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If $A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$ then $(A^{-1}) =$
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This specific matrix is an "Involutary Matrix." Any matrix that represents a reflection (like this one, which swaps the 1st and 3rd components) will always be its own inverse.
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Updated On:
Jul 1, 2026
$A$
$-A$
$-2A$
$0$
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The Correct Option is
A
Solution and Explanation
1. Calculate $A^2$ ($A \cdot A$):
$$\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$ Multiplying Row 1 by each column:
• $(0 \times 0) + (0 \times 0) + (1 \times 1) = 1$
• $(0 \times 0) + (0 \times 1) + (1 \times 0) = 0$
• $(0 \times 1) + (0 \times 0) + (1 \times 0) = 0$
Multiplying Row 2 by each column:
• $(0 \times 0) + (1 \times 0) + (0 \times 1) = 0$
• $(0 \times 0) + (1 \times 1) + (0 \times 0) = 1$
• $(0 \times 1) + (1 \times 0) + (0 \times 0) = 0$
Multiplying Row 3 by each column:
• $(1 \times 0) + (0 \times 0) + (0 \times 1) = 0$
• $(1 \times 0) + (0 \times 1) + (0 \times 0) = 0$
• $(1 \times 1) + (0 \times 0) + (0 \times 0) = 1$
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