Solve the matrix equation: A (3 4; -1 2; 2 1) = (0 15; 1 -2).

The matrix on the left has 3 rows and 2 columns and the right side is $2 \times 2$, so $A$ must be a $2 \times 3$ matrix. Write $A = \begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}$. Row 1 gives $3a - b + 2c = 0$ and $4a + 2b + c = 15$. Eliminating $c = 15 - 4a - 2b$ gives $a + b = 6$, so $b = 6 - a$ and $c = 3 - 2a$. Row 2 gives $3d - e + 2f = 1$ and $4d + 2e + f = -2$. Eliminating $f = -2 - 4d - 2e$ gives $d + e = -1$, so $e = -1 - d$ and $f = -2d$. There are 6 unknowns and only 4 equations, so the equation has no unique solution; the general solution is $A = \begin{pmatrix} a & 6 - a & 3 - 2a \\ d & -1 - d & -2d \end{pmatrix}$ for any real $a$, $d$ (for example $a = d = 0$ gives $A = \begin{pmatrix} 0 & 6 & 3 \\ 0 & -1 & 0 \end{pmatrix}$).