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3 and 4 .Determinants and Matrices
medium
$\left| {\,\begin{array}{*{20}{c}}{b + c}&{a - b}&a\\{c + a}&{b - c}&b\\{a + b}&{c - a}&c\end{array}\,} \right| = $
A
${a^3} + {b^3} + {c^3} - 3abc$
B
$3abc - {a^3} - {b^3} - {c^3}$
C
${a^3} + {b^3} + {c^3} - {a^2}b - {b^2}c - {c^2}a$
D
$(a+b+c)(a^2+b^2+c^2+ab+bc+ca)$
Solution
(b) $\Delta = \left| {\,\begin{array}{*{20}{c}}{2(a + b + c)}&0&{a + b + c}\\{c + a}&{b – c}&b\\{a + b}&{c – a}&c\end{array}\,} \right|$
by ${R_1} \to {R_1} + {R_2} + {R_3}$
$\Delta = (a + b + c)\,.\,\left| {\,\begin{array}{*{20}{c}}2&0&1\\{c + a}&{b – c}&b\\{a + b}&{c – a}&c\end{array}\,} \right|$
On expanding,
$ – (a + b + c)\,({a^2} + {b^2} + {c^2} – ab – bc – ca)$
= $ -(a^3 + b^3 + c^3 – 3abc) = 3abc – a^3 -b^3 – c^3$
Trick : Put $a = 1,\,b = 2,\,c = 3$ and check it.
Standard 12
Mathematics
