If `A=[[0, 1, 2], [1, 2, 3], [3, a, 1]] and A^(-1)=(-1)/(2)[[-1, 1, -1], [8, -6, -2k], [-5, 3, -1]]`, then

Options

A. `a=-1, k=1`
B. `a=1, k=-1`
C. `a=2, k=(-1)/(2)`
D. `a=(1)/(2), k=(1)/(2)`

Answer

To solve the problem, we need to find the values of \( a \) and \( k \) in the matrix \( A \) and its inverse \( A^{-1} \).

Given:
\[ A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix} \]
and
\[ A^{-1} = -\frac{1}{2} \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & -2k \\ -5 & 3 & -1 \end{bmatrix} \]

### Step 1: Calculate the Determinant of \( A \)

The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \( A \):
\[ \text{det}(A) = 0 \cdot (2 \cdot 1 - 3 \cdot a) - 1 \cdot (1 \cdot 1 - 3 \cdot 3) + 2 \cdot (1 \cdot a - 2 \cdot 3) \]
Calculating this:
\[ = 0 - 1 \cdot (1 - 9) + 2 \cdot (a - 6) \]
\[ = 8 + 2a - 12 \]
\[ = 2a - 4 \]

### Step 2: Calculate the Adjoint of \( A \)

The adjoint of a matrix is the transpose of the cofactor matrix. We will calculate the cofactors of each element in \( A \).

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = \begin{vmatrix} 2 & 3 \\ a & 1 \end{vmatrix} = 2 \cdot 1 - 3 \cdot a = 2 - 3a \]

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = -\begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} = - (1 \cdot 1 - 3 \cdot 3) = - (1 - 9) = 8 \]

3. **Cofactor \( C_{13} \)**:
\[ C_{13} = \begin{vmatrix} 1 & 2 \\ 3 & a \end{vmatrix} = 1 \cdot a - 2 \cdot 3 = a - 6 \]

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = -\begin{vmatrix} 1 & 2 \\ a & 1 \end{vmatrix} = - (1 \cdot 1 - 2 \cdot a) = - (1 - 2a) = 2a - 1 \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = \begin{vmatrix} 0 & 2 \\ 3 & 1 \end{vmatrix} = 0 \cdot 1 - 2 \cdot 3 = -6 \]

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = -\begin{vmatrix} 0 & 1 \\ 3 & a \end{vmatrix} = - (0 \cdot a - 1 \cdot 3) = 3 \]

7. **Cofactor \( C_{31} \)**:
\[ C_{31} = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = 1 \cdot 3 - 2 \cdot 2 = 3 - 4 = -1 \]

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = -\begin{vmatrix} 0 & 2 \\ 1 & 3 \end{vmatrix} = - (0 \cdot 3 - 2 \cdot 1) = 2 \]

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} = 0 \cdot 2 - 1 \cdot 1 = -1 \]

Thus, the cofactor matrix \( C \) is:
\[ C = \begin{bmatrix} 2 - 3a & 8 & a - 6 \\ 2a - 1 & -6 & 3 \\ -1 & 2 & -1 \end{bmatrix} \]

The adjoint \( \text{adj}(A) \) is the transpose of \( C \):
\[ \text{adj}(A) = \begin{bmatrix} 2 - 3a & 2a - 1 & -1 \\ 8 & -6 & 2 \\ a - 6 & 3 & -1 \end{bmatrix} \]

### Step 3: Calculate \( A^{-1} \)

Using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]
Substituting the values:
\[ A^{-1} = \frac{1}{2a - 4} \begin{bmatrix} 2 - 3a & 2a - 1 & -1 \\ 8 & -6 & 2 \\ a - 6 & 3 & -1 \end{bmatrix} \]

### Step 4: Set \( A^{-1} \) Equal to the Given Inverse

We have:
\[ -\frac{1}{2} \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & -2k \\ -5 & 3 & -1 \end{bmatrix} \]
Setting the two expressions for \( A^{-1} \) equal to each other, we can compare corresponding elements.

### Step 5: Solve for \( a \) and \( k \)

1. From the first element:
\[ \frac{2 - 3a}{2a - 4} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 2(2 - 3a) = (2a - 4) \implies 4 - 6a = 2a - 4 \implies 8 = 8a \implies a = 1 \]

2. From the second row, third column:
\[ \frac{2}{2a - 4} = -k \]
Substituting \( a = 1 \):
\[ \frac{2}{2(1) - 4} = -k \implies \frac{2}{2 - 4} = -k \implies \frac{2}{-2} = -k \implies -1 = -k \implies k = 1 \]

### Final Answer
Thus, the values are:
\[ a = 1, \quad k = 1 \]