1
$\begingroup$

Show that $(1-t, 2+3t, 3-2t)$ is a solution for all values of $t$ to following linear system:
$a + b + c = 6$
$a - b - 2c = -7$
$5a + b - c = 4$

I have found that these rows are linearly dependent but I do not know how to continue. Can you help me? Here is after Gaussian Elimination:

$$\begin{bmatrix}1 & 0 & -\dfrac{1}{2} & -\dfrac{1}{2} \\0 & 1 & \dfrac{3}{2} & \dfrac{13}{2} \\ 0 & 0 & 0 & 0 \end{bmatrix} $$

$\endgroup$
2
  • $\begingroup$ Just put in $(1-t)$ for $a$ and so on, and check if it fits $\endgroup$ Commented Jan 26, 2014 at 23:00
  • $\begingroup$ @amzoti yes ı have tried and found that these equations ae dependent. $\endgroup$ Commented Jan 26, 2014 at 23:01

2 Answers 2

2
$\begingroup$

Hint: One approach is to use Gaussian Elimination, you end up with:

$$\begin{bmatrix}1 & 0 & -\dfrac{1}{2} & -\dfrac{1}{2} \\0 & 1 & \dfrac{3}{2} & \dfrac{13}{2} \\ 0 & 0 & 0 & 0 \end{bmatrix} $$

Can you take it from here?

Update

We have:

  • $a = -\dfrac{1}{2} +\dfrac{1}{2}c$
  • $b = \dfrac{13}{2} -\dfrac{3}{2}c$

So, $c$ is a free variable and from the problem statement, they chose $c = 3-2t$.

Now, what are $a$ and $b$? Also, can you write this in another form?

$\endgroup$
1
  • $\begingroup$ Thank you very much,I was look for how can I write matrix, copied from yor answer :) I would be nice if you can show me what sould I do after elimination. $\endgroup$ Commented Jan 26, 2014 at 23:13
0
$\begingroup$

It is not asked to find all soltions for your system but only, if the given points are really solutions of your siystem, so just put them in your system, and see if it fits:

$$ 6\stackrel{!}{=} (1-t) + (2+3t)+(3-2t) = \ldots = 6 $$

so the first equation is true for all values of $t$.

For the 2nd and 3rd equation it is the same you have to do.

$\endgroup$
1
  • $\begingroup$ Thank you, I have did id so. $\endgroup$ Commented Jan 26, 2014 at 23:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .