Question:
If $A = \begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$ then $A(I + \text{adj } A) =$
If $A = \begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$ then $A(I + \text{adj } A) =$
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Property: $A \cdot \text{adj}(A) = |A|I$.
Updated On: Apr 30, 2026
- $\begin{bmatrix} 9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 11 \end{bmatrix}$
- $\begin{bmatrix} 8 & -2 & 2 \\ 0 & 9 & -3 \\ 3 & -2 & 10 \end{bmatrix}$
- $\begin{bmatrix} 9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 12 \end{bmatrix}$
- $\begin{bmatrix} 3 & 2 & -2 \\ 0 & 10 & 3 \\ -3 & 2 & 12 \end{bmatrix}$
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The Correct Option is A
Solution and Explanation
Step 1: Simplify the Expression
\[ A(I + \text{adj } A) = AI + A(\text{adj } A) = A + |A|I \]
Step 2: Calculate Determinant \( |A| \)
\[ |A| = 1(8 - 6) - (-2)(0 - (-9)) + 2(0 - 6) \]
\[ |A| = 1(2) + 2(9) + 2(-6) = 2 + 18 - 12 = 8 \]
Step 3: Calculate \( A + 8I \)
\[ \begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} + \begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix} = \begin{bmatrix} 9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 12 \end{bmatrix} \]
Final Answer: (C)
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