`A=[{:(x+y," "y),(" "x,x-y):}],B=[{:(3),(-2):}]and C=[{:(4),(-2):}]`. If AB = C, then what is `A^(2)` equal to ?

Options

A. `[{:(4,8),(-4,-16):}]`
B. `[{:(4,-4),(8,-16):}]`
C. `[{:(-4,-8),(4,12):}]`
D. `[{:(-4,-8),(8,12):}]`

Explanation

To solve the problem, we need to find \( A^2 \) given that \( AB = C \). Let's break this down step by step.

### Step 1: Define the Matrices
We have the matrices:
\[ A = \begin{pmatrix} x+y & y \\ x & x-y \end{pmatrix}, \quad B = \begin{pmatrix} 3 \\ -2 \end{pmatrix}, \quad C = \begin{pmatrix} 4 \\ -2 \end{pmatrix} \]

### Step 2: Multiply Matrices A and B
We need to calculate \( AB \):
\[ AB = \begin{pmatrix} x+y & y \\ x & x-y \end{pmatrix} \begin{pmatrix} 3 \\ -2 \end{pmatrix} \]

Calculating the elements:
- First row, first column:
\[ (x+y) \cdot 3 + y \cdot (-2) = 3(x+y) - 2y = 3x + 3y - 2y = 3x + y \]
- Second row, first column:
\[ x \cdot 3 + (x-y) \cdot (-2) = 3x - 2(x-y) = 3x - 2x + 2y = x + 2y \]

Thus, we have:
\[ AB = \begin{pmatrix} 3x + y \\ x + 2y \end{pmatrix} \]

### Step 3: Set Up the Equations
Since \( AB = C \):
\[ \begin{pmatrix} 3x + y \\ x + 2y \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} \]

This gives us two equations:
1. \( 3x + y = 4 \) (Equation 1)
2. \( x + 2y = -2 \) (Equation 2)

### Step 4: Solve the System of Equations
From Equation 1:
\[ y = 4 - 3x \]

Substituting \( y \) in Equation 2:
\[ x + 2(4 - 3x) = -2 \]
\[ x + 8 - 6x = -2 \]
\[ -5x + 8 = -2 \]
\[ -5x = -10 \implies x = 2 \]

Now substituting \( x = 2 \) back into Equation 1 to find \( y \):
\[ 3(2) + y = 4 \implies 6 + y = 4 \implies y = 4 - 6 = -2 \]

Thus, we have:
\[ x = 2, \quad y = -2 \]

### Step 5: Substitute Values Back into Matrix A
Now substituting \( x \) and \( y \) back into matrix \( A \):
\[ A = \begin{pmatrix} 2 + (-2) & -2 \\ 2 & 2 - (-2) \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ 2 & 4 \end{pmatrix} \]

### Step 6: Calculate \( A^2 \)
Now we need to calculate \( A^2 = A \cdot A \):
\[ A^2 = \begin{pmatrix} 0 & -2 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} 0 & -2 \\ 2 & 4 \end{pmatrix} \]

Calculating the elements:
- First row, first column:
\[ 0 \cdot 0 + (-2) \cdot 2 = 0 - 4 = -4 \]
- First row, second column:
\[ 0 \cdot (-2) + (-2) \cdot 4 = 0 - 8 = -8 \]
- Second row, first column:
\[ 2 \cdot 0 + 4 \cdot 2 = 0 + 8 = 8 \]
- Second row, second column:
\[ 2 \cdot (-2) + 4 \cdot 4 = -4 + 16 = 12 \]

Thus, we have:
\[ A^2 = \begin{pmatrix} -4 & -8 \\ 8 & 12 \end{pmatrix} \]

### Final Answer
\[ A^2 = \begin{pmatrix} -4 & -8 \\ 8 & 12 \end{pmatrix} \]

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