Question

a, b, c are distinct real numbers, not equal to one. If $ax+y+z=0$ , $x+by+z=0$ and $x+y+cz=0$ have a non-trivial solution, then the value of $\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}$ is equal to
(a) $-1$
(b) 1
(c) zero
(d) none of these

Answer 1

Hint: Here, first we will convert the given equation into matrix form and will equate with zero as it is told that equation have non-trivial solution i.e. given as $\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|=0$. Then we will apply the row transformation method in such a way that we will some terms become in the form of \[\left( 1-b \right)\] , \[\left( 1-c \right)\] and some terms to be zero. Then by finding the determinant of the matrix and on solving we will get the answer.

Complete step-by-step answer:
Here, we are given that equations $ax+y+z=0$ , $x+by+z=0$ , $x+y+cz=0$ have non-trivial solutions. It means that the determinant is equal to zero. Determinant means we have to convert this equation in matrix form.
Now, we will write the coefficient of all three equations in matrix form i.e. $\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|$ . So, we get as
$\left| \begin{matrix}
   a & 1 & 1 \\
   1 & b & 1 \\
   1 & 1 & c \\
\end{matrix} \right|=0$
We will use a row transformation method. We have to perform operations like addition, subtraction, multiplication, or division to make any of the coefficients equals to zero so it can be easier to solve.
So, here we are performing operation on Row1 i.e. $Row1=Row1-Row2$ .So, be doing this we will get matrix as
$\left| \begin{matrix}
   a-1 & 1-b & 1-1 \\
   1 & b & 1 \\
   1 & 1 & c \\
\end{matrix} \right|=0$
On solving, we get
$\left| \begin{matrix}
   a-1 & 1-b & 0 \\
   1 & b & 1 \\
   1 & 1 & c \\
\end{matrix} \right|=0$
Now, here we are performing operation on Row2 i.e. $Row2=Row2-Row3$ .So, be doing this we will get matrix as
$\left| \begin{matrix}
   a-1 & 1-b & 0 \\
   1-1 & b-1 & 1-c \\
   1 & 1 & c \\
\end{matrix} \right|=0$
On further solving, we get
$\left| \begin{matrix}
   a-1 & 1-b & 0 \\
   0 & b-1 & 1-c \\
   1 & 1 & c \\
\end{matrix} \right|=0$
Now, here we will find determinant of last row which we can find as: taking last row first element 1, we will get as
$\left| \begin{matrix}
   1-b & 0 \\
   b-1 & 1-c \\
\end{matrix} \right|$
So, on multiplying, we get the equation as $1\left[ \left( 1-b \right)\left( 1-c \right)-0 \right]$ .
Similarly, now taking second element of last row: 1, we will get as $\left| \begin{matrix}
   a-1 & 0 \\
   0 & 1-c \\
\end{matrix} \right|$ . So, on solving we get $-1\left[ \left( a-1 \right)\left( 1-c \right)-0 \right]$ . Here, minus sign is taken because there are rules for this i.e. $\left| \begin{matrix}
   + & - & + \\
   - & + & - \\
   + & - & + \\
\end{matrix} \right|$ . According to this, coefficient should be multiplied by appropriate signs. So, we have taken minus signs.
Now, we will take last element of last row: c, we will get as \[\left| \begin{matrix}
   a-1 & 1-b \\
   0 & b-1 \\
\end{matrix} \right|\] . On solving, we get \[c\left[ \left( a-1 \right)\left( b-1 \right)-0 \right]\] .
Now, on adding all the determinant values, we get as
\[\left( 1-b \right)\left( 1-c \right)-1\left( a-1 \right)\left( 1-c \right)+c\left( a-1 \right)\left( b-1 \right)=0\]
Now, as we have to find value of $\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}$ where terms are in form of \[\left( 1-a \right),\left( 1-b \right),\left( 1-c \right)\] so, in the equation we convert the required terms in the required form. So, we will get as
\[\left( 1-b \right)\left( 1-c \right)+\left( 1-a \right)\left( 1-c \right)+c\left( 1-a \right)\left( 1-b \right)=0\]
In the last two terms we have taken minus sign common and sign are changed accordingly.
Now, dividing the whole equation by \[\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)\] . We will get as
\[\dfrac{\left( 1-b \right)\left( 1-c \right)}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)}+\dfrac{\left( 1-a \right)\left( 1-c \right)}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)}+\dfrac{c\left( 1-a \right)\left( 1-b \right)}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)}=0\]
On cancelling the common terms, we will get
\[\dfrac{1}{\left( 1-a \right)}+\dfrac{1}{\left( 1-b \right)}+\dfrac{c}{\left( 1-c \right)}=0\]
We can see that the first two terms are in the proper form but in the last term we want 1 in the numerator so, we will add plus 1 and minus 1 in last term. We will get as
\[\dfrac{1}{\left( 1-a \right)}+\dfrac{1}{\left( 1-b \right)}+\dfrac{1-\left( 1-c \right)}{\left( 1-c \right)}=0\]
On further solving, we get
\[\dfrac{1}{\left( 1-a \right)}+\dfrac{1}{\left( 1-b \right)}+\dfrac{1}{\left( 1-c \right)}-1=0\]
So, we will get as \[\dfrac{1}{\left( 1-a \right)}+\dfrac{1}{\left( 1-b \right)}+\dfrac{1}{\left( 1-c \right)}=1\]
Hence, option (b) is the correct answer.

Note: Remember that we can make any row transformation applying proper operation. Here, we have made changes in row2 and row3, but we can also make changes in row1 and can find the answer. Just we have to try to make some terms in form of zero and \[\left( 1-a \right),\left( 1-b \right),\left( 1-c \right)\] which on solving is easier to find determinant and answer we get correct.