# Determine the value(s) of x such that [[x,2,1]][[-2,-1,2],[-1,0,2],[2,7,1]][[x],[-1],[2]] = 0

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### 2 Answers

Our given system is

`[[x,2,1]].[[-2,-1,2],[-1,0,2],[2,7,1]].[[x],[-1],[2]]=0`

Multiplying the first two matrices we get

`[[-2x-2+2,-x+7,2x+4+1]].[[x],[-1],[2]]=0`

or, `[[-2x,-x+7,2x+5]][[x],[-1],[2]]=0`

Multiplying the above matrices we get

`[-2x^2+x-7+4x+10]=0`

or, `[-2x^2+5x+3]=[0]`

as the element 0 can be written in matrix form `[0].`

Comparing the terms we get

`-2x^2+5x+3=0`

Which is a quadratic equation.

Solving the above equation we get

`x=-1/2` and `x=3.` Answer.

[[x,2,1]][[-2,-1,2],[-1,0,2],[2,7,1]][[x],[-1],[2]] = 0

`[[x,2,1]][[-2,-1,2],[-1,0,2],[2,7,1]][[x],[-1],[2]]=[0]`

`` `[[-2x-2+2,-x+2.0+7,2x+4+1]][[x],[-1],[2]]=[0]`

`[[-2x,-x+7,2x+5]][[x],[-1],[2]]=[0]`

`[-2x^2+x-7+4x+10]=[0]`

comparing element in both side of matrix ,we have

`-2x^2+5x+3=0`

`` `2x^2-6x+x-3=0`

`2x(x-3)+1(x-3)=0`

`(2x+1)(x-3)=0`

either 2x+1=0

or x-3=0

so x=-1/2 or x=3