Question:
If
\[
A=
\begin{bmatrix}
1 & -3 & 2\\
-2 & 1 & 3\\
3 & 2 & -1
\end{bmatrix}
\]
then \(A^2\operatorname{Adj}A=\)
If
\[
A=
\begin{bmatrix}
1 & -3 & 2\\
-2 & 1 & 3\\
3 & 2 & -1
\end{bmatrix}
\]
then \(A^2\operatorname{Adj}A=\)
Show Hint
Always remember the identity:
\[
A\operatorname{Adj}A=|A|I
\]
Using this, expressions like \(A^2\operatorname{Adj}A\) can be simplified as \(|A|A\).
Updated On: Jul 18, 2026
- \(21A\)
- \(-42A\)
- \(7A^{-1}\)
- \(14(\operatorname{Adj}A)\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Use the property of adjoint matrix.
We know that for any square matrix \(A\), \[ A\operatorname{Adj}A=|A|I \] Therefore, \[ A^2\operatorname{Adj}A=A(A\operatorname{Adj}A) \] So, \[ A^2\operatorname{Adj}A=A(|A|I) \] Hence, \[ A^2\operatorname{Adj}A=|A|A \]
Step 2: Find the determinant of \(A\).
\[ |A|= \begin{vmatrix} 1 & -3 & 2\\ -2 & 1 & 3\\ 3 & 2 & -1 \end{vmatrix} \] Expanding along the first row, \[ |A|=1 \begin{vmatrix} 1 & 3\\ 2 & -1 \end{vmatrix} -(-3) \begin{vmatrix} -2 & 3\\ 3 & -1 \end{vmatrix} +2 \begin{vmatrix} -2 & 1\\ 3 & 2 \end{vmatrix} \] Now, \[ \begin{vmatrix} 1 & 3\\ 2 & -1 \end{vmatrix} =1(-1)-3(2)=-1-6=-7 \] \[ \begin{vmatrix} -2 & 3\\ 3 & -1 \end{vmatrix} =(-2)(-1)-3(3)=2-9=-7 \] \[ \begin{vmatrix} -2 & 1\\ 3 & 2 \end{vmatrix} =(-2)(2)-1(3)=-4-3=-7 \] Therefore, \[ |A|=1(-7)+3(-7)+2(-7) \] \[ |A|=-7-21-14=-42 \]
Step 3: Substitute the determinant value.
Since, \[ A^2\operatorname{Adj}A=|A|A \] and \[ |A|=-42 \] we get \[ A^2\operatorname{Adj}A=-42A \]
Step 4: Final conclusion.
Therefore, \[ \boxed{-42A} \]
We know that for any square matrix \(A\), \[ A\operatorname{Adj}A=|A|I \] Therefore, \[ A^2\operatorname{Adj}A=A(A\operatorname{Adj}A) \] So, \[ A^2\operatorname{Adj}A=A(|A|I) \] Hence, \[ A^2\operatorname{Adj}A=|A|A \]
Step 2: Find the determinant of \(A\).
\[ |A|= \begin{vmatrix} 1 & -3 & 2\\ -2 & 1 & 3\\ 3 & 2 & -1 \end{vmatrix} \] Expanding along the first row, \[ |A|=1 \begin{vmatrix} 1 & 3\\ 2 & -1 \end{vmatrix} -(-3) \begin{vmatrix} -2 & 3\\ 3 & -1 \end{vmatrix} +2 \begin{vmatrix} -2 & 1\\ 3 & 2 \end{vmatrix} \] Now, \[ \begin{vmatrix} 1 & 3\\ 2 & -1 \end{vmatrix} =1(-1)-3(2)=-1-6=-7 \] \[ \begin{vmatrix} -2 & 3\\ 3 & -1 \end{vmatrix} =(-2)(-1)-3(3)=2-9=-7 \] \[ \begin{vmatrix} -2 & 1\\ 3 & 2 \end{vmatrix} =(-2)(2)-1(3)=-4-3=-7 \] Therefore, \[ |A|=1(-7)+3(-7)+2(-7) \] \[ |A|=-7-21-14=-42 \]
Step 3: Substitute the determinant value.
Since, \[ A^2\operatorname{Adj}A=|A|A \] and \[ |A|=-42 \] we get \[ A^2\operatorname{Adj}A=-42A \]
Step 4: Final conclusion.
Therefore, \[ \boxed{-42A} \]
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