Let A a b c d be a square matrix of size 4 det A 1 Find d

Let A = [a b c d] be a square matrix of size 4, det A = - 1. Find det [-a + b 2b + a + c c d + a].

Solution

|-a+b 2b+a+c c d+a|

= |-a 2b+a+c c d+a| + |b 2b+a+c c d+a|

= |-a 2b c d+a| + |-a a c d+a| +  |-a c c d+a| + |b 2b c d+a| + |b a c d+a| +|b c c d+a|

=|-a 2b c d| + |-a 2b c a| +  |-a a c d| +  |-a a c a| + |-a c c d| +|-a c c a| +|b 2b c d| +|b 2b c a| +  |b a c a| + |b a c d| + |b c c d| + |b c c a|

=(-2)*(-1) + (-2)*(-1) + (2)*(-1) + (2)*(-1) = 0

Here the terms with all a b c and d all are the ones for which we have the determinant and one column is multioplied by 2 so we multiply the determinant value by 2 for the four cases.

For the rest 8 cases (using the fact the column jumbling does not change the detreminant value) we see that the 4 are negative and corresppondingly the 4 are positive which makes the total to be 0 for them.