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 \( k = 15 \). 
 

Answer 2

Step 1: Given Matrix and Inverse Relationship
We are given that \( A = \begin{bmatrix} 3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{bmatrix} \) and the inverse of matrix \( A \), denoted as \( A^{-1} \), is related to the adjugate of \( A \) by the formula: \[ A^{-1} = \frac{1}{k} \text{adj}(A). \] We are asked to find the value of \( k \).

Step 2: Formula for Matrix Inverse
The formula for the inverse of a matrix is: \[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A). \] By comparing this formula with the given expression \( A^{-1} = \frac{1}{k} \text{adj}(A) \), we can deduce that: \[ \frac{1}{k} = \frac{1}{\text{det}(A)}. \] This implies: \[ k = \text{det}(A). \]

Step 3: Finding the Determinant of \( A \)
We will now calculate the determinant of \( A \). The matrix \( A \) is: \[ A = \begin{bmatrix} 3 & -2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{bmatrix}. \] The determinant of a 3x3 matrix is given by: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg), \] where \( a, b, c, d, e, f, g, h, i \) are the elements of the matrix: \[ A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}. \] Substituting the values from matrix \( A \): \[ \text{det}(A) = 3 \left( 2 \times 1 - (-1) \times 1 \right) - (-2) \left( 1 \times 1 - (-1) \times 0 \right) + 4 \left( 1 \times 1 - 2 \times 0 \right). \] Simplifying: \[ \text{det}(A) = 3 \left( 2 + 1 \right) + 2 \left( 1 + 0 \right) + 4 \left( 1 - 0 \right). \] \[ \text{det}(A) = 3 \times 3 + 2 \times 1 + 4 \times 1 = 9 + 2 + 4 = 15. \]

Step 4: Conclusion
Thus, the value of \( k \) is: \[ k = \text{det}(A) = 15. \]

The correct answer is: 15