Question: 5. Use matrices to solve for real numbers $x$ and $y$ if
$bx - ay = a + 2b$
$ax + by = 2a - b$.
1. **State the problem:** Solve the system of equations for real numbers $x$ and $y$:
$$bx - ay = a + 2b$$
$$ax + by = 2a - b$$
2. **Write the system in matrix form:**
$$\begin{pmatrix} b & -a \\ a & b \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a + 2b \\ 2a - b \end{pmatrix}$$
3. **Identify matrix $A$, vector $X$, and vector $B$:**
$$A = \begin{pmatrix} b & -a \\ a & b \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} a + 2b \\ 2a - b \end{pmatrix}$$
4. **Find the determinant of $A$ to check if $A$ is invertible:**
$$\det(A) = b \cdot b - (-a) \cdot a = b^2 + a^2$$
Since $a^2 + b^2 \neq 0$ for real $a,b$ not both zero, $A$ is invertible.
5. **Find the inverse of $A$:**
$$A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} b & a \\ -a & b \end{pmatrix} = \frac{1}{a^2 + b^2} \begin{pmatrix} b & a \\ -a & b \end{pmatrix}$$
6. **Solve for $X$ using $X = A^{-1} B$:**
$$X = \frac{1}{a^2 + b^2} \begin{pmatrix} b & a \\ -a & b \end{pmatrix} \begin{pmatrix} a + 2b \\ 2a - b \end{pmatrix}$$
7. **Multiply the matrices:**
$$\begin{aligned}
x &= \frac{1}{a^2 + b^2} \left( b(a + 2b) + a(2a - b) \right) \\
y &= \frac{1}{a^2 + b^2} \left( -a(a + 2b) + b(2a - b) \right)
\end{aligned}$$
8. **Simplify each expression:**
$$\begin{aligned}
x &= \frac{1}{a^2 + b^2} \left( ab + 2b^2 + 2a^2 - ab \right) = \frac{1}{a^2 + b^2} \left( 2a^2 + 2b^2 \right) = \frac{2(a^2 + b^2)}{a^2 + b^2} = 2 \\
y &= \frac{1}{a^2 + b^2} \left( -a^2 - 2ab + 2ab - b^2 \right) = \frac{1}{a^2 + b^2} \left( -a^2 - b^2 \right) = -1
\end{aligned}$$
9. **Final answer:**
$$\boxed{x = 2, \quad y = -1}$$