Left| {,begin{array}{*{20}{c}}0&{{b^3} - {a^3}}&{{c^3} - {a^3}}{{a^3}

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3 and 4 .Determinants and Matrices

easy

$\left| {\,\begin{array}{*{20}{c}}0&{{b^3} - {a^3}}&{{c^3} - {a^3}}\\{{a^3} - {b^3}}&0&{{c^3} - {b^3}}\\{{a^3} - {c^3}}&{{b^3} - {c^3}}&0\end{array}\,} \right| = . . $

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

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

$0$

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

Solution

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

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

$[{C_2} \to {C_2} – {C_1}$ and ${C_3} \to {C_3} – {C_1}]$ and then taking out common

$({b^2} – {a^3})$ from $II^{nd}$ column and ( ${c^3} – {a^3}$) from $III^{rd}$ column].

Standard 12

Mathematics

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)

સાબિત કરો કે, $\Delta=\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & m
\end{array}\right|$

medium

જો $\left| {\,\begin{array}{*{20}{c}}{y + z}&{x – z}&{x – y}\\{y – z}&{z – x}&{y – x}\\{z – y}&{z – x}&{x + y}\end{array}\,} \right| = k\,xyz$, તો $k$ મેળવો.

medium

$2\,\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2} – bc}&{{b^2} – ac}&{{c^2} – ab}\end{array}\,} \right| = $

hard

$\left|\begin{array}{ccc}102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|$ નું મૂલ્ય શોધો.