If {Delta _1} = left| {begin{array}{*{20}{c}} | vedclass

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(B) Given ${\Delta _1} = \left| {\begin{array}{*{20}{c}} x & b & b \ a & x & b \ a & a & x \end{array}} ight|$.Expanding the determinant ${\Delta

Vedclass · Accepted Answer

$\frac{d}{{dx}}({\Delta _1}) = 3{\Delta _2}$

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(B) Given ${\Delta _1} = \left| {\begin{array}{*{20}{c}} x & b & b \ a & x & b \ a & a & x \end{array}} ight|$.Expanding the determinant ${\Delta _1}$ along the first row:${\Delta _1} = x(x^2 - ab) - b(ax - ab) + b(a^2 - ax)$${\Delta _1} = x^3 - abx - abx + ab^2 + a^2b - abx$${\Delta _1} = x^3 - 3abx$.Now,differentiating ${\Delta _1}$ with respect to $x$:$\frac{d}{{dx}}({\Delta _1}) = \frac{d}{{dx}}(x^3 - 3abx) = 3x^2 - 3ab = 3(x^2 - ab)$.Given ${\Delta _2} = \left| {\begin{array}{*{20}{c}} x & b \ a & x \end{array}} ight| = x^2 - ab$.Substituting ${\Delta _2}$ into the derivative:$\frac{d}{{dx}}({\Delta _1}) = 3{\Delta _2}$.Thus,option $B$ is correct.

If ${\Delta _1} = \left| {\begin{array}{{20}{c}} x & b & b \\ a & x & b \\ a & a & x \end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{{20}{c}} x & b \\ a & x \end{array}} \right|$ are the given determinants,then:

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If ${\Delta _1} = \left| {\begin{array}{*{20}{c}} x & b & b \\ a & x & b \\ a & a & x \end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{*{20}{c}} x & b \\ a & x \end{array}} \right|$ are the given determinants,then: