For the matrix `A=[3 1 7 5]` , find `x` and `y` so that `A^2+x I=y Adot`

Explanation

To solve the equation \( A^2 + xI = yA \) for the matrix \( A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \), we will follow these steps:

### Step 1: Calculate \( A^2 \)
We need to multiply matrix \( A \) by itself:

\[ A^2 = A \cdot A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \cdot \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \]

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,

\[ A^2 = \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix} \]

### Step 2: Write the equation \( A^2 + xI = yA \)
The identity matrix \( I \) for a 2x2 matrix is:

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

So,

\[ xI = \begin{bmatrix} x & 0 \\ 0 & x \end{bmatrix} \]

Now substituting \( A^2 \) and \( xI \) into the equation:

\[ \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix} + \begin{bmatrix} x & 0 \\ 0 & x \end{bmatrix} = y \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} \]

### Step 3: Rewrite the equation
This gives us:

\[ \begin{bmatrix} 16 + x & 8 \\ 56 & 32 + x \end{bmatrix} = \begin{bmatrix} 3y & y \\ 7y & 5y \end{bmatrix} \]

### Step 4: Equate corresponding elements
Now we equate the corresponding elements from both matrices:

1. \( 16 + x = 3y \) (1)
2. \( 8 = y \) (2)
3. \( 56 = 7y \) (3)
4. \( 32 + x = 5y \) (4)

### Step 5: Solve for \( y \)
From equation (2), we directly find:

\[ y = 8 \]

### Step 6: Substitute \( y \) into other equations
Substituting \( y = 8 \) into equations (1), (3), and (4):

- From (1):
\[ 16 + x = 3(8) \implies 16 + x = 24 \implies x = 24 - 16 = 8 \]

- From (3):
\[ 56 = 7(8) \implies 56 = 56 \quad \text{(True)} \]

- From (4):
\[ 32 + x = 5(8) \implies 32 + x = 40 \implies x = 40 - 32 = 8 \]

### Final Answer
Thus, the values of \( x \) and \( y \) are:

\[ x = 8, \quad y = 8 \]