Dolly Robinson

2021-02-25

Determine if $\left\{\left[\begin{array}{c}2\\ -4\\ 1\end{array}\right],\left[\begin{array}{c}-3\\ 5\\ -1\end{array}\right]\right\}$ is a basis for
$Col\left[\begin{array}{c}1\\ -3\\ 1\end{array}\right],\left[\begin{array}{c}-3\\ 7\\ -2\end{array}\right]$

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

Since

are linearly independent.

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

From (3) ${C}_{1}={C}_{2}+1,$

From (1) $2{C}_{2}+2-3{C}_{2}=1⇒-{C}_{2}=-1⇒{C}_{2}=1$

Thus ${C}_{1}=2$

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

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

From (6) $x=y-2$

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

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

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

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

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

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

forms a basis of

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