1. Stating the problem: Consider matrix $A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$. We need to find determinants for $A$, $4A$, and $\begin{bmatrix} 11 & 1 \\ -9 & -1 \end{bmatrix}$. Then, use the classical adjoint to find $A^{-1}$, solve a system using $A^{-1}$, and verify using Cramer's Rule. 2. Finding determinants: (i) Determinant of $A$: $$\det(A) = (1)(-1) - (1)(1) = -1 - 1 = -2.$$ (ii) Determinant of $4A$: Since $4A = 4 \times A$, each element multiplied by 4 in a $2 \times 2$ matrix, determinant scales by $4^2=16$: $$\det(4A) = 16 \times \det(A) = 16 \times (-2) = -32.$$ (iii) Determinant of $\begin{bmatrix} 11 & 1 \\ -9 & -1 \end{bmatrix}$: $$\det = (11)(-1) - (1)(-9) = -11 + 9 = -2.$$ 3. Computing inverse using classical adjoint: The classical adjoint $\operatorname{adj}(A)$ is the transpose of the cofactor matrix. Cofactors: $$C_{11} = -1, C_{12} = -1, C_{21} = -1, C_{22} = 1.$$ Thus, $$\operatorname{adj}(A) = \begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix}.$$ The inverse: $$A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) = \frac{1}{-2} \begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}.$$ 4. Solving system using $A^{-1}$: System: $$\begin{cases}-x + y = 1 \\ x + y = 1\end{cases}$$ Matrix form: $AX = B$ where $$A=\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 1 \end{bmatrix}.$$ Multiply both sides by $A^{-1}$: $$X = A^{-1} B = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \times 1 + \frac{1}{2} \times 1 \\ \frac{1}{2} \times 1 - \frac{1}{2} \times 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}.$$ So, solution is $x=1$, $y=0$. 5. Verifying solution via Cramer's Rule: $$\det(A) = -2.$$ Replace first column of $A$ by $B$ to get $A_x$: $$A_x = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = A$$ Wait, original matrix $A$ same, but original A first column is $(1,1)$, $A_x$ replaces first column with $B=(1,1)$ so matrix is actually the same as $A$. So, $$\det(A_x) = \det(A) = -2.$$ Replace second column by $B$ for $A_y$: $$A_y = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix},$$ $$\det(A_y) = (1)(1) - (1)(1) = 0.$$ By Cramer's Rule: $$x = \frac{\det(A_x)}{\det(A)} = \frac{-2}{-2} = 1,$$ $$y = \frac{\det(A_y)}{\det(A)} = \frac{0}{-2} = 0.$$ This matches the solution found using inverse.