Question

A question about using the Gaussian method of Elimination to find a Determinant.

When subtracting/adding rows in the following form:

R2 → R2 - 2*R1

The determinant's value remains. However when I have to multiply the rows I'm changing, the value of the determinant changes completely. Why is that so?
Examples below:

Answer 1

\[\left[\begin{matrix}1 & 4 & 3 & 2 \\ 2 & 1 & 4 & 3 \\ 3 & 2 & 1 & 4 \\ 4 & 3 & 2 & 1\end{matrix}\right]\]
det = -120

\[R_2 \rightarrow R_2 -2R_1\]
\[R_3 \rightarrow R_3 -3R_1\]
\[R_4 \rightarrow R_4 -4R_1\]
\[\left[\begin{matrix}1 & 4 & 3 & 2 \\ 0 & -7 & -2 & -1 \\ 0 & -10 & -8 & -2 \\ 0 & -13 & -11 & -6\end{matrix}\right]\]
det = -120

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\[\left[\begin{matrix}2 & -3 & 0 & 1 \\ 1 & 5 & -1 & 2 \\ -1 & 0 & 4 & -2 \\ 3 & 2 & 7 & 6\end{matrix}\right]\]
det = 267

\[R_2 \rightarrow 2R_2 - R_1\]
\[R_3 \rightarrow 2R_3 + R_1\]
\[R_4 \rightarrow 2R_4 - 3R_1\]
\[\left[\begin{matrix}2 & -3 & 0 & 1 \\ 0 & 13 & -2 & 3 \\ 0 & -3 & 8 & -3 \\ 0 & 13 & 14 & 9\end{matrix}\right]\]
det = 2136

Answer 2

I triple checked the row elimination calculations, but the inconsistencies always seem to return when I'm multiplying the row I'm changing by a scalar.

Answer 3

Actually, when you multiply a row by something (i.e. when you replace \(R_j\) by \(c\cdot R_j\)), you should divide the determinant by \(c\), and when you switch row, multiply the determinant by -1.

When you add rows, however, you don't change the determinant.

Answer 4

Your steps should look like
\(\color{#000000}{ \displaystyle a\cdot R_j +R_i\rightarrow R_i }\)

Don't do,
\(\color{#000000}{ \displaystyle a\cdot R_j +bR_i\rightarrow R_i }\)

because you will get confused and forget to multiply the determinant by scalar.