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(B) Let $A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3 \end{bmatrix}$. The adjoint of $A$ is the transpose of the cofactor matrix,$adj(A)

Vedclass · Accepted Answer

$\begin{bmatrix} 3 & -4 & -5 \ -9 & 1 & 4 \ -5 & 3 & 1 \end{bmatrix}$

Vedclass · Answer

(B) Let $A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & -3 \ 2 & -1 & 3 \end{bmatrix}$. The adjoint of $A$ is the transpose of the cofactor matrix,$adj(A) = [C_{ij}]^T = \begin{bmatrix} C_{11} & C_{21} & C_{31} \ C_{12} & C_{22} & C_{32} \ C_{13} & C_{23} & C_{33} \end{bmatrix}$.Calculating the cofactors:$C_{11} = +\begin{vmatrix} 2 & -3 \ -1 & 3 \end{vmatrix} = (6 - 3) = 3$$C_{12} = -\begin{vmatrix} 1 & -3 \ 2 & 3 \end{vmatrix} = -(3 + 6) = -9$$C_{13} = +\begin{vmatrix} 1 & 2 \ 2 & -1 \end{vmatrix} = (-1 - 4) = -5$$C_{21} = -\begin{vmatrix} 1 & 1 \ -1 & 3 \end{vmatrix} = -(3 + 1) = -4$$C_{22} = +\begin{vmatrix} 1 & 1 \ 2 & 3 \end{vmatrix} = (3 - 2) = 1$$C_{23} = -\begin{vmatrix} 1 & 1 \ 2 & -1 \end{vmatrix} = -(-1 - 2) = 3$$C_{31} = +\begin{vmatrix} 1 & 1 \ 2 & -3 \end{vmatrix} = (-3 - 2) = -5$$C_{32} = -\begin{vmatrix} 1 & 1 \ 1 & -3 \end{vmatrix} = -(-3 - 1) = 4$$C_{33} = +\begin{vmatrix} 1 & 1 \ 1 & 2 \end{vmatrix} = (2 - 1) = 1$Thus,$adj(A) = \begin{bmatrix} 3 & -4 & -5 \ -9 & 1 & 4 \ -5 & 3 & 1 \end{bmatrix}$. The correct option is $B$.

The adjoint of $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$ is

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