InterviewSolution
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InterviewSolution
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If a, b, c are non zero complex numbers satisfying `a^(2) + b^(2) + c^(2) = 0 and |(b^(2) + c^(2),ab,ac),(ab,c^(2) + a^(2),bc),(ac,bc,a^(2) + b^(2))| = k a^(2) b^(2) c^(2)`, then k is equal toA. 3B. 2C. 4D. 1 |
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Answer» Correct Answer - C Let `Delta = |(b^(2) + c^(2),ab,ac),(ab,c^(2) + a^(2),bc),(ac,bc,a^(2) b^(2))|` Applying `R_(1) rarr R_(1) (a), R_(1) rarr R_(2) (b) and R_(3) (c)`, we get `Delta = (1)/(abc) |(a(b^(2) + c^(2)),a^(2)b,a^(2)c),(ab^(2),b(c^(2) + a^(2)),b^(2)c),(ac^(2),bc^(2),c(a^(2) + b^(2)))|` Taking a, b and c common from `C_(1), C_(2) and C_(3)` respectively, we get `Delta = (abc)/(abc) |(b^(2) + c^(2),a^(2),a^(2)),(b^(2),c^(2) + a^(2),b^(2)),(c^(2),c^(2),a^(2) + b^(2))|` `rArr Delta = |(2(b^(2) + c^(2)),2(c^(2) + a^(2)),2(a^(2) + b^(2))),(b^(2),c^(2) + a^(2),b^(2)),(c^(2),c^(2),a^(2) + b^(2))|` `rArr Delta = 2 |(b^(2) + c^(2),c^(2) + a^(2),a^(2) + b^(2)),(b^(2),c^(2) + a^(2),b^(2)),(c^(2),c^(2),a^(2) + b^(2))|` [Taking 2 common from `R_(1)`] Applying `R_(2) rarr R_(2) - R_(1) and R_(3) rarr R_(3) - R_(1)`, we get `Delta = 2 |(b^(2) + c^(2),c^(2) + a^(2),a^(2) + b^(2)),(-c^(2),0,-a^(2)),(-b^(2),-a^(2),0)|` `rArr Delta = 2 |(0,c^(2),b^(2)),(-c^(2),0,-a^(2)),(-b^(2),-a^(2),0)| " Applying "R_(1) rarr R_(1) + R_(2) + R_(3)` `rArr Delta = 2 {0 |(0,-a^(2)),(-a^(2),0)| -c^(2) |(-c^(2),-a^(2)),(-b^(2),0)|+ b^(2) |(-c^(2),0),(-b^(2),-a^(2))|}` `rArr Delta = 2 (a^(2) b^(2) c^(2) + a^(2) b^(2) c^(2)) = 4a^(2) b^(2) c^(2)` But, it is given that `Delta = ka^(2) b^(2) c^(2)` `:. k = 4` |
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