1. Given the matrix equation: $$\begin{pmatrix} 3 & 2 \\ 7 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 12 \\ 8 \end{pmatrix}$$ We want to find $x$ and $y$. 2. This corresponds to the system of linear equations: $$3x + 2y = 12$$ $$7x - 2y = 8$$ 3. Add the two equations to eliminate $y$: $$ (3x + 2y) + (7x - 2y) = 12 + 8 $$ $$ 3x + 7x + 2y - 2y = 20 $$ $$ 10x = 20 $$ $$ x = \frac{20}{10} = 2 $$ 4. Substitute $x = 2$ back into the first equation to find $y$: $$ 3(2) + 2y = 12 $$ $$ 6 + 2y = 12 $$ $$ 2y = 12 - 6 = 6 $$ $$ y = \frac{6}{2} = 3 $$ 5. So, the solution is: $$ x = 2, \quad y = 3 $$ 6. Next, find a $2 \times 3$ matrix $A = (a_{ij})$ where $a_{ij} = -1$ if $i = j$ and $a_{ij} = 1$ if $i \neq j$. 7. Since the matrix has 2 rows and 3 columns, index $i$ goes from 1 to 2, and $j$ from 1 to 3. 8. Fill the entries: - For row 1 ($i=1$): - $j=1$, $a_{11} = -1$ (since $i=j$) - $j=2$, $a_{12} = 1$ (since $i \neq j$) - $j=3$, $a_{13} = 1$ (since $i \neq j$) - For row 2 ($i=2$): - $j=1$, $a_{21} = 1$ (since $i \neq j$) - $j=2$, $a_{22} = -1$ (since $i=j$) - $j=3$, $a_{23} = 1$ (since $i \neq j$) 9. Thus the matrix is: $$ A = \begin{pmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \end{pmatrix} $$