Question

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix}$, then the value of the determinant of $A^{-1}$ is

• A. $-6$
• B. $-\frac{1}{6}$
• C. $\frac{1}{36}$
• D. $36$

Hint:

Never spend time calculating the adjoint or inverse matrix if the question only asks for the determinant of an inverse or power of a matrix. Always use determinant properties like $|A^{-1}| = |A|^{-1}$ or $|A^n| = |A|^n$ to save valuable time.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem provides a $3 \times 3$ square matrix $A$ and asks us to find the determinant of its inverse matrix, denoted as $|A^{-1}|$.

Step 2: Key Formula or Approach:
Instead of finding the full inverse matrix $A^{-1}$ explicitly, we can use a fundamental property of determinants: $$|A^{-1}| = \frac{1}{|A|}$$ Therefore, we only need to compute the determinant of the original matrix $A$.

Step 3: Detailed Explanation:
Let's find the determinant of matrix $A$: $$|A| = \begin{vmatrix} 1 & 0 & 1
0 & 2 & 3
1 & 2 & 1 \end{vmatrix}$$ Expanding the determinant along the first row: $$|A| = 1 \cdot \begin{vmatrix} 2 & 3
2 & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} 0 & 3
1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 2
1 & 2 \end{vmatrix}$$ $$|A| = 1 \cdot (2 \cdot 1 - 3 \cdot 2) - 0 + 1 \cdot (0 \cdot 2 - 2 \cdot 1)$$ $$|A| = 1 \cdot (2 - 6) + 1 \cdot (0 - 2)$$ $$|A| = -4 - 2 = -6$$ Now, applying the determinant property for inverse matrices: $$|A^{-1}| = \frac{1}{|A|} = \frac{1}{-6} = -\frac{1}{6}$$

Step 4: Final Answer:
The determinant of $A^{-1}$ is $-\frac{1}{6}$, matching option (B).