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If a, b, c are non - zero real numbers and if the system of equations

$(a-1)x=y+z,$

$(b-1)y=z+x,$

$(c-1)z=x+y,$

has a non-trivial solution, then ab+bc+ca equals :

Option 1)
a+b+c

Option 2)
abc

Option 3)
1

Option 4)
-1

Answers (1)

As we learnt in

Cramer's rule for solving system of linear equations -

When $\Delta =0$ and $\Delta _{1}=\Delta _{2}=\Delta _{3}=0$ ,

then the system of equations has infinite solutions.

- wherein

$a_{1}x+b_{1}y+c_{1}z=d_{1}$

$a_{2}x+b_{2}y+c_{2}z=d_{2}$

$a_{3}x+b_{3}y+c_{3}z=d_{3}$

and

$\Delta =\begin{vmatrix} a_{1} &b_{1} &c_{1} \\ a_{2} & b_{2} &c_{2} \\ a_{3}&b _{3} & c_{3} \end{vmatrix}$

$\Delta _{1},\Delta _{2},\Delta _{3}$ are obtained by replacing column 1,2,3 of $\Delta$ by $\left ( d_{1},d_{2},d_{3} \right )$ column

$\begin{vmatrix} a-1 &-1 &-1 \\ -1 & b-1 &-1 \\ -1&-1 &c-1 \end{vmatrix} = 0$

$\Rightarrow \left ( a-1 \right )\left [ 1+bc-b-c-1 \right ]+1\left [ -c+1-1 \right ]-1\left ( 1+b-1 \right )=0$

$\Rightarrow \left ( a-1 \right )\left [ bc-b-c \right ]-c-b=0$

$\Rightarrow abc-ab-bc-ac=0$

$\Rightarrow ab+bc+ac=abc$

Option 1)

a+b+c

This option is incorrect.

Option 2)

abc

This option is correct.

Option 3)

1

This option is incorrect.

Option 4)

-1

This option is incorrect.

Posted by

Sabhrant Ambastha

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