1. **Problem statement:** Given linear transformations $f: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $f(x,y) = (2x + 5y, 3x + 8y)$ and $g(x,y) = (x + 2y, 4x + 5y)$, find: i) The matrix of $f$ relative to the standard basis. ii) The matrix of $g$ relative to the standard basis. iii) The matrix of $f \circ g$ (composition $f(g(x,y))$) relative to the standard basis. iv) The matrix of $g \circ f$ (composition $g(f(x,y))$) relative to the standard basis. v) The inverse of $f$. vi) The inverse of $g$. 2. **Recall:** The matrix of a linear transformation $T(x,y) = (a_{11}x + a_{12}y, a_{21}x + a_{22}y)$ relative to the standard basis is $$ \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} $$ The composition $f \circ g$ corresponds to matrix multiplication $M_f M_g$. The inverse of a $2 \times 2$ matrix $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ is $$ \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} $$ provided $ad - bc \neq 0$. 3. **Find matrix of $f$:** From $f(x,y) = (2x + 5y, 3x + 8y)$, $$ M_f = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} $$ 4. **Find matrix of $g$:** From $g(x,y) = (x + 2y, 4x + 5y)$, $$ M_g = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} $$ 5. **Find matrix of $f \circ g$:** $$ M_{f \circ g} = M_f M_g = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 2 \times 1 + 5 \times 4 & 2 \times 2 + 5 \times 5 \\ 3 \times 1 + 8 \times 4 & 3 \times 2 + 8 \times 5 \end{bmatrix} = \begin{bmatrix} 2 + 20 & 4 + 25 \\ 3 + 32 & 6 + 40 \end{bmatrix} = \begin{bmatrix} 22 & 29 \\ 35 & 46 \end{bmatrix} $$ 6. **Find matrix of $g \circ f$:** $$ M_{g \circ f} = M_g M_f = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix} = \begin{bmatrix} 1 \times 2 + 2 \times 3 & 1 \times 5 + 2 \times 8 \\ 4 \times 2 + 5 \times 3 & 4 \times 5 + 5 \times 8 \end{bmatrix} = \begin{bmatrix} 2 + 6 & 5 + 16 \\ 8 + 15 & 20 + 40 \end{bmatrix} = \begin{bmatrix} 8 & 21 \\ 23 & 60 \end{bmatrix} $$ 7. **Find inverse of $f$:** Calculate determinant: $$ \det(M_f) = 2 \times 8 - 5 \times 3 = 16 - 15 = 1 $$ Since determinant is 1 (nonzero), inverse exists: $$ M_f^{-1} = \frac{1}{1} \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 8 & -5 \\ -3 & 2 \end{bmatrix} $$ So, $$ f^{-1}(x,y) = (8x - 5y, -3x + 2y) $$ 8. **Find inverse of $g$:** Calculate determinant: $$ \det(M_g) = 1 \times 5 - 2 \times 4 = 5 - 8 = -3 $$ Inverse exists since determinant $\neq 0$: $$ M_g^{-1} = \frac{1}{-3} \begin{bmatrix} 5 & -2 \\ -4 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{5}{3} & \frac{2}{3} \\ \frac{4}{3} & -\frac{1}{3} \end{bmatrix} $$ So, $$ g^{-1}(x,y) = \left(-\frac{5}{3}x + \frac{2}{3}y, \frac{4}{3}x - \frac{1}{3}y\right) $$