## Elementary Linear Algebra 7th Edition

$S$ is a basis for $M_{2,2}$.
The set $S=\left\{\left[\begin{array}{ll}{2} & {0} \\ {0} & {3}\end{array}\right],\left[\begin{array}{ll}{1} & {4} \\ {0} & {1}\end{array}\right],\left[\begin{array}{ll}{0} & {1} \\ {3} & {2}\end{array}\right],\left[\begin{array}{ll}{0} & {1} \\ {2} & {0}\end{array}\right]\right\}$ is a basis for $M_{2,2}$. Indeed, suppose the following linear combination $$a\left[\begin{array}{ll}{2} & {0} \\ {0} & {3}\end{array}\right]+b\left[\begin{array}{ll}{1} & {4} \\ {0} & {1}\end{array}\right]+c\left[\begin{array}{ll}{0} & {1} \\ {3} & {2}\end{array}\right]+d\left[\begin{array}{ll}{0} & {1} \\ {2} & {0}\end{array}\right]=\left[\begin{array}{cc}{0} & {0} \\ {0} & {0}\end{array}\right].$$ Comparing the entries of both sides of the above equation, we have \begin{align*} 2a+b&=0\\ 4b+ c+d&=0\\ 3c+2d&=0\\ 3a+b+2c&=0. \end{align*} The augmented matrix of the above system is given by $$\left[\begin{array}{cccc} {2}&{1}&{0}&{0}\\ {0}&{4}&{1}&{1}\\ {0}&{0}&{3}&{2}\\ {3}&{1}&{2}&{0}\\ \end{array}\right]$$ and its determinant is $-31$. Therefore, the above system has unique solution, that is, the trivial solution. Then $S$ is linearly independent set of vectors and since $M_{2,2}$ has dimension $4$, then by Theorem 4.12, $S$ is a basis for $M_{2,2}$.