Gauss Jordan Elemination Method (Stucked)

[farmd684]

Gauss Jordan Elemination Method (Stucked)!

Homework Statement

5x1+3x2+2x3=4
3x1+3x2+2x3=2
+x2+x3 =5
I have to solve this problem by calcutaing the inverse if this equation's augmented matrix

Homework Equations

I used this method to find its inverse
5 3 2 1 0 0
3 3 2 0 1 0
0 1 1 0 0 1

My goal is to make first matrix to Identity matrix like the second one

The Attempt at a Solution

So far here is my workings

1ST STEP: R1/5 R2/3
2ND STEP: R3*(3/5)
3RD STEP: R2=R1-R2
4TH STEP: R1=R1-R3 R3=R3-R1
5TH STEP: R2*(-5/2)
6TH STEP: R3*(1/5)
7TH STEP: R1=(1/5)R3+R1 AND R2=R2+(-2/3)R3
Then the first one becomes Identity matrix but the result is not right :(
Pls check the steps if i have done anything wrong or suggest me any techniques so that i can solve these type of problems with no confusions.I m having many problems with these maths.Pls help

 

[Defennder]

There isn't any need to find the inverse of the matrix. This approach works only if the system has exactly one solution only, which in this case it has, but not always so in general. Furthermore, it's time consuming since after finding the inverse you have to verify your calculations and THEN multiply the RHS with the inverse matrix.

You have expressed the system of linear equations in terms of an augmented matrix, so why not just performed row operations on the augmented matrix directly to reduce it to reduced row echelon form and from there simply read off the values of x1,x2,x3?

 

[farmd684]

There isn't any need to find the inverse of the matrix. This approach works only if the system has exactly one solution only, which in this case it has, but not always so in general. Furthermore, it's time consuming since after finding the inverse you have to verify your calculations and THEN multiply the RHS with the inverse matrix.

You have expressed the system of linear equations in terms of an augmented matrix, so why not just performed row operations on the augmented matrix directly to reduce it to reduced row echelon form and from there simply read off the values of x1,x2,x3?

Yeah i know that & i also know it would save my time.But i m particulary asked to solve this problem by finding the inverse of augmented matrix.I have also mentioned my workings sir.Have u cheked it?
I have tried many times but each time i failed.Let me know how could i get the inverse of such matrices as i have many problems to do.
Thanks

 

[Defennder]

Yeah i know that & i also know it would save my time.But i m particulary asked to solve this problem by finding the inverse of augmented matrix.I have also mentioned my workings sir.Have u cheked it?

Well I missed that part of the question.

1ST STEP: R1/5 R2/3
2ND STEP: R3*(3/5)
3RD STEP: R2=R1-R2
4TH STEP: R1=R1-R3 R3=R3-R1
5TH STEP: R2*(-5/2)
6TH STEP: R3*(1/5)
7TH STEP: R1=(1/5)R3+R1 AND R2=R2+(-2/3)R3

How exactly did you do these steps? And why? The matrix just before step 4 is:

$$\left ( \begin{array}{cccccc}1&\frac{3}{5}&\frac{2}{5}&\frac{1}{5}&0&-\frac{3}{5}\\0&\frac{2}{5}&\frac{4}{15}&-\frac{1}{5}&\frac{1}{3}&0\\0&\frac{3}{5}&\frac{3}{5}&0&0&\frac{3}{5} \end{array} \right )$$.

Doing R1=-R3+R1 will change the top row to:
1 0 -1/5 1/5 0 0

and it can be seen that the subsequent R3=R3-R1 will hence be unable to reduce the top row's 2nd entry from the left to 0. This may be your mistake. Row-operations cannot be performed "simulataneously". So you can't use the old R1 after it has been reduced itself to row reduce another row. I followed your steps right till the end and did not get the identity matrix on the left. So I'm assuming this is your mistake.