If `A=[[4, -1], [-1, k]]` such that `A^(2)-6A+7I=0 and k=`

Options

A. `1`
B. `3`
C. `2`
D. `4`

Answer

To solve the problem, we need to find the value of \( k \) such that the matrix \( A = \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix} \) satisfies the equation \( A^2 - 6A + 7I = 0 \).

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we multiply matrix \( A \) by itself:

\[ A^2 = A \cdot A = \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix} \cdot \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix} \]

Calculating the elements of \( A^2 \):

- First row, first column:
\[ 4 \cdot 4 + (-1) \cdot (-1) = 16 + 1 = 17 \]

- First row, second column:
\[ 4 \cdot (-1) + (-1) \cdot k = -4 - k \]

- Second row, first column:
\[ (-1) \cdot 4 + k \cdot (-1) = -4 - k \]

- Second row, second column:
\[ (-1) \cdot (-1) + k \cdot k = 1 + k^2 \]

Thus, we have:
\[ A^2 = \begin{bmatrix} 17 & -4 - k \\ -4 - k & 1 + k^2 \end{bmatrix} \]

### Step 2: Set up the equation \( A^2 - 6A + 7I = 0 \)

We know:
\[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
So,
\[ 7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \]

Now, substituting \( A \) and \( I \) into the equation:
\[ A^2 - 6A + 7I = 0 \]

Substituting \( A^2 \) and \( 6A \):
\[ \begin{bmatrix} 17 & -4 - k \\ -4 - k & 1 + k^2 \end{bmatrix} - 6 \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = 0 \]

Calculating \( 6A \):
\[ 6A = \begin{bmatrix} 24 & -6 \\ -6 & 6k \end{bmatrix} \]

### Step 3: Combine the matrices

Now, we combine:
\[ \begin{bmatrix} 17 & -4 - k \\ -4 - k & 1 + k^2 \end{bmatrix} - \begin{bmatrix} 24 & -6 \\ -6 & 6k \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = 0 \]

Calculating each element:
- First row, first column:
\[ 17 - 24 + 7 = 0 \]

- First row, second column:
\[ -4 - k + 6 = 2 - k \]

- Second row, first column:
\[ -4 - k + 6 = 2 - k \]

- Second row, second column:
\[ 1 + k^2 - 6k + 7 = k^2 - 6k + 8 \]

### Step 4: Set each element to zero

From the above, we have:
1. \( 0 = 0 \) (first element)
2. \( 2 - k = 0 \) (second element)
3. \( 2 - k = 0 \) (third element)
4. \( k^2 - 6k + 8 = 0 \) (fourth element)

From \( 2 - k = 0 \):
\[ k = 2 \]

### Step 5: Verify \( k = 2 \) in the quadratic equation

Substituting \( k = 2 \) into the quadratic equation:
\[ k^2 - 6k + 8 = 0 \rightarrow 2^2 - 6 \cdot 2 + 8 = 4 - 12 + 8 = 0 \]

Since both conditions are satisfied, we conclude:
\[ \boxed{2} \]