If left| {begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} {b

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3 and 4 .Determinants and Matrices

normal

If $\left| {\begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} \\ {b - c}&{c - a}&{a - b} \\ {c - a + 1}&{a - b}&{b - c} \end{array}} \right| = 0$ ,$\left( {a,b,c \in R - \left\{ 0 \right\}} \right),$ then

$a$,$b$ and $c$ must be equal

$a$,$b$ and $c$ may not be all equal

$a$,$b$ and $c$ must be in $G.P$.

$a$,$b$ and $c$ must be distinct

Solution

$\left|\begin{array}{lll}{a-b} & {b-c} & {c-a} \\ {b-c} & {c-a} & {a-b} \\ {c-a} & {a-b} & {b-c}\end{array}\right|+\left|\begin{array}{ccc}{0} & {b-c} & {c-a} \\ {0} & {c-a} & {a-b} \\ {1} & {a-b} & {b-c}\end{array}\right|=0$

$\Rightarrow 0+(a-b)(b-c)=(c-a)^{2}$

$\Rightarrow a b-a c-b^{2}+b c=c^{2}+a^{2}-2 a c$

$\Rightarrow a^{2}+b^{2}+c^{2}=a b+b c+c a$

$\Rightarrow a=b=c$

Standard 12

Mathematics

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

medium

Let $\alpha $, $\beta$ $\gamma$, $\delta$ are distinct imaginary roots of

$z^5=1$ then value of $\left| {\begin{array}{*{20}{c}}
{{e^\alpha }}&{{e^{2\alpha }}}&{{e^{3\alpha + 1}}}&{ – {e^{ – \delta }}} \\
{{e^\beta }}&{{e^{2\beta }}}&{{e^{3\beta + 1}}}&{ – {e^{ – \delta }}} \\
{{e^\gamma }}&{{e^{2\gamma }}}&{{e^{3\gamma + 1}}}&{ – {e^{ – \delta }}}
\end{array}} \right|$

normal

If $\left| {\,\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{{(c + a)}^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{{(a + b)}^2}}\end{array}\,} \right| = k\,abc{(a + b + c)^3}$, then the value of $k$ is

hard

If $\left| {\,\begin{array}{*{20}{c}}{y + z}&{x – z}&{x – y}\\{y – z}&{z – x}&{y – x}\\{z – y}&{z – x}&{x + y}\end{array}\,} \right| = k\,xyz$, then the value of $k $ is