Problem is taken from A First Course in Linear Algebra, Chapter M, Matrices

A:

Using Matrix Scalar Multiplication

\(\alpha\left[\begin{array}{cc}1 & 2\\4 & 1\end{array}\right] + \beta\left[\begin{array}{cc}2 & 1\\3 & 1\end{array}\right] = \left[\begin{array}{cc}-1 & 4\\6 & 1\end{array}\right]\)

Converts to \(\left[\begin{array}{cc}1\alpha & 2\alpha\\4\alpha & 1\alpha\end{array}\right] + \left[\begin{array}{cc}2\beta & 1\beta\\3\beta & 1\beta\end{array}\right] = \left[\begin{array}{cc}-1 & 4\\6 & 1\end{array}\right]\)

Using Matrix Addition

\(\left[\begin{array}{cc}1\alpha + 2\beta & 2\alpha + 1\beta\\4\alpha + 3\beta & 1\alpha + 1\beta\end{array}\right] = \left[\begin{array}{cc}-1 & 4\\6 & 1\end{array}\right]\)

This results in four equations \(1\alpha + 2\beta = -1\), \(2\alpha + 1\beta = 4\), \(4\alpha + 3\beta = 6\), \(1\alpha + 1\beta = 1\)

Solving two equations \(1\alpha + 2\beta = -1\), \(2\alpha + 1\beta = 4\)

\(-2*(1\alpha + 2\beta) = -1 * -2\), \(2\alpha + 1\beta = 4\)

\(-2\alpha - 4\beta = 2\), \(2\alpha + 1\beta = 4\)

\(-2\alpha - 4\beta = 2\), \(2\alpha + 1\beta = 4\)

\(-3\beta = 6\)

\(\beta = -2\)

Substituting beta value

\(2\alpha + 1\beta = 4\)

\(2\alpha + 1(-2) = 4\)

\(2(\alpha - 1) = 4\)

\((\alpha - 1) = 2\)

\(\alpha = 3\)

Substituting values of alpha and beta

\(\left[\begin{array}{cc}1(3) + 2(-2) & 2(3) + 1(-2)\\4(3) + 3(-2) & 1(3) + 1(-2)\end{array}\right] = \left[\begin{array}{cc}-1 & 4\\6 & 1\end{array}\right]\)

\(\left[\begin{array}{cc}(3 -4) & (6 -2)\\(12 - 6) & (3 - 2)\end{array}\right] = \left[\begin{array}{cc}-1 & 4\\6 & 1\end{array}\right]\)

\(\left[\begin{array}{cc}-1 & 4\\6 & 1\end{array}\right] = \left[\begin{array}{cc}-1 & 4\\6 & 1\end{array}\right]\)