Question

Given $A=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right];B=\left[\begin{array}{cc}4& 2\\ 1& 0\end{array}\right];C=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$

Find A(B+C)

• A. $\left[\begin{array}{cc}17& 6\\ 26& 4\end{array}\right]$
• B. $\left[\begin{array}{cc}17& 16\\ 26& 24\end{array}\right]$
• C. $\left[\begin{array}{cc}7& 16\\ 6& 24\end{array}\right]$
• D. $\left[\begin{array}{cc}7& 6\\ 6& 4\end{array}\right]$

Answer 1

The correct option is B

$\left[\begin{array}{cc}17& 16\\ 26& 24\end{array}\right]$

$B+C=\left[\begin{array}{cc}4& 2\\ 1& 0\end{array}\right]+\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}4+1& 2+2\\ 1+3& 0+4\end{array}\right]\Rightarrow \left[\begin{array}{cc}5& 4\\ 4& 4\end{array}\right]$

$A(B+C)=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]\left[\begin{array}{cc}5& 4\\ 4& 4\end{array}\right]\Rightarrow \left[\begin{array}{cc}1\times 5+3\times 4& 1\times 4+3\times 4\\ 2\times 5+4\times 4& 2\times 4+4\times 4\end{array}\right]$

$=\left[\begin{array}{cc}5+12& 4+12\\ 10+16& 8+16\end{array}\right]=\left[\begin{array}{cc}17& 16\\ 26& 24\end{array}\right]$