If for a matrix A, `absA=6and adj A=`{:[(1,-2,4),(4,1,1),(-1,k,0)]:}`, then k is equal to

Options

A. -1
B. 0
C. 1
D. 2

Explanation

To find the value of \( k \) in the given problem, we will follow these steps:

### Step 1: Write down the given information
We are given:
- \( \text{det}(A) = 6 \)
- \( \text{adj}(A) = \begin{pmatrix} 1 & -2 & 4 \\ 4 & 1 & 1 \\ -1 & k & 0 \end{pmatrix} \)

### Step 2: Use the property of determinants
The property of the adjoint of a matrix states that:
\[ \text{det}(\text{adj}(A)) = \text{det}(A)^{n-1} \]
where \( n \) is the order of the matrix. Here, \( n = 3 \) (since \( A \) is a \( 3 \times 3 \) matrix).

Thus, we have:
\[ \text{det}(\text{adj}(A)) = 6^{3-1} = 6^2 = 36 \]

### Step 3: Calculate the determinant of adjoint matrix
To find \( \text{det}(\text{adj}(A)) \), we will calculate the determinant of the matrix:
\[ \text{adj}(A) = \begin{pmatrix} 1 & -2 & 4 \\ 4 & 1 & 1 \\ -1 & k & 0 \end{pmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix:
\[ \text{det}(B) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( B = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).

Here:
- \( a = 1, b = -2, c = 4 \)
- \( d = 4, e = 1, f = 1 \)
- \( g = -1, h = k, i = 0 \)

Calculating the determinant:
\[ \text{det}(\text{adj}(A)) = 1 \cdot (1 \cdot 0 - 1 \cdot k) - (-2) \cdot (4 \cdot 0 - 1 \cdot -1) + 4 \cdot (4 \cdot k - 1 \cdot -1) \]
\[ = 1 \cdot (0 - k) + 2 \cdot (0 + 1) + 4 \cdot (4k + 1) \]
\[ = -k + 2 + 16k + 4 \]
\[ = 15k + 6 \]

### Step 4: Set the determinant equal to 36
Now we set the expression for the determinant equal to 36:
\[ 15k + 6 = 36 \]

### Step 5: Solve for \( k \)
Subtract 6 from both sides:
\[ 15k = 30 \]
Now divide by 15:
\[ k = 2 \]

### Final Answer
Thus, the value of \( k \) is \( \boxed{2} \).