Determine if left{ begin{bmatrix}2 -4 1 end{bmatrix},begin{bmatrix}-3 5 -1 end{bmatrix} right} is a basis for Col begin{bmatrix}1 -3 1 end{bmatrix},begin{bmatrix}-3 7 -2 end{bmatrix}

Question

Determine if

{[\begin{matrix} 2 \\ - 4 \\ 1 \end{matrix}], [\begin{matrix} - 3 \\ 5 \\ - 1 \end{matrix}]}

is a basis for

C o l [\begin{matrix} 1 \\ - 3 \\ 1 \end{matrix}], [\begin{matrix} - 3 \\ 7 \\ - 2 \end{matrix}]

Answer 1

${[\begin{matrix} 2 \\ - 4 \\ 1 \end{matrix}], [\begin{matrix} - 3 \\ 5 \\ - 1 \end{matrix}]}$

Since $⧸ \exists any C in R such that [\begin{matrix} + 2 \\ - 4 \\ 1 \end{matrix}], [\begin{matrix} - 3 \\ 5 \\ - 1 \end{matrix}]$

$So [\begin{matrix} 2 \\ - 4 \\ 1 \end{matrix}], [\begin{matrix} - 3 \\ 5 \\ - 1 \end{matrix}]$

are linearly independent.

Now for $[\begin{matrix} 1 \\ - 3 \\ 1 \end{matrix}], = C_{1} [\begin{matrix} 2 \\ - 4 \\ 1 \end{matrix}] + C_{2} [\begin{matrix} - 3 \\ 5 \\ - 1 \end{matrix}] \Rightarrow \begin{matrix} 2 C_{1} - 3 C_{2} = 1 - (1) \\ - 4 C_{1} + 5 C_{2} = - 3 - (2) \\ C_{1} - C_{2} = 1 - (3) \end{matrix}$

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

From (1) $2 C_{2} + 2 - 3 C_{2} = 1 \Rightarrow - C_{2} = - 1 \Rightarrow C_{2} = 1$

Thus $C_{1} = 2$

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

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

From (6) $x = y - 2$

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

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

So $[\begin{matrix} - 3 \\ 7 \\ 2 \end{matrix}] = (- 3) [\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}] + (- 1) [\begin{matrix} - 3 \\ 5 \\ - 1 \end{matrix}]$