Find the value of k so that `A^2=8A+kI` where `A=[(1,0),(-1,7)].`

Options

A. 7
B. -7
C. 1
D. -1

Answer

To find the value of \( k \) such that \( A^2 = 8A + kI \) for the matrix \( A = \begin{pmatrix} 1 & 0 \\ -1 & 7 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate \( A^2 \)
To find \( A^2 \), we multiply matrix \( A \) by itself.

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 \\ -1 & 7 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ -1 & 7 \end{pmatrix} \]

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

- First row, first column: \( 1 \cdot 1 + 0 \cdot (-1) = 1 \)
- First row, second column: \( 1 \cdot 0 + 0 \cdot 7 = 0 \)
- Second row, first column: \( -1 \cdot 1 + 7 \cdot (-1) = -1 - 7 = -8 \)
- Second row, second column: \( -1 \cdot 0 + 7 \cdot 7 = 49 \)

Thus,

\[ A^2 = \begin{pmatrix} 1 & 0 \\ -8 & 49 \end{pmatrix} \]

### Step 2: Calculate \( 8A \)
Next, we calculate \( 8A \):

\[ 8A = 8 \cdot \begin{pmatrix} 1 & 0 \\ -1 & 7 \end{pmatrix} = \begin{pmatrix} 8 & 0 \\ -8 & 56 \end{pmatrix} \]

### Step 3: Calculate \( kI \)
The identity matrix \( I \) for a \( 2 \times 2 \) matrix is:

\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Thus,

\[ kI = k \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} \]

### Step 4: Set up the equation \( A^2 = 8A + kI \)
Now we set up the equation:

\[ \begin{pmatrix} 1 & 0 \\ -8 & 49 \end{pmatrix} = \begin{pmatrix} 8 & 0 \\ -8 & 56 \end{pmatrix} + \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} \]

### Step 5: Combine the right-hand side
Combining the matrices on the right-hand side gives:

\[ \begin{pmatrix} 8 + k & 0 \\ -8 & 56 + k \end{pmatrix} \]

### Step 6: Equate the matrices
Now we equate the corresponding elements from both sides:

1. From the first row, first column:
\[ 1 = 8 + k \implies k = 1 - 8 = -7 \]

2. From the second row, second column:
\[ 49 = 56 + k \implies k = 49 - 56 = -7 \]

Both equations give us the same value for \( k \).

### Final Answer
Thus, the value of \( k \) is

\[ \boxed{-7} \]