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Calculate the determinant of a 3x3 matrix

  • Thread starter Cade
  • Start date
  • #1
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[Solved] Calculate the determinant of a 3x3 matrix

Homework Statement



Use elementary row operations to calculate the determinant of this 3x3 matrix.

Code:
1-a      1       1
   1   1-a       1
   1      1    1-a

Homework Equations



I think that the problem wants me to reduce this to an upper triangular matrix and multiply the diagonals to find the determinant. Scaling a row scales the determinant. Swapping a pair of rows/columns flips the sign of the determinant.

The Attempt at a Solution


I know how to calculate the determinant through cofactors, but I don't know how to approach the problem through row reduction.
 
Last edited:

Answers and Replies

  • #2
22,097
3,277
You want to obtain a row filled with zeroes except for possibly one element. So for example, you will want to make a zero in the first row at the first and second columns (so in places (1,1) and (1,2) if you understand what I mean).

So, let's try to make a zero at (1,2) (=first row, second column). We can do that by subtracting the third column. What do you obtain when you did that?
 
  • #3
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To get an upper triangle matrix, I'd have to deal with the bottom-left cells, not the top-right cells. I've tried reducing the rows like this, but I get stuck:

Code:
1-a      1       1
   1   1-a       1
   1      1    1-a

R3 = R3 - R2

1-a      1       1
   1   1-a       1
   0     a      -a

R2 = R2 + R3

1-a      1       1
   1     1       1 - a
   0     a      -a

R1 = R1 - R2
-a     0       a
 1     1       1 - a
 0     a      -a
 
  • #4
22,097
3,277
OK, first of all, reducing it to a triangular matrix is not always possible and not always the good thing to do. You can obviously do it if you like, but it isn't necessary. To calculate the determinant, you can make zeroes everywhere.

However, let's proceed your way. Your first step is good. A thing that immediately comes to mind after your first step is to make a zero at place (3,2). Can you think of a column operation that will make things work?
 
  • #5
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I could add column 3 to column 2, but I don't know if that's a permitted elementary operation. Also, my assignment sheet says that I can only use row operations.
 
  • #6
92
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Oh, I have it now.

Code:
1-a      1       1
   1   1-a       1
   1      1    1-a

R3 = R3 - R2

1-a      1       1
   1   1-a       1
   0     a      -a

R2 = R2 + R3

1-a      1       1
   1     1       1 - a
   0     a      -a

R1 = R1 - R2
-a     0       a
 1     1       1 - a
 0     a      -a

R2 = R2 * a
-a     0       a
 a     a       a - a^2
 0     a      -a

R2 = R1 + R2
-a     0       a
 0     a       2a - a^2
 0     a      -a

R3 = R3 - R2
-a     0       a
 0     a       2a - a2
 0     0      -a - 2a + a^2
The determinant is divided by a since I multiplied a row by a. The unsimplified determinant is (1/a)(-a)(a)(-a -2a + a^2)
 
  • #7
22,097
3,277
Yes, this is correct!!
 
  • #8
92
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I've just solved this in my previous post, thanks for your help. :)

I'm going to go learn LaTeX now.
 
  • #9
33,086
4,793
Click to see what I did.
[tex]\left|\begin{array} {c c c}1-a & 1 & 1\\
1 & 1-a & 1 \\
1 & 1 & 1-a \end{array}\right|[/tex]
 
  • #10
92
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Thanks, that helps.
 

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