Question

If A'= \(\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 &1 \end{bmatrix}\)\(\begin{bmatrix}  -1 & 2 & 1 \\ 1 &2 & 3\end{bmatrix}\) , then verify that 
(i) \((A+B)'=A'+B' \)
(ii) \((A-B)'=A'-B'\)

Answer 1

(i) It is known that A=(A')' 
Therefore, we have:
A= \(\begin{bmatrix} 3 & -1 &  0 \\ 4 & 2 & 1  \end{bmatrix}\) 

B'= \(\begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 &3 \end{bmatrix}\) 

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

\(\therefore (A+B)'=\)  \(\begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 &4 \end{bmatrix}\)

\(A'+B'=\) \(\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 &1 \end{bmatrix}\)+ \(\begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 &3 \end{bmatrix}\)= \(\begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 &4 \end{bmatrix}\)

Thus, we verified that:(A+B)'=A'+B' 

(ii) \(A-B\)= \(\begin{bmatrix} 3 & -1 &  0 \\ 4 & 2 & 1  \end{bmatrix}\)- \(\begin{bmatrix}  -1 & 2 & 1 \\ 1 &2 & 3\end{bmatrix}\) = \(\begin{bmatrix}  4 & -3 & -1 \\ 3 &0 & -2\end{bmatrix}\)

so\( (A-B)'\) = \(\begin{bmatrix} -4 & 3 \\ -3 & 0 \\ -1 &-2 \end{bmatrix}\)

A'-B'= \(\begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 &1 \end{bmatrix}\)- \(\begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 &3 \end{bmatrix}\)= \(\begin{bmatrix} -4 & 3 \\ -3 & 0 \\ -1 &-2 \end{bmatrix}\) 

Hence we verified that: \((A-B)'=A'-B'\)