse of F = $-$F = $\left[ \begin{matrix} -\sqrt{3} & -1 \\ 1 & -\sqrt{2} \\ \end{matrix} \right]$

Solution:

(i) A + $\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]$

= A + $\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1 & 2 \\ 2 & 1 \\ \end{matrix} \right]+~\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1+1 & 2+1 \\ 2+1 & 1+1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & 3 \\ 3 & 2 \\ \end{matrix} \right]$

So

A + $~\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]=~\left[ \begin{matrix} 0 & 3 \\ 3 & 2 \\ \end{matrix} \right]$

 

(ii) B + $\left[ \begin{matrix} -2 \\ 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 \\ -1 \\ \end{matrix} \right]$ + $\left[ \begin{matrix} -2 \\ 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+\left( -2 \right) \\ \left( -1 \right)+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1 \\ 2 \\ \end{matrix} \right]$

So

B + $\left[ \begin{matrix} 1 \\ -1 \\ \end{matrix} \right]=~\left[ \begin{matrix} -1 \\ 2 \\ \end{matrix} \right]$

 

(iii) C + $\left[ \begin{matrix} -2 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & -1 & 2 \\ \end{matrix} \right]$ + $\left[ \begin{matrix} -2 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+\left( -2 \right) & -1+1 & 2+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1 & 0 & 5 \\ \end{matrix} \right]$

So

C + $\left[ \begin{matrix} -2 & 1 & 3 \\ \end{matrix} \right]$ = $\left[ \begin{matrix} -1 & 0 & 5 \\ \end{matrix} \right]$

 

(iv) D + $\left[ \begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 2 & 3 \\ -1 & 0 & 2 \\ \end{matrix} \right]~+~\left[ \begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+0 & 2+1 & 3+0 \\ -1+2 & 0+0 & 2+1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 3 & 3 \\ 1 & 0 & 3 \\ \end{matrix} \right]$

So

D + $\left[ \begin{matrix} 0 & 1 & 0 \\ 2 & 0 & 1 \\ \end{matrix} \right]=~$ $\left[ \begin{matrix} 1 & 3 & 3 \\ 1 & 0 & 3 \\ \end{matrix} \right]$

 

(v) $2$A

= $2 \times \left[ \begin{matrix} -1 & 2 \\ 2 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2~\times \left( -1 \right) & 2~\times 2 \\ 2\times 2 & 2\times 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2 & 4 \\ 4 & 2 \\ \end{matrix} \right]$

So

$2$A = $\left[ \begin{matrix} -2 & 4 \\ 4 & 2 \\ \end{matrix} \right]$

 

(vi) $(-1)$B

= $(-1) \times \left[ \begin{matrix} 1 \\ -1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \left( -1 \right)\times 1 \\ \left( -1 \right)\times -1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1 \\ 1 \\ \end{matrix} \right]$

So

$(-1)$B = $\left[ \begin{matrix} -1 \\ 1 \\ \end{matrix} \right]$

 

(vii) $(-2)$C

= $(-2) \times \left[ \begin{matrix} 1 & -1 & 2 \\ \end{matrix} \right]$

= $\left[ \left( -2 \right)\times \begin{matrix} 1 & \left( -2 \right)\times -1 & \left( -2 \right)\times 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2 & 2 & -4 \\ \end{matrix} \right]$

So

$(-2)$C = $\left[ \begin{matrix} -2 & 2 & -4 \\ \end{matrix} \right]$

 

(viii) $3$D

= $3 \times \left[ \begin{matrix} 1 & 2 & 3 \\ -1 & 0 & 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3\times 1 & 3\times 2 & 3\times 3 \\ 3\times -1 & 3\times 0 & 3\times 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & 6 & 9 \\ -3 & 0 & 6 \\ \end{matrix} \right]$

So

$3$D = $\left[ \begin{matrix} 3 & 6 & 9 \\ -3 & 0 & 6 \\ \end{matrix} \right]$

 

(ix) $3$C

= $3 \times \left[ \begin{matrix} 1 & -1 & 2 \\ \end{matrix} \right]$

= $\left[ 3\times \begin{matrix} 1 & 3\times -1 & 3\times 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & -3 & 6 \\ \end{matrix} \right]$

So

$3$C = $\left[ \begin{matrix} 3 & -3 & 6 \\ \end{matrix} \right]$

Solution:

(i) $\left( \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]+~\left[ \begin{matrix} 0 & 2 \\ 3 & 0 \\ \end{matrix} \right] \right)+~\left[ \begin{matrix} 1 & 1 \\ 1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+0 & 0+2 \\ 0+3 & 1+0 \\ \end{matrix} \right]+~\left[ \begin{matrix} 1 & 1 \\ 1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 2 \\ 3 & 1 \\ \end{matrix} \right]+~\left[ \begin{matrix} 1 & 1 \\ 1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+1 & 2+1 \\ 3+1 & 1+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 3 \\ 4 & 1 \\ \end{matrix} \right]$

 

(ii) $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]+~\left( \left[ \begin{matrix} 0 & 2 \\ 3 & 0 \\ \end{matrix} \right]-~\left[ \begin{matrix} 1 & 1 \\ 1 & 0 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]+~\left[ \begin{matrix} 0-1 & 2-1 \\ 3-1 & 0-0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]+~\left[ \begin{matrix} -1 & 1 \\ 2 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+(-1) & 0+1 \\ 0+2 & 1+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & 1 \\ 2 & 1 \\ \end{matrix} \right]$

 

(iii) $\left[ \begin{matrix} 2 & 3 & 1 \\ \end{matrix} \right]+\left( \left[ \begin{matrix} 1 & 0 & 2 \\ \end{matrix} \right]-~\left[ \begin{matrix} 2 & 2 & 2 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} 2 & 3 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} 1-2 & 0-2 & 2-2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 3 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} -1 & -2 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2+(-1) & 3+(-2) & 1+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 1 & 1 \\ \end{matrix} \right]$

 

(iv) $\left[ \begin{matrix} 1 & 2 & 3 \\ -1 & -1 & -1 \\ 0 & 1 & 2 \\ \end{matrix} \right]$ $+\left[ \begin{matrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+1 & 2+1 & 3+1 \\ -1+2 & -1+2 & -1+2 \\ 0+3 & 1+3 & 2+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 3 & 4 \\ 1 & 1 & 1 \\ 3 & 4 & 5 \\ \end{matrix} \right]$

 

(v) $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 0 & -2 \\ -2 & -1 & 0 \\ 0 & 2 & -1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+1 & 2+0 & 3+\left( -2 \right) \\ 2+\left( -2 \right) & 3+\left( -1 \right) & 1+0 \\ 3+0 & 1+2 & 2+\left( -1 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 2 & 1 \\ 0 & 2 & 1 \\ 3 & 3 & 1 \\ \end{matrix} \right]$

 

(vi) $\left( \left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]+~\left[ \begin{matrix} 2 & 1 \\ 1 & 0 \\ \end{matrix} \right] \right)+~\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+2 & 2+1 \\ 0+1 & 1+0 \\ \end{matrix} \right]+~\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & 3 \\ 1 & 1 \\ \end{matrix} \right]+~\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3+1 & 3+1 \\ 1+1 & 1+1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & 4 \\ 2 & 2 \\ \end{matrix} \right]$

 

Solution:

(i) A + C = C + A

Taking Left Hand Side: A + C

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right] + \left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+\left( -1 \right) & 2+0 & 3+0 \\ 2+0 & 3+\left( -2 \right) & 1+3 \\ 1+1 & -1+1 & 0+2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & 2 & 3 \\ 2 & 1 & 4 \\ 2 & 0 & 2 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: C + A

= $\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1+1 & 0+2 & 0+3 \\ 0+2 & -2+3 & 3+1 \\ 1+1 & 1+\left( -1 \right) & 2+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & 2 & 3 \\ 2 & 1 & 4 \\ 2 & 0 & 2 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

From (1) and (2), it is proved that: A + C = C + A

 

(ii) A + B = B + A

Taking Left Hand Side: A + B

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+1 & 2+\left( -1 \right) & 3+1 \\ 2+2 & 3+\left( -2 \right) & 1+2 \\ 1+3 & -1+1 & 0+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 1 & 4 \\ 4 & 1 & 3 \\ 4 & 0 & 3 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: B + A

= $\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]~+~\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+1 & -1+2 & 1+3 \\ 2+2 & -2+3 & 2+1 \\ 3+1 & 1+\left( -1 \right) & 3+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 1 & 4 \\ 4 & 1 & 3 \\ 4 & 0 & 3 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

From (1) and (2), it is proved that: A + B = B + A

 

(iii) B + C = C + B

Taking Left Hand Side: B + C

= $\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right] + \left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+\left( -1 \right) & -1+0 & 1+0 \\ 2+0 & -2+\left( -2 \right) & 2+3 \\ 3+1 & 1+1 & 3+2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & -1 & 1 \\ 2 & -4 & 5 \\ 4 & 2 & 5 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: C + B

= $\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1+1 & 0+\left( -1 \right) & 0+1 \\ 0+2 & -2+\left( -2 \right) & 3+2 \\ 1+3 & 1+1 & 2+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & -1 & 1 \\ 2 & -4 & 5 \\ 4 & 2 & 5 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

From (1) and (2), it is proved that: B + C = C + B

 

(iv) A + (B + A) = 2A + B

Taking Left Hand Side: A + (B + A)

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right] + \left( \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]+~\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ } \right)$

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right] + \left[ \begin{matrix} 1+1 & -1+2 & 1+3 \\ 2+2 & -2+3 & 2+1 \\ 3+1 & 1+(-1) & 3+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right] + \left[ \begin{matrix} 2 & 1 & 4 \\ 4 & 1 & 3 \\ 4 & 0 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+2 & 2+1 & 3+4 \\ 2+4 & 3+1 & 1+3 \\ 1+4 & -1+0 & 0+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & 3 & 7 \\ 6 & 4 & 4 \\ 5 & -1 & 3 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: 2A + B

= $2\times \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+ \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2\times 1 & 2\times 2 & 2\times 3 \\ 2\times 2 & 2\times 3 & 2\times 1 \\ 2\times 1 & 2\times \left( -1 \right) & 2\times 0 \\ \end{matrix} \right]+ \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 4 & 6 \\ 4 & 6 & 2 \\ 2 & -2 & 0 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2+1 & 4+\left( -1 \right) & 6+1 \\ 4+2 & 6+\left( -2 \right) & 2+2 \\ 2+3 & -2+1 & 0+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & 3 & 7 \\ 6 & 4 & 4 \\ 5 & -1 & 3 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

From (1) and (2), it is proved that: A + (B + A) = 2A+B

 

(v) (C$~-$ B) + A = C + (A $-$ B)

Taking Left Hand Side: (C$-$B) + A

= $\left( \left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right] \right)+ \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-1 & 0-\left( -1 \right) & 0-1 \\ 0-2 & -2-\left( -2 \right) & 3-2 \\ 1-3 & 1-1 & 2-3 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2 & 1 & -1 \\ -2 & 0 & 1 \\ -2 & 0 & -1 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2+1 & 1+2 & -1+3 \\ -2+2 & 0+3 & 1+1 \\ -2+1 & 0+\left( -1 \right) & -1+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1 & 3 & 2 \\ 0 & 3 & 2 \\ -1 & -1 & -1 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: C + (A - B)

= $\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right] + \left( \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right ]-\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right] + \left[ \begin{matrix} 1-1 & 2-\left( -1 \right) & 3-1 \\ 2-2 & 3-\left( -2 \right) & 1-2 \\ 1-3 & -1-1 & 0-3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right] + \left[ \begin{matrix} 0 & 3 & 2 \\ 0 & 5 & -1 \\ -2 & -2 & -3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1+0 & 0+3 & 0+2 \\ 0+0 & -2+5 & 3+\left( -1 \right) \\ 1+\left( -2 \right) & 1+\left( -2 \right) & 2+\left( -3 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1 & 3 & 2 \\ 0 & 3 & 2 \\ -1 & -1 & -1 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

From (1) and (2), it is proved that: (C$~-$ B) + A = C + (A $-$ B)

 

(vi) 2A + B = A + (A + B)

Taking Left Hand Side: 2A + B

= $2 \times \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2\times 1 & 2\times 2 & 2\times 3 \\ 2\times 2 & 2\times 3 & 2\times 1 \\ 2\times 1 & 2\times \left( -1 \right) & 2\times 0 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 4 & 6 \\ 4 & 6 & 2 \\ 2 & -2 & 0 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2+1 & 4+\left( -1 \right) & 6+1 \\ 4+2 & 6+\left( -2 \right) & 2+2 \\ 2+3 & \left( -2 \right)+1 & 0+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & 3 & 7 \\ 6 & 4 & 4 \\ 5 & -1 & 3 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: A + (A + B)

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+ \left( \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 1+1 & 2+\left( -1 \right) & 3+1 \\ 2+2 & 3+\left( -2 \right) & 1+2 \\ 1+3 & -1+1 & 0+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 2 & 1 & 4 \\ 4 & 1 & 3 \\ 4 & 0 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+2 & 2+1 & 3+4 \\ 2+4 & 3+1 & 1+3 \\ 1+4 & -1+0 & 0+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & 3 & 7 \\ 6 & 4 & 4 \\ 5 & -1 & 3 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

From (1) and (2), it is proved that: 2A + B = A + (A + B)

 

(vii) (C $-$ B) $-$A = (C$-$A) B

Taking Left Hand Side: (C – B) – A

= $\left( \left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right] \right) - \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-1 & 0-\left( -1 \right) & 0-1 \\ 0-2 & -2-\left( -2 \right) & 3-2 \\ 1-3 & 1-1 & 2-3 \\ \end{matrix} \right] - \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2 & 1 & -1 \\ -2 & 0 & 1 \\ -2 & 0 & -1 \\ \end{matrix} \right] - \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2-1 & 1-2 & -1-3 \\ -2-2 & 0-3 & 1-1 \\ -2-1 & 0-\left( -1 \right) & -1-0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -3 & -1 & -4 \\ -4 & -3 & 0 \\ -3 & 1 & -1 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: (C – A) – B

= $\left( \left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right] - \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right ] \right)-\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -1-1 & 0-2 & 0-3 \\ 0-2 & -2-3 & 3-1 \\ 1-1 & 1-\left( -1 \right) & 2-0 \\ \end{matrix} \right] - \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2 & -2 & -3 \\ -2 & -5 & 2 \\ 0 & 2 & 2 \\ \end{matrix} \right] - \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2-1 & -2-\left( -1 \right) & -3-1 \\ -2-2 & -5-\left( -2 \right) & 2-2 \\ 0-3 & 2-1 & 2-3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -3 & -1 & -4 \\ -4 & -3 & 0 \\ -3 & 1 & -1 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

From (1) and (2), it is proved that: (C $-$ B) $-$A = (C$-$A) B

 

(viii) (A + B) + C = A + (B + C)

Taking Left Hand Side: (A + B) + C

= $\left( \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right] \right) + \left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+1 & 2+\left( -1 \right) & 3+1 \\ 2+2 & 3+\left( -2 \right) & 1+2 \\ 1+3 & -1+1 & 0+3 \\ \end{matrix} \right] + \left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 1 & 4 \\ 4 & 1 & 3 \\ 4 & 0 & 3 \\ \end{matrix} \right] + \left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2+\left( -1 \right) & 1+0 & 4+0 \\ 4+0 & 1+\left( -2 \right) & 3+3 \\ 4+1 & 0+1 & 3+2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 1 & 4 \\ 4 & -1 & 6 \\ 5 & 1 & 5 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: A + (B + C)

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+$ $\left( \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }+\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 1+\left( -1 \right) & -1+0 & 1+0 \\ 2+0 & -2+\left( -2 \right) & 2+3 \\ 3+1 & 1+1 & 3+2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & -1 & 1 \\ 2 & -4 & 5 \\ 4 & 2 & 5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+0 & 2+\left( -1 \right) & 3+1 \\ 2+2 & 3+\left( -4 \right) & 1+5 \\ 1+4 & -1+2 & 0+5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 1 & 4 \\ 4 & -1 & 6 \\ 5 & 1 & 5 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

From (1) and (2), it is proved that: (A + B) + C = A + (B + C)

 

(ix) A + (B$-$ C) = (A$-$ C) + B

Taking Left Hand Side: A + (B $-$ C)

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+$ $\left( \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right] - \left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 1-\left( -1 \right) & -1-0 & 1-0 \\ 2-0 & -2-\left( -2 \right) & 2-3 \\ 3-1 & 1-1 & 3-2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 2 & -1 & 1 \\ 2 & 0 & -1 \\ 2 & 0 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+2 & 2+\left( -1 \right) & 3+1 \\ 2+2 & 3+0 & 1+\left( -1 \right) \\ 1+2 & -1+0 & 0+1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & 1 & 4 \\ 4 & 3 & 0 \\ 3 & -1 & 1 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: (A$-$ C) + B

= $\left( \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]-\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -2 & 3 \\ 1 & 1 & 2 \\ \end{matrix} \right]\right) + \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1-\left( -1 \right) & 2-0 & 3-0 \\ 2-0 & 3-\left( -2 \right) & 1-3 \\ 1-1 & -1-1 & 0-2 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 2 & 3 \\ 2 & 5 & -2 \\ 0 & -2 & -2 \\ \end{matrix} \right] + \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2+1 & 2+\left( -1 \right) & 3+1 \\ 2+2 & 5+\left( -2 \right) & -2+2 \\ 0+3 & -2+1 & -2+3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & 1 & 4 \\ 4 & 3 & 0 \\ 3 & -1 & 1 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

From (1) and (2), it is proved that: A + (B $-$ C) = (A $-$ C) + B

 

(x) 2A + 2B = 2(A + B)

Taking Left Hand Side: 2A + 2B

= $\left( 2\times\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right] \right) +  \left( 2~\times \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right] \right)$

= $\left[ \begin{matrix} 2\times 1 & 2\times 2 & 2\times 3 \\ 2\times 2 & 2\times 3 & 2\times 1 \\ 2\times 1 & 2\times \left( -1 \right) & 2\times 0 \\ \end{matrix} \right] + \left[ \begin{matrix} 2\times 1 & 2\times -1 & 2\times 1 \\ 2\times 2 & 2\times -2 & 2\times 2 \\ 2\times 3 & 2\times 1 & 2\times 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 4 & 6 \\ 4 & 6 & 2 \\ 2 & -2 & 0 \\ \end{matrix} \right] + \left[ \begin{matrix} 2 & -2 & 2 \\ 4 & -4 & 4 \\ 6 & 2 & 6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2+2 & 4+\left( -2 \right) & 6+2 \\ 4+4 & 6+\left( -4 \right) & 2+4 \\ 2+6 & \left( -2 \right)+2 & 0+6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & 2 & 8 \\ 8 & 2 & 6 \\ 8 & 0 & 6 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side: 2(A + B)

= $2 \times \left( \left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }+\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & 1 & 3 \\ \end{matrix} \right] \right)$

= $2 \times\left[ \begin{matrix} 1+1 & 2+\left( -1 \right) & 3+1 \\ 2+2 & 3+\left( -2 \right) & 1+2 \\ 1+3 & -1+1 & 0+3 \\ \end{matrix} \right]$

= $2 \times\left[ \begin{matrix} 2 & 1 & 4 \\ 4 & 1 & 3 \\ 4 & 0 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2\times2 & 2\times1 & 2\times4 \\ 2\times4 & 2\times1 & 2\times3 \\ 2\times4 & 2\times0 & 2\times3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & 2 & 8 \\ 8 & 2 & 6 \\ 8 & 0 & 6 \\ \end{matrix} \right]\quad\quad\quad\quad\quad\quad...~(2)$

From (1) and (2), it is proved that: 2A + 2B = 2(A + B)

Solution:

(i) 3A $-$ 2B

= $3\times \left[ \begin{matrix} 1 & -2 \\ 3 & 4 \\ \end{matrix} \right] - 2\times \left[ \begin{matrix} 0 & 7 \\ -3 & 8 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3~\times 1 & 3\times -2 \\ 3\times 3 & 3\times 4 \\ \end{matrix} \right]$ $-$ $\left[ \begin{matrix} 2\times 0 & 2\times 7 \\ 2\times \left( -3 \right) & 2\times 8 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & -6 \\ 9 & 12 \\ \end{matrix} \right]$ $-$ $\left[ \begin{matrix} 0 & 14 \\ -6 & 16 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3-0 & -6-14 \\ 9-\left( -6 \right) & 12-16 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3 & -20 \\ 15 & -4 \\ \end{matrix} \right]$

 

So, 3A $-$ 2B = $\left[ \begin{matrix} 3 & -20 \\ 15 & -4 \\ \end{matrix} \right]$

 

(ii) 2A$^{t} -$ 3B$^{t}$

A = $\left[ \begin{matrix} 1 & -2 \\ 3 & 4 \\ \end{matrix} \right]$

A$^{t}$ = $\left[ \begin{matrix} 1 & 3 \\ -2 & 4 \\ \end{matrix} \right]$

 

B = $\left[ \begin{matrix} 0 & 7 \\ -3 & 8 \\ \end{matrix} \right]$

B$^{t}$ = $\left[ \begin{matrix} 0 & -3 \\ 7 & 8 \\ \end{matrix} \right]$

 

2A$^{t} -$ 3B$^{t}$

= $2\times \left[ \begin{matrix} 1 & 3 \\ -2 & 4 \\ \end{matrix} \right]-3\times \left[ \begin{matrix} 0 & -3 \\ 7 & 8 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2\times 1 & 2\times 3 \\ 2\times \left( -2 \right) & 2\times 4 \\ \end{matrix} \right]-\left[ \begin{matrix} 3\times 0 & 3\times \left( -3 \right) \\ 3\times 7 & 3\times 8 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 6 \\ -4 & 8 \\ \end{matrix} \right]-\left[ \begin{matrix} 0 & -9 \\ 21 & 24 \\ \end{matrix} \right]$

=$\left[ \begin{matrix} 2-0 & 6-\left( -9 \right) \\ -4-21 & 8-24 \\ \end{matrix} \right]$

=$\left[ \begin{matrix} 2 & 15 \\ -25 & -16 \\ \end{matrix} \right]$

 

So, 2A$^{t} -$ 3B$^{t}$ = $\left[ \begin{matrix} 2 & 15 \\ -25 & -16 \\ \end{matrix} \right]$

Solution:

The given equation is

$2\left[ \begin{matrix} 2 & 4 \\ -3 & a \\ \end{matrix} \right]+3\left[ \begin{matrix} 1 & b \\ 8 & -4 \\ \end{matrix} \right]=\left[ \begin{matrix} 7 & 10 \\ 18 & 1 \\ \end{matrix} \right]$

 

Taking Left Hand Side

= $2\left[ \begin{matrix} 2 & 4 \\ -3 & a \\ \end{matrix} \right]+3\left[ \begin{matrix} 1 & b \\ 8 & -4 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2\times 2 & 2\times 4 \\ 2\times \left( -3 \right) & 2\times a \\ \end{matrix} \right]+\left[ \begin{matrix} 3\times 1 & 3\times b \\ 3\times 8 & 3\times \left( -4 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & 8 \\ -6 & 2a \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & 3b \\ 24 & -12 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4+3 & 8+3b \\ \left( -6 \right)+24 & 2a+\left( -12 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 7 & 8+3b \\ 18 & 2a-12 \\ \end{matrix} \right]$

 

By equating it with the Right Hand Side, we have:

$\left[ \begin{matrix} 7 & 8+3b \\ 18 & 2a-12 \\ \end{matrix} \right]=~\left[ \begin{matrix} 7 & 10 \\ 18 & 1 \\ \end{matrix} \right]$

 

By comparing corresponding elements:

$8+3b=10$

$3b=10-8$

$3b=2$

$b=\frac{2}{3}\quad\quad\quad\quad\quad ...~(1)$

 

$2a-12=1$

$2a=1+12$

$2a=13$

$a=\frac{13}{2} \quad\quad\quad\quad\quad ...~(2)$

 

From equations $(1)$ and $(2)$, we get $a=\frac{13}{2}$ and $b=\frac{2}{3}$

Solution:

(i) (A $+$ B)$^{t}$ = A$^{t}+$ B$^{t}$

Taking Left Hand Side

= (A $+$ B)$^{t}$

= ${{\left( \left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }+\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ } \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} 1+1 & 2+1 \\ 0+2 & 1+0 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ } \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} 2 & 3 \\ 2 & 1 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ } \right)}^{t}}$

= $\left[ \begin{matrix} 2 & 2 \\ 3 & 1 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side

= A$^{t}+$ B$^{t}$

= ${{\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]}^{t}}+{{\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} 1 & 0 \\ 2 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 2 \\ 1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+1 & 0+2 \\ 2+1 & 1+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 2 \\ 3 & 1 \\ \end{matrix} \right]\quad\quad\quad\quad\quad...~(2)$

 

From (1) and (2), it is proved that: (A $+$ B)$^{t}$ = A$^{t}+$ B$^{t}$

 

(ii) (A $-$ B)$^{t}$ = A$^{t}-$ B$^{t}$ 

Taking Left Hand Side

= (A $-$ B)$^{t}$

= ${{\left( \left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]-\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ } \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} 1-1 & 2-1 \\ 0-2 & 1-0 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ } \right)}^{t}}$

= ${{\left( \left[ \begin{matrix} 0 & 1 \\ -2 & 1 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ } \right)}^{t}}$

= $\left[ \begin{matrix} 0 & -2 \\ 1 & 1 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

 

Taking Right Hand Side

= A$^{t}-$ B$^{t}$

= ${{\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]}^{t}}-{{\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} 1 & 0 \\ 2 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 2 \\ 1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1-1 & 0-2 \\ 2-1 & 1-0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & -2 \\ 1 & 1 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

 

From $(1)$ and $(2)$, it is proved that: (A $-$ B)$^{t}$ = A$^{t}-$ B$^{t}$

 

(iii) To prove A $+$ A$^{t}$ is symmetric

A $+$ A$^{t}$

= $\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]+{{\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 0 \\ 2 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+1 & 2+0 \\ 0+2 & 1+1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 2 \\ 2 & 2 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

 

Now we will take transpose of A$ + $A$^{t}$

(A $+$ A$^{t}$)$^{t}$

= ${{\left[ \begin{matrix} 2 & 2 \\ 2 & 2 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} 2 & 2 \\ 2 & 2 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

 

From $1$ and $2$, it is proved that: A $+$ A$^{t}$ = (A $+$ A$^{t}$)$^{t}$

So, it is symmetric.

 

(iv) To prove A $-$ A$^{t}$ is skew-symmetric

A $-$ A$^{t}$

= $\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]-{{\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 0 \\ 2 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1-1 & 2-0 \\ 0-2 & 1-1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & 2 \\ -2 & 0 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

 

Now we will take transpose of A $-$ A$^{t}$

(A $-$ A$^{t}$)$^{t}$

= ${{\left[ \begin{matrix} 0 & 2 \\ -2 & 0 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} 0 & -2 \\ 2 & 0 \\ \end{matrix} \right]$

= $\left( -1 \right)\left[ \begin{matrix} 0 & 2 \\ -2 & 0 \\ \end{matrix} \right]$

= $-\left[ \begin{matrix} 0 & 2 \\ -2 & 0 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

= $-$(A $-$ A$^{t}$)

 

From $(1)$ and $(2)$, it is proved that: (A $-$ A$^{t}$)$^{t}$ = $-$(A $-$ A$^{t}$)

So A $-$ A$^{t}$ is skew-symmetric

 

(v) To prove B $+$ B$^{t}$ is symmetric

B $+$ B$^{t}$

= $\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]+{{\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 2 \\ 1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1+1 & 1+2 \\ 2+1 & 0+0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2 & 3 \\ 3 & 0 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

 

Now we will take transpose of B $+$ B$^{t}$

(B $+$ B$^{t}$)$^{t}$

= ${{\left[ \begin{matrix} 2 & 3 \\ 3 & 0 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} 2 & 3 \\ 3 & 0 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

 

From $(1)$ and $(2)$, it is proved that: B $+$ B$^{t}$ = (B $+$ B$^{t}$)$^{t}$

So, it is symmetric.

 

(vi) To prove B $-$ B$^{t}$ is skew-symmetric

B $-$ B$^{t}$

= $\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]-{{\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 2 \\ 1 & 0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1-1 & 1-2 \\ 2-1 & 0-0 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

 

Now we will take transpose of B $-$ B$^{t}$

(B $-$B $^{t}$)$^{t}$

= ${{\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]}^{t}}$

= $\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \\ \end{matrix} \right]$

= $\left( -1 \right)\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]$

= $-\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

= $-$(B $-$ B$^{t}$)

 

From $(1)$ and $(2)$, it is proved that: (B $-$ B$^{t}$)$^{t}$ = $-$(B $-$ B$^{t}$)

So, it is skew-symmetric.