Question

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

• (A) $\left[\begin{array}{c}-2\\ 4\end{array}\right]$
• (B) $\left[\begin{array}{cc}-2& 4\end{array}\right]$
• (C) $\left[\begin{array}{c}2\\ -4\end{array}\right]$
• (D) $\left[\begin{array}{ll}-2& -4\end{array}\right]$

Answer 1

The Correct Option is B

Explanation:
Given,$A=\left[\begin{array}{ccc}1& -1& 0\\ 3& 2& -1\end{array}\right]$ and $B=\left[\begin{array}{l}1\\ 3\\ 5\end{array}\right]$$\Rightarrow AB=\left[\begin{array}{ccc}1& -1& 0\\ 3& 2& -1\end{array}\right]\times \left[\begin{array}{l}1\\ 3\\ 5\end{array}\right]$$\Rightarrow AB=\left[\begin{array}{c}1-3+0\\ 3+6-5\end{array}\right]$$\Rightarrow AB=\left[\begin{array}{c}-2\\ 4\end{array}\right]$As we know,The new matrix obtained by interchanging the rows and columns of the original matrix is called as the transpose of the matrix. It is denoted by $A^{\prime }$ or $A^T$.$\therefore (AB{)}^T=\left[\begin{array}{ll}-2& 4\end{array}\right]$Hence, the correct option is (B).