Question:
If
\(A = \begin{bmatrix} 1 & -2 & 2 0 & 2 & -3 3 & -2 & 4 \end{bmatrix}\),
find the value of the matrix expression \(A(I + \text{adj } A)\), where \(I\) is the identity matrix of the same order as \(A\).
If
\(A = \begin{bmatrix} 1 & -2 & 2 0 & 2 & -3 3 & -2 & 4 \end{bmatrix}\),
find the value of the matrix expression \(A(I + \text{adj } A)\), where \(I\) is the identity matrix of the same order as \(A\).
Show Hint
Always replace \(A\cdot \text{adj}(A)\) with \(|A|I\). It saves time and avoids the heavy computation of cofactors in higher-order matrices.
Updated On: May 19, 2026
- \( \begin{bmatrix} 8 & -2 & 2 0 & 8 & -3 3 & -2 & 8 \end{bmatrix} \)
- \( \begin{bmatrix} 9 & -2 & 2 0 & 10 & -3 3 & -2 & 12 \end{bmatrix} \)
- \( \begin{bmatrix} 1 & -16 & 16 0 & 16 & -24 24 & -16 & 32 \end{bmatrix} \)
- \( \begin{bmatrix} 9 & 0 & 0 0 & 10 & 0 0 & 0 & 12 \end{bmatrix} \)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
We use key matrix identities instead of directly computing the adjoint:
• Distributive law: \(A(B + C) = AB + AC\)
• Identity property: \(AI = A\)
• Adjoint identity: \(A \cdot \text{adj}(A) = |A|I\) So, \[ A(I + \text{adj}A) = A + A\cdot \text{adj}A = A + |A|I \]
Step 1: Simplifying expression.
\[ A(I + \text{adj}A) = AI + A\cdot \text{adj}A \] \[ = A + |A|I \] So we only need determinant of \(A\).
Step 2: Finding determinant of \(A\).
\[ A = \begin{bmatrix} 1 & -2 & 2 0 & 2 & -3 3 & -2 & 4 \end{bmatrix} \] Expand along first column: \[ |A| = 1 \begin{vmatrix} 2 & -3 -2 & 4 \end{vmatrix} + 3 \begin{vmatrix} -2 & 2 2 & -3 \end{vmatrix} \] Compute minors: \[ \begin{vmatrix} 2 & -3 -2 & 4 \end{vmatrix} = (2)(4) - (-3)(-2) = 8 - 6 = 2 \] \[ \begin{vmatrix} -2 & 2 2 & -3 \end{vmatrix} = (-2)(-3) - (2)(2) = 6 - 4 = 2 \] So, \[ |A| = 1(2) + 3(2) = 2 + 6 = 8 \]
Step 3: Compute final expression.
\[ A + 8I \] \[ 8I = \begin{bmatrix} 8 & 0 & 0 0 & 8 & 0 0 & 0 & 8 \end{bmatrix} \] \[ A + 8I = \begin{bmatrix} 1+8 & -2 & 2 0 & 2+8 & -3 3 & -2 & 4+8 \end{bmatrix} \] \[ = \begin{bmatrix} 9 & -2 & 2 0 & 10 & -3 3 & -2 & 12 \end{bmatrix} \]
• Distributive law: \(A(B + C) = AB + AC\)
• Identity property: \(AI = A\)
• Adjoint identity: \(A \cdot \text{adj}(A) = |A|I\) So, \[ A(I + \text{adj}A) = A + A\cdot \text{adj}A = A + |A|I \]
Step 1: Simplifying expression.
\[ A(I + \text{adj}A) = AI + A\cdot \text{adj}A \] \[ = A + |A|I \] So we only need determinant of \(A\).
Step 2: Finding determinant of \(A\).
\[ A = \begin{bmatrix} 1 & -2 & 2 0 & 2 & -3 3 & -2 & 4 \end{bmatrix} \] Expand along first column: \[ |A| = 1 \begin{vmatrix} 2 & -3 -2 & 4 \end{vmatrix} + 3 \begin{vmatrix} -2 & 2 2 & -3 \end{vmatrix} \] Compute minors: \[ \begin{vmatrix} 2 & -3 -2 & 4 \end{vmatrix} = (2)(4) - (-3)(-2) = 8 - 6 = 2 \] \[ \begin{vmatrix} -2 & 2 2 & -3 \end{vmatrix} = (-2)(-3) - (2)(2) = 6 - 4 = 2 \] So, \[ |A| = 1(2) + 3(2) = 2 + 6 = 8 \]
Step 3: Compute final expression.
\[ A + 8I \] \[ 8I = \begin{bmatrix} 8 & 0 & 0 0 & 8 & 0 0 & 0 & 8 \end{bmatrix} \] \[ A + 8I = \begin{bmatrix} 1+8 & -2 & 2 0 & 2+8 & -3 3 & -2 & 4+8 \end{bmatrix} \] \[ = \begin{bmatrix} 9 & -2 & 2 0 & 10 & -3 3 & -2 & 12 \end{bmatrix} \]
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