How to show that Det(A) = 0 without directly evaluating the determinant ? A= first row ( -2 8 1 4 ) second row ( 3 2 5 1 ) third row ( 1 10 6 5 ) and fourth row ( 4 -6 4 -3 )

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Answer 1 · 2015-09-16T10:09:32

Standard Properties of Determinant .

By Theorem DRCMA Determinant for Row or Column Multiples and Addition of Standard Properties of Determinant.

$RR_1 ->RR_1 +RR_2$ value of $det$ is unchanged here.
$RR_1$ becomes ( 1 10 6 5 )
$RR_3$ is ( 1 10 6 5 )

We can see here that corresponding elements in the $RR_1 and RR_3$ are equal.Hence the rows are equal.

By theorem DERC of the above link if two rows or column are equal
then value of determinant is $0$ .

Therefore , $det(A)=0$

How to show that Det(A) = 0￼ without directly evaluating the determinant ? A= first row ( -2 8 1 4 ) second row ( 3 2 5 1 ) third row ( 1 10 6 5 ) and fourth row ( 4 -6 4 -3 )