If det a 1 d b 1 e c 1 f 2 and det a 1 d b 2 e c 3 f 2 Solut

If det

a 1 d b 1 e c 1 f

=2 and det

a 1 d b 2 e c 3 f

=2,

Solution

In the case of constant multiplication factor,

Det(new matrix) = K * Det(old matrix)

=> 7 * Det(a,1,d;b,1,e;c,1,f)

=> 7 * -2

=> -14

Hence the answer to the first part is equal to -14

The second matrix is given by

2 * Det(a,1,d;b,2,e;c,3,f) + 2 * Det(a,1,d;b,1,e;c,1,f)

=> 2 * (-2) + 2 * (-2)

=> -8

multiplying second matrix column by -2 and first column by -1, we get it as

now adding 2 * (first matrix) by multiplying the first column and adding the two matrices to get the desired matrix