1. **Problem statement:** Given matrices $$X=\begin{pmatrix}1 & 2 \\ 3 & 0\end{pmatrix}, Y=\begin{pmatrix}4 & 1 \\ 2 & 5\end{pmatrix}, Z=\begin{pmatrix}0 & 3 \\ 1 & 2\end{pmatrix}$$ Compute: (a) $3X - Z$ (b) $ZY$ (c) $X^{-1}$ 2. **Formula and rules:** - Scalar multiplication: multiply each element by the scalar. - Matrix subtraction: subtract corresponding elements. - Matrix multiplication: multiply rows of first matrix by columns of second matrix. - Inverse of 2x2 matrix $A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ is $A^{-1} = \frac{1}{ad - bc} \begin{pmatrix}d & -b \\ -c & a\end{pmatrix}$ if $ad - bc \neq 0$. 3. **Calculate (a) $3X - Z$:** - $3X = 3 \times \begin{pmatrix}1 & 2 \\ 3 & 0\end{pmatrix} = \begin{pmatrix}3 & 6 \\ 9 & 0\end{pmatrix}$ - $3X - Z = \begin{pmatrix}3 & 6 \\ 9 & 0\end{pmatrix} - \begin{pmatrix}0 & 3 \\ 1 & 2\end{pmatrix} = \begin{pmatrix}3-0 & 6-3 \\ 9-1 & 0-2\end{pmatrix} = \begin{pmatrix}3 & 3 \\ 8 & -2\end{pmatrix}$ 4. **Calculate (b) $ZY$:** - Multiply $Z=\begin{pmatrix}0 & 3 \\ 1 & 2\end{pmatrix}$ by $Y=\begin{pmatrix}4 & 1 \\ 2 & 5\end{pmatrix}$: $$ZY = \begin{pmatrix}0\times4 + 3\times2 & 0\times1 + 3\times5 \\ 1\times4 + 2\times2 & 1\times1 + 2\times5\end{pmatrix} = \begin{pmatrix}6 & 15 \\ 8 & 11\end{pmatrix}$$ 5. **Calculate (c) $X^{-1}$:** - For $X=\begin{pmatrix}1 & 2 \\ 3 & 0\end{pmatrix}$, determinant $det = 1\times0 - 3\times2 = 0 - 6 = -6$ - Since $det \neq 0$, inverse exists. - $X^{-1} = \frac{1}{-6} \begin{pmatrix}0 & -2 \\ -3 & 1\end{pmatrix} = \begin{pmatrix}0 & \frac{1}{3} \\ \frac{1}{2} & -\frac{1}{6}\end{pmatrix}$ **Final answers:** (a) $3X - Z = \begin{pmatrix}3 & 3 \\ 8 & -2\end{pmatrix}$ (b) $ZY = \begin{pmatrix}6 & 15 \\ 8 & 11\end{pmatrix}$ (c) $X^{-1} = \begin{pmatrix}0 & \frac{1}{3} \\ \frac{1}{2} & -\frac{1}{6}\end{pmatrix}$