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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 to
A. \[4\lambda abc\]
B. \[ - 4\lambda abc\]
C. \[4{\lambda ^2}\]
D. \[ - 4{\lambda ^2}\]

Answer Verified Verified
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.
\[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  {pd}&{pe}&{pf} \\
  g&h&i
\end{array}} \right| = p\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right|\]

Complete step-by-step answer:
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
\[ \Rightarrow 4{\lambda ^3}\left| {\begin{array}{*{20}{c}}
  {{a^2}}&{{b^2}}&{{c^2}} \\
  a&b&c \\
  1&1&1
\end{array}} \right|\] … (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, don’t directly subtract in the first step, always use the formula for opening the squares and then subtract.