## Elementary Linear Algebra 7th Edition

$S$ is is linearly independent .
Consider the combination $$a \left[\begin{array}{ll}{1} & {-1} \\ {4} & {5}\end{array}\right]+b\left[\begin{array}{cc}{4} & {3} \\ {-2} & {3}\end{array}\right]+c\left[\begin{array}{ll}{1} & {-8} \\ {22} & {23}\end{array}\right]= \left[\begin{array}{ll}{0} & {0} \\ {0} & {0}\end{array}\right], \quad a,b,c\in R.$$ Which yields the following system of equations \begin{align*} a+4b+c&=0\\ -a+3 b-8c&=0\\ 4a-2b+22c&=0\\ 5a+3b+23c&=0. \end{align*} The coefficient matrix of the above system is given by $$\left[ \begin {array}{cccc} 1&4&1\\ -1&3&-8 \\ 4&-2&22\\ 5&3&23 \end {array} \right].$$ Using Gauss elimination, the coefficient matrix has the row-echelon form $$\left[ \begin {array}{ccc} 1&4&1\\ 0&7&-7 \\ 0&0&1\\ 0&0&0\end {array} \right] .$$ Hence the system has unique solution, that is, $$a=0, \quad b=0, \quad c=0.$$ Consequently, $S$ is is linearly independent .