# Finding Solutions for Linear Algebra: k Values for No/Unique/Infinite Solutions

• rafehi
In summary, the conversation revolved around finding the values of k for which a system of linear equations has no solution, a unique solution, or more than one solution. The suggested method was to reduce the system to row echelon form and then use the determinant of the coefficient matrix to determine the number of solutions. It was also mentioned that Cramer's rule could be used in certain cases, but the preferred method was Gauss triangularization. The conversation ended with a positive note about the success of the poster in their linear algebra exam.
rafehi
First time poster. Have attempted the problem, but keep hitting dead ends and have no idea how to proceed.

## Homework Statement

Determine the values of k for which the system of linear equations has (i) no solution vector, (ii) a unique solution vector, (iii) more than one solution vector (x,y,z):
kx + y + z = 1
x + ky + z = 1
x + y + kz = 1

2. The attempt at a solution

The only approach I can think of is reducing it to row echelon form, but given the k's I'm finding it impossible to do so. Putting it in augmented matrix form and reducing it as best I can, I get:
[k 1 1 | 1]
[0 (k-1) (1-k) | 0]
[1 1 k | 1]

The above is getting me nowhere. I've tried taking out factors of k, but I'm not sure if that's the way to go because the end result is overly complicated.

Any help?

You need to reduce it before you can do anything and yes that can be messy.

Once, you have reasonably reduced refer to when a matrix has no solution, unique solution, or infinite.

I've been trying that but I just can't seem to reduce to anything doable. I've got it down to row echelon form, but I'm left with the bottom line of [0 0 -k^2+2k+3|-k-3}

I don't know if I've done something wrong, but with that last line I just can't solve it.

Can someone help me out here or at least give me a push in the right direction? This question is doing my head in...

Do the determinant of the coefficient matrix, and you get a polynomial. Case by case, replace in the original matrix with the roots of the polynomial and solve. You'll get 0 or infinite solutions depending on the root.

Thanks mate, worked great.

If you don't mind me asking, what's the theory behind the method?

Also, would there be a way to extend it to find a unique solution (in an example which has one), or would row reduction be possible in such a situation?

Let's call A the coefficient matrix of a system of linear equations.
If det(A) = 0, that means at least one of the rows is a linear combination of the other rows. That's why we look for the roots in the determinant.
Now one row is linear combination of two others, for example, row 3 = row 1 + row 2 (for the coefficients). Now if the constants in row 1 and 2 add to the constant in row 3, some variables are free (infinite solutions). If the constants in row 1 and 2 don't add to the constant in row 3, there's a contradiction and there's no solution.

Now if AX = C (where X, C are vectors ; X is the unknown vector and C is the constant vector) and det(A) is not 0...
- you can solve the system with Gauss triangularization
- you can find A-1 (since det(A) is not 0) and X = A-1C
- you can use Cramer's rule to do it only with determinants
Unless you're solving a system where many coefficients are symbols, the best is Gauss.

Cramer's rule would be useful if you had to solve something like:
ax + 2by + 3az = 1
2ax + ay + 5cz = d
2x + 3cy + 5bz = 2d
Because you can do it with Gauss, but... ahem...
x = (5ab - 15 c2 - 6a2d - 10b2d + 9acd + 20bc d)/(-6 aLet's call A the coefficient matrix of a system of linear equations.
If det(A) = 0, that means at least one of the rows is a linear combination of the other rows. That's why we look for the roots in the determinant.
Now one row is linear combination of two others, for example, row 3 = row 1 + row 2 (for the coefficients). Now if the constants in row 1 and 2 add to the constant in row 3, some variables are free (infinite solutions). If the constants in row 1 and 2 don't add to the constant in row 3, there's a contradiction and there's no solution.

Now if AX = C (where X, C are vectors ; X is the unknown vector and C is the constant vector) and det(A) is not 0...
- you can solve the system with Gauss triangularization
- you can find A-1 (since det(A) is not 0) and X = A-1C
- you can use Cramer's rule to do it only with determinants
Unless you're solving a system where many coefficients are symbols, the best is Gauss.

Cramer's rule would be useful if you had to solve something like:
ax + 2by + 3az = 1
2ax + ay + 5cz = d
2x + 3cy + 5bz = 2d
Because you can do it with Gauss, but if det(A) is not 0... ahem...

$$x = \frac{5 a b - 15 c^2 - 6 a^2 d - 10 b^2 d + 9 a c d + 20 b c d}{-6 a^2 + 5 a^2 b - 20 a b^2 + 18 a^2 c + 20 b c - 15 a c^2}$$

$$y = \frac{-10 a b + 10 c - 6 a d + 12 a^2 d + 5 a b d - 10 a c d}{-6 a^2 + 5 a^2 b - 20 a b^2 + 18 a^2 c + 20 b c - 15 a c^2}$$

$$z = \frac{-2 a + 6 a c + 2 a^2 d + 4 b d - 8 a b d - 3 a c d}{-6 a^2 + 5 a^2 b - 20 a b^2 + 18 a^2 c + 20 b c - 15 a c^2}$$

So it would be kind of easy to get lost in the process.

PS: I had this semester's linear algebra exam today... it went damn well :)

Last edited:

Thanks heaps for that springo.

## 1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations and their representations in vector spaces. It involves the use of matrices, vectors, and systems of equations to analyze and solve various mathematical problems.

## 2. What are k values in linear algebra?

K values, also known as constants or coefficients, are numerical values that are used to represent the variables in a linear equation. They can affect the slope and position of a line or the values of a matrix, and are essential in solving systems of linear equations.

## 3. How do you determine the number of solutions for a linear algebra problem?

The number of solutions for a linear algebra problem depends on the values of the coefficients in the equations. If the coefficients are consistent and independent, there will be a unique solution. If one or more equations are redundant or contradictory, there will be no solutions. If there are infinitely many solutions, the equations will be consistent but dependent on each other.

## 4. What are some methods for finding solutions in linear algebra?

There are several methods for finding solutions in linear algebra, including substitution, elimination, and using matrices to solve systems of equations. These methods involve manipulating the equations to isolate the variables and solve for their values.

## 5. How can I use linear algebra in real-life applications?

Linear algebra has many real-life applications, such as in computer graphics, data analysis, and engineering. It can be used to solve optimization problems, analyze networks and systems, and create mathematical models for various phenomena in the physical world.

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