# Matrix Solutions: Find Linearly Independent Solutions

• UOAMCBURGER
In summary, the textbook says that the two solutions (1,-1,0) and (1,0,-1) are the only solutions to the equation x1+x2+x3=0, but the equation is "rigged" to make these solutions easy to spot.

## Homework Statement

given the matrix
https://scontent.fhlz2-1.fna.fbcdn.net/v/t1.15752-9/40882602_313421129235428_2500668957757800448_n.png?_nc_cat=0&oh=7f5d6372c263996c6a11969b072d1349&oe=5BF8F2B7

in RREF
we see solution to this system is x1+x2+x3 = 0
in the textbook it says which solutions are spanned by the vectors f1 = (−1, 1, 0) T and f2 = (−1, 0, 1) T

## The Attempt at a Solution

[/B]
I don't really understand how this is the case? Is the only way to see this to find all possible solutions to this system and then make sure each solution is linearly independent ? Or set x1 = 1 (since x1 is the only leading variable) and then clearly the system is only satisfied when (x1, x2, x3) = (1, -1, 0) or (1, 0, -1), but then why in the textbook would they set the pivot to -1 and not just 1 ?

Last edited by a moderator:
UOAMCBURGER said:

## Homework Statement

given the matrix
https://scontent.fhlz2-1.fna.fbcdn.net/v/t1.15752-9/40882602_313421129235428_2500668957757800448_n.png?_nc_cat=0&oh=7f5d6372c263996c6a11969b072d1349&oe=5BF8F2B7

in RREF
we see solution to this system is x1+x2+x3 = 0
in the textbook it says which solutions are spanned by the vectors f1 = (−1, 1, 0) T and f2 = (−1, 0, 1) T[/B]

## The Attempt at a Solution

I don't really understand how this is the case? Is the only way to see this to find all possible solutions to this system and then make sure each solution is linearly independent ? Or set x1 = 1 (since x1 is the only leading variable) and then clearly the system is only satisfied when (x1, x2, x3) = (1, -1, 0) or (1, 0, -1), but then why in the textbook would they set the pivot to -1 and not just 1 ?[/B]
What problem are you trying to solve? Finding the kernel of the given matrix?

UOAMCBURGER said:

## Homework Statement

given the matrix
https://scontent.fhlz2-1.fna.fbcdn.net/v/t1.15752-9/40882602_313421129235428_2500668957757800448_n.png?_nc_cat=0&oh=7f5d6372c263996c6a11969b072d1349&oe=5BF8F2B7

in RREF
we see solution to this system is x1+x2+x3 = 0
in the textbook it says which solutions are spanned by the vectors f1 = (−1, 1, 0) T and f2 = (−1, 0, 1) T[/B]

## The Attempt at a Solution

I don't really understand how this is the case? Is the only way to see this to find all possible solutions to this system and then make sure each solution is linearly independent ? Or set x1 = 1 (since x1 is the only leading variable) and then clearly the system is only satisfied when (x1, x2, x3) = (1, -1, 0) or (1, 0, -1), but then why in the textbook would they set the pivot to -1 and not just 1 ?[/B]
(1) Turn off the bold font!
(2) The equation ##x_1+x_2+x_3=0## describes a 2-dimensional plane in 3 dimensions. You can express any vector in the plane as a linear combination of two linearly-independent vectors, so that for any point ##{\bf p} = (x_1,x_2,x_3) ## in the plane we can write ##{\bf p} = c_1 {\bf p}_1 + c_2 {\bf p}_2##. As we vary ##(c_1,c_2)## we sweep out the whole plane.

The solution is merely giving you two possible linearly-independent vectors ##{\bf p}_1## and ##{\bf p}_2## in the plane; there are infinitely many other possible choices.

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UOAMCBURGER
Three variables and one linear constraint should tell you that there is a two-dimensional solution space. The solutions (1,-1,0) and (1,0,-1) are clearly two linearly independent solutions. The problem is "rigged" to make those solutions easy to spot. Any other solution, like (0,1,-1), can be obtained by a linear combination of the two: (0,1,-1) = -1(1,-1,0)+1(1,0,-1).
(Your statement that (1,-1,0) and (1,0,-1) are the only solutions is not correct. They form a basis for the entire space of solutions.)

Ray Vickson said:
(1) Turn off the bold font!
Fixed.

## 1. What are matrix solutions?

Matrix solutions are a set of values that satisfy a linear system of equations. In other words, they are the values of the variables that make all the equations in the system true.

## 2. How do you find linearly independent solutions?

To find linearly independent solutions, you need to first solve the linear system of equations using various methods such as substitution or elimination. Then, you can check if the solutions are linearly independent by forming a matrix using the coefficients of the variables and checking its determinant. If the determinant is non-zero, the solutions are linearly independent.

## 3. What is the importance of finding linearly independent solutions?

Finding linearly independent solutions is important because they form a basis for the solution space of a linear system of equations. This means that any other solution can be written as a linear combination of these linearly independent solutions. Additionally, linearly independent solutions help in finding the general solution to a linear system of equations.

## 4. Can a linear system of equations have more than one set of linearly independent solutions?

Yes, a linear system of equations can have more than one set of linearly independent solutions. This is because there can be multiple ways to solve a linear system of equations, and each method might result in a different set of linearly independent solutions.

## 5. How does the number of linearly independent solutions relate to the number of variables in a linear system of equations?

The number of linearly independent solutions is related to the number of variables in a linear system of equations by the rank-nullity theorem. This theorem states that the number of linearly independent solutions is equal to the number of variables minus the rank of the coefficient matrix. In other words, the more variables a linear system has, the more linearly independent solutions it can have.