Question

Show that using the properties of matrices
\[\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & xy & zx \\
   xy & {{\left( x+z \right)}^{2}} & yz \\
   xz & yz & {{\left( x+y \right)}^{2}} \\
\end{matrix} \right|=2xyz{{\left( x+y+z \right)}^{3}}\]

Answer 1

Hint: We solve this problem by using the determinant properties by taking the LHS of the given equation. That is we assume that
\[L=\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & xy & zx \\
   xy & {{\left( x+z \right)}^{2}} & yz \\
   xz & yz & {{\left( x+y \right)}^{2}} \\
\end{matrix} \right|\]
The properties of determinant are simple, that is nothing but changing the rows and columns by subtracting or adding the columns or rows with respect to other columns and rows. By using the suitable transformations we get RHS from LHS we need to get two values as ‘0’ either in one row or column so that we can expand the determinant easily.

Complete step-by-step solution:
We are given to prove that
\[\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & xy & zx \\
   xy & {{\left( x+z \right)}^{2}} & yz \\
   xz & yz & {{\left( x+y \right)}^{2}} \\
\end{matrix} \right|=2xyz{{\left( x+y+z \right)}^{3}}\]
Let us take the LHS from the above equation as
\[\Rightarrow L=\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & xy & zx \\
   xy & {{\left( x+z \right)}^{2}} & yz \\
   xz & yz & {{\left( x+y \right)}^{2}} \\
\end{matrix} \right|\]
Now, by multiplying and dividing with \['xyz'\] outside the determinant we get
\[\begin{align}
  & \Rightarrow L=\dfrac{xyz}{xyz}\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & xy & zx \\
   xy & {{\left( x+z \right)}^{2}} & yz \\
   xz & yz & {{\left( x+y \right)}^{2}} \\
\end{matrix} \right| \\
 & \Rightarrow L=\dfrac{1}{xyz}.x.y.z\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & xy & zx \\
   xy & {{\left( x+z \right)}^{2}} & yz \\
   xz & yz & {{\left( x+y \right)}^{2}} \\
\end{matrix} \right| \\
\end{align}\]
Now, let us multiply the first row with \['x'\], second row with \['y'\] and third row with \['z'\] then we get
\[\Rightarrow L=\dfrac{1}{xyz}\left| \begin{matrix}
   x{{\left( y+z \right)}^{2}} & {{x}^{2}}y & z{{x}^{2}} \\
   x{{y}^{2}} & y{{\left( x+z \right)}^{2}} & {{y}^{2}}z \\
   x{{z}^{2}} & y{{z}^{2}} & z{{\left( x+y \right)}^{2}} \\
\end{matrix} \right|\]
Now, let us take out common term from each column that is let us take out \['x'\] from first column and \['y'\] from second column and \['z'\] from third column, then we get
\[\begin{align}
  & \Rightarrow L=\dfrac{1}{xyz}\left( xyz \right)\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & {{x}^{2}} & {{x}^{2}} \\
   {{y}^{2}} & {{\left( x+z \right)}^{2}} & {{y}^{2}} \\
   {{z}^{2}} & {{z}^{2}} & {{\left( x+y \right)}^{2}} \\
\end{matrix} \right| \\
 & \Rightarrow L=\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & {{x}^{2}} & {{x}^{2}} \\
   {{y}^{2}} & {{\left( x+z \right)}^{2}} & {{y}^{2}} \\
   {{z}^{2}} & {{z}^{2}} & {{\left( x+y \right)}^{2}} \\
\end{matrix} \right| \\
\end{align}\]
Now, let us use the column transformations that is let us use the transformation \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] then we get
\[\begin{align}
  & \Rightarrow L=\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & {{x}^{2}}-{{\left( y+z \right)}^{2}} & {{x}^{2}} \\
   {{y}^{2}} & {{\left( x+z \right)}^{2}}-{{y}^{2}} & {{y}^{2}} \\
   {{z}^{2}} & {{z}^{2}}-{{z}^{2}} & {{\left( x+y \right)}^{2}} \\
\end{matrix} \right| \\
 & \Rightarrow L=\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & \left( x-y-z \right)\left( x+y+z \right) & {{x}^{2}} \\
   {{y}^{2}} & \left( x+z-y \right)\left( x+z+y \right) & {{y}^{2}} \\
   {{z}^{2}} & 0 & {{\left( x+y \right)}^{2}} \\
\end{matrix} \right| \\
\end{align}\]
Now, similarly let us use the column transformations to the third column that is let us use \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\] then we get
\[\begin{align}
  & \Rightarrow L=\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & \left( x-y-z \right)\left( x+y+z \right) & {{x}^{2}}-{{\left( y+z \right)}^{2}} \\
   {{y}^{2}} & \left( x+z-y \right)\left( x+z+y \right) & {{y}^{2}}-{{y}^{2}} \\
   {{z}^{2}} & 0 & {{\left( x+y \right)}^{2}}-{{z}^{2}} \\
\end{matrix} \right| \\
 & \Rightarrow L=\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & \left( x-y-z \right)\left( x+y+z \right) & \left( x-y-z \right)\left( x+y+z \right) \\
   {{y}^{2}} & \left( x+z-y \right)\left( x+z+y \right) & 0 \\
   {{z}^{2}} & 0 & \left( x+y-z \right)\left( x+y+z \right) \\
\end{matrix} \right| \\
\end{align}\]
Now, let us take the term \[\left( x+y+z \right)\] common out from second and third column we get
\[\Rightarrow L={{\left( x+y+z \right)}^{2}}\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & \left( x-y-z \right) & \left( x-y-z \right) \\
   {{y}^{2}} & \left( x+z-y \right) & 0 \\
   {{z}^{2}} & 0 & \left( x+y-z \right) \\
\end{matrix} \right|\]
Now, let us use the row transformation that is \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}-{{R}_{3}}\] then we get
\[\Rightarrow L={{\left( x+y+z \right)}^{2}}\left| \begin{matrix}
   2yz & -2z & -2y \\
   {{y}^{2}} & \left( x+z-y \right) & 0 \\
   {{z}^{2}} & 0 & \left( x+y-z \right) \\
\end{matrix} \right|\]
Now, let us take the number ‘2’ common out from first row and by applying the column transformation \[{{C}_{2}}\to {{C}_{2}}+\dfrac{1}{y}{{C}_{1}}\] then we get
\[\begin{align}
  & \Rightarrow L=2{{\left( x+y+z \right)}^{2}}\left| \begin{matrix}
   yz & -z+\dfrac{yz}{y} & -y \\
   {{y}^{2}} & \left( x+z-y \right)+\dfrac{{{y}^{2}}}{y} & 0 \\
   {{z}^{2}} & 0+\dfrac{{{z}^{2}}}{y} & \left( x+y-z \right) \\
\end{matrix} \right| \\
 & \Rightarrow L=2{{\left( x+y+z \right)}^{2}}\left| \begin{matrix}
   yz & 0 & -y \\
   {{y}^{2}} & \left( x+z \right) & 0 \\
   {{z}^{2}} & 0+\dfrac{{{z}^{2}}}{y} & \left( x+y-z \right) \\
\end{matrix} \right| \\
\end{align}\]
Now, let us use the column transformation for third column that is let us use the transformation \[{{C}_{3}}\to {{C}_{3}}+\dfrac{1}{z}{{C}_{1}}\] then we get
\[\begin{align}
  & \Rightarrow L=2{{\left( x+y+z \right)}^{2}}\left| \begin{matrix}
   yz & 0 & -y+\dfrac{yz}{z} \\
   {{y}^{2}} & \left( x+z \right) & 0+\dfrac{{{y}^{2}}}{z} \\
   {{z}^{2}} & 0+\dfrac{{{z}^{2}}}{y} & \left( x+y-z \right)+\dfrac{{{z}^{2}}}{z} \\
\end{matrix} \right| \\
 & \Rightarrow L=2{{\left( x+y+z \right)}^{2}}\left| \begin{matrix}
   yz & 0 & 0 \\
   {{y}^{2}} & \left( x+z \right) & \dfrac{{{y}^{2}}}{z} \\
   {{z}^{2}} & 0+\dfrac{{{z}^{2}}}{y} & \left( x+y \right) \\
\end{matrix} \right| \\
\end{align}\]
Here, we can see that the first row has two values equal to ‘0’. So, let us expand the determinant along first row then we will get
\[\Rightarrow L=2{{\left( x+y+z \right)}^{2}}yz\left[ \left( x+z \right)\left( x+y \right)-\left( \dfrac{{{z}^{2}}}{y} \right)\left( \dfrac{{{y}^{2}}}{z} \right) \right]\]
Now, by multiplying the terms in the above equation we get
\[\begin{align}
  & \Rightarrow L=2{{\left( x+y+z \right)}^{2}}yz\left[ {{x}^{2}}+xy+zx+zy-zy \right] \\
 & \Rightarrow L=2{{\left( x+y+z \right)}^{2}}yz\left[ x\left( x+y+z \right) \right] \\
 & \Rightarrow L=2xyz{{\left( x+y+z \right)}^{3}} \\
\end{align}\]
Here, we can see that the value of LHS is equal to the value of RHS.
So we can say that the required result has been proved.

Note: In this problem we can apply the two transformations at a time to reduce the length of problem solving. Since, the given determinant is \[3\times 3\] matric we can apply two transformations at a time but those transformations need to be the same, that is either row or column transformations. Here, we have
\[\Rightarrow L=\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & {{x}^{2}} & {{x}^{2}} \\
   {{y}^{2}} & {{\left( x+z \right)}^{2}} & {{y}^{2}} \\
   {{z}^{2}} & {{z}^{2}} & {{\left( x+y \right)}^{2}} \\
\end{matrix} \right|\]
We can apply two column transformations at a time that is we can apply \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] to get the following result directly.
\[\Rightarrow L={{\left( x+y+z \right)}^{2}}\left| \begin{matrix}
   {{\left( y+z \right)}^{2}} & \left( x-y-z \right) & \left( x-y-z \right) \\
   {{y}^{2}} & \left( x+z-y \right) & 0 \\
   {{z}^{2}} & 0 & \left( x+y-z \right) \\
\end{matrix} \right|\]
This is the shortcut to reduce the length and time of problem solving. Similarly we can do for the rest of the solution.