Inverse of a matrix with an X

  • Thread starter rey242
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  • #1
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Hi everyone!

Homework Statement


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?
 

Answers and Replies

  • #2
D H
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Keep pushing through. What conditions would make the "pushing through" invalid?
 
  • #3
Dick
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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
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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
Dick
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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
D H
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Are you saying that x-3 does not have a multiplicative inverse?
 
  • #7
D H
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What happened to the x in the second row?
She swapped the second and third rows.
 
  • #8
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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
D H
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Should I declare this not invertible?
No.

What is the multiplicative inverse of x-3?
Under what conditions does x-3 not have a multiplicative inverse?
 
  • #10
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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
Dick
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She swapped the second and third rows.

Oh, yeah. I see it now. Continue with your excellent help. And thanks for the exposition rey242.
 
  • #12
D H
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Should I multiply row 3 by the multiplicative inverse ?
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
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Thanks for all your help

I understand now.

I really appreciate it!
 

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