Answer
(a) incompatible dimensions.
(b) incompatible dimensions.
(c) $\begin{bmatrix} 6&10\\3&-2\\-3&9 \end{bmatrix}$
(d) $\begin{bmatrix} 36&58\\0&-3\\18&28 \end{bmatrix}$
(e) $\begin{bmatrix} 2&-\frac{3}{2}\\-1&1 \end{bmatrix}$
(f) not a square matrix.
(g) not a square matrix.
(h) $-3$
Work Step by Step
The following matrices are given:
$\begin{array} \\A= \\ \end{array}
\begin{bmatrix} 2&3\\2&4 \end{bmatrix},
\begin{array} \\B= \\ \end{array}
\begin{bmatrix} 2&4\\-1&1\\3&0 \end{bmatrix},
\begin{array} \\C= \\ \end{array}
\begin{bmatrix} 1&0&4\\-1&1&2\\0&1&3 \end{bmatrix}$
(a) $\begin{array} \\A+B= \\ \end{array}
\begin{bmatrix} 2&3\\2&4 \end{bmatrix}
\begin{array} \\+ \\ \end{array}
\begin{bmatrix} 2&4\\-1&1\\3&0 \end{bmatrix}$
This can not be performed as the two matrices are of incompatible dimensions.
(b) $\begin{array} \\AB= \\ \end{array}
\begin{bmatrix} 2&3\\2&4 \end{bmatrix}
\begin{bmatrix} 2&4\\-1&1\\3&0 \end{bmatrix}$
The product of $2\times2$ and $3\times2$ can not be performed due to incompatible dimensions.
(c) $\begin{array} \\BA= \\ \end{array}
\begin{bmatrix} 2&4\\-1&1\\3&0 \end{bmatrix}
\begin{bmatrix} 2&3\\2&4 \end{bmatrix}=\begin{bmatrix} 12&22\\0&1\\6&9 \end{bmatrix}$
$\begin{array} \\BA-3B= \\ \end{array}
=\begin{bmatrix} 12&22\\0&1\\6&9 \end{bmatrix}-3\begin{bmatrix} 2&4\\-1&1\\3&0 \end{bmatrix}
=\begin{bmatrix} 6&10\\3&-2\\-3&9 \end{bmatrix}$
(d) $\begin{array} \\CB= \\ \end{array}
\begin{bmatrix} 1&0&4\\-1&1&2\\0&1&3 \end{bmatrix}\begin{bmatrix} 2&4\\-1&1\\3&0 \end{bmatrix}=\begin{bmatrix} 14&4\\3&-3\\8&1 \end{bmatrix}$
$\begin{array} \\CBA= \\ \end{array}
\begin{bmatrix} 14&4\\3&-3\\8&1 \end{bmatrix}\begin{bmatrix} 2&3\\2&4 \end{bmatrix}
=\begin{bmatrix} 36&58\\0&-3\\18&28 \end{bmatrix}$
(e) Use the inverse formula for a 2x2 matrix:
$\begin{array} \\A^{-1}=\frac{1}{8-6} \\ \end{array}
\begin{bmatrix} 4&-3\\-2&2 \end{bmatrix}
=\begin{bmatrix} 2&-\frac{3}{2}\\-1&1 \end{bmatrix}$
(f) $B^{-1}$ can not be derived as it is not a square matrix.
(g) $det(B)$ can not be evaluated as it is not a square matrix.
(h) Expand with row-1:
$\begin{array} \\det(C)= \\ \end{array}
\begin{vmatrix} 1&0&4\\-1&1&2\\0&1&3 \end{vmatrix}=1(3-2)+4(-1-0)=-3$
