If the matrix $$1$$−$$1$$ $$x1$$ x $$1x$$−$$1$$ $$1$$ has no inverse then the number of real values of x is

If $$1-1x1x1x-11$$ has no inverse, then $$1-1x1x1x-11=0$$ ⇒$$1x+1+11-x+x-1-x2=0$$ ⇒$$x3+x-2=0$$ ⇒$$x-1x2+x+2=0$$ ⇒$$x=1$$ is the only real root of the equation. ∴number of real roots is $$1$$.

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If the matrix $[\begin{matrix} 1 & - 1 & x \\ 1 & x & 1 \\ x & - 1 & 1 \end{matrix}]$ has no inverse then the number of real values of x is

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$I f [\begin{matrix} 1 & - 1 & x \\ 1 & x & 1 \\ x & - 1 & 1 \end{matrix}] h a s n o i n v e r s e, t h e n |\begin{matrix} 1 & - 1 & x \\ 1 & x & 1 \\ x & - 1 & 1 \end{matrix}| = 0 \Rightarrow 1 (x + 1) + 1 (1 - x) + x (- 1 - x^{2}) = 0 \Rightarrow x^{3} + x - 2 = 0 \Rightarrow (x - 1) (x^{2} + x + 2) = 0 \Rightarrow x = 1 i s t h e o n l y r e a l r o o t o f t h e e q u a t i o n . ∴ n u m b e r o f r e a l r o o t s i s 1 .$

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Applications of determinant

Remember concepts with our Masterclasses.

If the matrix 1−1 x1 x 1x−1 1 has no inverse then the number of real values of x is

If 1-1x1x1x-11 has no inverse, then 1-1x1x1x-11=0 ⇒1x+1+11-x+x-1-x2=0 ⇒x3+x-2=0 ⇒x-1x2+x+2=0 ⇒x=1 is the only real root of the equation. ∴number of real roots is 1.

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If the matrix $[\begin{matrix} 1 & - 1 & x \\ 1 & x & 1 \\ x & - 1 & 1 \end{matrix}]$ has no inverse then the number of real values of x is