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# Compute the indicated products: $(vi)\;\begin{bmatrix}3 & -1 &3\\-1 & 0 &2\end{bmatrix}\begin{bmatrix}2 & -3\\1 & 0\\3 & 1\end{bmatrix}$

This question has 6 parts and each part has been answered separately here.

Toolbox:
• Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix.
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B:
• $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
$\begin{bmatrix}3 & -1 &3\\-1 & 0 &2\end{bmatrix}\begin{bmatrix}2 & -3\\1 & 0\\3 & 1\end{bmatrix} = \begin{bmatrix}3\times 2+-1\times 1+3\times 3 & 3\times -3+-1\times 0+3\times 1 \\-1\times 2+0\times 1 +2\times 3 & -1\times -3+0\times 0+2\times 1\end{bmatrix}$
$\begin{bmatrix}3 & -1 &3\\-1 & 0 &2\end{bmatrix}\begin{bmatrix}2 & -3\\1 & 0\\3 & 1\end{bmatrix} = \begin{bmatrix}6-1+9 & -9+0+3 \\-2+0+6 & 3+0+2\end{bmatrix}$
$\begin{bmatrix}3 & -1 &3\\-1 & 0 &2\end{bmatrix}\begin{bmatrix}2 & -3\\1 & 0\\3 & 1\end{bmatrix} = \begin{bmatrix}14 & -6\\4 & 5\end {bmatrix}$