Thread: 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

Thank you for your help!

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.

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.

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: .

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

displays:

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:



clearing out the first column leaves us with:



dividing row 2 by 6:



clearing out the second column:



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



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:



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:



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).

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!

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:



I divide by 3 giving:




Then I subtract from giving me:



Then I multiply by 2 and then subtract from giving me:



At this point we can notice that and are the same, so I can disregard yielding:



Then I can divide by 2 giving:



I then assume that is different than 0 and therefore I can multiply by k and subtract that from giving me:



Rewriting by multiplying through and factoring gives me:



Now to get a 1 in the 3rd row and 3rd column I must multiply 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:



So now I know that = -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 by (k+3)/2 and subtract that from and then get:



Then I multiply by 3 and add it to getting:




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 , , ?

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! :)

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

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

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":

there are a couple problems.

first, k times the third entry in the second row is , so you should have:



this affects what you have after the |: it becomes

, which is what should be.

this means that

finally, .

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.

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!