Question

Using properties of determinants, prove that
 $ \left| {\begin{array}{*{20}{c}}
  {{{\left( {x + y} \right)}^2}}&{zx}&{zy} \\
  {zx}&{{{\left( {z + y} \right)}^2}}&{xy} \\
  {zy}&{xy}&{{{\left( {z + x} \right)}^2}}
\end{array}} \right| = 2xyz{\left( {x + y + z} \right)^3} $

Answer 1

Hint: In this particular question use the concept of multiplication a single variable into matrix for example in this question multiply z, x, and y in first second and third row respectively and multiply by (1/xyz) outside the determinant so that overall no change, then take z, x, and y common from first, second and third column respectively so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given identity
 $ \left| {\begin{array}{*{20}{c}}
  {{{\left( {x + y} \right)}^2}}&{zx}&{zy} \\
  {zx}&{{{\left( {z + y} \right)}^2}}&{xy} \\
  {zy}&{xy}&{{{\left( {z + x} \right)}^2}}
\end{array}} \right| = 2xyz{\left( {x + y + z} \right)^3} $
Consider the LHS of the given identity we have,
 $ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {{{\left( {x + y} \right)}^2}}&{zx}&{zy} \\
  {zx}&{{{\left( {z + y} \right)}^2}}&{xy} \\
  {zy}&{xy}&{{{\left( {z + x} \right)}^2}}
\end{array}} \right| $
Now multiply z, x, and y in first second and third row respectively and multiply by (1/xyz) outside the determinant so that overall no change, so we have,
 $ \Rightarrow \dfrac{1}{{zxy}}\left| {\begin{array}{*{20}{c}}
  {z{{\left( {x + y} \right)}^2}}&{{z^2}x}&{{z^2}y} \\
  {z{x^2}}&{x{{\left( {z + y} \right)}^2}}&{{x^2}y} \\
  {z{y^2}}&{x{y^2}}&{y{{\left( {z + x} \right)}^2}}
\end{array}} \right| $
Now take z, x, and y common from first, second and third column respectively we have,
 $ \Rightarrow \dfrac{{zxy}}{{zxy}}\left| {\begin{array}{*{20}{c}}
  {{{\left( {x + y} \right)}^2}}&{{z^2}}&{{z^2}} \\
  {{x^2}}&{{{\left( {z + y} \right)}^2}}&{{x^2}} \\
  {{y^2}}&{{y^2}}&{{{\left( {z + x} \right)}^2}}
\end{array}} \right| $
 $ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {{{\left( {x + y} \right)}^2}}&{{z^2}}&{{z^2}} \\
  {{x^2}}&{{{\left( {z + y} \right)}^2}}&{{x^2}} \\
  {{y^2}}&{{y^2}}&{{{\left( {z + x} \right)}^2}}
\end{array}} \right| $
Now apply determinant properties i.e. apply $ {C_1} \to {C_1} - {C_3},{C_2} \to {C_2} - {C_3} $ we have,
 $ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {{{\left( {x + y} \right)}^2} - {z^2}}&0&{{z^2}} \\
  0&{{{\left( {z + y} \right)}^2} - {x^2}}&{{x^2}} \\
  {{y^2} - {{\left( {z + x} \right)}^2}}&{{y^2} - {{\left( {z + x} \right)}^2}}&{{{\left( {z + x} \right)}^2}}
\end{array}} \right| $
Now as we know that $ \left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) $ so use this property in the above determinant we have,
 $ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {\left( {x + y + z} \right)\left( {x + y - z} \right)}&0&{{z^2}} \\
  0&{\left( {x + y + z} \right)\left( {z + y - x} \right)}&{{x^2}} \\
  {\left( {x + y + z} \right)\left( {y - z - x} \right)}&{\left( {x + y + z} \right)\left( {y - z - x} \right)}&{{{\left( {z + x} \right)}^2}}
\end{array}} \right| $
Now take (x + y + z) common from first and second column respectively we have,
 $ \Rightarrow {\left( {x + y + z} \right)^2}\left| {\begin{array}{*{20}{c}}
  {\left( {x + y - z} \right)}&0&{{z^2}} \\
  0&{\left( {z + y - x} \right)}&{{x^2}} \\
  {\left( {y - z - x} \right)}&{\left( {y - z - x} \right)}&{{{\left( {z + x} \right)}^2}}
\end{array}} \right| $
Now apply, $ {R_3} \to {R_3} - \left( {{R_1} + {R_2}} \right) $ we have,
 $ \Rightarrow {\left( {x + y + z} \right)^2}\left| {\begin{array}{*{20}{c}}
  {\left( {x + y - z} \right)}&0&{{z^2}} \\
  0&{\left( {z + y - x} \right)}&{{x^2}} \\
  {\left( {y - z - x} \right) - \left( {x + y - z} \right)}&{\left( {y - z - x} \right) - \left( {z + y - x} \right)}&{{{\left( {z + x} \right)}^2} - {z^2} - {x^2}}
\end{array}} \right| $
Now simplify the determinant we have,
 $ \Rightarrow {\left( {x + y + z} \right)^2}\left| {\begin{array}{*{20}{c}}
  {\left( {x + y - z} \right)}&0&{{z^2}} \\
  0&{\left( {z + y - x} \right)}&{{x^2}} \\
  { - 2x}&{ - 2z}&{2zx}
\end{array}} \right| $ , $ \left[ {\because {{\left( {z + x} \right)}^2} = {z^2} + {x^2} + 2zx} \right] $
Now apply, $ {C_1} \to {C_1} + \dfrac{{{C_3}}}{z},{C_2} \to {C_2} + \dfrac{{{C_3}}}{x} $ we have,
 $ \Rightarrow {\left( {x + y + z} \right)^2}\left| {\begin{array}{*{20}{c}}
  {\left( {x + y - z} \right) + z}&{\dfrac{{{z^2}}}{x}}&{{z^2}} \\
  {\dfrac{{{x^2}}}{z}}&{\left( {z + y - x} \right) + x}&{{x^2}} \\
  { - 2x + 2x}&{ - 2z + 2z}&{2zx}
\end{array}} \right| $
 $ \Rightarrow {\left( {x + y + z} \right)^2}\left| {\begin{array}{*{20}{c}}
  {\left( {x + y} \right)}&{\dfrac{{{z^2}}}{x}}&{{z^2}} \\
  {\dfrac{{{x^2}}}{z}}&{\left( {z + y} \right)}&{{x^2}} \\
  0&0&{2zx}
\end{array}} \right| $
Now expand the determinant w.r.t third row we have,
 $ \Rightarrow {\left( {x + y + z} \right)^2}\left[ {2zx\left| {\begin{array}{*{20}{c}}
  {x + y}&{\dfrac{{{z^2}}}{x}} \\
  {\dfrac{{{x^2}}}{z}}&{z + y}
\end{array}} \right| - 0 + 0} \right] $
Now expand the mini determinant we have,
 $ \Rightarrow {\left( {x + y + z} \right)^2}\left[ {2zx\left( {\left( {x + y} \right)\left( {z + y} \right) - \dfrac{{{z^2}}}{x}.\dfrac{{{x^2}}}{z}} \right)} \right] $
Now simplify we have,
 $ \Rightarrow {\left( {x + y + z} \right)^2}\left[ {2zx\left( {xz + xy + yz + {y^2} - xz} \right)} \right] $
 $ \Rightarrow {\left( {x + y + z} \right)^2}\left[ {2zx\left( {xy + yz + {y^2}} \right)} \right] $
 $ \Rightarrow {\left( {x + y + z} \right)^2}\left[ {2zxy\left( {x + z + y} \right)} \right] $
 $ \Rightarrow 2xyz{\left( {x + y + z} \right)^3} $
= RHS
Hence proved.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall how to apply the determinant rule (i.e. if we apply determinant in any particular column, the variables of this column is changed according to the variables in the other column we cannot change the variables of any particular column according to the variables of the row).