Suppose we have the following matrix: $$A= \begin{bmatrix} x+\dfrac12&8&-\dfrac{27}2\\ -\dfrac12&x+10& -\dfrac{27}{2}\\ -\dfrac12&8&x-\dfrac{21}2 \end{bmatrix}$$
My goal is to find the determinant of matrix A. One can see the matrix A as: $$ xI-B=A $$ So finding the determinant of A is the same as finding the characterist polynomial of B. I know the result is: $$det(A)=x^3 $$
Which can be determined. However, the calculation of such determinant takes a long time.
My question is if there is any way to simplify the calculation and save some time.
Thank you in advance.
If you add the second row of $A$ to the first one and if you subtract the second line from the third one, the determinant stays the same and the matrix becomes$$\begin{pmatrix}x&x+18&-27\\-\frac12&x+10&-\frac{27}2\\0&-x-2&x+3\end{pmatrix}.$$Now, if you add to the first line the second one times $2x$, again the determinant stays the same and you get$$\begin{pmatrix}0&2x^2+21x+18&-27x-27\\-\frac12&x+10&-\frac{27}2\\0&-x-2&x+3\end{pmatrix}.$$Now, thanks to thos two $0$'s, it's quite easy to see that the determinant is $x^3$.
You can also check that $B^3$ is the null matrix. Therefore, its characteristic polynomial is $x^3$.