Question

Answer the following question:

Evaluate determinant along second column

`|(1, -1, 2),(3, 2, -2),(0, 1, -2)|`

Answer 1

Let D = `|(1, -1, 2),(3, 2, -2),(0, 1, -2)|`

The expansion of the determinant D along second column is

D = – 1 (cofactor of – 1) + 2 (cofactor of 2) + 1 (cofactor of 1)

= `-1 xx (-1)^(1 + 2) |(3, -2),(0, -2)| + 2 xx (-1)^(2 + 2)|(1, 2),(0, -2)| + 1 xx (-1)^(2 + 3)|(1, 2),(3, -2)|`

= –1( –1) ( – 6 + 0) + 2 (1)( – 2 – 0) +1 ( –1)( –2 –  6)

= 1 ( – 6) + 2( – 2) – 1( – 8)

= –6 – 4 + 8

= – 2.