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If a, b, c are non-zero real numbers and if the system of equations
$\begin{align}
  & \left( a-1 \right)x=y+z \\
 & \left( b-1 \right)y=z+x \\
 & \left( c-1 \right)z=x+y \\
\end{align}$
has a non-trivial solution, then $ab+bc+ca$ is equal to
A. a + b + c
B. abc
C. 1
D. -1

Answer
VerifiedVerified
156.6k+ views
Hint: Make the 3 equations into matrix form. It will become a 3 x 3 matrix. Now, find its determinant. If the matrix is taken as A, then determinant of A is $\left| A \right|=0$. Solve and get the answer.

Complete step-by-step answer:
A solution or example that is not trivial, if the solution is non-zero. Solution/examples that involve the number zero are considered as trivial.
For example the equation x + 5y = 0 has trivial solution (0, 0).
Now-trivial solutions include (5, -1) and (2, 0.4).
Consider the 3 equations
$\begin{align}
  & \left( a-1 \right)x=y+z\Rightarrow \left( a-1 \right)x-y-z=0 \\
 & \left( b-1 \right)y=z+x\Rightarrow -x+\left( b-1 \right)y-z=0 \\
 & \left( c-1 \right)z=x+y\Rightarrow -x-y-\left( c-1 \right)z=0 \\
\end{align}$
These 3 equations can be considered in 3 x 3 matrix form
A 3 x 3 matrix is of the form \[\left[ \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & {{z}_{1}} \\
   {{x}_{2}} & {{y}_{2}} & {{z}_{2}} \\
   {{x}_{3}} & {{y}_{3}} & {{z}_{3}} \\
\end{matrix} \right]\]
Similarly determinant is of form \[\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & {{z}_{1}} \\
   {{x}_{2}} & {{y}_{2}} & {{z}_{2}} \\
   {{x}_{3}} & {{y}_{3}} & {{z}_{3}} \\
\end{matrix} \right|\]
Let us consider $A=\left[ \begin{matrix}
   a-1 & -1 & -1 \\
   -1 & b-1 & -1 \\
   -1 & -1 & c-1 \\
\end{matrix} \right]$
$\therefore \left| A \right|=\left| \begin{matrix}
   a-1 & -1 & -1 \\
   -1 & b-1 & -1 \\
   -1 & -1 & c-1 \\
\end{matrix} \right|\begin{matrix}
   \to Row1\left( {{R}_{1}} \right) \\
   \to Row2\left( {{R}_{2}} \right) \\
   \to Row3\left( {{R}_{3}} \right) \\
\end{matrix}$
Do ${{R}_{3}}\to {{R}_{3}}\to {{R}_{2}}$
$\left| \begin{matrix}
   a-1 & -1 & -1 \\
   -1 & b-1 & -1 \\
   \left( -1+1 \right) & \left( -1-b+1 \right) & \left( c-1+1 \right) \\
\end{matrix} \right|=\left| \begin{matrix}
   a-1 & -1 & -1 \\
   -1 & b-1 & -1 \\
   0 & -b & c \\
\end{matrix} \right|$
Now do ${{R}_{2}}\to {{R}_{2}}\to {{R}_{1}}$
\[\left| \begin{matrix}
   a-1 & -1 & -1 \\
   \left( -1-a+1 \right) & \left( b-1+1 \right) & \left( -1+1 \right) \\
   0 & -b & c \\
\end{matrix} \right|=\left| \begin{matrix}
   a-1 & -1 & -1 \\
   -a & b & 0 \\
   0 & -b & c \\
\end{matrix} \right|\]
We know that $\left| A \right|=0$
$\begin{align}
  & \left| A \right|\Rightarrow \left( a-1 \right)\left[ bc \right]+1\left( -ac \right)-1\left( ab \right) \\
 & =\left( a-1 \right)bc-ac-ab \\
 & =abc-bc-ac-ab \\
 & \left| A \right|=0 \\
 & \Rightarrow abc-bc-ac-ab=0 \\
 & \Rightarrow ab+bc+ac=abc \\
\end{align}$
Therefore, the correct answer is option C.

Note: Simplify the determinant A before equating it to zero or else the answer will become complex.One must be aware of the rows and columns operations which helps in simplifying the determinant.