Thread: Gauss-Jordan complete elimination

1. Gauss-Jordan complete elimination

Use Gauss-Jordan complete elimination to solve this linear system in four variables.

$\displaystyle \begin{bmatrix} 0&0&1&0\\ 0&1&0&0\\ 1&0&0&2\\ 0&0&2&0\end{bmatrix}$

I've tried this question but can't get really far ... can anyone show a few steps on how to tackle this question

2. Originally Posted by tsal15
Use Gauss-Jordan complete elimination to solve this linear system in four variables.

$\displaystyle \begin{bmatrix} 0&0&1&0\\ 0&1&0&0\\ 1&0&0&2\\ 0&0&2&0\end{bmatrix}$

I've tried this question but can't get really far ... can anyone show a few steps on how to tackle this question
If you have an equation $\displaystyle A\mathbf{x} = \mathbf{b}$ where A is a square matrix and x and b are column vectors, you should be able to solve the system of equations.

But you haven't given x or b...

3. Originally Posted by Prove It
If you have an equation $\displaystyle A\mathbf{x} = \mathbf{b}$ where A is a square matrix and x and b are column vectors, you should be able to solve the system of equations.

But you haven't given x or b...
oh yes yes you're absolutely correct...i haven't given you guys an x or b...i will now sorry
b = $\displaystyle \begin{bmatrix}16\\4\\15\\0\end{bmatrix}$
x = $\displaystyle \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}$

but um i've tried using row operations...still i can't get to a final answer

Thanks for your continued assistance

4. Originally Posted by tsal15
oh yes yes you're absolutely correct...i haven't given you guys an x or b...i will now sorry
b = $\displaystyle \begin{bmatrix}16\\4\\15\\0\end{bmatrix}$
x = $\displaystyle \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}$

but um i've tried using row operations...still i can't get to a final answer

Thanks for your continued assistance
This system of equations is nonsense.

Multiply A by x you get...

$\displaystyle A\mathbf{x}=\begin{bmatrix}x_3\\x_2\\x_1 + 2x_4\\x_3\end{bmatrix}$.

So you have $\displaystyle \begin{bmatrix}x_3\\x_2\\x_1 + 2x_4\\x_3\end{bmatrix} = \begin{bmatrix}16\\4\\15\\0\end{bmatrix}$.

How can $\displaystyle x_3 = 16$ AND $\displaystyle 0$???

5. Originally Posted by Prove It
This system of equations is nonsense.

Multiply A by x you get...

$\displaystyle A\mathbf{x}=\begin{bmatrix}x_3\\x_2\\x_1 + 2x_4\\x_3\end{bmatrix}$.

So you have $\displaystyle \begin{bmatrix}x_3\\x_2\\x_1 + 2x_4\\x_3\end{bmatrix} = \begin{bmatrix}16\\4\\15\\0\end{bmatrix}$.

How can $\displaystyle x_3 = 16$ AND $\displaystyle 0$???
Do you mean that the matrix has infinitely many solutions? also how did you get your answer? also why did you not choose the method of reducing the matrix to a row echelon form? and how can you solve for $\displaystyle x_1$ and $\displaystyle x_4$ if $\displaystyle x_2$ nor $\displaystyle x_3$ is part of that equation?

Thanks again for your continued support

6. Originally Posted by tsal15
Do you mean that the matrix has infinitely many solutions? also how did you get your answer? also why did you not choose the method of reducing the matrix to a row echelon form? and how can you solve for $\displaystyle x_1$ and $\displaystyle x_4$ if $\displaystyle x_2$ nor $\displaystyle x_3$ is part of that equation?

Thanks again for your continued support
This is a system of the form $\displaystyle A\mathbf{x} = \mathbf{b}$.

I could see that there were a lot of 0's in A, so just by using matrix multiplication, I could multiply A by x and get a vector that is equal to b.

This system is not only unsolvable, it's nonsense. $\displaystyle x_1$ and $\displaystyle x_4$ can not be solved, and $\displaystyle x_3$ has to equal two different things at once in order for the system to make any sense at all.

Are you sure you copied the matrix A down correctly?

7. Originally Posted by Prove It
This is a system of the form $\displaystyle A\mathbf{x} = \mathbf{b}$.

I could see that there were a lot of 0's in A, so just by using matrix multiplication, I could multiply A by x and get a vector that is equal to b.

This system is not only unsolvable, it's nonsense. $\displaystyle x_1$ and $\displaystyle x_4$ can not be solved, and $\displaystyle x_3$ has to equal two different things at once in order for the system to make any sense at all.

Are you sure you copied the matrix A down correctly?
Oh my god you are absolutely correct... that matrix was for another question sorry to have wasted your time but this is the true matrix A:

$\displaystyle \begin{bmatrix}1&2&4&8\\1&-1&-2&2\\0&3&6&9\\-1&2&4&0\end{bmatrix}$

Now, i've checked and double checked and so far this matrix A is correctly copied. also, x and b still are the same.

Sorry to have caused such trouble

8. Originally Posted by Prove It
This is a system of the form $\displaystyle A\mathbf{x} = \mathbf{b}$.

I could see that there were a lot of 0's in A, so just by using matrix multiplication, I could multiply A by x and get a vector that is equal to b.

This system is not only unsolvable, it's nonsense. $\displaystyle x_1$ and $\displaystyle x_4$ can not be solved, and $\displaystyle x_3$ has to equal two different things at once in order for the system to make any sense at all.

Are you sure you copied the matrix A down correctly?
Sorry but can you still help? thanks Prove It

9. $\displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & 4 & 8 & 16\\ 1 & -1 & -2 & 2 & 4\\ 0 & 3 & 6 & 9 & 15\\ -1 & 2 & 4 & 0 & 0\end{array}\right]$

Do [tex]R_2 - R_1[tex] and $\displaystyle R_4 + R_1$

$\displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & 4 & 8 & 16\\ 0 & -3 & -6 & -6 & -12\\ 0 & 3 & 6 & 9 & 15\\ 0 & 4 & 8 & 8 & 16\end{array}\right]$

Do $\displaystyle -\frac{1}{3}R_2$

$\displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & 4 & 8 & 16\\ 0 & 1 & 2 & 2 & 4\\ 0 & 3 & 6 & 9 & 15\\ 0 & 4 & 8 & 8 & 16\end{array}\right]$

Do $\displaystyle R_3 - 3R_2$ and $\displaystyle R_4 - 4R_2$

$\displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & 4 & 8 & 16\\ 0 & 1 & 2 & 2 & 4\\ 0 & 0 & 0 & 3 & 3\\ 0 & 0 & 0 & 0 & 0\end{array}\right]$.

So we have...

$\displaystyle 3x_4 = 3$ so $\displaystyle x_4 = 1$.

Notice that we can not solve for $\displaystyle x_3$. So we call $\displaystyle x_3 = t$, an arbitrary parameter.

Back substituting gives

$\displaystyle x_2 + 2x_3 + 2x_4 = 4$

$\displaystyle x_2 + 2t + 2(1) = 4$

$\displaystyle x_2 = 2 - 2t$

$\displaystyle x_1 + 2x_2 + 4x_3 + 8x_4 = 16$

$\displaystyle x_1 + 2(2 - 2t) + 4t + 8(1) = 16$

$\displaystyle x_1 + 4 - 4t + 4t + 8 = 16$

$\displaystyle x_1 = 4$

So $\displaystyle \mathbf{x} = \left[ \begin{array}{c} 1\\ 2 - 2t\\ t\\ 1\end{array}\right]$.

10. Originally Posted by Prove It
$\displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & 4 & 8 & 16\\ 1 & -1 & -2 & 2 & 4\\ 0 & 3 & 6 & 9 & 15\\ -1 & 2 & 4 & 0 & 0\end{array}\right]$

Do [tex]R_2 - R_1[tex] and $\displaystyle R_4 + R_1$

$\displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & 4 & 8 & 16\\ 0 & -3 & -6 & -6 & -12\\ 0 & 3 & 6 & 9 & 15\\ 0 & 4 & 8 & 8 & 16\end{array}\right]$

Do $\displaystyle -\frac{1}{3}R_2$

$\displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & 4 & 8 & 16\\ 0 & 1 & 2 & 2 & 4\\ 0 & 3 & 6 & 9 & 15\\ 0 & 4 & 8 & 8 & 16\end{array}\right]$

Do $\displaystyle R_3 - 3R_2$ and $\displaystyle R_4 - 4R_2$

$\displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & 4 & 8 & 16\\ 0 & 1 & 2 & 2 & 4\\ 0 & 0 & 0 & 3 & 3\\ 0 & 0 & 0 & 0 & 0\end{array}\right]$.

So we have...

$\displaystyle 3x_4 = 3$ so $\displaystyle x_4 = 1$.

Notice that we can not solve for $\displaystyle x_3$. So we call $\displaystyle x_3 = t$, an arbitrary parameter.

Back substituting gives

$\displaystyle x_2 + 2x_3 + 2x_4 = 4$

$\displaystyle x_2 + 2t + 2(1) = 4$

$\displaystyle x_2 = 2 - 2t$

$\displaystyle x_1 + 2x_2 + 4x_3 + 8x_4 = 16$

$\displaystyle x_1 + 2(2 - 2t) + 4t + 8(1) = 16$

$\displaystyle x_1 + 4 - 4t + 4t + 8 = 16$

$\displaystyle x_1 = 4$

So $\displaystyle \mathbf{x} = \left[ \begin{array}{c} 1\\ 2 - 2t\\ t\\ 1\end{array}\right]$.
hey, that is sooooo helpful, if only my textbook could be as helpful as you are thanks for your help Prove it

11. Originally Posted by Prove It
$\displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & 4 & 8 & 16\\ 1 & -1 & -2 & 2 & 4\\ 0 & 3 & 6 & 9 & 15\\ -1 & 2 & 4 & 0 & 0\end{array}\right]$

Do [tex]R_2 - R_1[tex] and $\displaystyle R_4 + R_1$

$\displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & 4 & 8 & 16\\ 0 & -3 & -6 & -6 & -12\\ 0 & 3 & 6 & 9 & 15\\ 0 & 4 & 8 & 8 & 16\end{array}\right]$

Do $\displaystyle -\frac{1}{3}R_2$

$\displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & 4 & 8 & 16\\ 0 & 1 & 2 & 2 & 4\\ 0 & 3 & 6 & 9 & 15\\ 0 & 4 & 8 & 8 & 16\end{array}\right]$

Do $\displaystyle R_3 - 3R_2$ and $\displaystyle R_4 - 4R_2$

$\displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & 4 & 8 & 16\\ 0 & 1 & 2 & 2 & 4\\ 0 & 0 & 0 & 3 & 3\\ 0 & 0 & 0 & 0 & 0\end{array}\right]$.

So we have...

$\displaystyle 3x_4 = 3$ so $\displaystyle x_4 = 1$.

Notice that we can not solve for $\displaystyle x_3$. So we call $\displaystyle x_3 = t$, an arbitrary parameter.

Back substituting gives

$\displaystyle x_2 + 2x_3 + 2x_4 = 4$

$\displaystyle x_2 + 2t + 2(1) = 4$

$\displaystyle x_2 = 2 - 2t$

$\displaystyle x_1 + 2x_2 + 4x_3 + 8x_4 = 16$

$\displaystyle x_1 + 2(2 - 2t) + 4t + 8(1) = 16$

$\displaystyle x_1 + 4 - 4t + 4t + 8 = 16$

$\displaystyle x_1 = 4$

So $\displaystyle \mathbf{x} = \left[ \begin{array}{c} 1\\ 2 - 2t\\ t\\ 1\end{array}\right]$.
I did all the reduction parts but that was where i got stuck now i got something i can learn from So, how did you know we cannot solve for $\displaystyle x_3$? that is what has been holding me up from completing this question.

Thanks again

12. Originally Posted by tsal15
I did all the reduction parts but that was where i got stuck now i got something i can learn from So, how did you know we cannot solve for $\displaystyle x_3$? that is what has been holding me up from completing this question.

Thanks again
Because there isn't an equation we can back-substitute $\displaystyle x_4$ into that can give us JUST $\displaystyle x_3$.

When we back substitute, we get an equation that has both $\displaystyle x_2$ and $\displaystyle x_3$.

So the best we can do is write one of them in terms of the other.

In other words, we can let $\displaystyle x_3$ be whatever we like, it won't change the system. The only thing that will change is $\displaystyle x_2$, because it depends on $\displaystyle x_3$.

13. Originally Posted by Prove It
Because there isn't an equation we can back-substitute $\displaystyle x_4$ into that can give us JUST $\displaystyle x_3$.

When we back substitute, we get an equation that has both $\displaystyle x_2$ and $\displaystyle x_3$.

So the best we can do is write one of them in terms of the other.

In other words, we can let $\displaystyle x_3$ be whatever we like, it won't change the system. The only thing that will change is $\displaystyle x_2$, because it depends on $\displaystyle x_3$.
uhuh so does that mean we could've let $\displaystyle x_2$=t?

14. Originally Posted by tsal15
uhuh so does that mean we could've let $\displaystyle x_2$=t?
Yep. Either of them can be kept arbitrary. We just take note that one depends on the other.

15. Originally Posted by Prove It
Yep. Either of them can be kept arbitrary. We just take note that one depends on the other.
You have been very helpful thank you

I don't know if i'm asking for too much here, but i've got another thread "Gaussian - Jordan Elimination" - would you be able to have some input on that? thanks again Prove It