Linear Algebra Help: Reduce Matrix & Find Error

[JohnPrior3]

So I had a quiz on Wednesday and got the problem wrong but don't know why. The question is:

Use elementary row operations to reduce the matrix

A= 3 1 -1
2 3 1
-4 0 2

to upper-triangular form. Each of these elementary row operations should have the form Ri :=Ri-likRk where lil is an appropriate multiplier. Show the values of each of the ilk's and the order in which each of the elementary row operations is applied. (For this exercise, do not interchange any rows.)

Hint: When you are done, the product of the three pivots should be -2.

I had l21= 2/3 and did R2:= R3=2/3R1 and and then did l31=-4/3 R3:=-R3-(-4/3)R1

My matrix at the end was
3 1 -1
0 7/3 5/3
0 0 2/3

Which is upper triangular form, but incorrect. What did I do wrong?

 

[mathman]

Shouldn't the determinants be the same? They are not! In any case you didn't finish. Following what you did the third row is (0,4/3,2/3).

 

[JohnPrior3]

Shouldn't the determinants be the same? They are not! In any case you didn't finish. Following what you did the third row is (0,4/3,2/3).

Then how would I get the upper triangular system? I know I am supposed to have three pivots, but I don't know how to take three such that the product would be -2.

 

[HallsofIvy]

A= \begin{bmatrix}3 & 1 & -1 \\ 2 & 3 & 1 \\ -4 & 0 & 2 \end{bmatrix}

You want "0"s below the "3" in the first column so subtract 2/3 times the first row from the second row and add 4/3 times the first row from the third row. That gives
\begin{bmatrix}3 & 1 & -1 \\ 0 & 7/3 & 5/3 \\ 0 & 4/3 & 2/3 \end{bmatrix}

Now you want a "0" below the "7/3" in the second column so subtract (4/3)/(7/3)= 4/7 times the second row from the third row. 2/3- (4/7)(5/3)= 14/21- 20/21= -6/21= -2/7, not 2/3. It looks like you just forgot to do that final subtraction!

 

[JohnPrior3]

A= \begin{bmatrix}3 & 1 & -1 \\ 2 & 3 & 1 \\ -4 & 0 & 2 \end{bmatrix}

You want "0"s below the "3" in the first column so subtract 2/3 times the first row from the second row and add 4/3 times the first row from the third row. That gives
\begin{bmatrix}3 & 1 & -1 \\ 0 & 7/3 & 5/3 \\ 0 & 4/3 & 2/3 \end{bmatrix}

Now you want a "0" below the "7/3" in the second column so subtract (4/3)/(7/3)= 4/7 times the second row from the third row. 2/3- (4/7)(5/3)= 14/21- 20/21= -6/21= -2/7, not 2/3. It looks like you just forgot to do that final subtraction!

Thanks! I am really happy you say that. I had gotten to that conclusion yesterday and I thought it was wrong. So how does my revision look?