1. **State the problem:** Find the minor and cofactor of entry $a_{11}$ in matrix $$A = \begin{bmatrix} 5 & -3 & 2 \\ -9 & 5 & -4 \\ -3 & 1 & -2 \end{bmatrix}$$ Then find the determinant of $A$ and determine if $A$ is singular or nonsingular. 2. **Formula and rules:** - The minor $M_{ij}$ of entry $a_{ij}$ is the determinant of the matrix formed by deleting the $i$-th row and $j$-th column. - The cofactor $C_{ij} = (-1)^{i+j} M_{ij}$. - The determinant of a $3 \times 3$ matrix $A$ with entries $a_{ij}$ is $$\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}.$$ - A matrix is singular if its determinant is zero; otherwise, it is nonsingular. 3. **Find the minor $M_{11}$:** Delete row 1 and column 1: $$\begin{bmatrix} 5 & -4 \\ 1 & -2 \end{bmatrix}$$ Calculate its determinant: $$M_{11} = (5)(-2) - (-4)(1) = -10 + 4 = -6.$$ 4. **Find the cofactor $C_{11}$:** $$C_{11} = (-1)^{1+1} M_{11} = (+1)(-6) = -6.$$ 5. **Find the determinant of $A$ using the first row:** Calculate cofactors $C_{12}$ and $C_{13}$: - Minor $M_{12}$: delete row 1, column 2: $$\begin{bmatrix} -9 & -4 \\ -3 & -2 \end{bmatrix}$$ $$M_{12} = (-9)(-2) - (-4)(-3) = 18 - 12 = 6.$$ $$C_{12} = (-1)^{1+2} M_{12} = -6.$$ - Minor $M_{13}$: delete row 1, column 3: $$\begin{bmatrix} -9 & 5 \\ -3 & 1 \end{bmatrix}$$ $$M_{13} = (-9)(1) - (5)(-3) = -9 + 15 = 6.$$ $$C_{13} = (-1)^{1+3} M_{13} = (+1)(6) = 6.$$ 6. **Calculate determinant:** $$\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} = 5(-6) + (-3)(-6) + 2(6) = -30 + 18 + 12 = 0.$$ 7. **Conclusion:** Since $$\det(A) = 0,$$ matrix $A$ is singular. **Final answers:** - Minor of $a_{11}$: $-6$ - Cofactor of $a_{11}$: $-6$ - Determinant of $A$: $0$ - Matrix $A$ is singular.