If `3A=[-1-2-2 2 1-2x-2y]` such that `A.A^T=I` , then which of the following are correct? `x+2y=4` (b) `x-y=1` `x^2+y^2=-3` (d) `x^2+y^2=5`

Options

A. `-3`
B. `-2`
C. `-1`
D. 0

Answer

To solve the given problem, we start with the equation provided:

Given:
\[ 3A = \begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ x & -2 & y \end{bmatrix} \]

### Step 1: Find Matrix A
To find matrix \( A \), we divide the given matrix by 3:
\[ A = \frac{1}{3} \begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ x & -2 & y \end{bmatrix} = \begin{bmatrix} -\frac{1}{3} & -\frac{2}{3} & -\frac{2}{3} \\ \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ \frac{x}{3} & -\frac{2}{3} & \frac{y}{3} \end{bmatrix} \]

### Step 2: Find A Transpose
Now, we find the transpose of matrix \( A \):
\[ A^T = \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} & \frac{x}{3} \\ -\frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ -\frac{2}{3} & -\frac{2}{3} & \frac{y}{3} \end{bmatrix} \]

### Step 3: Calculate \( A \cdot A^T \)
Next, we calculate the product \( A \cdot A^T \):
\[ A \cdot A^T = \begin{bmatrix} -\frac{1}{3} & -\frac{2}{3} & -\frac{2}{3} \\ \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ \frac{x}{3} & -\frac{2}{3} & \frac{y}{3} \end{bmatrix} \cdot \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} & \frac{x}{3} \\ -\frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ -\frac{2}{3} & -\frac{2}{3} & \frac{y}{3} \end{bmatrix} \]

Calculating each element of the resulting matrix:
1. First row, first column:
\[ \left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right) + \left(-\frac{2}{3}\right)\left(-\frac{2}{3}\right) + \left(-\frac{2}{3}\right)\left(-\frac{2}{3}\right) = \frac{1}{9} + \frac{4}{9} + \frac{4}{9} = \frac{9}{9} = 1 \]
2. First row, second column:
\[ \left(-\frac{1}{3}\right)\left(\frac{2}{3}\right) + \left(-\frac{2}{3}\right)\left(\frac{1}{3}\right) + \left(-\frac{2}{3}\right)\left(-\frac{2}{3}\right) = -\frac{2}{9} - \frac{2}{9} + \frac{4}{9} = 0 \]
3. First row, third column:
\[ \left(-\frac{1}{3}\right)\left(\frac{x}{3}\right) + \left(-\frac{2}{3}\right)\left(-\frac{2}{3}\right) + \left(-\frac{2}{3}\right)\left(\frac{y}{3}\right) = -\frac{x}{9} + \frac{4}{9} - \frac{2y}{9} = 0 \]
This gives us the equation:
\[ -x + 4 - 2y = 0 \implies x + 2y = 4 \quad \text{(1)} \]

4. Second row, second column:
\[ \left(\frac{2}{3}\right)\left(-\frac{2}{3}\right) + \left(\frac{1}{3}\right)\left(\frac{1}{3}\right) + \left(-\frac{2}{3}\right)\left(-\frac{2}{3}\right) = -\frac{4}{9} + \frac{1}{9} + \frac{4}{9} = 1 \]

5. Second row, third column:
\[ \left(\frac{2}{3}\right)\left(\frac{x}{3}\right) + \left(\frac{1}{3}\right)\left(-\frac{2}{3}\right) + \left(-\frac{2}{3}\right)\left(\frac{y}{3}\right) = \frac{2x}{9} - \frac{2}{9} - \frac{2y}{9} = 0 \]
This gives us the equation:
\[ 2x - 2 - 2y = 0 \implies x - y = 1 \quad \text{(2)} \]

6. Third row, third column:
\[ \left(\frac{x}{3}\right)\left(\frac{x}{3}\right) + \left(-\frac{2}{3}\right)\left(-\frac{2}{3}\right) + \left(\frac{y}{3}\right)\left(\frac{y}{3}\right) = \frac{x^2}{9} + \frac{4}{9} + \frac{y^2}{9} = 1 \]
This gives us the equation:
\[ x^2 + 4 + y^2 = 9 \implies x^2 + y^2 = 5 \quad \text{(3)} \]

### Step 4: Solve the Equations
Now we have the following equations:
1. \( x + 2y = 4 \)
2. \( x - y = 1 \)
3. \( x^2 + y^2 = 5 \)

From equation (2), we can express \( x \) in terms of \( y \):
\[ x = y + 1 \]

Substituting \( x \) in equation (1):
\[ (y + 1) + 2y = 4 \implies 3y + 1 = 4 \implies 3y = 3 \implies y = 1 \]

Substituting \( y = 1 \) back into \( x = y + 1 \):
\[ x = 1 + 1 = 2 \]

### Step 5: Verify the Options
Now we have \( x = 2 \) and \( y = 1 \). Let's check the options:
- (a) \( x + 2y = 4 \): \( 2 + 2(1) = 4 \) (Correct)
- (b) \( x - y = 1 \): \( 2 - 1 = 1 \) (Correct)
- (c) \( x^2 + y^2 = -3 \): \( 2^2 + 1^2 = 4 + 1 = 5 \) (Incorrect)
- (d) \( x^2 + y^2 = 5 \): \( 2^2 + 1^2 = 4 + 1 = 5 \) (Correct)

### Final Answer
The correct options are (a), (b), and (d).