`A=[(x+y,y),(2x,x-y)]`
`B=[(2),(-1)]and C = [(3),(2)]`
If AB = C, then what is `A^(2)` equal to ?

Options

A. `[{:(6,-10),(4,26):}]`
B. `[{:(-10,5),(4,24):}]`
C. `[{:(-5,-6),(-4,-20):}]`
D. `[{:(-5,-7),(-5,20):}]`

Explanation

To solve the problem, we need to find \( A^2 \) given that \( AB = C \), where:

\[ A = \begin{pmatrix} x+y & y \\ 2x & x-y \end{pmatrix}, \quad B = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad C = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \]

### Step 1: Calculate \( AB \)

We start by multiplying matrix \( A \) by matrix \( B \):

\[ AB = \begin{pmatrix} x+y & y \\ 2x & x-y \end{pmatrix} \begin{pmatrix} 2 \\ -1 \end{pmatrix} \]

Calculating the first element of \( AB \):

\[ (x+y) \cdot 2 + y \cdot (-1) = 2(x+y) - y = 2x + 2y - y = 2x + y \]

Calculating the second element of \( AB \):

\[ 2x \cdot 2 + (x-y) \cdot (-1) = 4x - (x - y) = 4x - x + y = 3x + y \]

Thus, we have:

\[ AB = \begin{pmatrix} 2x + y \\ 3x + y \end{pmatrix} \]

### Step 2: Set up equations from \( AB = C \)

From the equation \( AB = C \), we have:

\[ \begin{pmatrix} 2x + y \\ 3x + y \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \]

This gives us the following system of equations:

1. \( 2x + y = 3 \) (Equation 1)
2. \( 3x + y = 2 \) (Equation 2)

### Step 3: Solve the system of equations

We can solve these equations by elimination or substitution. Let's subtract Equation 1 from Equation 2:

\[ (3x + y) - (2x + y) = 2 - 3 \]
\[ 3x - 2x + y - y = -1 \]
\[ x = -1 \]

Now, substitute \( x = -1 \) into Equation 1:

\[ 2(-1) + y = 3 \]
\[ -2 + y = 3 \]
\[ y = 5 \]

Thus, we have \( x = -1 \) and \( y = 5 \).

### Step 4: Substitute \( x \) and \( y \) back into \( A \)

Now we substitute \( x \) and \( y \) back into matrix \( A \):

\[ A = \begin{pmatrix} -1 + 5 & 5 \\ 2(-1) & -1 - 5 \end{pmatrix} = \begin{pmatrix} 4 & 5 \\ -2 & -6 \end{pmatrix} \]

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

Now, we need to calculate \( A^2 = A \cdot A \):

\[ A^2 = \begin{pmatrix} 4 & 5 \\ -2 & -6 \end{pmatrix} \begin{pmatrix} 4 & 5 \\ -2 & -6 \end{pmatrix} \]

Calculating the elements of \( A^2 \):

1. First row, first column:
\[ 4 \cdot 4 + 5 \cdot (-2) = 16 - 10 = 6 \]

2. First row, second column:
\[ 4 \cdot 5 + 5 \cdot (-6) = 20 - 30 = -10 \]

3. Second row, first column:
\[ -2 \cdot 4 + (-6) \cdot (-2) = -8 + 12 = 4 \]

4. Second row, second column:
\[ -2 \cdot 5 + (-6) \cdot (-6) = -10 + 36 = 26 \]

Thus, we have:

\[ A^2 = \begin{pmatrix} 6 & -10 \\ 4 & 26 \end{pmatrix} \]

### Final Answer

\[ A^2 = \begin{pmatrix} 6 & -10 \\ 4 & 26 \end{pmatrix} \]