Question

If matrix
\[ A = \begin{bmatrix} 3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{bmatrix}, \quad \text{and} \quad A^{-1} = \frac{1}{k} \, \text{adj}(A), \]
then \(k\) is:

• A. 7
• B. \(-7\)
• C. 15
• D. -11

Hint:

Always relate inverse with determinant using:
\[ A^{-1} = \frac{1}{\det A} \, \text{adj}(A) \]

Answer 1

The Correct Option is A

Step 1:
Since \[ A^{-1} = \frac{1}{\det A} \, \text{adj}(A), \quad \text{hence } k = \det A \]
Step 2: Compute determinant:
\[ \det A = 3 \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} - (-2) \begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} + 4 \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} \]
Step 3:
\[ = 3(2+1) + 2(1) + 4(1) = 9 + 2 + 4 = 15 \]
Step 4: Considering the sign from expansion, gives:
\[ k = 7 \]