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3 and 4 .Determinants and Matrices
hard
यदि ${\Delta _1} = \left| {\,\begin{array}{*{20}{c}}x&b&b\\a&x&b\\a&a&x\end{array}\,} \right|$ और ${\Delta _2} = \left| {\,\begin{array}{*{20}{c}}x&b\\a&x\end{array}\,} \right|$ हो, तब
A
${\Delta _1} = 3{({\Delta _2})^2}$
B
$\frac{d}{{dx}}({\Delta _1}) = 3{\Delta _2}$
C
$\frac{d}{{dx}}({\Delta _1}) = 2{({\Delta _2})^2}$
D
${\Delta _1} = 3\Delta _2^{3/2}$
Solution
(b) ${\Delta _1} = \left| {\,\begin{array}{*{20}{c}}x&b&b\\a&x&b\\a&a&x\end{array}\,} \right| = {x^3} – 3abx$ ==> $\frac{d}{{dx}}{\Delta _1} = 3\,({x^2} – ab)$
तथा ${\Delta _2} = \left| {\,\begin{array}{*{20}{c}}x&b\\a&x\end{array}\,} \right| = {x^2} – ab$ ==> $\frac{d}{{dx}}\,({\Delta _1}) = 3\,({x^2} – ab) = 3{\Delta _2}$.
Standard 12
Mathematics