# saqifriends's Question from Math Help Forum

#### Sudharaka

##### Well-known member
MHB Math Helper
Title: Find a standard basis vector

saqifriends said:
Find a standard basis vector that can be added to the set {v1, v2} to produce a basis for R3 where;
v1 = (-1, 2, 3), v2 = (1, -2, -2)
Hi saqifriends,

We can use the determinants to see which standard basis vectors are linearly independent with the given two vectors.

$\begin{vmatrix} 1 & 0 & 0\\-1 & 2 & 3\\1 & -2 & -2 \end{vmatrix}\neq 0$

$\begin{vmatrix} 0 & 1 & 0\\-1 & 2 & 3\\1 & -2 & -2 \end{vmatrix}\neq 0$

$\begin{vmatrix} 0 & 0 & 1\\-1 & 2 & 3\\1 & -2 & -2 \end{vmatrix}=0$

Therefore, $$\{(1,0,0),\,(-1, 2, 3),\,(1, -2, -2)\}\mbox{ and }\{(0,1,0),\,(-1, 2, 3),\,(1, -2, -2)\}$$ are linearly independent sets.

Now we shall show that, $$\{(1,0,0),\,(-1, 2, 3),\,(1, -2, -2)\}$$ spans $$\Re^{3}$$.

Take any, $$(x,y,z)\in\Re^{3}$$.

$(x,y,z)=\alpha(1,0,0)+\beta(-1, 2, 3)+\gamma(1, -2, -2)$

$\Rightarrow \alpha=x+\frac{y}{2},\,\beta=z-y,\,\gamma=z-\frac{3\,y}{2}$

$\Rightarrow \alpha,\,\beta,\,\gamma\in\Re$

Therefore, $$\{(1,0,0),\,(-1, 2, 3),\,(1, -2, -2)\}$$ spans $$\Re^{3}$$.

Similarly it could be shown that, $$\{(0,1,0),\,(-1, 2, 3),\,(1, -2, -2)\}$$ spans $$\Re^{3}$$.

Hence both $$\{(1,0,0),\,(-1, 2, 3),\,(1, -2, -2)\}\mbox{ and }\{(0,1,0),\,(-1, 2, 3),\,(1, -2, -2)\}$$ are bases of $$\Re^{3}$$.

Kind Regards,
Sudharaka.