Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Did you try Gaussian Elimination to reduce the system and see if there are values of $t$ of interest? – Amzoti Jun 30 '13 at 3:56
I'm not there yet to do that. any suggestions without matrix methods? – Sarunas Jun 30 '13 at 3:58
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. – Ataraxia Jun 30 '13 at 4:30
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? – DJohnM Jun 30 '13 at 4:38
$\displaystyle \left(\begin{array}[c]\\ 3\\-1\\1\end{array}\right)= 2\displaystyle \left(\begin{array}[c]\\ 2\\1\\1\end{array}\right)- \displaystyle \left(\begin{array}[c]\\ 1\\3\\1\end{array}\right)$ – Angela Richardson Jun 30 '13 at 5:22
up vote 3 down vote accepted

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.

share|cite|improve this answer
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. – copper.hat Jun 30 '13 at 5:43
Fixed! Sorry about the noise. – copper.hat Jun 30 '13 at 5:45
I would have done pretty much the same thing, but I've missed the dependence of one of the vectors. Thank you! – 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'.

share|cite|improve this answer
Yes, but OP in the comments asks for answers not using matrix methods. – Gerry Myerson Jun 30 '13 at 5:26
I guess you are referring to the use of the rank nullity theorem? – copper.hat Jun 30 '13 at 5:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.