Question

If $A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$ then $(A^{-1}) =$

• A. $A$
• B. $-A$
• C. $-2A$
• D. $0$

Hint:

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.

Answer 1

The Correct Option is A

To find $A^{-1}$, we can use the property that if $A \cdot A = I$ (where $I$ is the identity matrix), then $A$ is its own inverse ($A = A^{-1}$). 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:

• $(1 \times 0) + (0 \times 0) + (0 \times 1) = 0$

• $(1 \times 0) + (0 \times 1) + (0 \times 0) = 0$

• $(1 \times 1) + (0 \times 0) + (0 \times 0) = 1$

2. Conclusion: The resulting matrix is: $$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$$ Since $A \cdot A = I$, by definition of an inverse matrix, $A^{-1} = A$.