Mashaal Masha
Question:
 
If A = $\left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]$, B = $\left[ \begin{matrix} 5 & -4 \\ -2 & -1 \\ \end{matrix} \right]$, then find the following.
 
(i) 2A $+$ 3B
(ii) $-$3A $+$ 2B
(iii) $-$3(A $+$ 2B)
(iv) $\frac{2}{3}$(2A $-$ 3B)
Difficulty: Easy

Solution:

(i) 2A $+$ 3B

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

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

= $\left[ \begin{matrix} 4 & 6 \\ 2 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 15 & -12 \\ -6 & -3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4+15 & 6-12 \\ 2-6 & 0-3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 19 & -6 \\ -4 & -3 \\ \end{matrix} \right]$

 

So, 2A $+$ 3B = $\left[ \begin{matrix} 19 & -6 \\ -4 & -3 \\ \end{matrix} \right]$

 

(ii) $-$3A $+$ 2B

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

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

= $\left[ \begin{matrix} -6 & -9 \\ -3 & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 10 & -8 \\ -4 & -2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \left( -6 \right)+10 & \left( -9 \right)+\left( -8 \right) \\ -3+\left( -4 \right) & 0+(-2) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 4 & -17 \\ -7 & -2 \\ \end{matrix} \right]$

 

So, $-$3A $+$ 2B  = $\left[ \begin{matrix} 4 & -17 \\ -7 & -2 \\ \end{matrix} \right]$

 

(iii) $-$3(A $+$ 2B)

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

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

= $-3\left( \left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]\text{ }\!\!~\!\!\text{ }+\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 10 & -8 \\ -4 & -2 \\ \end{matrix} \right] \right)$

= $-3\left[ \begin{matrix} 2+10 & 3-8 \\ 1-4 & 0+-2 \\ \end{matrix} \right]$

= $-3\left[ \begin{matrix} 12 & -5 \\ -3 & -2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -3~\times 12 & -3~\times \left( -5 \right) \\ -3~\times \left( -3 \right) & -3~\times \left( -2 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -36 & 15 \\ 9 & 6 \\ \end{matrix} \right]$

 

So, $-$3(A $+$ 2B) = $\left[ \begin{matrix} -36 & 15 \\ 9 & 6 \\ \end{matrix} \right]$

 

(iv) $\frac{2}{3}$(2A $-$ 3B)

= $\frac{2}{3}\left( 2\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 2 & 3 \\ 1 & 0 \\ \end{matrix} \right]-\text{ }\!\!~\!\!\text{ }3\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 5 & -4 \\ -2 & -1 \\ \end{matrix} \right] \right)$

= $\frac{2}{3}\left( \left[ \begin{matrix} 2~\times 2 & 2\times 3 \\ 2\times 1 & 2\times 0 \\ \end{matrix} \right]-\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 3\times 5 & 3\times \left( -4 \right) \\ 3\times \left( -2 \right) & 3\times \left( -1 \right) \\ \end{matrix} \right] \right)$

= $\frac{2}{3}\left( \left[ \begin{matrix} 4 & 6 \\ 2 & 0 \\ \end{matrix} \right]-\text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 15 & -12 \\ -6 & -3 \\ \end{matrix} \right] \right)$

= $\frac{2}{3}\left( \left[ \begin{matrix} 4-15 & 6+12 \\ 2+6 & 0+3 \\ \end{matrix} \right] \right)$

= $\frac{2}{3}\times \left[ \begin{matrix} -11 & 18 \\ 8 & 3 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -\frac{22}{3} & \frac{36}{3} \\ \frac{16}{3} & \frac{6}{3} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -\frac{22}{3} & 12 \\ \frac{16}{3} & 2 \\ \end{matrix} \right]$

 

So, $\frac{2}{3}$(2A $-$ 3B) = $\left[ \begin{matrix} -\frac{22}{3} & 12 \\ \frac{16}{3} & 2 \\ \end{matrix} \right]$