Question

If $A=\left[\begin{array}{ccc}1& 3& 0\\ -1& 2& 1\\ 0& 0& 2\end{array}\right],B=\left[\begin{array}{ccc}2& 3& 4\\ 1& 2& 3\\ -1& 1& 2\end{array}\right]$ then $AB=$

• A. $\left[\begin{array}{ccc}5& 3& 11\\ 1& 2& 2\\ 1& 3& 5\end{array}\right]$
• B. $\left[\begin{array}{ccc}5& 9& 13\\ -1& 2& 4\\ -2& 2& 4\end{array}\right]$
• C. $\left[\begin{array}{ccc}5& 8& 11\\ 1& 2& 3\\ 2& 2& -3\end{array}\right]$
• D. $\left[\begin{array}{ccc}5& 3& 13\\ -1& 2& 4\\ -2& 3& -5\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{ccc}5& 9& 13\\ -1& 2& 4\\ -2& 2& 4\end{array}\right]$

Given,

$A=\left[\begin{array}{ccc}1& 3& 0\\ -1& 2& 1\\ 0& 0& 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}2& 3& 4\\ 1& 2& 3\\ -1& 1& 2\end{array}\right]$

Then, $AB=\left[\begin{array}{ccc}1& 3& 0\\ -1& 2& 1\\ 0& 0& 2\end{array}\right]\left[\begin{array}{ccc}2& 3& 4\\ 1& 2& 3\\ -1& 1& 2\end{array}\right]$

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

$=\left[\begin{array}{ccc}5& 9& 13\\ -1& 2& 4\\ -2& 2& 4\end{array}\right]$

Hence, the correct option is $\left(B\right)$