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
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
•
Total views: 325.2k
•
Views today: 5.84k
Recently Updated Pages
If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?
