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

Let $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 3 & 4\end{array}\right], B=\left[\begin{array}{ccc}4 & 0 & -3 \\ -1 & -2 & -3\end{array}\right]$ and $C=\left[\begin{array}{cccc}2 & -3 & 0 & 1 \\ 5 & -1 & -4 & 2 \\ -1 & 0 & 0 & 3\end{array}\right]$, what is $A^T B$ ?
  1. $\left[\begin{array}{ccc}4 & 0 & -3 \\ -7 & -6 & -6 \\ 4 & -8 & -18\end{array}\right]$
  2. $A^{\top} B$ is not defined
  3. $\left[\begin{array}{ccc}4 & -7 & 4 \\ 0 & -6 & -8 \\ -3 & 12 & 6\end{array}\right]$
  4. $A^T B=0$

Solution

$A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 3 & 4\end{array}\right], B=\left[\begin{array}{ccc}4 & 0 & -3 \\ -1 & -2 & -3\end{array}\right]$ and $\quad C=\left[\begin{array}{cccc}2 & -3 & 0 & 1 \\ 5 & -1 & -4 & 2 \\ -1 & 0 & 0 & 3\end{array}\right]$ Now, $ A^T=\left[\begin{array}{cc} 1 & 0 \\ -1 & 3 \\ 2 & 4 \end{array}\right], B=\left[\begin{array}{ccc} 4 & 0 & -3 \\ -1 & -2 & -3 \end{array}\right] $ $\Rightarrow A^T B=\left[\begin{array}{cc}1 & 0 \\ -1 & 3 \\ 2 & 4\end{array}\right]\left[\begin{array}{ccc}4 & 0 & -3 \\ -1 & -2 & -3\end{array}\right]$ $=\left[\begin{array}{ccc}4+0 & 0+0 & -3+0 \\ -4-3 & 0-6 & 3-9 \\ 8-4 & 0-8 & -6-12\end{array}\right]$ $=\left[\begin{array}{ccc}4 & 0 & -3 \\ -7 & -6 & -6 \\ 4 & -8 & -18\end{array}\right]$

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

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