Question

If \( A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} \), then \( A^{-1} = \)

• A. \(\begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}\)
• B. \(\begin{bmatrix} \frac{1}{2} & -1 & \frac{5}{2} \\ 1 & -6 & 3 \\ 1 & 2 & -1 \end{bmatrix}\)
• C. \(\begin{bmatrix} 1 & -1 & -1 \\ -8 & 6 & -2 \\ 5 & -3 & 1 \end{bmatrix}\)
• D. \(\begin{bmatrix} 1 & -1 & -1 \\ -8 & 6 & -2 \\ 5 & -3 & 1 \end{bmatrix}\)

Hint:

When computing the inverse, double‑check the sign pattern for cofactors: \((-1)^{i+j}\). The adjugate is the transpose of the cofactor matrix. Verify by multiplying \(A \times A^{-1}\) to get the identity matrix.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a \(3 \times 3\) matrix \(A\) and need to find its inverse \(A^{-1}\).

Step 2: Key Formula or Approach:
For a matrix \(A\), \(A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)\). Compute the determinant and then the adjugate (transpose of cofactor matrix).

Step 3: Detailed Explanation: First, compute \(\det(A)\): \[ \det(A) = 0 \cdot (2\cdot1 - 3\cdot1) - 1 \cdot (1\cdot1 - 3\cdot3) + 2 \cdot (1\cdot1 - 2\cdot3) \] \[ = 0 - 1 \cdot (1 - 9) + 2 \cdot (1 - 6) = -1 \cdot (-8) + 2 \cdot (-5) = 8 - 10 = -2. \] Now find the cofactor matrix. \(C_{11} = +\begin{vmatrix}2&3\\1&1\end{vmatrix} = 2\cdot1 - 3\cdot1 = -1\) \(C_{12} = -\begin{vmatrix}1&3\\3&1\end{vmatrix} = -(1\cdot1 - 3\cdot3) = -(1-9) = 8\) \(C_{13} = +\begin{vmatrix}1&2\\3&1\end{vmatrix} = 1\cdot1 - 2\cdot3 = 1-6 = -5\) \(C_{21} = -\begin{vmatrix}1&2\\1&1\end{vmatrix} = -(1\cdot1 - 2\cdot1) = -(1-2)=1\) \(C_{22} = +\begin{vmatrix}0&2\\3&1\end{vmatrix} = 0\cdot1 - 2\cdot3 = -6\) \(C_{23} = -\begin{vmatrix}0&1\\3&1\end{vmatrix} = -(0\cdot1 - 1\cdot3) = -(-3)=3\) \(C_{31} = +\begin{vmatrix}1&2\\2&3\end{vmatrix} = 1\cdot3 - 2\cdot2 = 3-4 = -1\) \(C_{32} = -\begin{vmatrix}0&2\\1&3\end{vmatrix} = -(0\cdot3 - 2\cdot1) = -(-2)=2\) \(C_{33} = +\begin{vmatrix}0&1\\1&2\end{vmatrix} = 0\cdot2 - 1\cdot1 = -1\) Thus cofactor matrix = \(\begin{bmatrix} -1 & 8 & -5 \\ 1 & -6 & 3 \\ -1 & 2 & -1 \end{bmatrix}\). Adjugate = transpose = \(\begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}\). Finally, \(A^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}\).

Step 4: Final Answer:
This matches option (A).