Find all values of $t$ for which the system of equations

$$\begin{array} 22x_1 + x_2 + 4x_3 + 3x_4 = 1\\ x_1 + 3x_2 + 2x_3 − x_4 = 3t\\ x_1 + x_2 + 2x_3 + x_4 = t^2 \end{array}$$

has a solution?

I was given a theorem, that system has a solution, when column vector of RHS lies in the subspace spanned by column vectors of LHS. If we take respective column vectors, and notice that third is a scalar multiple of the first column, we get three linearly independent vectors. What I don't get is, why should there be particular $t$'s, for which the system doesn't have a solution, as if we have three linearly independent (column) vectors, they should span $\mathbb{R^3}$, and thus we could find solution for any set of $t$'s.

I suppose I'm wrong, but where's the mistake, and how should I check then, which $t$'s suffice?

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    $\begingroup$ Did you try Gaussian Elimination to reduce the system and see if there are values of $t$ of interest? $\endgroup$ – Amzoti Jun 30 '13 at 3:56
  • $\begingroup$ I'm not there yet to do that. any suggestions without matrix methods? $\endgroup$ – Sarunas Jun 30 '13 at 3:58
  • $\begingroup$ I don't think you could even begin to approach this without any matrix methods...especially since this is an underdefined system. I could be wrong though. $\endgroup$ – Ataraxia Jun 30 '13 at 4:30
  • $\begingroup$ Agreed; I could pick any value I want for $x_4$ and t, and have three equations in three unknowns. Or am I missing something really basic? $\endgroup$ – DJohnM Jun 30 '13 at 4:38
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    $\begingroup$ $\displaystyle \left(\begin{array}[c]\\ 3\\-1\\1\end{array}\right)= 2\displaystyle \left(\begin{array}[c]\\ 2\\1\\1\end{array}\right)- \displaystyle \left(\begin{array}[c]\\ 1\\3\\1\end{array}\right)$ $\endgroup$ – Angela Pretorius Jun 30 '13 at 5:22

Using Angela's observation, the system reduces to $$2x_1+x_2=1\\x_1+3x_2=3t\\x_1+x_2=t^2$$

Subtracting the third line from the first, we get $x_1=1-t^2$. The second line then gives $3x_2=3t-(1-t^2)=t^2+3t-1$, while the third line gives $x_2=t^2-(1-t^2)=2t^2-1$. Hence for the system to have a solution these must agree, i.e. $\frac{1}{3}(t^2+3t-1)=2t^2-1$ or $t^2+3t-1=6t^2-3$. This is a quadratic equation $5t^2-3t-2=0$ with two solutions, and copper.hat kindly found them explicitly.

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  • $\begingroup$ Hmm, I seem to have unintentionally & accidentally downvoted you, and seem to be unable to reverse my vote. Sorry about that. I think if you edit slightly I can reverse my vote. Actually, maybe I can edit it. $\endgroup$ – copper.hat Jun 30 '13 at 5:43
  • $\begingroup$ Fixed! Sorry about the noise. $\endgroup$ – copper.hat Jun 30 '13 at 5:45
  • $\begingroup$ I would have done pretty much the same thing, but I've missed the dependence of one of the vectors. Thank you! $\endgroup$ – Sarunas Jun 30 '13 at 12:46

Let $A$ be the matrix of the left hand side. Notice that $A (-2, 0 ,1, 0)^T = 0$ and $A(0, -1, 1, -1)^T = 0$. Also, $ \operatorname{sp} \{ (2,1,1)^T, (1 -2, 0)^T\} \subset {\cal R} A $, hence using the rank nullity theorem, we have $\dim {\cal R} A = 2$.

Aside: Consider $f(t) = (1,3t,t^2)^T$. We have $f(0), f(1), f(2)$ are linearly independent, hence it is impossible to find solutions for all $t$.

So, you need to solve $f(t) \in {\cal R} A = \operatorname{sp} \{ (2,1,1)^T, (1 -2, 0)^T\}$. Explicitly, find $x,y,t$ such that $f(t) = (1,3t,t^2)^T =x (2,1,1)^T + y(1 -2, 0)^T $. We see that $x=t^2$, $y = \frac{1}{2} t (t-3)$, and the first equation then gives $2 t^2 + \frac{1}{2} t(t-3) = 1$, which simplifies to $5 t^2-3t-2 = 0$.

This gives $t = 1$ or $t=-\frac{2}{5}$.

Addendum: Let me eliminate matrix methods from the above. Note that the third column is $-2$ times the first. So we need only worry about one of these, I will pick the first $(2,1,1)^T$. Notice that the third column equals the second plus the fourth column, so we need only worry about one of these. However, for computational simplicity, I want a zero in the vector, so instead of using the second, I will use the first minus the second $(1 -2, 0)^T$. The range of possible right hand sides is then given by $x (2,1,1)^T + y(1 -2, 0)^T $, where $x,y$ range over the real numbers. The remainder of the argument above is 'matrix free'.

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  • $\begingroup$ Yes, but OP in the comments asks for answers not using matrix methods. $\endgroup$ – Gerry Myerson Jun 30 '13 at 5:26
  • $\begingroup$ I guess you are referring to the use of the rank nullity theorem? $\endgroup$ – copper.hat Jun 30 '13 at 5:29

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