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Can I quickly check this:

I'm asked to find the value of the parameter $a$ so that the set of vectors: $\{(1,-2,3),\;(0,1,-2),\;(3,-2,a)\}$ does not span $\mathbb R^3$. What method to use to determine this?

(Am I correct that $a = 1$?)

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Make each vector a column vector in a $3 \times 3$ matrix.

$$\begin{bmatrix} 1 & 0 & 3\\ -2 & 1 & -2\\ 3&-2& a\end{bmatrix}$$

Calculate the determinant (it will be a function of $a$) and put the determinant equal to $0$, and then solve for $a$.

That value of $a$ will yield a set of vectors that are not linearly independent, and hence, those three vectors cannot span all of $\mathbb R^3$.

If you do this, you'll see that yes, indeed, $a$ must equal $1$ in order to ensure the three given vectors do not span $\mathbb R^3$.

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@ amWhy : does this method work for any n by n matrix? – user134785 May 7 '14 at 19:59
Yes, it certainly does. – amWhy May 7 '14 at 20:02

if you take 3 times the first, then the result will agree with the third vector in the first coordinate.

3 times the first gives -6 in the second coordinate and you can now add multiples of the second vector (which will have zero effect on the first coordinate) to get -2, that is you need to add 4 times the second vector to 3 times the first vector to get the first 2 coordinates to agree with (3, -2, a).

All that remains is to fix a so that when taking 4 times the second vector and 3 times the first vector you actually equal the third vector in all 3 coordinates. 4(-2) + 3(3) = 1. So yes, a = 1.

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