Matrix/Determinant/Inverse Q's

[dcl]

c = \left[ {\begin{array}{*{20}c}
{2 - x} & 5 & 1 \\
{ - 3} & 0 & x \\
{ - 2} & 1 & 2 \\
\end{array}} \right]

a) Calculate det(C).
My answer was x^2 - 12x + 27.

b) Calculate det(2C).
Umm, would this just be 2*det(C)?
Couldn't find anything more helpful in my notes.

c) State the values for 'x' for which C is not invertible.
I believe the value for 'x' that would make this non invertable would be the solution that det(c) = 0. (A matrix has no inverse when the determinant = 0 yeh?)
which would be x = 9 or 3
is this correct?

[Janitor]

I get the same answer as you on (a) and (c). I'm thinking that det(aC)=a^N det(c) where a is a constant and N is the "dimension" (wrong terminology? NxN matrix). If I am right about that, then for a 3x3 matrix, det(2C) = 8 det(C).

You may be thinking about the trace of a matrix: Tr(2C) = 2 Tr(C).

[dcl]

Hmm, yeh, what you're saying would make more sense.
Thanks for confirming the other answers.

[Icarus]

Janitor is correct about (b). To see it, just consider what happens if C is the identity: 2I has 2 along the diagonal and 0 everywhere else, so the determinant is det(2I) = 8 = 23det(I).