# Linear algebra determinant

## Homework Statement

I think I have broken maths. I am reducing a matrix to row echelon form to find the determinant. The matrix I will show is nearly in the desired form

1 -3 -2 1
0 1 2 -1
0 0 1 -0.8
0 0 1 0

## The Attempt at a Solution

As it stands, the determinant is 0.8. However subtracting the 4th row from the 3rd row changes the determinant to -0.8. I thought adding multiples of one row to another left the determinant unchanged. So why hasn't it? Thanks for answers.

Doing that leaves the determinant at +0.8.

Det(The matrix)=det(The matrix without the first row and first column)=det(The matrix without the first and second row and the first and second column).

I arrive at that from expansion along the first column twice.

So you arrive at det(The matrix)=(1)(0)-(1)(-0.8)=0.8

Subtracting the row like you said just zaps that first 1 to 0 making it (0)(0)-(1)(-0.8)=0.8.

Doing that leaves the determinant at +0.8.

Det(The matrix)=det(The matrix without the first row and first column)=det(The matrix without the first and second row and the first and second column).

I arrive at that from expansion along the first column twice.

So you arrive at det(The matrix)=(1)(0)-(1)(-0.8)=0.8

Subtracting the row like you said just zaps that first 1 to 0 making it (0)(0)-(1)(-0.8)=0.8.

Am I correct in thinking that the matrix obtained doing this calculation (3rd row - 4th row) is this
1 | -3 | -2 | 1
0 | 1 | 2 | -1
0 | 0 | 1 | -0.8
0 | 0 | 0 | -0.8

Since this is a triangular matrix, the determinant is the product of the entries on the leading diagonal, -0.8. Where have I gone wrong?

You originally said subtracting the 4th from the 3rd, which would make the bottom corner

0 -0.8
1 0

That is how I would usually interpret subtracting the 4th from the 3rd row, anyway.

So you mean what I would call subtracting the 3rd from the 4th row. You have done the calculation incorrectly, the bottom right entry should be +0.8 because it is 0-(-0.8).