Question:
If
\[
A=
\begin{bmatrix}
2 & -3 & 1\\
-3 & 1 & -2\\
1 & -2 & 3
\end{bmatrix},
\]
then \(\operatorname{Adj}(A)=\)
If
\[
A=
\begin{bmatrix}
2 & -3 & 1\\
-3 & 1 & -2\\
1 & -2 & 3
\end{bmatrix},
\]
then \(\operatorname{Adj}(A)=\)
Show Hint
To find the adjoint of a \(3\times3\) matrix, first compute all nine cofactors carefully, paying attention to the signs \((+,-,+;\,-,+,-;\,+,-,+)\), and then take the transpose of the cofactor matrix.
Updated On: Jul 29, 2026
- \[ \begin{bmatrix} 1 & -7 & 5\\ -7 & 5 & -1\\ 5 & -1 & 7 \end{bmatrix} \]
- \[ \begin{bmatrix} 1 & 7 & -5\\ 7 & 5 & 1\\ -5 & 1 & 7 \end{bmatrix} \]
- \[ \begin{bmatrix} -1 & 7 & 5\\ 7 & 5 & 1\\ 5 & 1 & -7 \end{bmatrix} \]
- \[ \begin{bmatrix} -1 & 7 & -5\\ 7 & -5 & 1\\ -5 & 1 & -7 \end{bmatrix} \]
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
The adjoint of a matrix is the transpose of its cofactor matrix.
\[
\operatorname{Adj}(A)=\left[C_{ij}\right]^T
\]
where \(C_{ij}\) denotes the cofactor of the element \(a_{ij}\).
Step 1: Find the cofactors of the first row. \[ C_{11}= \begin{vmatrix} 1 & -2\\ -2 & 3 \end{vmatrix} =3-4=-1 \] \[ C_{12}=- \begin{vmatrix} -3 & -2\\ 1 & 3 \end{vmatrix} =-(-9+2)=7 \] \[ C_{13}= \begin{vmatrix} -3 & 1\\ 1 & -2 \end{vmatrix} =6-1=5 \]
Step 2: Find the cofactors of the second row. \[ C_{21}=- \begin{vmatrix} -3 & 1\\ -2 & 3 \end{vmatrix} =-(-9+2)=7 \] \[ C_{22}= \begin{vmatrix} 2 & 1\\ 1 & 3 \end{vmatrix} =6-1=5 \] \[ C_{23}=- \begin{vmatrix} 2 & -3\\ 1 & -2 \end{vmatrix} =-(-4+3)=1 \]
Step 3: Find the cofactors of the third row. \[ C_{31}= \begin{vmatrix} -3 & 1\\ 1 & -2 \end{vmatrix} =6-1=5 \] \[ C_{32}=- \begin{vmatrix} 2 & 1\\ -3 & -2 \end{vmatrix} =-(-4+3)=1 \] \[ C_{33}= \begin{vmatrix} 2 & -3\\ -3 & 1 \end{vmatrix} =2-9=-7 \]
Step 4: Form the cofactor matrix and hence the adjoint. The cofactor matrix is \[ \begin{bmatrix} -1 & 7 & 5\\ 7 & 5 & 1\\ 5 & 1 & -7 \end{bmatrix}. \] Since this matrix is symmetric, its transpose is the same matrix. Therefore, \[ \operatorname{Adj}(A)= \begin{bmatrix} -1 & 7 & 5\\ 7 & 5 & 1\\ 5 & 1 & -7 \end{bmatrix}. \] \[ \boxed{ \operatorname{Adj}(A)= \begin{bmatrix} -1 & 7 & 5\\ 7 & 5 & 1\\ 5 & 1 & -7 \end{bmatrix} } \]
Step 1: Find the cofactors of the first row. \[ C_{11}= \begin{vmatrix} 1 & -2\\ -2 & 3 \end{vmatrix} =3-4=-1 \] \[ C_{12}=- \begin{vmatrix} -3 & -2\\ 1 & 3 \end{vmatrix} =-(-9+2)=7 \] \[ C_{13}= \begin{vmatrix} -3 & 1\\ 1 & -2 \end{vmatrix} =6-1=5 \]
Step 2: Find the cofactors of the second row. \[ C_{21}=- \begin{vmatrix} -3 & 1\\ -2 & 3 \end{vmatrix} =-(-9+2)=7 \] \[ C_{22}= \begin{vmatrix} 2 & 1\\ 1 & 3 \end{vmatrix} =6-1=5 \] \[ C_{23}=- \begin{vmatrix} 2 & -3\\ 1 & -2 \end{vmatrix} =-(-4+3)=1 \]
Step 3: Find the cofactors of the third row. \[ C_{31}= \begin{vmatrix} -3 & 1\\ 1 & -2 \end{vmatrix} =6-1=5 \] \[ C_{32}=- \begin{vmatrix} 2 & 1\\ -3 & -2 \end{vmatrix} =-(-4+3)=1 \] \[ C_{33}= \begin{vmatrix} 2 & -3\\ -3 & 1 \end{vmatrix} =2-9=-7 \]
Step 4: Form the cofactor matrix and hence the adjoint. The cofactor matrix is \[ \begin{bmatrix} -1 & 7 & 5\\ 7 & 5 & 1\\ 5 & 1 & -7 \end{bmatrix}. \] Since this matrix is symmetric, its transpose is the same matrix. Therefore, \[ \operatorname{Adj}(A)= \begin{bmatrix} -1 & 7 & 5\\ 7 & 5 & 1\\ 5 & 1 & -7 \end{bmatrix}. \] \[ \boxed{ \operatorname{Adj}(A)= \begin{bmatrix} -1 & 7 & 5\\ 7 & 5 & 1\\ 5 & 1 & -7 \end{bmatrix} } \]
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