Question:
If
\[
A=
\begin{bmatrix}
1 & -2 & 2 \\
2 & -6 & 5 \\
5 & 0 & 4
\end{bmatrix},
\]
then \(\operatorname{Adj}(A)=\)
If
\[
A=
\begin{bmatrix}
1 & -2 & 2 \\
2 & -6 & 5 \\
5 & 0 & 4
\end{bmatrix},
\]
then \(\operatorname{Adj}(A)=\)
Show Hint
To find the adjoint of a matrix, first compute all cofactors, form the cofactor matrix, and then take its transpose.
Updated On: Jun 26, 2026
- \[ \begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & -1 \\ 30 & -10 & 2 \end{bmatrix} \]
- \[ \begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & 1 \\ -30 & 10 & -2 \end{bmatrix} \]
- \[ \begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & -1 \\ 30 & -10 & -2 \end{bmatrix} \]
- \[ \begin{bmatrix} 24 & -8 & 2 \\ -17 & -6 & 1 \\ 30 & -10 & -2 \end{bmatrix} \]
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Find the cofactors of the first row.
\[ C_{11}= \begin{vmatrix} -6 & 5 \\ 0 & 4 \end{vmatrix} =-24 \] \[ C_{12} =-\begin{vmatrix} 2 & 5 \\ 5 & 4 \end{vmatrix} =-(8-25)=17 \] \[ C_{13} = \begin{vmatrix} 2 & -6 \\ 5 & 0 \end{vmatrix} =30 \]
Step 2: Find the cofactors of the second row.
\[ C_{21} =-\begin{vmatrix} -2 & 2 \\ 0 & 4 \end{vmatrix} =8 \] \[ C_{22} = \begin{vmatrix} 1 & 2 \\ 5 & 4 \end{vmatrix} =4-10=-6 \] \[ C_{23} =-\begin{vmatrix} 1 & -2 \\ 5 & 0 \end{vmatrix} =-10 \]
Step 3: Find the cofactors of the third row.
\[ C_{31} = \begin{vmatrix} -2 & 2 \\ -6 & 5 \end{vmatrix} =-10+12=2 \] \[ C_{32} =-\begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} =-(5-4)=-1 \] \[ C_{33} = \begin{vmatrix} 1 & -2 \\ 2 & -6 \end{vmatrix} =-6+4=-2 \]
Step 4: Form the cofactor matrix and take transpose.
The cofactor matrix is \[ C= \begin{bmatrix} -24 & 17 & 30 \\ 8 & -6 & -10 \\ 2 & -1 & -2 \end{bmatrix} \] Hence, \[ \operatorname{Adj}(A)=C^T = \begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & -1 \\ 30 & -10 & -2 \end{bmatrix} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{ \operatorname{Adj}(A)= \begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & -1 \\ 30 & -10 & -2 \end{bmatrix} } \] Hence, the correct option is \[ \boxed{(3)} \]
\[ C_{11}= \begin{vmatrix} -6 & 5 \\ 0 & 4 \end{vmatrix} =-24 \] \[ C_{12} =-\begin{vmatrix} 2 & 5 \\ 5 & 4 \end{vmatrix} =-(8-25)=17 \] \[ C_{13} = \begin{vmatrix} 2 & -6 \\ 5 & 0 \end{vmatrix} =30 \]
Step 2: Find the cofactors of the second row.
\[ C_{21} =-\begin{vmatrix} -2 & 2 \\ 0 & 4 \end{vmatrix} =8 \] \[ C_{22} = \begin{vmatrix} 1 & 2 \\ 5 & 4 \end{vmatrix} =4-10=-6 \] \[ C_{23} =-\begin{vmatrix} 1 & -2 \\ 5 & 0 \end{vmatrix} =-10 \]
Step 3: Find the cofactors of the third row.
\[ C_{31} = \begin{vmatrix} -2 & 2 \\ -6 & 5 \end{vmatrix} =-10+12=2 \] \[ C_{32} =-\begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} =-(5-4)=-1 \] \[ C_{33} = \begin{vmatrix} 1 & -2 \\ 2 & -6 \end{vmatrix} =-6+4=-2 \]
Step 4: Form the cofactor matrix and take transpose.
The cofactor matrix is \[ C= \begin{bmatrix} -24 & 17 & 30 \\ 8 & -6 & -10 \\ 2 & -1 & -2 \end{bmatrix} \] Hence, \[ \operatorname{Adj}(A)=C^T = \begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & -1 \\ 30 & -10 & -2 \end{bmatrix} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{ \operatorname{Adj}(A)= \begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & -1 \\ 30 & -10 & -2 \end{bmatrix} } \] Hence, the correct option is \[ \boxed{(3)} \]
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