Question

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

• A. 7
• B. -7
• C. 15
• D. -11

Hint:

Remember the formula \( A^{-1} = \frac{1}{det(A)} adj(A) \).

Answer 1

The Correct Option is C

Step 1: {Recall the relationship between \( A^{-1} \) and \( adj(A) \)}
The inverse of a matrix \( A \) is given by \( A^{-1} = \frac{1}{det(A)} adj(A) \), where \( det(A) \) is the determinant of \( A \) and \( adj(A) \) is the adjugate of \( A \). 
Step 2: {Compare with the given equation}
We are given that \( A^{-1} = \frac{1}{k} adj(A) \). Comparing this with the general formula, we see that \( k = det(A) \). 
Step 3: {Calculate the determinant of \( A \)}
\[ \begin{vmatrix} 3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{vmatrix} = 3 \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} - (-2) \begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} + 4 \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} \] \[ = 3(2(1) - (-1)(1)) + 2(1(1) - (-1)(0)) + 4(1(1) - 2(0)) \] \[ = 3(2 + 1) + 2(1 - 0) + 4(1 - 0) \] \[ = 3(3) + 2(1) + 4(1) \] \[ = 9 + 2 + 4 = 15 \] 
Step 4: {Identify the value of \( k \)}
Since \( k = det(A) \), we have \( k = 15 \).