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When looking for the eigenvalue of A, doing the determinant I get terms in cube which I don't know how to factorise. I know how to do them when I can expand my DET along a row/column with 2 zeroes so that I only have a cube to factorise but here I am stuck... How would you proceed?

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    see method in http://math.stackexchange.com/questions/2103936/minimal-polynomial-of-a-left-beginsmallmatrix-7-2-1-2-10-2/2103943#2103943 JUST IN CASE: do you know the determinant of $A?$ Do you know the determinant of $B?$ – 2017-01-21
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    Could I simplify the matrix B before doing the detrminant of B (Row operations to get zeroes where I want to expand my det. ? Can I do that or will it change my matrix? – 2017-01-21

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For example with $\;A\;$ :

$$\det(tI-A)=\begin{vmatrix} t-2&0&-8\\ -1&t-4&4\\ -1&-2&t+3\end{vmatrix}=(t-2)(t^2-t-12)-16+8(t-2)-8(t-4)=$$$${}$$

$$=t^3-3t^2-10t+24=(t+3)(t^2-6t+8)=(t+3)(t-2)(t-4)$$

and thus you have three distinct eigenvalues for a $\;3\times3\;$ matrix, so it is...

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    so it is diagonalizable. Thanks! I really have a hard time factoring the characteristic polynomial of degree 3... Do you have a special technique for that? – 2017-01-21
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    @SoHCahToha I tried to show it in the process of evaluating that determinant. You first look for rational roots, which must be integer ones as the characteristic polynomial is monic. Thus, integer roots, if there are any at all, must be divisors of the free coefficient, which is $\;\pm\;$ the matrix's determinant...etc. – 2017-01-21