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If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are non-zero real numbers and if the equations $(a-1) x=y+z,(b-1) y=z+x,(c-1) z=x+y$ have a non-trivial solution, then $a b+b c+c a=$
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Verified Answer
The correct answer is:
$a b c$
We have system of equation,
$$
(a-1) x-y-z=0 ; x-(b-1) y+z=0
$$
and $x+y-(c-1) z=0$
It is a homogeneous system of equations.
Now, for non-trivial solution.
$$
\begin{aligned}
& \left|\begin{array}{ccc}
a-1 & -1 & -1 \\
1 & -(b-1) & 1 \\
1 & 1 & -(c-1)
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
a-1 & -1 & -1 \\
1 & 1-b & 1 \\
1 & 1 & 1-c
\end{array}\right|=0 \\
& (a-1)[(1-b)(1-c)-1]+1[1-c-1] \\
& \quad-[1-(1-b)]=0 \\
& (a-1)(1-b)(1-c)-a+1-c-b=0 \\
& (a-1)(1-c-b+b c)-a-b-c+1=0 \\
& a-a c-a b+a b c-1+c+b-b c-a-b-c+1=0 \\
& \therefore \quad a b+b c+c a=a b c
\end{aligned}
$$
$$
(a-1) x-y-z=0 ; x-(b-1) y+z=0
$$
and $x+y-(c-1) z=0$
It is a homogeneous system of equations.
Now, for non-trivial solution.
$$
\begin{aligned}
& \left|\begin{array}{ccc}
a-1 & -1 & -1 \\
1 & -(b-1) & 1 \\
1 & 1 & -(c-1)
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
a-1 & -1 & -1 \\
1 & 1-b & 1 \\
1 & 1 & 1-c
\end{array}\right|=0 \\
& (a-1)[(1-b)(1-c)-1]+1[1-c-1] \\
& \quad-[1-(1-b)]=0 \\
& (a-1)(1-b)(1-c)-a+1-c-b=0 \\
& (a-1)(1-c-b+b c)-a-b-c+1=0 \\
& a-a c-a b+a b c-1+c+b-b c-a-b-c+1=0 \\
& \therefore \quad a b+b c+c a=a b c
\end{aligned}
$$
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