# Linear algebra - matrix equations

• karnten07
In summary, the conversation discusses solving a linear algebra problem involving matrix equations. The first part of the problem shows that the given equation has no solutions in R^3. The second part asks to find two linearly independent vectors b in R^3 such that the equation Ax=b has a solution. The conversation then explores different techniques for solving the problem, including row reduction and substitution. The final solution includes two linearly independent vectors (1,0,-1) and (0,1,1) that satisfy the equation.

#### karnten07

[SOLVED] linear algebra - matrix equations

## Homework Statement

Consider the matrix A =
1 1 1
2 1 0
1 0 -1

a.)
Show that the equation Ax =
1
2
3
has no solutions in R^3, where R is the set of real numbers.

b.) Find two linearly independant vectors b in R^3 such that the equations Ax=b has a solution in R^3.

## The Attempt at a Solution

For part a.) i inserted a vector x of the form
a
b
c
into the equation and then multiplied this x by A. I then had 3 equations in a b and c:

(1) a + b + c = 1
(2) 2a + b =2
(3) a - c = 3

Rearranging (3) i get c = a - 3 and substituting this into (1), it becomes 2a + b = 4
But this contradicts (2) so i conclude that the equation has no solution in R^3.

I am stuck on part b.), if i change the original vector(1,2,3) to b equalling (x,y,z) i can simply replace the terms in the original set of euqations.

In doing this i find that 2a+b=x+z meaning that y = x+z

Is this the right way to do this, and how do i find the numbers in vector b? Could i use x = 1, y=3 and z=2 as one vecotr b as this satisfies my equations? The other being the same numbers but each is negative?

Any help is greatly appreciated, thanks

It's strange you're not using the techniques of linear algebra to do the problem for a). I suspect that you're probably unfamiliar with the theory behind linear algebra for Ax = b to have solutions.

I don't know how much you have learned about linear algebra so far but have you learned anything about column spaces yet? It'll help if you do.

The "technique" Defennnder is talking about is, I think, "row reduction" and can be used to solve both of your problems in terms of matrix manipulations. However, what you have done is perfectly correct and is really the "basic" algebra version of those row reductions.

What you are saying, for (b), is that in order to have a solution, your <x, y, z> vector must satisfy y= x+ z. Choose values for two of those and solve for the third.
Could i use x = 1, y=3 and z=2 as one vecotr b as this satisfies my equations? The other being the same numbers but each is negative?
Yes, <1, 3, 2> works- in fact the equation then has an infinite number of solutions. It is also true that <-1, -3, -2> will also work but those two vectors are obviously not independent and you are asked for two linearly independent vectors that have that property.

I like to use "1" and "0" just because they are easy! Since, as you say, z= x+ y, if x= 1 and y= 0, z= 1. <1, 0, 1> is such a vector. If x= 0 and y= 1, then z= 1 so <0, 1, 1> is also such a vector. Can you see that they are obviously independent?

HallsofIvy said:
The "technique" Defennnder is talking about is, I think, "row reduction" and can be used to solve both of your problems in terms of matrix manipulations. However, what you have done is perfectly correct and is really the "basic" algebra version of those row reductions.

What you are saying, for (b), is that in order to have a solution, your <x, y, z> vector must satisfy y= x+ z. Choose values for two of those and solve for the third.

Yes, <1, 3, 2> works- in fact the equation then has an infinite number of solutions. It is also true that <-1, -3, -2> will also work but those two vectors are obviously not independent and you are asked for two linearly independent vectors that have that property.

I like to use "1" and "0" just because they are easy! Since, as you say, z= x+ y, if x= 1 and y= 0, z= 1. <1, 0, 1> is such a vector. If x= 0 and y= 1, then z= 1 so <0, 1, 1> is also such a vector. Can you see that they are obviously independent?

Thanks for your help guys, I've been looking over my course sylabus and i don't see row reduction in it but i think it is similar to Guassian elimination which is covered in my course. So i have used that to get it to row echelon form to show that in part a.) the matrix is inconsistent.

The row echelon form i get is:
1 0.5 0¦ b2/2
0 1 2 ¦ 2b1-b2
0 0 0 ¦ b1+b3-b2\right][/tex]

By rearranging and substitution i get:
(x1+x2+x3) = b1
(2x1+x2) = b2
(x1-x3) = b3

So using b1+b3 = b2 i see that there are solutions of b as (1, 0, -1), (0,1,1), (1,1,0) and the null space solution (0,0,0).

From this i get solutions of Ax=b for these particular b values as:
x = (-1,2,0), (0,1,-1), (0,1,0) and the null space solution (1,-2,1)

So i am asked for 2 linearly independant solutions. Which ones do i choose, and with each one do i include the nullspace solution as a part of each? Many thanks

p.s i just reread the question and i am only asked to find 2 linearly independant vectors, b, that produce a solution in R^3. So 1,0,-1 and 0,1,1 are independent. I might just go for those. Thanks again

Last edited:

## 1. What is a matrix equation?

A matrix equation is an equation that involves matrices and vectors. It is used to represent systems of linear equations in a concise and organized manner.

## 2. How do you solve a matrix equation?

To solve a matrix equation, one can use various methods such as Gaussian elimination, Cramer's rule, or matrix inversion. The method used depends on the size and complexity of the equation.

## 3. What is the purpose of using matrix equations?

Matrix equations are used to represent and solve systems of linear equations, which are commonly used in fields such as physics, engineering, and economics. They also have applications in computer graphics, data analysis, and machine learning.

## 4. Can matrix equations have multiple solutions?

Yes, matrix equations can have multiple solutions. This is known as an overdetermined system, where there are more equations than unknowns. In this case, the solutions form a subspace.

## 5. What is the difference between a matrix equation and a linear equation?

A linear equation involves only variables and constants, while a matrix equation involves matrices and vectors. Additionally, a linear equation has only one variable, while a matrix equation can have multiple variables represented by different matrices and vectors.

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