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If

then k is equal to:

  • Option 1)

    4\lambda abc

  • Option 2)

    -4\lambda \: abc

  • Option 3)

    4\lambda ^{2}

  • Option 4)

    -4\lambda ^{2}

 

Answers (1)

best_answer

As we learnt in 

Value of determinants of order 3 -

-

 

\begin{vmatrix} a^{2} &b^{2} &c^{2} \\ \left ( a+\lambda \right )^{2}& \left ( b+\lambda \right )^{2} &\left ( c+\lambda ^{2} \right ) \\ \left ( a-\lambda \right )^{2} & \left ( b-\lambda \right )^{2} & \left ( c-\lambda \right )^{2} \end{vmatrix}

\Rightarrow R_{2}\rightarrow R_{2}-R_{3}

\begin{vmatrix} a^{2} & b^{2} &c^{2} \\ 4a\lambda &4b\lambda & 4c\lambda \\ \left ( a-\lambda \right )^{2} & \left ( b-\lambda \right )^{2} & \left ( c-\lambda \right )^{2} \end{vmatrix} =4\lambda \begin{vmatrix} a^{2} & b^{2} &c^{2} \\ a &b & c \\ \left ( a-\lambda \right )^{2} & \left ( b-\lambda \right )^{2} & \left ( c-\lambda \right )^{2} \end{vmatrix}

\Rightarrow 4\lambda \begin{vmatrix} a^{2} & b^{2} &c^{2} \\ a &b &c \\ a^{2} & b^{2} &c^{2} \end{vmatrix} - 4\lambda . 2\lambda \begin{vmatrix} a^{2} &b^{2} &c^{2} \\ a & b&c \\ a &b & c \end{vmatrix}+ 4\lambda.\lambda ^{2}\begin{vmatrix} a^{a} &b^{2} & c^{2}\\ a &b &c \\ 1&1 &1 \end{vmatrix}

=0-0+4\lambda ^{3}\begin{vmatrix} a^{2} & b^{2} &c^{2} \\ a&b &c \\ 1 &1 &1 \end{vmatrix}

\therefore K=4\lambda ^{2}

 


Option 1)

4\lambda abc

This option is incorrect.

Option 2)

-4\lambda \: abc

This option is incorrect.

Option 3)

4\lambda ^{2}

This option is correct.

Option 4)

-4\lambda ^{2}

This option is incorrect.

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Aadil

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