- Home
- Standard 12
- Mathematics
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