Question

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Answer 1

1)  $R_{1}$->$R_{1}$- $R_{2}$   , $R_{2}$->$R_{2}$-$R_{3}$

=>I x      -x         0 I

   I 0       x        -x I    =  0

   I a       b      x+c I

=>$C_{1}$->$C_{1}$+$C_{2}$ , $C_{2}$->$C_{2}$+$C_{3}$

=>I 0         -x               0 I

   I x          0               -x I    =  0

   I a+b      x+b+c     x+c I

=>  -(-x) ($x^{2}$+xc+ax+bx)=0......1stequation

So by this equation.......

=>   (x=0) ,    $x^{2}$+x(a+b+c)=0

                     x(x+a+b+c)=0 .........2th equation

                     

We already have x = 0

So from 2nd equation we get.....

=>x= -(a+b+c)

Ans is ......x= -(a+b+c) and x = 0