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To find a standard basis vector that can be added to the set {v1, v2} to create a basis for R3, the determinants of matrices formed with standard basis vectors and the given vectors are calculated. The determinants for the vectors (1,0,0) and (0,1,0) show they are linearly independent from {v1, v2}, while the determinant for (0,0,1) is zero, indicating linear dependence. Both sets { (1,0,0), v1, v2 } and { (0,1,0), v1, v2 } span R3, confirming they are valid bases. Thus, either (1,0,0) or (0,1,0) can be added to the set to form a basis for R3.
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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.
 
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Sudharaka said:
Title: Find a standard basis vector
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.
Correct! one has to first solve the simultaneous equation:
##x=α-β+ϒ##
##y=2β-2ϒ##
##z=3β-2ϒ##

##x=α-β+ϒ⇒α=x+β-ϒ##
and from
##y=2β-2ϒ##⇒
##\dfrac{y}{2}##=##β-ϒ##
on substituting into
##α=x+β-ϒ##
##α=x+\dfrac{y}{2}## as shown...the other constants can be found in a similar approach (by use of elimination method).
 
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