Question

# If $\left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ {{{(a + \lambda )}^2}}&{{{(b + \lambda )}^2}}&{{{(c + \lambda )}^2}} \\ {{{(a - \lambda )}^2}}&{{{(b - \lambda )}^2}}&{{{(c - \lambda )}^2}} \end{array}} \right| = k\lambda \left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ a&b&c \\ 1&1&1 \end{array}} \right|$ , $\lambda \ne 0$, then k is equal toA. $4\lambda abc$B. $- 4\lambda abc$C. $4{\lambda ^2}$D. $- 4{\lambda ^2}$

Hint: Here we use the concept of row transformations to convert the matrix on the left side as the matrix on the right hand side. Use the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab,{(a - b)^2} = {a^2} + {b^2} - 2ab$ to open up the values in the determinant. We bring out the constants from each row as we convert the row to a similar row on the right hand side.
* If a row in the matrix contains elements which all have a common factor say p, then we can bring out the factor from the matrix.
$\left| {\begin{array}{*{20}{c}} a&b&c \\ {pd}&{pe}&{pf} \\ g&h&i \end{array}} \right| = p\left| {\begin{array}{*{20}{c}} a&b&c \\ d&e&f \\ g&h&i \end{array}} \right|$

Here we name each row as a variable
First row is ${R_1}$
Second row is ${R_2}$
Third row is ${R_3}$
We have the matrix
$\left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ {{{(a + \lambda )}^2}}&{{{(b + \lambda )}^2}}&{{{(c + \lambda )}^2}} \\ {{{(a - \lambda )}^2}}&{{{(b - \lambda )}^2}}&{{{(c - \lambda )}^2}} \end{array}} \right|$
Opening up the values in squares in the second row and the third row using the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and ${(a - b)^2} = {a^2} + {b^2} - 2ab$
$\Rightarrow \left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ {{a^2} + {\lambda ^2} + 2a\lambda }&{{b^2} + {\lambda ^2} + 2b\lambda }&{{c^2} + {\lambda ^2} + 2c\lambda } \\ {{a^2} + {\lambda ^2} - 2a\lambda }&{{b^2} + {\lambda ^2} - 2b\lambda }&{{c^2} + {\lambda ^2} - 2c\lambda } \end{array}} \right|$
Now we use the row transformation ${R_2} \to {R_2} - {R_1};{R_3} \to {R_3} - {R_1}$
$\Rightarrow \left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ {{\lambda ^2} + 2a\lambda }&{{\lambda ^2} + 2b\lambda }&{{\lambda ^2} + 2c\lambda } \\ {{\lambda ^2} - 2a\lambda }&{{\lambda ^2} - 2b\lambda }&{{\lambda ^2} - 2c\lambda } \end{array}} \right|$
Now we use the row transformation ${R_2} \to {R_2} - {R_3}$
$\Rightarrow \left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ {4a\lambda }&{4b\lambda }&{4c\lambda } \\ {{\lambda ^2} - 2a\lambda }&{{\lambda ^2} - 2b\lambda }&{{\lambda ^2} - 2c\lambda } \end{array}} \right|$
Now we take 2 common from ${R_2}$
$\Rightarrow 2\left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ {2a\lambda }&{2b\lambda }&{2c\lambda } \\ {{\lambda ^2} - 2a\lambda }&{{\lambda ^2} - 2b\lambda }&{{\lambda ^2} - 2c\lambda } \end{array}} \right|$
Now we use row transformation ${R_3} \to {R_3} + {R_2}$
$\Rightarrow 2\left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ {2a\lambda }&{2b\lambda }&{2c\lambda } \\ {{\lambda ^2}}&{{\lambda ^2}}&{{\lambda ^2}} \end{array}} \right|$
Taking $2\lambda$common from ${R_2}$and taking ${\lambda ^2}$common from ${R_3}$
$\Rightarrow 2({\lambda ^2})(2\lambda )\left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ a&b&c \\ 1&1&1 \end{array}} \right|$
Multiplying the terms outside the matrix
$\Rightarrow 4{\lambda ^3}\left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ a&b&c \\ 1&1&1 \end{array}} \right|$ â€¦ (1)
No we compare the equation (1) with RHS of the question
$\Rightarrow 4{\lambda ^3}\left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ a&b&c \\ 1&1&1 \end{array}} \right| = k\lambda \left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ a&b&c \\ 1&1&1 \end{array}} \right|$
Equating the constant terms outside the matrix
$\Rightarrow 4{\lambda ^3} = k\lambda$
Divide both sides of the equation by $\lambda$
$\Rightarrow \dfrac{{4{\lambda ^3}}}{\lambda } = \dfrac{{k\lambda }}{\lambda }$
Cancel the same terms from numerator and denominator from both sides of the equation.
$k = 4{\lambda ^2}$
So, option C is correct.

Note: Students many times start applying any row transformations to make the matrix on the left hand side easy, but remember that we apply row transformations in such a way that our matrix becomes similar to the matrix on the right side. Also, donâ€™t directly subtract in the first step, always use the formula for opening the squares and then subtract.