# Thread: System of linear equations with parameters

1. ## System of linear equations with parameters

Hi,

I am wondering if anyone could help me with this system or linear equations with parameter k. I've tried using reduced row echelon form, but I'm not sure about my answer. This is the problem:

x1 - x3 = -3
3x1 + 6x2 + 3kx3= 3
kx2 + 5x3= 4
2x1 + kx2 - x3= -2

2. ## Re: System of linear equations with parameters

can you show your work? you can use row-reduction for expressions involving parameters just like "normal" row-reduction, BUT:

if you wind up with values of k that make denominators of fractional expressions 0, these values of k have to be investigated separately.

3. ## Re: System of linear equations with parameters

my solutions at the end turn out to be x1 = 3x3 - 3; x2 = [(k+3)/2 ]x3 + x2; x3 = (2-k)/(k-5)(k+2). Therefore x1 and x2 are my variables and my x3 is my free variable. Is there anything else I should be mentioning for my k value other than the following?

When k = -2 my system is not consistent because I get 0x1 + 0x2 + x3= -3 which is impossible. Similarly when k = 5 my system is still not consistent. I am not sure in the other cases, when k is different than -2 and 5.

I did all the work out, but I am not able to write a matrix on this page so I am unable to show you my work. I'm sorry. I hope I didn't make any adding/subtracting errors.

4. ## Re: System of linear equations with parameters

i will share with you what little i've learned about using latex on this forum:

to display special math formatting use $$and$$ as "start" and "end" tags.

for example: $$x_1$$ displays as: $\displaystyle x_1$.

and $$\begin{bmatrix}a&b&c\\d&e&f\\g&h&j \end{bmatrix}$$

displays: $\displaystyle \begin{bmatrix}a&b&c\\d&e&f\\g&h&k \end{bmatrix}$

i suspect you may have made an error in arithmetic, but the only way i can be sure is to row-reduce it myself (i hate doing this, by the way).

we have the augmented matrix:

$\displaystyle \begin{bmatrix}1&0&-1&|&-3\\3&6&3k&|&3\\0&k&5&|&4\\2&k&-1&|&-2 \end{bmatrix}$

clearing out the first column leaves us with:

$\displaystyle \begin{bmatrix}1&0&-1&|&-3\\0&6&3k+3&|&12\\0&k&5&|&4\\0&k&1&|&4 \end{bmatrix}$

dividing row 2 by 6:

$\displaystyle \begin{bmatrix}1&0&-1&|&-3\\0&1&\frac{k+1}{2}&|&2\\0&k&5&|&4\\0&k&1&|&4 \end{bmatrix}$

clearing out the second column:

$\displaystyle \begin{bmatrix}1&0&-1&|&-3\\0&1&\frac{k+1}{2}&|&2\\0&0&5-\frac{k(k+1)}{2}&|&4-2k\\0&0&1-\frac{k(k+1)}{2}&|&4-2k \end{bmatrix}$

now if we subtract row 3 from row 4, and then multiply row 4 by -1/4, we get:

$\displaystyle \begin{bmatrix}1&0&-1&|&-3\\0&1&\frac{k+1}{2}&|&2\\0&0&5-\frac{k(k+1)}{2}&|&4-2k\\0&0&1&|&0 \end{bmatrix}$

this tells us that x3 = 0. the only way for this to be consistent with row 3 is if:

4 - 2k = 0, so k = 2.

then row 2 tells us that x2 = 2, and row 1 tells us that x1 = -3.

let's verify this is a solution:

-3 - 0 = 3 (check)
3(-3) + 6(2) + 6(0) = -9 + 12 = 3 (check)
2(2) + 5(0) = 4 (check)
2(-3) + 2(2) - (0) = -6 + 4 = -2 (check)

is this the ONLY solution?

well if x is a solution to Ax = b, and x0 is any solution to Ax0 = 0, then

A(x+x0) = A(x) + A(x0) = b + 0 = b.

similarly, if y is also a solution to Ax = b, so that Ay = b, then A(y-x) = A(y) - A(x) = b - b = 0, so the difference of any two solutions to Ax = b is a solution of Ax = 0.

so to find ALL solutions, we need to find the null space of the matrix:

$\displaystyle A = \begin{bmatrix}1&0&-1\\3&6&3k\\0&k&5\\2&k&-1 \end{bmatrix}$

row-reducing like before (and then since row 3 is clearly a multiple of row 4, subtracting the appropriate multiple of row 4 from row 3, and switching them), we get:

$\displaystyle \begin{bmatrix}1&0&-1\\0&1&\frac{k(k+1)}{2}\\0&0&1\\0&0&0 \end{bmatrix}$

which is enough to tell us that the null space is just {(0,0,0)}. so we have indeed found every solution.

(a clue that we must have some "special value" for k should have been we have FOUR equations, but only THREE unknowns. so k must be some value that makes the 4th equation a linear combination of the other three. it is not hard to see that 2 times equation 1 plus equation 3 minus equation 4 gives: 4x3 = 0, and thus equation 3 quickly gives us:

k = 4/x2

equation 1 gives us: x1 = -3, and then equation 2 gives us x2 = 2, so that k = 4/2 = 2).

5. ## Re: System of linear equations with parameters

Thank you so much! I did make an arithmetical mistake. My work was not similar to the one that we had done in class so I thought I had done something wrong or the problem was totally different.

Thank you so much for helping me! I really do appreciate it. I'll be helping out on the forums as much as I can to reciprocate.

Have a wonderful day!

6. ## Re: System of linear equations with parameters

Hi again,

I actually copied my initial problem incorrectly. Here is the work I've done. Would you mind looking at my work?

I begin with the augmented matrix:

$\displaystyle \begin{bmatrix}1&0&-3&|-3\\3&6&3k&| 3\\0&k&5&| 4\\2&k&-1&|-2 \end{bmatrix}$

I divide $\displaystyle R_2$ by 3 giving:

$\displaystyle \begin{bmatrix}1&0&-3&|-3\\1&2&k&| 1\\0&k&5&| 4\\2&k&-1&|-2 \end{bmatrix}$

Then I subtract $\displaystyle R_1$ from $\displaystyle R_2$ giving me:

$\displaystyle \begin{bmatrix}1&0&-3&|-3\\0&2&k+3&| 4\\0&k&5&| 4\\2&k&-1&|-2 \end{bmatrix}$

Then I multiply $\displaystyle R_1$ by 2 and then subtract from $\displaystyle R_4$ giving me:

$\displaystyle \begin{bmatrix}1&0&-3&|-3\\0&2&k+3&| 4\\0&k&5&| 4\\0&k&5&|4\end{bmatrix}$

At this point we can notice that $\displaystyle R_3$ and $\displaystyle R_4$ are the same, so I can disregard $\displaystyle R_4$ yielding:

$\displaystyle \begin{bmatrix}1&0&-3&|-3\\0&2&k+3&| 4\\0&k&5&| 4\end{bmatrix}$

Then I can divide $\displaystyle R_2$ by 2 giving:

$\displaystyle \begin{bmatrix}1&0&-3&|-3\\0&1&(k+3)/2&| 2\\0&k&5&| 4\end{bmatrix}$

I then assume that $\displaystyle k$ is different than 0 and therefore I can multiply $\displaystyle R_2$ by k and subtract that from $\displaystyle R_3$ giving me:

$\displaystyle \begin{bmatrix}1&0&-3&|-3\\0&1&(k+3)/2&| 2\\0&0&5 - (k^2 - 3k) /2&| 4-2k\end{bmatrix}$

Rewriting $\displaystyle R_3$ by multiplying through and factoring gives me:

$\displaystyle \begin{bmatrix}1&0&-3&|-3\\0&1&(k+3)/2&| 2\\0&0& (k-5)(k+2)/2 &| -2(k-2)\end{bmatrix}$

Now to get a 1 in the 3rd row and 3rd column I must multiply $\displaystyle R_3$ by 2/(k-5)(k+2). Therefor I must eliminate the values of k that make that fraction 0; k must be different than -2 and k must be different than 5. I then get:

$\displaystyle \begin{bmatrix}1&0&-3&|-3\\0&1&(k+3)/2&| 2\\0&0& 1 &| -4(k-2)/(k-5)(k+2)\end{bmatrix}$

So now I know that $\displaystyle x_3$ = -4(k-2)/(k-5)(k+2).

Now I have to go ahead and get 0s above all of my leading variables. Is that correct?

I'd have to state that k must be different than -3, and then multiply $\displaystyle R_3$ by (k+3)/2 and subtract that from $\displaystyle R_2$ and then get:

$\displaystyle \begin{bmatrix}1&0&-3&|-3\\0&1& 0 &| (2(k-5)(k+2) + 2(k-2)(k+3)) / (k-5)(k+2)\\0&0& 1 &| -4(k-2)/(k-5)(k+2)\end{bmatrix}$

Then I multiply $\displaystyle R_3$ by 3 and add it to $\displaystyle R_1$ getting:

$\displaystyle \begin{bmatrix}1&0&0&|-3 - (-12 (k-2))/(k-5)(k+2)\\0&1& 0 &| (2(k-5)(k+2) + 2(k-2)(k+3)) / (k-5)(k+2)\\0&0& 1 &| -4(k-2)/(k-5)(k+2)\end{bmatrix}$

At this point, should I go back and figure out what happens when k is equal to the values I eliminated? And if so, do I do that case my case. Meaning case 1, k=5, I'll plug 5 in where I have all my ks and figure out what I get for values of $\displaystyle x_1$, $\displaystyle x_2$, $\displaystyle x_3$?

Thank you for your help. I'm terribly sorry I made a mistake in copying my initial problem. This way I did get a chance to figure out and create a matrix. I'm sorry that my fractions are a bit weird. I don't know how to write a fraction.

Thanks again! :)

7. ## Re: System of linear equations with parameters

to write a fraction use: \frac{numerator}{denominator}

for example \frac{3}{k+4} produces: $\displaystyle \frac{3}{k+4}$

also, the "|" symbol will line up better if you put ampersands on both sides of it as if it were a number entry in your augmented matrices.

looking over your work when you "simplify":

$\displaystyle 5 - \frac{k^2 - 3k}{2}$ there are a couple problems.

first, k times the third entry in the second row is $\displaystyle \frac{k^2 + 3k}{2}$, so you should have:

$\displaystyle 5 - \frac{k^2 + 3k}{2} = \frac{10 - 3k - k^2}{2} = \frac{(5+k)(2-k)}{2}$

this affects what you have after the |: it becomes

$\displaystyle \frac{4(2-k)}{(5+k)(2-k)} = \frac{4}{k+5}$, which is what $\displaystyle x_3$ should be.

this means that $\displaystyle x_2 = 2 - \left(\frac{k+3}{2}\right)\left(\frac{4}{k+5} \right ) = \frac{2k+10}{k+5} - \frac{2k+6}{k+5} = \frac{4}{k+5}$

finally, $\displaystyle x_1 = \frac{12}{k+5} - 3 = \frac{12}{k+5} - \frac{3k+15}{k+5} = \frac{-3(k+1)}{k+5}$.

obviously, you will have to check the values k = 0, k = 2, and k = -5 separately (because of the steps you took).

i think you should find:

k = 0 does not affect the solution.
k = 2 gives a solution with one free variable,
k = -5 is inconsistent.

8. ## Re: System of linear equations with parameters

thank you for writing to me. I'll practice typing in fractions in other posts. I always make little mistakes. I've loved math and I love helping other people with it, but I always get frustrated when I make mistakes like that. I figured something was wrong, because my answers were a bit "odd".

thanks again for all the help you've given me! I truly appreciate it!