Determine if {[2−41],[−35−1]} is a basis for Col[1−31],[−37−2]

$\{\left[\begin{array}{c}2\\ -4\\ 1\end{array}\right],\left[\begin{array}{c}-3\\ 5\\ -1\end{array}\right]\}$

Since $\text{⧸}\mathrm{\exists }\text{any C in R such that}\left[\begin{array}{c}+2\\ -4\\ 1\end{array}\right],\left[\begin{array}{c}-3\\ 5\\ -1\end{array}\right]$

$\text{So}\left[\begin{array}{c}2\\ -4\\ 1\end{array}\right],\left[\begin{array}{c}-3\\ 5\\ -1\end{array}\right]$

are linearly independent.

Now for $\left[\begin{array}{c}1\\ -3\\ 1\end{array}\right],=C_1\left[\begin{array}{c}2\\ -4\\ 1\end{array}\right]+C_2\left[\begin{array}{c}-3\\ 5\\ -1\end{array}\right]\Rightarrow \begin{array}{c}2C_1-3C_2=1-(1)\\ -4C_1+5C_2=-3-(2)\\ C_1-C_2=1-(3)\end{array}$

From (3) $C_1=C_2+1,$

From (1) $2C_2+2-3C_2=1\Rightarrow -C_2=-1\Rightarrow C_2=1$

Thus $C_1=2$

So $\left[\begin{array}{c}1\\ -3\\ 1\end{array}\right]=2\left[\begin{array}{c}2\\ -4\\ 1\end{array}\right]+(1)x\left[\begin{array}{c}-3\\ 5\\ -1\end{array}\right]$

Against for $\left[\begin{array}{c}-3\\ 7\\ -2\end{array}\right]=x\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]+y\left[\begin{array}{c}-3\\ 5\\ -1\end{array}\right]\Rightarrow \begin{array}{c}2x-3y=-3-(4)\\ -4x+5y=+7-(5)\\ x-y=2-(6)\end{array}$

From (6) $x=y-2$

From (4) $2y-4-3y=-3\Rightarrow -y=1\Rightarrow y=-1$
$\Rightarrow x=-1-2=-3$

From (5) $-4x(-3)+5(-1)=12-5=7$

So $\left[\begin{array}{c}-3\\ 7\\ 2\end{array}\right]=(-3)\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]+(-1)\left[\begin{array}{c}-3\\ 5\\ -1\end{array}\right]$

So $\{\left[\begin{array}{c}2\\ -4\\ 1\end{array}\right],\left[\begin{array}{c}-3\\ 5\\ -1\end{array}\right]\}$

generate $col\left[\begin{array}{cc}1& -3\\ -3& 7\\ 1& 2\end{array}\right]$ .

So $\{\left[\begin{array}{c}2\\ -4\\ 1\end{array}\right],\left[\begin{array}{c}-3\\ 5\\ -1\end{array}\right]\}$

forms a basis of