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- Feb 5, 2012

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**Title: Find a standard basis vector**

Hi saqifriends,saqifriends said:Find a standard basis vector that can be added to the set {,v_{1}} to produce a basis forv_{2}where;R^{3}

= (-1, 2, 3),v_{1}= (1, -2, -2)v_{2}

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.