If: = A, find A.
$\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]$
Given, $\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]_{1 \times 3}\left[\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]_{3 \times 3}\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]_{3 \times 1}=A$
So, $\quad\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]_{1 \times 3}\left[\begin{array}{c}-1+0+1 \\ -1+0+0 \\ 0+0-1\end{array}\right]_{3 \times 1}=A$
$\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]_{1 \times 3}\left[\begin{array}{c}0 \\ -1 \\ -1\end{array}\right]_{3 \times 1}=A$
$[0-1-3]=A$
Thus, $A=[-4]$
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