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.

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