If `A=[(3,1),(-1,2)]` and `I=[(1,0),(0,1)]` find 'k' so that `A^(2)=5A+kI`.

Answer

To solve the problem, we need to find the value of \( k \) such that \( A^2 = 5A + kI \), where \( A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \) and \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).

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

We start by calculating \( A^2 \) which is \( A \cdot A \).

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

Calculating the elements:

- First row, first column: \( 3 \cdot 3 + 1 \cdot (-1) = 9 - 1 = 8 \)
- First row, second column: \( 3 \cdot 1 + 1 \cdot 2 = 3 + 2 = 5 \)
- Second row, first column: \( -1 \cdot 3 + 2 \cdot (-1) = -3 - 2 = -5 \)
- Second row, second column: \( -1 \cdot 1 + 2 \cdot 2 = -1 + 4 = 3 \)

Thus,

\[ A^2 = \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} \]

### Step 2: Calculate \( 5A \)

Next, we calculate \( 5A \):

\[ 5A = 5 \cdot \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 15 & 5 \\ -5 & 10 \end{pmatrix} \]

### Step 3: Calculate \( kI \)

Now, we calculate \( kI \):

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

### Step 4: Set up the equation \( A^2 = 5A + kI \)

We now have:

\[ A^2 = \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} \]
\[ 5A + kI = \begin{pmatrix} 15 + k & 5 \\ -5 & 10 + k \end{pmatrix} \]

Setting these equal:

\[ \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 15 + k & 5 \\ -5 & 10 + k \end{pmatrix} \]

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

From the equality of matrices, we can set up the following equations:

1. \( 8 = 15 + k \)
2. \( 3 = 10 + k \)

From the first equation:

\[ k = 8 - 15 = -7 \]

From the second equation:

\[ k = 3 - 10 = -7 \]

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

### Conclusion

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

\[ \boxed{-7} \]