$$[[1, 3], [7, 1]][[-2, 6], [-9, 3]]=$$

When multiplying matrices, we take the dot product of the rows and the columns. As a reminder, the dot product operates as follows:

$\mathit{a}\mathbf{·}\mathit{b}\mathbf{ }\mathbf{=}(a_1 x b_1)+(a_2 x b_2)+ ...$

For a 2x2 times 2x2 matrix multiplication, the following process is used

$\left.open\begin{array}{cc}a& b\\ c& d\end{array}\right]\left.open\begin{array}{cc}e& f\\ g& h\end{array}\right]=\left.open\begin{array}{cc}ae+bg& af+bh\\ ce+dg& cf+dh\end{array}\right]$

We can use this to solve our question:

$$\begin{gathered} \left.open\begin{array}{cc}1& 3\\ 7& 1\end{array}\right]\left.open\begin{array}{cc}-2& 6\\ -9& 3\end{array}\right]=\left.open\begin{array}{cc}\left(1\right)(-2)+\left(3\right)(-9)& \left(1\right)\left(6\right)+\left(3\right)\left(3\right)\\ \left(7\right)(-2)+\left(1\right)(-9)& \left(7\right)\left(6\right)+\left(1\right)\left(3\right)\end{array}\right] \\ \left.open\begin{array}{cc}1& 3\\ 7& 1\end{array}\right]\left.open\begin{array}{cc}-2& 6\\ -9& 3\end{array}\right] = \left.open\begin{array}{cc}-29& 15\\ -23& 45\end{array}\right] \end{gathered}$$