matrix row echelon form

• May 17th 2008, 02:20 PM
elizabeth
matrix row echelon form
I have never done this before and I am not sure I am doing this right.
I am taking college algebra online and I am lost in this matrix stuff. I have had several folks try to help me but no one has come up with the correct answer. The sheet I am doing is multiple choice. The two sets of answers that I come up with over and over are not one of the choices. This is what the problem said.

Use elementary row operations to write the matrix below in row echelon form.

1 -4 6 9
3 -11 8 1
2 -5 -3 -15
• May 17th 2008, 02:59 PM
Reckoner
Quote:

Originally Posted by elizabeth
The two answers I come up with are not the choices I have. It is a multiple choice question.

What answers do you come up with? If you show your work, we can see what you have done wrong, if anything.

Note that while a matrix will have a unique reduced row echelon form, there are many different ways you can put it into row echelon form. What are the choices that you are given?
• May 17th 2008, 02:59 PM
o_O
Conditions for a matrix in row echelon form:
1. If there are any rows of all zeros then they are at the bottom of the matrix.
2. If a row does not consist of all zeros then its first non-zero entry (i.e. the left most non-zero entry) is a 1. This 1 is called a leading 1.
3. In any two successive rows, neither of which consists of all zeroes, the leading 1 of the lower row is to the right of the leading 1 of the higher row.

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Edit: If it helps, what I'm trying to do with your matrix is get it into the form of: $\left[ \begin{array}{cccc} 1 & * & * & * \\ 0 & 1 & * & * \\ 0 & 0 & 1 & * \end{array}\right]$

where the *'s are the entries that come about after modifing the matrix. Note that this form will satisfy the conditions above for a matrix in row echelon form. Each leading 1 is to the right of the one above it.
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$\left[ \begin{array}{cccc} 1 & -4 & 6 & 9 \\ {\color{blue}3} & -11 & 8 & 1 \\ {\color{blue}2} & -5 & -3 & -15 \end{array} \right]$

Now with these conditions, we can use the 1 in the first row as the lead 1 of its column. Thus any leading 1's of the other rows must be to the right of it as suggested by condition 3. So we focus on getting the blue numbers on becoming 0's.

For the second row, we can subtract 3 times the first row from it ( $R_{2} - 3R_{1}$):
$\left[ \begin{array}{cccc} {\color{red}1} & {\color{red}-4} & {\color{red}6} & {\color{red}9} \\ {\color{blue}3}-3({\color{red}1}) & -11-3({\color{red}-4}) & 8 -3({\color{red}6}) & 1 -3({\color{red}9}) \\ {\color{blue}2} & -5 & -3 & -15 \end{array} \right]$ ${\color{white}.} \: \: = \: \: \left[ \begin{array}{cccc} 1 & -4 & 6 & 9 \\ 0 & 1 & -10 & -26 \\ {\color{blue}2} & -5 & -3 & -15 \end{array} \right]$

Use the first row again to get the blue 2 to become 0 by subtracting 2 times the first row from the 3rd:
$\left[ \begin{array}{cccc} {\color{magenta}1} & {\color{magenta}-4} & {\color{magenta}6} & {\color{magenta}9} \\ 0 & 1 & -10 & -26 \\ {\color{blue}2}-3({\color{magenta}1}) & -5-3({\color{magenta}-4}) & -3 - 3({\color{magenta}6}) & -15 - 3({\color{magenta}9}) \end{array} \right]$ ${\color{white}.} \: \: = \: \: \left[ \begin{array}{cccc} 1 & -4 & 6 & 9 \\ 0 & 1 & -10 & -26 \\ 0 & 7 & -21 & -42 \end{array} \right]$

Now all that's left is getting the 7 in the third row to become a 0 and the entry beside it to be the leading 1. Hopefully you can see what to do.

There may be some computational errors. Row-reducing was always a pain.
• May 17th 2008, 03:33 PM
Soroban
Hello, Elizabeth!

Quote:

Use elementary row operations to write the matrix below in row echelon form.

. . $\left[\begin{array}{ccc|c}
1 & \text{-}4 & 6 & 9 \\ 3 & \text{-}11 & 8 & 1 \\ 2 & \text{-}5 & \text{-}3 & \text{-}15 \end{array}\right]$

$\begin{array}{c} \\ R_2-3R_1 \\ R_3-2R_1\end{array}\left[\begin{array}{ccc|c} 1 & \text{-}4 & 6 & 9 \\ 0 & 1 & \text{-}10 & \text{-}26 \\ 0 & 3 & \text{-}15 & \text{-}33 \end{array}\right]$

$\begin{array}{c}R_1 + 4R_2 \\ \\ R_3-3R_2\end{array}\left[\begin{array}{ccc|c}1 & 0 & \text{-}34 & \text{-}95 \\ 0 & 1 & \text{-}10 & \text{-}26 \\ 0 & 0 & 15 & 45 \end{array}\right]$

. . . $\begin{array}{c}\\ \\ \frac{1}{15}R_3\end{array}\left[\begin{array}{ccc|c}1 & 0 & \text{-}34 & \text{-}95 \\ 0 & 1 & \text{-}10 & \text{-}26 \\ 0 & 0 & 1 & 3 \end{array}\right]$

$\begin{array}{c}R_1 + 34R_3 \\ R_2 + 10R_3 \\ \\ \end{array}\left[\begin{array}{ccc|c}1 & 0 & 0 & 7 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 3 \end{array}\right]$

• May 17th 2008, 03:49 PM
elizabeth
these are the choices
1 3 -2 1
0 1 -5 -3
0 0 1 -3

1 -4 6 9
0 1 -3 -5
0 0 1 3

1 3 1 -2
0 1 -5 -3
0 0 1 -3

1 3 -2 1
0 1 -3 -5
0 0 1 -3

1 -4 6 9
0 1 -5 -3
0 0 1 3

these are the choices I had I think I had it down to the 2nd and 4th one.
• May 17th 2008, 05:23 PM
Reckoner
Quote:

Originally Posted by elizabeth
these are the choices I had I think I had it down to the 2nd and 4th one.

The result that you came up with was probably in correct row echelon form, but as I said, the answers are not unique. So, you need to figure out which of the given choices are actually row-equivalent to your original matrix (or to your answer).

The easiest thing to do would be to take the answer you got and try to manipulate it with row operations to form one of the choices. Alternatively, you could convert the original matrix to reduced row echelon form, and then systematically convert all of the choices to reduced row echelon form and see which ones are equivalent, but that might be more work.