For a non - zero, real a, b and c left| {begin{array}{*{20}{c}}{frac{{{a^2} + {b^2}}}{c}}&amp

English

Gujarati

Hindi

3 and 4 .Determinants and Matrices

normal

For a non - zero, real $a, b$ and $c$ $\left| {\begin{array}{*{20}{c}}{\frac{{{a^2} + {b^2}}}{c}}&c&c\\a&{\frac{{{b^2} + {c^2}}}{a}}&a\\b&b&{\frac{{{c^2} + {a^2}}}{b}} \end{array}} \right|$ $= \alpha \, abc$, then the values of $\alpha$ is

$-4$

$0$

$2$

$4$

Solution

$\frac{1}{{abc}}\,\left| {\,\,\begin{array}{*{20}{c}}{{a^2} + {b^2}}&{{c^2}}&{{c^2}}\\ {{a^2}}&{{b^2} + {c^2}}&{{a^2}}\\{{b^2}}&{{b^2}}&{{c^2} + {a^2}}\end{array}\,\,} \right|$

use $R_1 \rightarrow R_1 – (R_2 + R_3)$

$\frac{1}{{abc}}\,\left| {\,\,\begin{array}{*{20}{c}}0&{ – 2{b^2}}&{ – 2{a^2}}\\{{a^2}}&{{b^2} + {c^2}}&{{a^2}}\\{{b^2}}&{{b^2}}&{{c^2} + {a^2}} \end{array}\,\,} \right|$

$R_2 \rightarrow R_2 + 1/2R_1$ and $R_3 \rightarrow R_3 + 1/2 R_1$

$\frac{1}{{abc}}\,\left| {\,\,\begin{array}{*{20}{c}}0&{ – 2{b^2}}&{ – 2{a^2}}\\ {{a^2}}&{{c^2}}&0\\{{b^2}}&0&{{c^2}}\end{array}\,\,} \right|$
$\frac{1}{{abc}}\,$ [$ 2b^2 (a^2 c^2) – 2a^2 (- b^2 c^2) $] =$\frac{{4{a^2}\,{b^2}\,{c^2}}}{{abc}}\, = \,4abc$

Standard 12

Mathematics

$\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}\,} \right| = $

medium

If $\omega $ is a complex cube root of unity, then the determinant $\left| {\,\begin{array}{*{20}{c}}2&{2\omega }&{ – {\omega ^2}}\\1&1&1\\1&{ – 1}&0\end{array}\,} \right| = $

easy

$\left| {\,\begin{array}{*{20}{c}}{a – b – c}&{2a}&{2a}\\{2b}&{b – c – a}&{2b}\\{2c}&{2c}&{c – a – b}\end{array}\,} \right| = $

hard

If $ab + bc + ca = 0$ and $\left| {\,\begin{array}{*{20}{c}}{a – x}&c&b\\c&{b – x}&a\\b&a&{c – x}\end{array}\,} \right| = 0$, then one of the value of $x$ is