Expert: Paul Klarreich - 6/25/2006

Question
Thanks so much.
I have last question.

How to solve this matrix operation:

1) Given 1 3 2   , find A^T

        2 1 -3

Thanks,
-------------------------

Followup To

Question -
1) How is this is Solved using Gaussian elimination:

x + 2 Y + z =0
-3x + 3 y +2z = -7
4x -2y -3 z =2

2) how to solve this using Cramer's rule:

2x + 4y +6z =2
x + 0y + 2z =0
2x + 3y -z = -5

Thanks,

Answer -
Hi, Eddie,

You wrote:
Subject:  Gaussian and Cramer's Rule
Question:  1) How is this is Solved using Gaussian elimination:

x + 2y + z =0
-3x + 3y +2z = -7
4x -2y -3 z =2

2) how to solve this using Cramer's rule:

2x + 4y +6z =2
x + 0y + 2z =0
2x + 3y -z = -5

Thanks,
--------------------------------------
WARNING!  VIEW THIS IN A FIXED-SIZE FONT.

1) When you do G-E, you try to use row operations to change your matrix:

1   2   1   0
-3   3   2  -7
4  -2  -3   2

into one that has this form:

1   ?   ?   ?
0   1   ?   ?
0   0   1   ?

Then you can just solve the last equation and back-substitute.  Now sometimes things go badly, like getting all zeroes in one row, but your textbook will explain that.

You could press the work further, and manipulate things to get:

1   0   0   ?
0   1   0   ?
0   0   1   ?

In which case you just read off the solution from the right side.  It's not clear which one is more work.

In your case:

1   2   1   0
-3   3   2  -7
4  -2  -3   2

Use the top row, since it already has a '1' in the first spot, to kill the numbers below it.  3 times the first row, added to the second will do that one, then -4 times the first added to the third will work.

1   2   1   0
0   9   5  -7
0 -10  -7   2

Now I would like to get a nice '1' in the second column.  A good move would be to add row 2 to row 3:

1   2   1   0
0   9   5  -7
0  -1  -2  -5

and flip the signs in row 3:

1   2   1   0
0   9   5  -7
0   1   2   5

and then make it row 2.

1   2   1   0
0   1   2   5
0   9   5  -7

Now use the '1' in row 2 to kill the numbers in column 2.

1   0  -3 -10
0   1   2   5
0   0 -13 -52

Simplify row 3:

1   0  -3 -10
0   1   2   5
0   0   1   4

Now that last row really says  z = 4.  You can now either
(1) back substitute or
(2) use that last '1' in the (3,3) position to kill the numbers above it.  I'll leave that much to you.

2) how to solve this using Cramer's rule:

2x + 4y +6z =2
x + 0y + 2z =0
2x + 3y -z = -5

First, remember that Cramer was an Italian, so you will always pronounce his name: Crah-maire.

Write four determinants, each of which you will compute.
1. D is the determinant of the coefficients.  Easy to write.

    | 2   4   6 |
D  = | 1   0   2 |
    | 2   3  -1 |

2. Dx is D, with the x-column replaced by the right-hand side.

    |-2   4   6 |
Dx = | 0   0   2 |
    |-5   3  -1 |

3. Dy likewise.
4. Dz likewise.

Now the solution is simple:

x = Dx/D
y = Dy/D
z = Dz/D

I'll leave the hackwork to you.


Answer
Hi, Eddie,

Subject:  Gaussian and Cramer's Rule
Question:  Thanks so much.
I have last question.

How to solve this matrix operation:

1) Given

1 3  2   , find A^T
2 1 -3
----------------------
I'm not sure about your notation, but this usually means the 'transpose' of the matrix, which is nothing more than the result of exchanging rows and columns:

1  2
3  1
2 -3