Let A=left[begin{array}{ccc}1 & -1 & 2 0 & 3 | vedclass

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(A) Given matrices are $A=\left[\begin{array}{ccc}1 & -1 & 2 \ 0 & 3 & 4\end{array}ight]$ and $B=\left[\begin{array}{ccc}4 & 0 & -3 \ -1 & -2 & -3

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$\left[\begin{array}{ccc}4 & 0 & -3 \ -7 & -6 & -6 \ 4 & -8 & -18\end{array}ight]$

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(A) Given matrices are $A=\left[\begin{array}{ccc}1 & -1 & 2 \ 0 & 3 & 4\end{array}ight]$ and $B=\left[\begin{array}{ccc}4 & 0 & -3 \ -1 & -2 & -3\end{array}ight]$.First,we find the transpose of matrix $A$,denoted as $A^T$:$A^T = \left[\begin{array}{cc}1 & 0 \ -1 & 3 \ 2 & 4\end{array}ight]$.Now,we compute the product $A^T B$:$A^T B = \left[\begin{array}{cc}1 & 0 \ -1 & 3 \ 2 & 4\end{array}ight] \left[\begin{array}{ccc}4 & 0 & -3 \ -1 & -2 & -3\end{array}ight]$.Performing matrix multiplication:$A^T B = \left[\begin{array}{ccc}(1)(4)+(0)(-1) & (1)(0)+(0)(-2) & (1)(-3)+(0)(-3) \ (-1)(4)+(3)(-1) & (-1)(0)+(3)(-2) & (-1)(-3)+(3)(-3) \ (2)(4)+(4)(-1) & (2)(0)+(4)(-2) & (2)(-3)+(4)(-3)\end{array}ight]$.$A^T B = \left[\begin{array}{ccc}4+0 & 0+0 & -3+0 \ -4-3 & 0-6 & 3-9 \ 8-4 & 0-8 & -6-12\end{array}ight]$.$A^T B = \left[\begin{array}{ccc}4 & 0 & -3 \ -7 & -6 & -6 \ 4 & -8 & -18\end{array}ight]$.

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 $A^T B$ ?

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