When is Matrix C Not Invertible?

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The discussion centers on the conditions under which the matrix C is not invertible, specifically when its determinant equals zero. The determinant of matrix C, defined as c = [[2 - x, 5, 1], [-3, 0, x], [-2, 1, 2]], is calculated to be x^2 - 12x + 27. The values of x that render C non-invertible are confirmed to be x = 3 and x = 9. Additionally, it is established that for a scalar multiplication of a matrix, the determinant of 2C is computed as 8 times the determinant of C, given that C is a 3x3 matrix.

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<br /> c = \left[ {\begin{array}{*{20}c}<br /> {2 - x} &amp; 5 &amp; 1 \\<br /> { - 3} &amp; 0 &amp; x \\<br /> { - 2} &amp; 1 &amp; 2 \\<br /> \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?
 
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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).
 
Last edited:
Hmm, yeh, what you're saying would make more sense.
Thanks for confirming the other answers.
 
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).
 

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