Question 1032073
Why is the matrix {{{(matrix(2,2,x-1,-1,-3,x-5))}}} 
not invertible for all real numbers x?
<pre>
A square matrix whose determinant is 0 is not invertible

{{{abs(matrix(2,2,x-1,-1,-3,x-5))}}}{{{""=""}}}{{{(x-1)(x-5)-(-1)(-3)}}}{{{""=""}}}{{{x^2-6x+5-3}}}{{{""=""}}}{{{x^2-6x+2}}}

If {{{x^2-6x+2}}}{{{""=""}}}{{{0}}} then the matrix is not invertible.

{{{x}}}{{{""=""}}}{{{(-(-6) +- sqrt((-6)^2-4(1)(2) ))/(2(1)) }}}

{{{x}}}{{{""=""}}}{{{(6 +- sqrt(36-8))/2 }}}

{{{x}}}{{{""=""}}}{{{(6 +- sqrt(28))/2 }}}

{{{x}}}{{{""=""}}}{{{(6 +- sqrt(4*7))/2 }}}

{{{x}}}{{{""=""}}}{{{(6 +- 2sqrt(7))/2 }}}

{{{x}}}{{{""=""}}}{{{6/2 +- 2sqrt(7)/2 }}}

{{{x}}}{{{""=""}}}{{{3 +- sqrt(7) }}}

The matrix is not invertible if {{{x}}}{{{""=""}}}{{{3 + sqrt(7) }}} or {{{x}}}{{{""=""}}}{{{3 - sqrt(7) }}}.

Edwin</pre>