A store sold 25 textbooks and 35 exercise books for 13500. The next day, it sold 21 textbooks and 38 exercise books for 12200. Set up and solve a matrix equation to find the price of each item.

The coefficients of $t$ and $e$ form the coefficient matrix.
The variables $t$ and $e$ form the variable matrix.
The constant terms form the constant matrix.

The matrix equation is:

$$\boxed{\begin{bmatrix} 25 & 35 \\ 21 & 38 \end{bmatrix} \begin{bmatrix} t \\ e \end{bmatrix} = \begin{bmatrix} 13500 \\ 12200 \end{bmatrix}}$$

Let $A = \begin{bmatrix} 25 & 35 \\ 21 & 38 \end{bmatrix}$.
The determinant of $A$ is $det(A) = (25)(38) - (35)(21)$.

$$det(A) = 950 - 735$$ $$det(A) = 215$$

The inverse of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

$$A^{-1} = \frac{1}{215} \begin{bmatrix} 38 & -35 \\ -21 & 25 \end{bmatrix}$$

$$\begin{bmatrix} t \\ e \end{bmatrix} = A^{-1} \begin{bmatrix} 13500 \\ 12200 \end{bmatrix}$$ $$\begin{bmatrix} t \\ e \end{bmatrix} = \frac{1}{215} \begin{bmatrix} 38 & -35 \\ -21 & 25 \end{bmatrix} \begin{bmatrix} 13500 \\ 12200 \end{bmatrix}$$

Perform the matrix multiplication:

$$\begin{bmatrix} t \\ e \end{bmatrix} = \frac{1}{215} \begin{bmatrix} (38)(13500) + (-35)(12200) \\ (-21)(13500) + (25)(12200) \end{bmatrix}$$ $$\begin{bmatrix} t \\ e \end{bmatrix} = \frac{1}{215} \begin{bmatrix} 513000 - 427000 \\ -283500 + 305000 \end{bmatrix}$$ $$\begin{bmatrix} t \\ e \end{bmatrix} = \frac{1}{215} \begin{bmatrix} 86000 \\ 21500 \end{bmatrix}$$ $$\begin{bmatrix} t \\ e \end{bmatrix} = \begin{bmatrix} \frac{86000}{215} \\ \frac{21500}{215} \end{bmatrix}$$ $$\begin{bmatrix} t \\ e \end{bmatrix} = \begin{bmatrix} 400 \\ 100 \end{bmatrix}$$

The cost of one textbook ($t$) is sh. 400.
The cost of one exercise book ($e$) is sh. 100.

The price of one textbook is $\boxed{\text{sh. } 400}$.
The price of one exercise book is $\boxed{\text{sh. } 100}$.