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AP EAPCET 25-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 15 of 160

Marks: +1, -0

\begin{vmatrix} x & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & x & 0 & 0 \\ 2 & 0 & x-1 & 0 \end{vmatrix}

-\begin{vmatrix} 0 & x & 0 \\ 0 & 0 & x-1 \\ 2 & 2 & 0 \end{vmatrix}=0

Solution: 👈: Video Solution

\begin{vmatrix} x & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & x & 0 & 0 \\ 2 & 0 & x-1 & 0 \end{vmatrix}

-\begin{vmatrix} 0 & x & 0 \\ 0 & 0 & x-1 \\ 2 & 2 & 0 \end{vmatrix}=0

On expanding first determinant along

R_{2}

and second determinant along

R_{1}

1[x^{2}(x-1)]-(-x)[-2(x-1)]=0

x^{3}-x^{2}+x(-2x+2)=0

x^{3}-x^{2}-2x^{2}+2x=0

x^{3}-3x^{2}+2x=0

x(x^{2}-3x+2)=0

x(x-1)(x-2)=0

⇒

x=0,1

and 2

∴ Sum of the roots

=0+1+2=3

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