For the matrix `A=[{:(3,1),(7,5):}]`. Find 'x' and 'y' so that `A^(2)+xI=yA`. Hence find `A^(-1)`.

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) such that \( A^2 + xI = yA \), and then we will find the inverse of matrix \( A \).

Given the matrix:
\[ A = \begin{pmatrix} 3 & 1 \\ 7 & 5 \end{pmatrix} \]

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

To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 3 & 1 \\ 7 & 5 \end{pmatrix} \cdot \begin{pmatrix} 3 & 1 \\ 7 & 5 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 3 \cdot 3 + 1 \cdot 7 = 9 + 7 = 16 \)
- First row, second column: \( 3 \cdot 1 + 1 \cdot 5 = 3 + 5 = 8 \)
- Second row, first column: \( 7 \cdot 3 + 5 \cdot 7 = 21 + 35 = 56 \)
- Second row, second column: \( 7 \cdot 1 + 5 \cdot 5 = 7 + 25 = 32 \)

Thus, we have:
\[ A^2 = \begin{pmatrix} 16 & 8 \\ 56 & 32 \end{pmatrix} \]

### Step 2: Set up the equation \( A^2 + xI = yA \)

The identity matrix \( I \) for a \( 2 \times 2 \) matrix is:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
So,
\[ xI = \begin{pmatrix} x & 0 \\ 0 & x \end{pmatrix} \]

Now, substituting into the equation:
\[ A^2 + xI = yA \]
becomes:
\[ \begin{pmatrix} 16 & 8 \\ 56 & 32 \end{pmatrix} + \begin{pmatrix} x & 0 \\ 0 & x \end{pmatrix} = y \begin{pmatrix} 3 & 1 \\ 7 & 5 \end{pmatrix} \]

This simplifies to:
\[ \begin{pmatrix} 16 + x & 8 \\ 56 & 32 + x \end{pmatrix} = \begin{pmatrix} 3y & y \\ 7y & 5y \end{pmatrix} \]

### Step 3: Equate the elements

From the matrices, we can set up the following equations:
1. \( 16 + x = 3y \)
2. \( 8 = y \)
3. \( 56 = 7y \)
4. \( 32 + x = 5y \)

### Step 4: Solve for \( y \)

From equation 2:
\[ y = 8 \]

### Step 5: Substitute \( y \) into the other equations

Substituting \( y = 8 \) into the first equation:
\[ 16 + x = 3(8) \implies 16 + x = 24 \implies x = 24 - 16 = 8 \]

### Step 6: Verify with the other equations

Substituting \( y = 8 \) into the third equation:
\[ 56 = 7(8) \implies 56 = 56 \quad \text{(True)} \]

Substituting \( y = 8 \) into the fourth equation:
\[ 32 + x = 5(8) \implies 32 + x = 40 \implies x = 40 - 32 = 8 \quad \text{(True)} \]

Thus, we have:
\[ x = 8, \quad y = 8 \]

### Step 7: Find \( A^{-1} \)

To find the inverse of \( A \), we use the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]

First, calculate the determinant of \( A \):
\[ \text{det}(A) = 3 \cdot 5 - 1 \cdot 7 = 15 - 7 = 8 \]

Next, find the adjugate of \( A \):
\[ \text{adj}(A) = \begin{pmatrix} 5 & -1 \\ -7 & 3 \end{pmatrix} \]

Thus,
\[ A^{-1} = \frac{1}{8} \begin{pmatrix} 5 & -1 \\ -7 & 3 \end{pmatrix} = \begin{pmatrix} \frac{5}{8} & -\frac{1}{8} \\ -\frac{7}{8} & \frac{3}{8} \end{pmatrix} \]

### Final Answer

The values are:
\[ x = 8, \quad y = 8 \]
And the inverse of \( A \) is:
\[ A^{-1} = \begin{pmatrix} \frac{5}{8} & -\frac{1}{8} \\ -\frac{7}{8} & \frac{3}{8} \end{pmatrix} \]