Solution MO.C14 Chris Black

The given equation \begin{align*} \begin{bmatrix} -1 & 4 \\ 6 & 1 \end{bmatrix} &= \alpha \begin{bmatrix} 1 & 2 \\ 4 & 1 \end{bmatrix} + \beta \begin{bmatrix} 2 & 1 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} \alpha + 2\beta & 2\alpha + \beta \\ 4\alpha + 3\beta & \alpha + \beta \end{bmatrix} \end{align*} gives a system of four equations in two variables \begin{align*} \alpha + 2\beta &= -1\\ 2\alpha + \beta &= 4\\ 4\alpha + 3\beta &= 6\\ \alpha + \beta &= 1. \end{align*} Solving this linear system by row-reducing the augmnented matrix shows that $\alpha = 3$, $\beta = -2$ is the only solution.