Inverse of a matrix with an X

1. Oct 21, 2009

rey242

Hi everyone!

1. The problem statement, all variables and given/known data
Find the inverse of the following using row reduction. If it does not exist, indicate clearly why.

3, 0, 6;
1, -2, x;
1, 2, 1;

2. The attempt at a solution

I started by augmenting with a 3 by 3 Identity matrix:
3, 0, 6, 1, 0, 0;
1, -2, x, 0, 1, 0;
1, 2, 1, 0, 0, 1;

Then I started using Gaussian Elimination until I hit a road bump:

3, 0, 6, 1, 0, 0;
0, -2, x-3, 1/3, 1, 0;
0, 0, x-5, 0, 0, 1;

Should I just keep on pushing though? Or should I try to figure X out? Or is there some sort of theorem out there than can help me?

2. Oct 21, 2009

D H

Staff Emeritus
Keep pushing through. What conditions would make the "pushing through" invalid?

3. Oct 21, 2009

Dick

I'm going to guess you meant to start by taking (1/3) times the first row and subtracting it from the second row. Something has gone wrong already.

4. Oct 21, 2009

rey242

Ok, I reworked it and I got this:
3, 0, 6, 1, 0, 0;
0, -2, 1, 1/3, 0, -1;
0, 0, (X-3), -2/3, 1, 1;

So would this matrix not have an inverse since I can never get the Identity on the left? It seems like it I'll keep getting number when I do row reduction after this...

5. Oct 21, 2009

Dick

It's still wrong. Can you break this out one step at a time and say what you are doing? What happened to the x in the second row?

6. Oct 21, 2009

D H

Staff Emeritus
Are you saying that x-3 does not have a multiplicative inverse?

7. Oct 21, 2009

D H

Staff Emeritus
She swapped the second and third rows.

8. Oct 21, 2009

rey242

Here is my process:
3, 0, 6, 1, 0, 0;
1, -2, x, 0, 1, 0;
1, 2, 1, 0, 0, 1;

I multiplied -1/3 to r1 and added to r2 and got
3, 0, 6, 1, 0, 0;
0, -2, x-2, -1/3, 1, 0;
1, 2, 1, 0, 0, 1;

I then did the same and added to r3 and got
3, 0, 6, 1, 0, 0;
0, -2, x-2, -1/3, 1, 0;
0, 2, -1, -1/3, 0, 1;

I then added r2 to r3 and got
3, 0, 6, 1, 0, 0;
0, -2, x-2, -1/3, 1, 0;
0, 0, x-3, -2/3, 1, 1;

Then I added -r3 to r2 to get:
3, 0, 6, 1, 0, 0;
0, -2, 1, 1/3, 0, -1;
0, 0, x-3, -2/3, 1, 1;

So I got rid of the x in this step. Should I declare this not invertible?

EDIT: It's He by the way, lol

9. Oct 21, 2009

D H

Staff Emeritus
No.

What is the multiplicative inverse of x-3?
Under what conditions does x-3 not have a multiplicative inverse?

10. Oct 21, 2009

rey242

The multiplicative inverse is 1/(x-3), this does not apply when x=3...
Hmm... I was thinking about this before but I felt it wasn't right. Should I multiply row 3 by the multiplicative inverse ?

11. Oct 21, 2009

Dick

Oh, yeah. I see it now. Continue with your excellent help. And thanks for the exposition rey242.

12. Oct 21, 2009

D H

Staff Emeritus
Give that man a prize!

With that you will have the third row in the desired form. The rest is just tedious elementary algebra. It's easy to make a mistake in all this tedious work. I suggest that you check your work at the end by multiplying the original matrix and its supposed inverse to make sure it really is the inverse.

13. Oct 21, 2009