Question
If A' = $\left[\begin{array}{cc} {-2} & {3} \\ {1} & {2} \end{array}\right] \text { and } \mathbf{B}=\left[\begin{array}{rr} {-1} & {0} \\ {1} & {2} \end{array}\right]$, then find (A + 2B)′

Answer

Here, 
$A^\prime$ = $\left[\begin{array}{cc} {-2} & {1} \\ {3} & {2} \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{cc} {-1} & {0} \\ {1} & {2} \end{array}\right] $
$\Rightarrow A=\left[\begin{array}{cc} {-2} & {3} \\ {1} & {2} \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{cc} {-1} & {0} \\ {1} & {2} \end{array}\right]$
So, $A+2 B=\left[\begin{array}{rr} {-2} & {1} \\ {3} & {2} \end{array}\right]+2\left[\begin{array}{cc} {-1} & {0} \\ {1} & {2} \end{array}\right]$
$\Rightarrow A+2 B=\left[\begin{array}{cc} {-2} & {1} \\ {3} & {2} \end{array}\right]+\left[\begin{array}{cc} {-2} & {0} \\ {2} & {4} \end{array}\right]$
$\Rightarrow A+2 B=\left[\begin{array}{rr} {-4} & {1} \\ {5} & {6} \end{array}\right]$
Now, (A+2B)' = transpose of A+2B
$\Rightarrow(A+2 B)^\prime=\left[\begin{array}{cc} {-4} & {5} \\ {1} & {6} \end{array}\right]$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free