Answer
The resulting matrix has 3 pivot rows and hence it is invertible by the matrix theorem.
this shows that the columns form a basis
Work Step by Step
From the given Vectors;
$\begin{bmatrix}2\\-2\\1\end{bmatrix},\begin{bmatrix}1\\-3\\2\end{bmatrix},\begin{bmatrix}-7\\5\\4\end{bmatrix}$
We need to inspect the vectors to determine which sets in bases for $R^3$
We can row reduce the matrix by combining the vectors to form a Matrix $M$
by forming the Matrix $\mathbf{M}$
$\mathbf{M}=\begin{bmatrix}2&1&-7\\-2&-3&5\\1&2&4\end{bmatrix}$
$\mathbf{M}=\begin{bmatrix}2&1&{ - 7}\\0&{ - 2}&{ - 2}\\0&3&{15}\end{bmatrix}$ ; $R_{2}=(R_{2}+R_{1})$ and $R_{3}=(2R_{3}-R_{1})$
$\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}2&1&{ - 7}\\0&1&1\\0&1&5\end{array}} \right]\,\,\,;\,\,\,\left\{ \begin{array}{l}{R_2} = \dfrac{{{R_2}}}{{ - 2}}\\{R_3} = \dfrac{{{R_3}}}{3}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}2&0&{ - 8}\\0&1&1\\0&0&4\end{array}} \right]\,\,\,;\,\,\,\left\{ \begin{array}{l}{R_1} = {R_1} - {R_2}\\{R_3} = {R_3} - {R_2}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 4}\\0&1&1\\0&0&1\end{array}} \right]\,;\,\,\,\left\{ \begin{array}{l}{R_2} = \dfrac{{{R_2}}}{2}\\{R_3} = \dfrac{{{R_3}}}{4}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\,;\,\,\,\left\{ \begin{array}{l}{R_1} = {R_1} + 4{R_3}\\{R_2} = {R_2} - {R_3}\end{array} \right\}\end{array}$
The resulting matrix has 3 pivot rows and hence it is invertible by the matrix theorem.
this shows that the columns form a basis
