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Question: If a, b, c are non-zero real numbers and if the system of equations \(\begin{aligned} & \left(...

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

Explanation

Solution

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 ∣A∣=0\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
(a−1)x=y+z⇒(a−1)x−y−z=0 (b−1)y=z+x⇒−x+(b−1)y−z=0 (c−1)z=x+y⇒−x−y−(c−1)z=0 \begin{aligned} & \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{aligned}
These 3 equations can be considered in 3 x 3 matrix form
A 3 x 3 matrix is of the form [x1y1z1 x2y2z2 x3y3z3 ]\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 ∣x1y1z1 x2y2z2 x3y3z3 ∣\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=[a−1−1−1 −1b−1−1 −1−1c−1 ]A=\left[ \begin{matrix} a-1 & -1 & -1 \\\ -1 & b-1 & -1 \\\ -1 & -1 & c-1 \\\ \end{matrix} \right]
∴∣A∣=∣a−1−1−1 −1b−1−1 −1−1c−1 ∣→Row1(R1) →Row2(R2) →Row3(R3) \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 R3→R3→R2{{R}_{3}}\to {{R}_{3}}\to {{R}_{2}}
∣a−1−1−1 −1b−1−1 (−1+1)(−1−b+1)(c−1+1) ∣=∣a−1−1−1 −1b−1−1 0−bc ∣\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 R2→R2→R1{{R}_{2}}\to {{R}_{2}}\to {{R}_{1}}

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 $\left| A \right|=0$ $\begin{aligned} & \left| A \right|\Rightarrow \left( a-1 \right)\left[ bc \right]+1\left( -ac \right)-1\left( ab \right) \\\ & =\left( a-1 \right)bc-ac-ab \\\ & =abc-bc-ac-ab \\\ & \left| A \right|=0 \\\ & \Rightarrow abc-bc-ac-ab=0 \\\ & \Rightarrow ab+bc+ac=abc \\\ \end{aligned}$ 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.