જો a,b,c એ અસમાન હોય અને left| {,begin{array}{*{20}{c}}a&{{a^2}}&amp

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3 and 4 .Determinants and Matrices

medium

જો $a,b,c$ એ અસમાન હોય અને $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} - 1}\\b&{{b^2}}&{{b^3} - 1}\\c&{{c^2}}&{{c^3} - 1}\end{array}\,} \right| = 0$ તો . . .

$a + b + c = 0$

$abc = 1$

$a + b + c = 1$

$ab + bc + ca = 0$

Solution

(b) $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} – 1}\\b&{{b^2}}&{{b^3} – 1}\\c&{{c^2}}&{{c^3} – 1}\end{array}\,} \right| = 0$

==> $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3}}\\b&{{b^2}}&{{b^3}}\\c&{{c^2}}&{{c^3}}\end{array}\,} \right| – \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&1\\b&{{b^2}}&1\\c&{{c^2}}&1\end{array}\,} \right| = 0$

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

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

Since $a,b,c$ are different, so $\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| \ne 0$

Hence $abc – 1 = 0$ i.e., $abc = 1$.

Standard 12

Mathematics

સાબિત કરો કે $\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|=0$

easy

સમીકરણ $\left|\begin{array}{*{20}{c}}
{\cos \,\,x}&{\sin \,\,x}&{\sin \,\,x}\\
{\sin \,\,x}&{\cos \,\,x}&{\sin \,\,x}\\
{\sin \,\,x}&{\sin \,\,x}&{\cos \,\,x}
\end{array}\right|\,\, = \,\,0$ ના કેટલા ભિન્ન વાસ્તવિક બીજ એ $\left[ { – \frac{\pi }{4},\frac{\pi }{4}} \right]$ અંતરાલ માં હશે ?

hard

(JEE MAIN-2016)

જો સમીકરણ સંહતિ $ax + y + z = 0$, $x + by + z = 0$ અને $x + y + cz = 0 $ ને શૂન્યતર ઉકેલ હોય તો $\frac{1}{{1 – a}} + \frac{1}{{1 – b}} + \frac{1}{{1 – c}} = . .. . $ (કે જ્યાં $a,b,c \ne 1$ )

medium

$\left| {\,\begin{array}{*{20}{c}}4&{ – 6}&1\\{ – 1}&{ – 1}&1\\{ – 4}&{11}&{ – 1\,}\end{array}} \right|= . . . $