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AP EAMCET Engineering 21 Sep 2020 Shift 2 Paper

Section: Mathematics

Question : 1 of 160

Marks: +1, -0

\begin{bmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{bmatrix}

x

Solution:

(d) It is given that matrix

A = \begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}

have no inverse,

|A| = 0\; \Rightarrow \begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix} = 0

\Rightarrow \;\; 1(x+1)+1(1-x)+x(-1-x^{2})=0

\Rightarrow \;\; x+1+1-x-x-x^{3}=0

\Rightarrow \;\; x^{3}+x-2=0 \Rightarrow (x-1)(x^{2}+x+2)=0

Either

x=1

x^{2}+x+2=0

But discriminant of quadratic equation

x^{2}+x+2=0

is negative so no real roots.

\therefore |A| = 0 \Rightarrow x=1

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