3 and 4 .Determinants and Matrices
medium

यदि $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right| = k(a + b + c)({a^2} + {b^2} + {c^2}$ $ - bc - ca - ab)$, तो  $k =$

A

$1$

B

$2$

C

$-1$

D

$-2$

Solution

(c) $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right| = a(bc – {a^2}) – b({b^2} – ca) + c(ab – {c^2})$

= $ – {a^3} – {b^3} – {c^3} + 3abc$ = $ – 1\,[{a^3} + {b^3} + {c^3} – 3abc]$

= $ – [(a + b + c)\,({a^2} + {b^2} + {c^2} – ab – bc – ca)]$

==> $k = – 1$.

Standard 12
Mathematics

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