Question:medium

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.
Updated On: Jul 1, 2026
  • $A$
  • $-A$
  • $-2A$
  • $0$
Show Solution

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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