open vertical bar table row cell left parenthesis straight b plus straight c right parenthesis hat 2 end cell ab ca row ab cell left parenthesis straight a plus straight c right parenthesis hat 2 end cell bc row ac bc cell left parenthesis straight a plus straight b right parenthesis hat 2 end cell end table close vertical bar equals 2 abc left parenthesis straight a plus straight b plus straight c right parenthesis hat 3

Asked by indupremachandran | 26th May, 2019, 01:08: PM

Expert Answer:

Follow given steps to prove the question which you have posted.
 
 
i) Multiplying R1, R2 and R3 by a, b, c respectively and dividing by abc 
ii) Take a, b, c common from C1, C2 and C3 respectively.
iii) begin mathsize 16px style straight C subscript 3 rightwards arrow straight C subscript 3 minus straight C subscript 1 space and space straight C subscript 2 rightwards arrow straight C subscript 2 minus straight C subscript 1 end style
iv) (a + b + c) common from C2 and C3
v) R1 →R- (R2 + R3)
vi) begin mathsize 16px style straight C subscript 2 rightwards arrow straight C subscript 2 plus 1 over straight b straight C subscript 1 space and space straight C subscript 3 rightwards arrow straight C subscript 3 plus 1 over straight c straight C subscript 1 end style
Now solve using determinant.

Answered by Sneha shidid | 27th May, 2019, 10:25: AM