Question

If $A=\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}3& 1& 2\\ 2& 0& 5\\ 1& 2& 0\end{array}\right]$, find $AB$.

• A. $\left[\begin{array}{ccc}5& 6& -9\\ 4& 3& -9\\ -1& 3& -2\end{array}\right]$
• B. $\left[\begin{array}{ccc}2& 4& 7\\ 1& 5& 3\\ -1& 0& -5\end{array}\right]$
• C. $\left[\begin{array}{ccc}7& 11& -9\\ 4& 10& -11\\ -1& 2& -5\end{array}\right]$
• D. $\left[\begin{array}{ccc}7& 8& -19\\ 4& 1& -5\\ -1& 12& -5\end{array}\right]$

Answer 1

The correct option is C $\left[\begin{array}{ccc}7& 11& -9\\ 4& 10& -11\\ -1& 2& -5\end{array}\right]$

$A=\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]B=\left[\begin{array}{ccc}3& 1& 2\\ 2& 0& 5\\ 1& 2& 0\end{array}\right]AB=\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]\times \left[\begin{array}{ccc}3& 1& 2\\ 2& 0& 5\\ 1& 2& 0\end{array}\right]=\left[\begin{array}{ccc}9-6+4& 3+0+8& 6-15+0\\ 6-6+4& 2+0+8& 4-15+0\\ 0-2+1& 0+0+2& 0-5+0\end{array}\right]=\left[\begin{array}{ccc}7& 11& -9\\ 4& 10& -11\\ -1& 2& -5\end{array}\right]$