E1.4b Determinant with zero column

[karush]

$$\left[\begin{array}{rrrrr}
1 &0 &2 &1\\
1 &1 &0 &1\\
1 &3 &4 &1\\
-1 &-3 &-4 &-1
\end{array}\right]=\color{red}{0}$$Answer (red) via W|Aok I did not do any operations on this
Since by observation the 4th column can become all zero'showever didn't see anything in the book to support this
only that the co-factor expansion would result in multiplying zeros throughoutany suggestions?

 

[HOI]

Surely in your textbook there is the rule "If two rows or columns of a matrix are equal" (or the more general "if one row or column is a multiple of another") "then the determinant is 0". One way to evaluate the determinant of matrix is to use "row operations" to reduce the matrix to a simpler matrix. The row operations are "multiply an entire row by a constant", "swap two rows", and "add a multiple of on row to another". The first multiplies the determinant by that constant. The second multiplies the determinant by -1. The third does not change the determinant.

In particular, if two rows of a matrix are the same, adding -1 times one of those rows to the other gives a matrix having all "0"s in one row. "Expanding" on that row gives 0 as the determinant 0.

(Of course you can replace "row" by "column" thoughout.)