If `A=[(1,-3), (2,k)]` and `A^(2) - 4A + 10I = A`, then k is equal to

Options

A. 0
B. `-4`
C. 4
D. 1 or 4

Answer

To solve the problem, we need to find the value of \( k \) such that the equation \( A^2 - 4A + 10I = A \) holds true for the matrix \( A = \begin{pmatrix} 1 & -3 \\ 2 & k \end{pmatrix} \).

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

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

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & -3 \\ 2 & k \end{pmatrix} \cdot \begin{pmatrix} 1 & -3 \\ 2 & k \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( 1 \cdot 1 + (-3) \cdot 2 = 1 - 6 = -5 \)
- First row, second column: \( 1 \cdot (-3) + (-3) \cdot k = -3 - 3k \)
- Second row, first column: \( 2 \cdot 1 + k \cdot 2 = 2 + 2k \)
- Second row, second column: \( 2 \cdot (-3) + k \cdot k = -6 + k^2 \)

Thus,

\[ A^2 = \begin{pmatrix} -5 & -3 - 3k \\ 2 + 2k & k^2 - 6 \end{pmatrix} \]

**Hint for Step 1:** Remember to multiply the rows of the first matrix by the columns of the second matrix and sum the products to get each element.

### Step 2: Calculate \( 4A \)

Next, we calculate \( 4A \):

\[ 4A = 4 \cdot \begin{pmatrix} 1 & -3 \\ 2 & k \end{pmatrix} = \begin{pmatrix} 4 & -12 \\ 8 & 4k \end{pmatrix} \]

**Hint for Step 2:** Simply multiply each element of matrix \( A \) by 4.

### Step 3: Calculate \( 10I \)

The identity matrix \( I \) of order 2 is:

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

Thus,

\[ 10I = 10 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 10 & 0 \\ 0 & 10 \end{pmatrix} \]

**Hint for Step 3:** The identity matrix has 1s on the diagonal and 0s elsewhere. Multiply it by 10.

### Step 4: Set up the equation

Now we substitute \( A^2 \), \( 4A \), and \( 10I \) into the equation \( A^2 - 4A + 10I = A \):

\[ \begin{pmatrix} -5 & -3 - 3k \\ 2 + 2k & k^2 - 6 \end{pmatrix} - \begin{pmatrix} 4 & -12 \\ 8 & 4k \end{pmatrix} + \begin{pmatrix} 10 & 0 \\ 0 & 10 \end{pmatrix} = \begin{pmatrix} 1 & -3 \\ 2 & k \end{pmatrix} \]

### Step 5: Simplify the left side

Calculating the left side:

\[ \begin{pmatrix} -5 - 4 + 10 & (-3 - 3k) + 12 \\ (2 + 2k) - 8 & (k^2 - 6) - 4k + 10 \end{pmatrix} = \begin{pmatrix} 1 & 9 - 3k \\ -6 + 2k & k^2 - 4k + 4 \end{pmatrix} \]

### Step 6: Set equal to \( A \)

Now we set the left side equal to \( A \):

\[ \begin{pmatrix} 1 & 9 - 3k \\ -6 + 2k & k^2 - 4k + 4 \end{pmatrix} = \begin{pmatrix} 1 & -3 \\ 2 & k \end{pmatrix} \]

### Step 7: Solve for \( k \)

From the first row, second column:

\[ 9 - 3k = -3 \implies 3k = 12 \implies k = 4 \]

From the second row, first column:

\[ -6 + 2k = 2 \implies 2k = 8 \implies k = 4 \]

Both equations give us \( k = 4 \).

### Final Answer

Thus, the value of \( k \) is:

\[ \boxed{4} \]