Solution MO.C13 Chris Black

The given equation \begin{align*} \begin{bmatrix} 2 & 1 \\ 1 & -2 \\ 2 & 6 \end{bmatrix} &= \alpha\begin{bmatrix} 3 & 1 \\ 2 & 0 \\ 1 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 1 \\ 3 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3\alpha - 4 & \alpha - 1 \\ 2\alpha - 3 & -2 \\ \alpha & 4\alpha - 1 \end{bmatrix} \end{align*} gives a system of six equations in $\alpha$: \begin{align*} 3\alpha - 4 &= 2\\ \alpha - 1 &= 1 \\ 2\alpha - 3&= 1\\ -2 &= -2\\ \alpha &= 2\\ 4\alpha - 1 &= 6. \end{align*} Solving each of these equations, we see that the first three and the fifth all lead to the solution $\alpha = 2$, the fourth equation is true no matter what the value of $\alpha$, but the last equation is only solved by $\alpha = 7/4$. Thus, the system has no solution, and the original matrix equation also has no solution.