)
2) So, let's begin with the lower-left product of matrices:
[tex]\begin{gathered} \begin{pmatrix}-2 & 12 \\ -1 & -3\end{pmatrix}\times\begin{pmatrix}2 & 1 \\ 4 & 2\end{pmatrix}= \\ \begin{pmatrix}\left(-2\right)\cdot \:2+12\cdot \:4&\left(-2\right)\cdot \:1+12\cdot \:2\\ \left(-1\right)\cdot \:2+\left(-3\right)\cdot \:4&\left(-1\right)\cdot \:1+\left(-3\right)\cdot \:2\end{pmatrix} \\ \begin{pmatrix}44&22\\ -14&-7\end{pmatrix} \end{gathered}[/tex]Now for the next pair of matrices in the middle:
[tex]\begin{gathered} \begin{pmatrix}2 & 0 \\ 1 & 0\end{pmatrix}\times\begin{pmatrix}3 & -1 \\ 2 & 2\end{pmatrix}= \\ \begin{pmatrix}2\cdot \:3+0\cdot \:2&2\left(-1\right)+0\cdot \:2\\ 1\cdot \:3+0\cdot \:2&1\cdot \left(-1\right)+0\cdot \:2\end{pmatrix} \\ \begin{pmatrix}6&-2\\ 3&-1\end{pmatrix} \end{gathered}[/tex]Notice that each row is multiplied by each correspondent column.
And finally, the pair on the lower-right:
[tex]\begin{gathered} \begin{pmatrix}5&-2\\ \:\:\:4\:&-3\end{pmatrix}\times \begin{pmatrix}2&-1\\ \:\:3&1\end{pmatrix} \\ \begin{pmatrix}5\cdot \:2+\left(-2\right)\cdot \:3&5\left(-1\right)+\left(-2\right)\cdot \:1\\ 4\cdot \:2+\left(-3\right)\cdot \:3&4\left(-1\right)+\left(-3\right)\cdot \:1\end{pmatrix} \\ \begin{pmatrix}4&-7\\ -1&-7\end{pmatrix} \end{gathered}[/tex]) Thus, the se are theanswers .
