Solution MO.C12 Chris Black

The given equation \begin{align*} \begin{bmatrix} 7 & 12 & 6\\ 6 & 4 & -2 \end{bmatrix} &= \alpha\begin{bmatrix} 1 & 3 & 4 \\ 2 & 1 & -1 \end{bmatrix} + \begin{bmatrix} 4 & 3 & -6 \\ 0 & 1 & 1 \end{bmatrix}\\ &= \begin{bmatrix} \alpha & 3\alpha & 4\alpha \\ 2\alpha & \alpha & -\alpha \end{bmatrix} + \begin{bmatrix} 4 & 3 & -6 \\ 0 & 1 & 1 \end{bmatrix}\\ &= \begin{bmatrix} 4 + \alpha & 3 + 3\alpha & -6 + 4\alpha \\ 2\alpha & 1 + \alpha & 1 - \alpha \end{bmatrix} \end{align*} leads to the 6 equations in $\alpha$: \begin{align*} 4 + \alpha &= 7\\ 3 + 3\alpha &= 12\\ -6 + 4\alpha &= 6\\ 2\alpha &= 6\\ 1 + \alpha &= 4\\ 1 - \alpha &= -2. \end{align*} The only value that solves all 6 equations is $\alpha = 3$, which is the solution to the original matrix equation.