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: State the plan.
To invert $A$, use $A^{-1} = \dfrac{1}{\det A}\,\text{adj}(A)$, where the adjugate is the transpose of the cofactor matrix.
Step 2: Find the determinant.
Expand along the first row of $A = \begin{bmatrix} 0&1&2\\1&2&3\\3&1&1 \end{bmatrix}$: \[ \det A = 0 - 1(1\cdot1 - 3\cdot3) + 2(1\cdot1 - 2\cdot3) = -1(-8) + 2(-5) = 8 - 10 = -2 \]
Step 3: Build the cofactors (row 1 and 2).
$C_{11} = (2-3) = -1$, $C_{12} = -(1-9) = 8$, $C_{13} = (1-6) = -5$, $C_{21} = -(1-2) = 1$, $C_{22} = (0-6) = -6$, $C_{23} = -(0-3) = 3$.
Step 4: Build the cofactors (row 3).
$C_{31} = (3-4) = -1$, $C_{32} = -(0-2) = 2$, $C_{33} = (0-1) = -1$.
Step 5: Transpose to get the adjugate.
\[ \text{adj}(A) = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix} \]
Step 6: Divide by the determinant.
\[ A^{-1} = \frac{1}{-2}\,\text{adj}(A) = \begin{bmatrix} \frac12 & -\frac12 & \frac12 \\ -4 & 3 & -1 \\ \frac52 & -\frac32 & \frac12 \end{bmatrix} \] \[ \boxed{\begin{bmatrix} \frac12 & -\frac12 & \frac12 \\ -4 & 3 & -1 \\ \frac52 & -\frac32 & \frac12 \end{bmatrix}} \]