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I should find the values x for which the matrix is non-singular.

Given matrix is

I already know that the matrix is singular when det(A) is 0. So basically the matrix will be non-singular when the determinant is something else than 0. I know how to find a determinant but I guess this should be solved "other way around"?

It's a homework so hint would be appreciated.

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  • $\begingroup$ Reduce it to reduced row echelon form and check when the leading term is $0$. $\endgroup$ – A---B Oct 4 '17 at 19:39
  • $\begingroup$ You can calculate $\det A$ outright. But you can do row operations to $A$ to perhaps make it easier to work with. $\endgroup$ – Doug M Oct 4 '17 at 19:45
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Hint: Calculate the determinant of this matrix. You get an expression that depends on $x$. When is this expression zero?

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  • $\begingroup$ OH! Now I get the quadratic equation! So does it mean that the solution of quadratic equation is the solution for x as well? $\endgroup$ – MiMaKo Oct 4 '17 at 19:50
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    $\begingroup$ Yes, the determinant is zero if, and only if, the quadratic equation you get is zero. Therefore the matrix is singular if $x$ is a solution to that equation and non-singular otherwise. $\endgroup$ – Epiousios Oct 4 '17 at 20:16

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