## Elementary Linear Algebra 7th Edition

$S$ is not a basis for $M_{2,2}$.
The set $S=\left\{\left[\begin{array}{rr}{1} & {2} \\ {-5} & {4}\end{array}\right],\left[\begin{array}{rr}{2} & {-7} \\ {6} & {2}\end{array}\right],\left[\begin{array}{rr}{4} & {-9} \\ {11} & {12}\end{array}\right],\left[\begin{array}{rr}{12} & {-16} \\ {17} & {42}\end{array}\right]\right\}$ is not a basis for $M_{2,2}$. Indeed, suppose the following linear combination $$a\left[\begin{array}{rr}{1} & {2} \\ {-5} & {4}\end{array}\right]+b\left[\begin{array}{rr}{2} & {-7} \\ {6} & {2}\end{array}\right]+c\left[\begin{array}{rr}{4} & {-9} \\ {11} & {12}\end{array}\right]+d\left[\begin{array}{rr}{12} & {-16} \\ {17} & {42}\end{array}\right]=\left[\begin{array}{cc}{0} & {0} \\ {0} & {0}\end{array}\right].$$ Comparing the entries of both sides of the above equation, we have \begin{align*} a+2b+4c+12d&=0\\ 2a-7b-9c-16d&=0\\ -5a+6b+11c+17d&=0\\ 4a+2b+12c+42d&=0. \end{align*} The augmented matrix of the above system is given by $$\left[\begin{array}{cccc} 1&2&4&12\\ 2&-7&-9&-16 \\ -5&6&11&17\\ 4&2&12&42 \\ \end{array}\right]$$ and its determinant is $0$. Therefore, the above system has infinite number of solutions, that is, there is lots of non trivial solution. Then $S$ is linearly dependent set of vectors and then $S$ is not a basis for $M_{2,2}$.