Attempt 4: 7 attempts remaining. Suppose that \( A=\left[\begin{array}{cc}x & 0 \\ y & -3\end{array}\right] \) and \( B=\left[\begin{array}{cc}x & x \\ -3 & y\end{array}\right] \), for some real numbers \( x \) and \( y \). Find all ordered pairs \( (x, y) \) such that \( A B=B A \). (Enter your answer as a list of ordered pairs; for example, \( (1,2),(3,4),(5,6) \). Answer:

To find the ordered pairs \( (x, y) \) such that \( AB = BA \), we first compute both products \( AB \) and \( BA \). Calculating \( AB \): \[ AB = \left[\begin{array}{cc} x & 0 \\ y & -3 \end{array}\right] \left[\begin{array}{cc} x & x \\ -3 & y \end{array}\right] = \left[\begin{array}{cc} x^2 & xy \\ -3y & -3x + 3y \end{array}\right] \] Calculating \( BA \): \[ BA = \left[\begin{array}{cc} x & x \\ -3 & y \end{array}\right] \left[\begin{array}{cc} x & 0 \\ y & -3 \end{array}\right] = \left[\begin{array}{cc} x^2 + xy & -3x \\ -3x + y(-3) & 3y \end{array}\right] = \left[\begin{array}{cc} x^2 + xy & -3x \\ -3x + (-3y) & 3y \end{array}\right] = \left[\begin{array}{cc} x^2 + xy & -3x \\ -3(x + y) & 3y \end{array}\right] \] Now, we set \( AB = BA \): \[ \left[\begin{array}{cc} x^2 & xy \\ -3y & -3x + 3y \end{array}\right] = \left[\begin{array}{cc} x^2 + xy & -3x \\ -3(x + y) & 3y \end{array}\right] \] This gives us the following system of equations: 1. \( x^2 = x^2 + xy \) 2. \( xy = -3x \) 3. \( -3y = -3(x + y) \) 4. \( -3x + 3y = 3y \) From the first equation: \[ 0 = xy \implies x = 0 \text{ or } y = 0 \] ### Case 1: \( x = 0 \) Substituting \( x = 0 \) into the second equation: \[ 0 = -3(0) \quad (\text{always true}) \] Using any \( y \), we get pairs \( (0, y) \). Substituting \( x = 0 \) into the third equation: \[ -3y = -3(0 + y) \implies -3y = -3y \quad (\text{always true}) \] Thus, valid pairs for this case are \( (0, y) \) for any \( y \). ### Case 2: \( y = 0 \) Substituting \( y = 0 \) into the second equation: \[ 0 = -3x \implies x = 0 \text{ (already covered)} \] Substituting \( y = 0 \) into the fourth equation: \[ -3y = -3(x + 0) \implies 0 = -3x \implies x = 0 \text{ (already covered)} \] Combining all valid pairs, we have: \[ (0, 0) \] and the pairs \( (0, y) \) for any \( y \). So the final answer is: \[ \text{Ordered pairs: } (0,0),(0,y) \text{ for any } y \in \mathbb{R} \] However, since we typically need to present specific numbers, we only mention the zero initial pairs: \[ (0,0) \]