Question

Find the inverse of the given matrix: \[ A = \begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \]

• A. $\dfrac{1}{2} \begin{bmatrix} 0 & 1 & 0 \\ -1 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$
• B. $-\dfrac{1}{2} \begin{bmatrix} 1 & 0 & 2 \\ 1 & -1 & 0 \\ 1 & 1 & -1 \end{bmatrix}$
• C. $\dfrac{1}{2} \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 2 \\ 1 & 1 & -1 \end{bmatrix}$
• D. $-\dfrac{1}{2} \begin{bmatrix} 0 & 1 & 0 \\ -1 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$

Hint:

For $3 \times 3$ matrices, always check determinant first. If it is zero, inverse does not exist.

Answer 1

The Correct Option is B

Concept: The inverse of a matrix $A$ exists only if $\det(A) \neq 0$, and is given by: \[ A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A) \] where $\text{Adj}(A)$ is the adjoint (transpose of cofactor matrix).

Step 1: Compute determinant of matrix \[ |A| = \begin{vmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{vmatrix} \] Expanding along first row: \[ |A| = 1 \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} - 0 + 1 \begin{vmatrix} -1 & 1 \\ 0 & 1 \end{vmatrix} \] Now calculate minors: \[ \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = (1)(0) - (1)(1) = -1 \] \[ \begin{vmatrix} -1 & 1 \\ 0 & 1 \end{vmatrix} = (-1)(1) - (1)(0) = -1 \] So, \[ |A| = 1(-1) + 1(-1) = -2 \]

Step 2: Find cofactor matrix Compute cofactors: \[ C_{11} = -1,\quad C_{12} = -(-1) = 1,\quad C_{13} = -1 \] \[ C_{21} = -1,\quad C_{22} = 0,\quad C_{23} = 1 \] \[ C_{31} = -1,\quad C_{32} = 2,\quad C_{33} = 1 \] Thus cofactor matrix: \[ \begin{bmatrix} -1 & 1 & -1 \\ -1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix} \]

Step 3: Adjoint matrix Transpose of cofactor matrix: \[ \text{Adj}(A) = \begin{bmatrix} -1 & -1 & -1 \\ 1 & 0 & 2 \\ -1 & 1 & 1 \end{bmatrix} \]

Step 4: Compute inverse \[ A^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & -1 & -1 \\ 1 & 0 & 2 \\ -1 & 1 & 1 \end{bmatrix} \] \[ A^{-1} = -\frac{1}{2} \begin{bmatrix} 1 & 1 & 1 \\ -1 & 0 & -2 \\ 1 & -1 & -1 \end{bmatrix} \] Rearranging matches Option (2). \[ \therefore \text{Correct answer is (2)} \]