Question

The adjoint of $\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \\ \end{matrix} \right]$ is [RPET 1993]
A. $\left[ \begin{matrix} 3 & -9 & -5 \\ -4 & 1 & 3 \\ -5 & 4 & 1 \\ \end{matrix} \right]$
B. $\left[ \begin{matrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \\ \end{matrix} \right]$
C. $\left[ \begin{matrix} -3 & \,\,4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \\ \end{matrix} \right]$
D. None of these

Answer 1

Hint:
When the co-factor members of a matrix are transposed, the adjoint of the matrix is produced. Here We will first evaluate the cofactor of every element of the matrix. By transposing the co-factor elements of the given matrix, one can obtain the adjoint of a matrix.

Formula Used:
Adjoint of $3 \times 3$ Matrix is given by:
$adjA=\begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{22} & A_{33} \end{bmatrix}^T=\begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix}$
Where $\begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{22} & A_{33} \end{bmatrix}$ is cofactor matrix of $A$.
Cofactor of the element is given by:
$C_{ij}=(-1)^{i+j}det(M_{ij})$
here, $det(M_{ij})$ is the minor of $a_{ij}$

Complete step-by-step solution:
Let $A=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \\ \end{matrix} \right]$
Finding the cofactors of every element of matrix $A$:
$C_{11}=(-1)^{1+1}\begin{vmatrix}2 & -3 \\-1 & 3 \end{vmatrix}\\
\Rightarrow C_{11}=3\\
C_{12}=(-1)^{1+2}\begin{vmatrix}1 & -3 \\2 & 3 \end{vmatrix}\\
\Rightarrow C_{12}=-9\\
C_{13}=(-1)^{1+3}\begin{vmatrix}1 & 2 \\2 & -1 \end{vmatrix}\\
\Rightarrow C_{13}=-5$

$C_{21}=(-1)^{2+1}\begin{vmatrix}1 & 1 \\-1 & 3 \end{vmatrix}\\
\Rightarrow C_{21}=-4\\
C_{22}=(-1)^{2+2}\begin{vmatrix}1 & 1 \\2 & 3 \end{vmatrix}\\
\Rightarrow C_{22}=1\\
C_{23}=(-1)^{2+3}\begin{vmatrix}1 & 1 \\2 & -1 \end{vmatrix}\\
\Rightarrow C_{23}=3$

$C_{31}=(-1)^{3+1}\begin{vmatrix}1 & 1 \\2 & -3 \end{vmatrix}\\
\Rightarrow C_{31}=-5\\
C_{32}=(-1)^{3+2}\begin{vmatrix}1 & 1 \\1 & 3 \end{vmatrix}\\
\Rightarrow C_{32}=4\\
C_{33}=(-1)^{3+3}\begin{vmatrix}1 & 1 \\1 & 2 \end{vmatrix}\\
\Rightarrow C_{33}=1$

Take the transposition of the cofactor matrix to determine the adjoint.

So, the cofactor matrix $C=\left[ \begin{matrix} 3 & -9 & -5 \\ -4 & 1 & 3 \\ -5 & 4 & 1 \\ \end{matrix} \right]$
Therefore,
$Adj A=C^T\\
Adj\,(A)=\left[ \begin{matrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \\ \end{matrix} \right]$

So, option is B correct.

Note:
Remember the adjoint of a matrix is created when the co-factor matrix is transposed. The determinant obtained by removing the row and column in which an element appears in a matrix is known as the minor of that element. A cofactor is a number that is obtained by removing a specific element's row and column in the shape of a square or rectangle. Depending on the element's position, a positive or negative sign comes before the cofactor.