- #1

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i gotta show that the determinant

|..1.......1......-1...|

|..a......-b.......c...| = (a+b)(a+c)(c-b).

|a^2...b^2...-c^2.|

however, i cannot quite get the answer to this. there is definitely one of these questions on my exam on friday and i just cannot ever reach the finish line with these ones.

here is as far as i get with this particular one:

|..1.......1.......-1..|

|..a......-b........c..|

|a^2...b^2...-c^2|

=

|..1-1............1...........1-1....|

|..a+b...........-b..........a+c....| (took column 2 from column 1 - determinant unchanged)

|a^2+b^2....b^2....a^2-c^2..|

=

|......0.............1.............0......|

|.....a+b..........-b...........a+c....| (neatened things up a bit)

|(a+b)(a-b).....b^2....(a+c)(a-c)|

=

|..0......1......0...|

1/(a+b)(a+c)|..1......-b.....1...| (multiplied column 1 by 1/(a+b) & column 2 by 1/(a+c)

|a-b...b^2...a-c.| - therefore multiplied determinant by same amount)

= (b-c)/(a+b)(a+c) when that simplified determinant is calculated.

ive tried messing around a few ways but this is the closest i have got to the answer, as in this attempt i at least got the (a+b) and (a+c) parts.

can somebody show me the right way to do this please, and most importantly explain it?

thanks.