Augmented matrices?

Question

We're working on solving systems of equations by using augmented matrices in school. Is there someone who could try to explain it to me (again) so that I get it? It seems like a always do something wrong. If it'll help, you can im me on aol ([e-mail]) or msn ([e-mail]) I just really want to understand this...

Answer 1

Hi,

Enter your equations coefficients into your matrix that is "n" rows and "n + 1" columns.. Your goal is to multiply rows and add them together to get the first "n" columns to have 1 in each position on the diagonal starting in the top left corner and 0 in every other position in those "n" columns.

If your matrix for variables x, y, and z became:
[1...0...0...8] then x = 8
[0...1...0..-2] then y = -2
[0...0...1...5] then z = 5

To do this, get a 1 in the first spot in the first row by dividing every term on that line by that first number.

Now, use the 1 in that row to get zeros as the first term in all the rows underneath it. If the next row started with a 5, multiply the first row times -5 and add these results to row 2. DO NOT really change the first line. Now your first row still starts with a 1 and your second row starts with a zero. If your next row starts with -18. Do that by multiplying the first row times 18 and add these results to row 3. DO NOT really change the first line. Now your first row still starts with a 1 and your second and third rows start with a zero. Continue this pattern for all rows until the first column has a 1 followed by zeros to the bottom.

Now look at the second column. You need to get a 1 on the second row in the second column, so divide every term on that line by the number in the second column of the second row. This is the 1 you will keep forever on this line.

Now, use the 1 in that row to get zeros as the second term in all the other rows in the matrix. If a row had 7 in the second column, multiply the second row times -7 and add these results to the other row. DO NOT really change the second line. Now your second row still has a 1 in its second column and your other row has a zero in its second column. If another row has 4 in its second column, multiplying the second row times -4 and add these results to the other row. DO NOT really change the second line. Now your first row still starts with a 1 and your second row has a 1 in its second column. Continue this pattern for all rows until the second column has a 1 only on the second row and zeros on every other row in the second column.

Repeat this process for every other column to get the desired matrix.

If you want a quick way to check what your final answer matrix should be, enter your matrix, say as [A]. Then, on the TI83, press MATRIX, MATH, and choose rref( which will find you the reduced row echelon form of the matrix. After rref( is on your screen, go to MATRIX and select matrix [A] so rref([A]) is on your screen. Press ENTER and you will get a matrix like:

[1...0...0...8]
[0...1...0..-2]
[0...0...1...5]

These are the answers to your matrix in the final form of your matrix.