InterviewSolution
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InterviewSolution
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Usingproperties of determinants, prove that`|b+c q+r y+z c+a r+p z+x c+b p+q x+y|=2 |a p x b q y c r z|` |
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Answer» We have, `LHS = |[b+c,c+a, a+b],[q+r, r+p, p+q],[y+z, z+x,x+y]|` `=2 |[a+b+c,c+a, a+b],[p+q+r, r+p, p+q],[x+y+z, z+x,x+y]|["applying"C_(1) to (C_(1) + C_(2) + C_(3))"and taking out 2 common from"C_(1)]` `=2 |[a+b+c,-b, -c],[p+q+r, -q, -r],[x+y+z, -y,-z]|[C_(2) to (C_(2) -C_(1)), C_(3) to (C_(3) -C_(1))]` `=2(-1)(-1) * |[a+b+c,b, c],[p+q+r, q, r],[x+y+z, y,z]|["taking out(-1) common from each one of "C_(2) "and"C_(3)]` `=2 |[a,b, c],[p, q, r],[x, y,z]| = RHS ["applying "C_(1) to C_(1) -(C_(2) + C_(3))]` Hence, LHS=RHS. |
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