# Factoring out multiples to determine determinants. Linear algebra

I'm reading this text and I'm a bit confused about how they determined the determinant of this matrix:

The factoring out of the -7 and the -3 is because of rule 2 in theoreum 3.3 right?

What is the intuition behind these rules? Aren't row operations supposed to keep matrices equal? Aren't matrices that result from the elementary row operations supposed to be equivalent? So why would that change the determinants if you say interchange two rows?

• Row operations do not keep matrices "equal" but they do keep them "equivalent." There are many similarities between two matrices which are equivalent, such as dimension of the row space, dimension of the column space, dimension of the kernel, invertability, etc... but of course equality is not one of them. – JMoravitz Aug 6 '18 at 21:09
• Perhaps a clearer explanation of what is happening is in pointing out the property that for two square matrices $A$ and $B$ of the same shape, one has that $\det(AB)=\det(A)\cdot \det(B)$. Then remember that performing elementary row operations is like "factoring out elementary matrices." (note: to be fair, the properties mentioned in the image are usually used to prove the property I mention here in my comment) – JMoravitz Aug 6 '18 at 21:11
• Indeed, in some treatments those rules are taken as defining properties of the determinant. – amd Aug 6 '18 at 22:07
• @Jwan622 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – gimusi Sep 6 '18 at 23:11

For a simple intuition and check of the Properties of the determinant let consider

$$\begin{vmatrix}1&0\\0&1\end{vmatrix}=1$$

but for interchanging rows/colums

$$\begin{vmatrix}0&1\\1&0\end{vmatrix}=-1$$

multiply a row/column by a scalar

$$\begin{vmatrix}2&0\\0&1\end{vmatrix}=2$$

add first row to second row

$$\begin{vmatrix}1&0\\1&1\end{vmatrix}=1$$