Matrices

Concept:The adjoint of a matrix is the transpose of its cofactor matrix.Explanation:First, compute the cofactor for each element of $A$. For element $a_{11}=1$, minor = $\det\begin{bmatrix}5 & 6 \\ 8 & 9\end{bmatrix} = 45-48 = -3$, cofactor $C_{11}=(-1)^{1+1}(-3) = -3$. For $a_{12}=2$, minor = $\det\begin{bmatrix}4 & 6 \\ 7 & 9\end{bmatrix} = 36-42 = -6$, $C_{12}=(-1)^{1+2}(-6)=6$. For $a_{13}=3$, minor = $\det\begin{bmatrix}4 & 5 \\ 7 & 8\end{bmatrix} = 32-35 = -3$, $C_{13}=(-1)^{1+3}(-3)=-3$. For $a_{21}=4$, minor = $\det\begin{bmatrix}2 & 3 \\ 8 & 9\end{bmatrix} = 18-24 = -6$, $C_{21}=(-1)^{2+1}(-6)=6$. For $a_{22}=5$, minor = $\det\begin{bmatrix}1 & 3 \\ 7 & 9\end{bmatrix} = 9-21 = -12$, $C_{22}=(-1)^{2+2}(-12)=-12$. For $a_{23}=6$, minor = $\det\begin{bmatrix}1 & 2 \\ 7 & 8\end{bmatrix} = 8-14 = -6$, $C_{23}=(-1)^{2+3}(-6)=6$. For $a_{31}=7$, minor = $\det\begin{bmatrix}2 & 3 \\ 5 & 6\end{bmatrix} = 12-15 = -3$, $C_{31}=(-1)^{3+1}(-3)=-3$. For $a_{32}=8$, minor = $\det\begin{bmatrix}1 & 3 \\ 4 & 6\end{bmatrix} = 6-12 = -6$, $C_{32}=(-1)^{3+2}(-6)=6$. For $a_{33}=9$, minor = $\det\begin{bmatrix}1 & 2 \\ 4 & 5\end{bmatrix} = 5-8 = -3$, $C_{33}=(-1)^{3+3}(-3)=-3$. Thus the cofactor matrix is $\begin{bmatrix} -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 \end{bmatrix}$. The adjoint is the transpose of this matrix, which remains the same because it is symmetric.Answer:$\operatorname{adj} A = \begin{bmatrix} -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 \end{bmatrix}$, which matches option C.