MHB What is the determinant of the given matrix and why do the scalers change sign?

[karush]

Compute the determinant of the following matrix
$$\left| \begin{array}{cccc}
2 & 1 & 0 & 2 \\1 & 2 & 1 & 2 \\-1 & 1 & -3 & 2 \\1 & -1 & 1 & 0
\end{array} \right|
\sim
\left| \begin{array}{cccc}
2 & 1 & 0 & 2 \\1 & 2 & 1 & 2 \\-1 & 1 & -3 & 2 \\0 & 0 & -2 & 2
\end{array} \right|
\sim
\left| \begin{array}{cccc}
2 & 1 & 0 & 2 \\1 & 2 & 1 & 3 \\-1 & 1 & -3 & -1 \\0 & 0 & -2 & 0
\end{array} \right|$$
then
$$2\left| \begin{array}{cccc}
2 & 1 & 2 \\1 & 2 & 3 \\-1 & 1 & -1
\end{array} \right|
\sim
2\left| \begin{array}{cccc}
0 & 3 & 0 \\1 & 2 & 3 \\-1 & 1 & -1
\end{array} \right|$$
then
$$(-2)(3)\left| \begin{array}{cccc}
1 & 3 \\-1 & -1
\end{array} \right|$$
finally
$$=-6[((1)\cdot(-1))-((-1)\cdot(3))]=-12$$

ok prob some typos but why do the scalers change sign?

https://www.physicsforums.com/attachments/8756

 

[Evgeny.Makarov]

If the only nonzero element in $i$th row is in $j$th column, then this element is taken out and multiplied by $(-1)^{i+j}$ and the minor obtained by erasing $i$th row and $j$th column. For example, in line 3 of your computation 3 is in line 1, row 2, so the $2\times2$ determinant is multiplied by $(-1)^{1+2}\cdot 3=-3$.

 

[topsquark]

@karush

A notation comment: Why are you using " ~ " instead of " = "?

-Dan

 

[Wilmer]

@karush

A notation comment: Why are you using " ~ " instead of " = "?

-Dan

He's being nice and waving at you :)

 

[karush]

@karush

A notation comment: Why are you using " ~ " instead of " = "?

-Dan

The teacher uses it $\sim$ meas similar

 

[topsquark]

The teacher uses it $\sim$ meas similar

That's kind of what I was wondering. The matrices are similar, the determinants of those matrices are equal. You wrote those as determinants so you should be using the " = ".

-Dan