Question

Find the number of positive integral solutions of the equation $\left| \begin{matrix}
   {{y}^{3}}+1 & {{y}^{2}}z & {{y}^{2}}x \\
   y{{z}^{2}} & {{z}^{3}}+1 & {{z}^{2}}x \\
   y{{x}^{2}} & {{x}^{2}}z & {{x}^{3}}+1 \\
\end{matrix} \right|=11$.

Answer 1

Hint: We only need to find the determinant value. If start is by row and column operation to simplify the given form. Then we use the multiplication form to change the determinant. In the end, we break the determinant to get the cubic equation of the variables. We solve it to find the solution.

Complete step-by-step solution:
We try to find the determinant value of the given by using operations.
$\left| \begin{matrix}
   {{y}^{3}}+1 & {{y}^{2}}z & {{y}^{2}}x \\
   y{{z}^{2}} & {{z}^{3}}+1 & {{z}^{2}}x \\
   y{{x}^{2}} & {{x}^{2}}z & {{x}^{3}}+1 \\
\end{matrix} \right|=11$
The determinant has y, z, x in every column.
Taking x, y, and z common from every column.
$\left[ {{C}_{1}}^{'}=\dfrac{{{C}_{1}}}{y},{{C}_{2}}^{'}=\dfrac{{{C}_{2}}}{z},{{C}_{3}}^{'}=\dfrac{{{C}_{3}}}{x} \right]$
So, \[\left| \begin{matrix}
   {{y}^{3}}+1 & {{y}^{2}}z & {{y}^{2}}x \\
   y{{z}^{2}} & {{z}^{3}}+1 & {{z}^{2}}x \\
   y{{x}^{2}} & {{x}^{2}}z & {{x}^{3}}+1 \\
\end{matrix} \right|=11\Rightarrow xyz\left| \begin{matrix}
   \dfrac{{{y}^{3}}+1}{y} & {{y}^{2}} & {{y}^{2}} \\
   {{z}^{2}} & \dfrac{{{z}^{3}}+1}{z} & {{z}^{2}} \\
   {{x}^{2}} & {{x}^{2}} & \dfrac{{{x}^{3}}+1}{x} \\
\end{matrix} \right|=11\],
Now we multiply y, z, x in the 1st, 2nd, 3rd row respectively.
$\left[ {{R}_{1}}^{'}=y{{R}_{1}},{{R}_{2}}^{'}=z{{R}_{2}},{{R}_{3}}^{'}=x{{R}_{3}} \right]$
\[xyz\left| \begin{matrix}
   \dfrac{{{y}^{3}}+1}{y} & {{y}^{2}} & {{y}^{2}} \\
   {{z}^{2}} & \dfrac{{{z}^{3}}+1}{z} & {{z}^{2}} \\
   {{x}^{2}} & {{x}^{2}} & \dfrac{{{x}^{3}}+1}{x} \\
\end{matrix} \right|=11\Rightarrow \left| \begin{matrix}
   {{y}^{3}}+1 & {{y}^{3}} & {{y}^{3}} \\
   {{z}^{3}} & {{z}^{3}}+1 & {{z}^{3}} \\
   {{x}^{3}} & {{x}^{3}} & {{x}^{3}}+1 \\
\end{matrix} \right|=11\],
Now we apply some operations(Substrcting column I from column II and III ) to get value 0 in some elements
\[\begin{align}
  & \left| \begin{matrix}
   {{y}^{3}}+1 & {{y}^{3}} & {{y}^{3}} \\
   {{z}^{3}} & {{z}^{3}}+1 & {{z}^{3}} \\
   {{x}^{3}} & {{x}^{3}} & {{x}^{3}}+1 \\
\end{matrix} \right|=11 \\
 & \Rightarrow \left| \begin{matrix}
   {{y}^{3}}+1 & -1 & -1 \\
   {{z}^{3}} & 1 & 0 \\
   {{x}^{3}} & 0 & 1 \\
\end{matrix} \right|=11 \\
\end{align}\]
Now we take the determinant value
\[\begin{align}
  & \left| \begin{matrix}
   {{y}^{3}}+1 & -1 & -1 \\
   {{z}^{3}} & 1 & 0 \\
   {{x}^{3}} & 0 & 1 \\
\end{matrix} \right|=11 \\
 & \Rightarrow \left( {{y}^{3}}+1 \right)-{{z}^{3}}\left( -1 \right)+{{x}^{3}}=11 \\
 & \Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=10 \\
\end{align}\]
We need to find positive integral solutions for x, y, z. So, $x,y,z>0$.
So, possible solutions are of the sum of cubes of three numbers are only 1, 1, 2.
So, we can arrange 3 digits for three unknown numbers in $\dfrac{3!}{2!}=3$ ways.
The solutions are $\left( x,y,z \right)=\left( 1,1,2 \right),\left( 1,2,1 \right),\left( 2,1,1 \right)$.

Note: We need to find the determinant value. But if we start it with breaking the determinant from the very first step then the problem will become more chaotic. In this type of question, the properties of determinants are of utmost importance as they tend to make the problem much easier.