3 and 4 .Determinants and Matrices
medium

$\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right| = $

A

$3abc + {a^3} + {b^3} + {c^3}$

B

$3abc - {a^3} - {b^3} - {c^3}$

C

$abc - {a^3} + {b^3} + {c^3}$

D

$abc + {a^3} - {b^3} - {c^3}$

Solution

(b) $\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right| = \left| {\,\begin{array}{*{20}{c}}{a + b + c}&{a + b + c}&{a + b + c}\\b&c&a\\c&a&b\end{array}\,} \right|$,

                                                       $({R_1} \to {R_1} + {R_2} + {R_3})$

=$(a + b + c)$ $\left| {\,\begin{array}{*{20}{c}}{1}&{1}&{1}\\b&c&a\\c&a&b\end{array}\,} \right|$  =$(a + b + c)$  $\left| {\,\begin{array}{*{20}{c}}   1&1&1 \\    b&c&a \\    c&a&b  \end{array}\,} \right|$  =$(a + b + c)$    $\left| {\,\begin{array}{*{20}{c}}   1&0&0 \\    b&{b – c}&{c – a} \\    c&{c – a}&{a – b}  \end{array}\,} \right|$

= $3abc – {a^3} – {b^3} – {c^3}$,                    (After simplification).

Standard 12
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.