Given
[tex]A = \begin{bmatrix}-1&2\\1&4\end{bmatrix} \text{ and } B = \begin{bmatrix}1&3\\3&2\end{bmatrix} \text{ and } C = \begin{bmatrix}2&-1\\2&0\end{bmatrix}[/tex]
we have
(1)
[tex]AB = \begin{bmatrix}-1&2\\1&4\end{bmatrix}\begin{bmatrix}1&3\\3&2\end{bmatrix} \\\\ AB = \begin{bmatrix}-1\times1+2\times3 & -1\times3+2\times2 \\ 1\times1+4\times3 & 1\times3 + 4\times2\end{bmatrix} \\\\ AB = \begin{bmatrix}5&1\\13&11\end{bmatrix}[/tex]
[tex]BA = \begin{bmatrix}1&3\\3&2\end{bmatrix}\begin{bmatrix}-1&2\\1&4\end{bmatrix} \\\\ BA = \begin{bmatrix}1\times(-1)+3\times1 & 1\times2+3\times4 \\ 3\times(-1)+2\times1 & 3\times2+2\times4\end{bmatrix} \\\\ BA = \begin{bmatrix}2&14\\-1&14\end{bmatrix}[/tex]
So AB ≠ BA. This is generally true for arbitrary matrices A and B, since matrix multplication is not commutative.
(2)
[tex]2B \cdot 2C = 2^2 \cdot BC = 4 BC[/tex]
so we have 2BC = 2B•2C = 4BC if and only if BC is the zero matrix.
[tex]BC = \begin{bmatrix}1&3\\3&2\end{bmatrix}\begin{bmatrix}2&-1\\2&0\end{bmatrix} \\\\ BC = \begin{bmatrix}1\times3+3\times2 & 1\times(-1)+3\times0\\3\times3+2\times2&3\times(-1)+2\times0\end{bmatrix} \\\\ BC = \begin{bmatrix}9&-1\\13&-3\end{bmatrix}[/tex]
So 2BC ≠ 2B•2C.
(3) Yes, (AB)C = A(BC) because matrix multplication is associative. You don't need to compute this, but just to confirm this result:
[tex](AB)C = \begin{bmatrix}5&1\\13&11\end{bmatrix}\begin{bmatrix}2&-1\\2&0\end{bmatrix} \\\\ (AB)C = \begin{bmatrix}5\times3+1\times2&5\times(-1)+1\times0\\13\times3+11\times2&13\times(-1)+11\times0\end{bmatrix} \\\\ (AB)C = \begin{bmatrix}17&-5\\61&-13\end{bmatrix}[/tex]
[tex]A(BC) = \begin{bmatrix}-1&2\\1&4\end{bmatrix}\begin{bmatrix}9&-1\\13&-3\end{bmatrix} \\\\ A(BC) = \begin{bmatrix}(-1)\times9+2\times13 & (-1)\times(-1)+2\times(-3)\\1\times9+4\times13&1\times(-1) + 4\times(-3)\end{bmatrix} \\\\ A(BC) = \begin{bmatrix}17&-5\\61&-13\end{bmatrix}[/tex]
