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:

x_{1 } - x_{3 }= -3

3x_{1} + 6x_{2} + 3kx_{3}= 3

kx_{2} + 5x_{3}= 4

2x_{1} + kx_{2} - x_{3}= -2

Thank you for your help!

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.

Re: System of linear equations with parameters

my solutions at the end turn out to be x_{1} = 3x_{3} - 3; x_{2} = [(k+3)/2 ]x_{3} + x_{2}; x_{3} = (2-k)/(k-5)(k+2). Therefore x_{1} and x_{2} are my variables and my x_{3} 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 0x_{1} + 0x_{2} + x_{3}= -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.

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 [tex] and [/tex] as "start" and "end" tags.

for example: [tex]x_1[/tex] displays as: $\displaystyle x_1$.

and [tex]\begin{bmatrix}a&b&c\\d&e&f\\g&h&j \end{bmatrix}[/tex]

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 x_{3} = 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 x_{2} = 2, and row 1 tells us that x_{1} = -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 x_{0} is any solution to Ax_{0} = 0, then

A(x+x_{0}) = A(x) + A(x_{0}) = 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: 4x_{3} = 0, and thus equation 3 quickly gives us:

k = 4/x_{2}

equation 1 gives us: x_{1} = -3, and then equation 2 gives us x_{2} = 2, so that k = 4/2 = 2).

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! :)

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! :)

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.

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!