Given `A=[(x, 3),(y, 3)]`
If `A^(2)=3I`, where I is the identity matrix of order 2, find x and y.

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) given the matrix \( A = \begin{pmatrix} x & 3 \\ y & 3 \end{pmatrix} \) and the condition \( A^2 = 3I \), where \( I \) is the identity matrix of order 2.

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

First, we calculate \( A^2 \) by multiplying matrix \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} x & 3 \\ y & 3 \end{pmatrix} \cdot \begin{pmatrix} x & 3 \\ y & 3 \end{pmatrix} \]

### Step 2: Perform the multiplication

Now we perform the multiplication:

1. For the element at (1,1):
\[ x \cdot x + 3 \cdot y = x^2 + 3y \]

2. For the element at (1,2):
\[ x \cdot 3 + 3 \cdot 3 = 3x + 9 \]

3. For the element at (2,1):
\[ y \cdot x + 3 \cdot y = xy + 3y \]

4. For the element at (2,2):
\[ y \cdot 3 + 3 \cdot 3 = 3y + 9 \]

Thus, we have:

\[ A^2 = \begin{pmatrix} x^2 + 3y & 3x + 9 \\ xy + 3y & 3y + 9 \end{pmatrix} \]

### Step 3: Set \( A^2 \) equal to \( 3I \)

Since \( A^2 = 3I \), we have:

\[ 3I = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]

This gives us the equations:

1. \( x^2 + 3y = 3 \) (from the (1,1) entry)
2. \( 3x + 9 = 0 \) (from the (1,2) entry)
3. \( xy + 3y = 0 \) (from the (2,1) entry)
4. \( 3y + 9 = 3 \) (from the (2,2) entry)

### Step 4: Solve the equations

From the second equation \( 3x + 9 = 0 \):

\[ 3x = -9 \implies x = -3 \]

From the fourth equation \( 3y + 9 = 3 \):

\[ 3y = 3 - 9 \implies 3y = -6 \implies y = -2 \]

### Step 5: Verify with the first and third equations

Now, we can verify if these values satisfy the first and third equations:

1. For \( x^2 + 3y = 3 \):
\[ (-3)^2 + 3(-2) = 9 - 6 = 3 \quad \text{(True)} \]

2. For \( xy + 3y = 0 \):
\[ (-3)(-2) + 3(-2) = 6 - 6 = 0 \quad \text{(True)} \]

Thus, the values we found are correct.

### Final Answer

The values of \( x \) and \( y \) are:

\[ x = -3, \quad y = -2 \]