# 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

Verified

325.2k+ 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.

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.

Last updated date: 28th May 2023

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