calculating the characteristic polynomial I have the following matrix: $$A=\begin{pmatrix}
    -9 & 7 & 4 \\
    -9 & 7 & 5\\
    -8 & 6 & 2
\end{pmatrix}$$
And I need to find the characteristic polynomial so I use det(xI-A) which is $$\begin{vmatrix}
    x+9 & -7 & -4 \\
    9 & x-7 & -5\\
    8 & -6 & x-2
\end{vmatrix}$$
Is there a way to calculate the determinate faster or is way is:
$$(x+9)\cdot\begin{vmatrix}
    x-7 & -5  \\
    -6 & x-2 \\
\end{vmatrix}+7\cdot\begin{vmatrix}
    9 & -5  \\
    8 & x-2 \\
\end{vmatrix} -4\begin{vmatrix}
    9 & x-7  \\
    8 & -6 \\
\end{vmatrix}=$$
$$=(x+9)[(x-7)(x-2)-30]+7[9x-18+40]-4[54-8x+56]=(x+9)[x^2-9x-16]+7[9x+22]-4[-8x+2]=x^3-2x+2$$

I am not sure if this method is "faster", but it does seem to involve less numbers in the actual calculation: 
We know from a different theorem that two similar matrices share the same characteristic polynomial, see this post Elegant proofs that similar matrices have the same characteristic polynomial?. 
From this fact, we can put $A$ in row echelon form. That is $A = \begin{bmatrix}
    {-9} & {7} & {4} \\
    {-9} & {7} & {5} \\
    {-8} & {6} & {2}
\end{bmatrix}$ and for some invertable matrix $E, F = E*A*E^{-1} = \begin{bmatrix}
    {1} & {-7/9} & {-4/9} \\
    {0} & {-2/9} & {-14/9} \\
    {0} & {0} & {1}
\end{bmatrix}$, that is $A$ is similar to $F$. From the previous theorem, it is clear that $det(F-I*x)=det(A-I*x)=(1-x)^2(\frac{-2}{9}-x)$.  
But what you did is correct. 
