System of equations and solving for an unknown

[Kolika28]

Homework Statement

I need help with this problem. I have tried everything and get different answers all the time. Given the three equations under, I'm supposed to find out for which value of k do we have


a) No soultion b) A unique solution c) Infinitely many solutions

Relevant Equations

My teacher has told me the following is true if the matrix is in row echelon form :

1. If there are one or more columns with no pivot-number, there are infinitely many solutions
2. If every column contains a pivot number, its a unique solution.
3. If every element in the row is zero, except for the argument column, there is no solution.

The first thing I do is making the argumented matrix:

Then I try to rearrange to make the row echelon form. But maybe that's what confusses me the most. I have tried different ways of doing it, for example changing the order of the equations. I always end up with $k+number$ expression in the second column in the third row that I try to make zero. I have tried to different ways to make the row echelon form now:

1.
Rearrange R1 and R3:

Then taking -2R1+R2 and -R1+R3

The last thing I did was multiplying R3 with $\frac{1}{k-10}$ and afterwards took $\frac{1}{21}$R2+R3:

From rule number 3 there must be no solution when k=-5 because then row 3 will be 0 - 0 - 0 - (8/105). And if k=3, the last row will be 0 - 0 - 0 - 0. From rule number one, there will be infinite many solutions.

2. If I use this calculator: Calculator, I get that the row echelon form is:

This is a very different matrix then what I calculated. But if I assume that this is the right answer and given the rules my teacher has taught me, there will be no solution when k=-5 because then the last row will be 0 - 0 - 0 - (2/3). And if k=3, the last row will be 0 - 0 - 0 - 0. From rule number one, there will be infinite many solutions.So with both row echelon forms of the matrix, I find the numbers -5 and 3. But I don't manage to find out for which k we have a unique solution. What confuses me even more are the fractions I get containing k in both examples. k=10 is not valid in example 1 and k=-1/2 is not valid in example 2. Why are these values different in both examples? A friend of me told me that she got 1/2 as the unique solution, but I don't see how. Thank you if you have taken the time reading everything. I really need some help with this one, because I have been stuck on this problem for three days now.

 

[docnet]

If you performed the matrix operations correctly, the two resulting echelon form matrices will represent the same system of equations. I believe your explanation regarding $k=-5$ and $k=3$ is correct.

Regarding $k=-.5$ and $k=10$, you could try plugging them into the equations to see if there is indeed a unique solution for each

 

[Kolika28]

If you performed the matrix operations correctly, the two resulting echelon form matrices will represent the same system of equations. I believe your explanation regarding $k=-5$ and $k=3$ is correct.

Regarding $k=-.5$ and $k=10$, you could try plugging them into the equations to see if there is indeed a unique solution for each

So $k=-0.5$ and $k=10$ can still be a valid solution for k even though they make the denominator zero? And if $k=-5$ and $k=3$ gives no solution and infinite many solutions, does that mean that every other value will give a unique solution? Thank you so much for your reply by the way

 

[docnet]

So $k=-0.5$ and $k=10$ can still be a valid solution for k even though they make the denominator zero? And if $k=-5$ and $k=3$ gives no solution and infinite many solutions, does that mean that every other value will give a unique solution? Thank you so much for your reply by the way

Yes, and you're welcome

 

[Mark44]

So with both row echelon forms of the matrix, I find the numbers -5 and 3. But I don't manage to find out for which k we have a unique solution. What confuses me even more are the fractions I get containing k in both examples. k=10 is not valid in example 1 and k=-1/2 is not valid in example 2. Why are these values different in both examples?

k - 10 doesn't really enter into the decision.
If, instead of dividing by k - 10 as you did, you multiply row 2 by k - 10 and add it to 21 times row 3, you get this augmented matrix:
$$\begin{bmatrix}1&10&-6&|&1 \\ 0&-20&k+12&|&3 \\ 0 & 0 & k^2 + 2k - 15 &|&3k - 9 \end{bmatrix}$$

Pretty clearly, if k = 3 or k = -5, the first three entries in the bottom row are all 0. If k = 3, the entire row consists of 0 values, so there are an infinite number of solutions. If k = -5, the fourth entry in the bottom row is nonzero, so the system is inconsistent (no solutions). If k is any value other than 3 or -5, the system is consistent and there will be a unique solution of the system for each value of k.

 

[docnet]

If, instead of dividing by k - 10 as you did, you multiply row 2 by k - 10 and add it to 21 times row 3, you get this augmented matrix:

This is a valid point. $k$ is treated as a constant while performing row operations. we can multiply any row of a matrix by a constant to give an equivalent matrix. For example, we multiply 3rd row of op's reduced echelon matrix

$$\begin{bmatrix}1&10&-6&|&1 \\ 0&-20&k+12&|&3 \\ 0 & 0 & \frac{(-k^2 + 2k - 15)}{21k-210} &|& \frac{k - 3}{7k-70} \end{bmatrix}$$

by $-(21k-210)$, which is a constant, and turn it into

$$\begin{bmatrix}1&10&-6&|&1 \\ 0&-20&k+12&|&3 \\ 0 & 0 & k^2 + 2k - 15 &|& 9-3k \end{bmatrix}$$

 

[Kolika28]

k - 10 doesn't really enter into the decision.
If, instead of dividing by k - 10 as you did, you multiply row 2 by k - 10 and add it to 21 times row 3, you get this augmented matrix:
$$\begin{bmatrix}1&10&-6&|&1 \\ 0&-20&k+12&|&3 \\ 0 & 0 & k^2 + 2k - 15 &|&3k - 9 \end{bmatrix}$$

Pretty clearly, if k = 3 or k = -5, the first three entries in the bottom row are all 0. If k = 3, the entire row consists of 0 values, so there are an infinite number of solutions. If k = -5, the fourth entry in the bottom row is nonzero, so the system is inconsistent (no solutions). If k is any value other than 3 or -5, the system is consistent and there will be a unique solution of the system for each value of k.

Thank you so much! I never thought about multiplying row 2 by k - 10. That gives a lot of sense and make the matrix look a bit more "tidier". Appreciate your help! ♥

 

[Kolika28]

This is a valid point. $k$ is treated as a constant while performing row operations. we can multiply any row of a matrix by a constant to give an equivalent matrix. For example, we multiply 3rd row of op's reduced echelon matrix

$$\begin{bmatrix}1&10&-6&|&1 \\ 0&-20&k+12&|&3 \\ 0 & 0 & \frac{(-k^2 + 2k - 15)}{21k-210} &|& \frac{k - 3}{7k-70} \end{bmatrix}$$

by $-(21k-210)$, which is a constant, and turn it into

$$\begin{bmatrix}1&10&-6&|&1 \\ 0&-20&k+12&|&3 \\ 0 & 0 & k^2 + 2k - 15 &|& 9-3k \end{bmatrix}$$

Thank you so much, that was a good explanation!