calculating the characteristic polynomial

Question

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:

$$(x+9)\cdot\begin{vmatrix} x-7 & -5 \\ -6 & x-2 \\ \end{vmatrix}+7\cdot\begin{vmatrix} 9 & -5 \\ 8 & x-2 \\ \end{vmatrix} -4\begin{vmatrix} 9 & x-7 \\ 8 & -6 \\ \end{vmatrix}=$$

$$=(x+9)[(x-7)(x-2)-30]+7[9x-18+40]-4[54-8x+56]=(x+9)[x^2-9x-16]+7[9x+22]-4[-8x+2]=x^3-2x+2$$

You could use that since it's going to be a 3rd degree polynomial, you only need to calculate it for 4 different values of x and then use Lagrange interpolation. You can then choose 4 values for which the determinant is easy to calculate. — Count Iblis, Aug 27 '15 at 17:52
This is the correct way. The last step should be $x^3-2x+2$ though. — KittyL, Aug 27 '15 at 17:52

score 2 · Accepted Answer · answered Aug 27 '15 at 22:27

You could use $\displaystyle\begin{vmatrix}x+9&-7&-4\\9&x-7&-5\\8&-6&x-2\end{vmatrix}=\begin{vmatrix}x+2&-7&-4\\x+2&x-7&-5\\2&-6&x-2\end{vmatrix}$ $\;\;\;$(adding C2 to C1)

$\displaystyle\hspace{2.6 in}=\begin{vmatrix}0&-x&1\\x+2&x-7&-5\\2&-6&x-2\end{vmatrix}$$\;\;\;$(subtracting R2 from R1)

$\displaystyle\hspace{2.6 in}=\begin{vmatrix}0&0&1\\x+2&-4x-7&-5\\2&x^2-2x-6&x-2\end{vmatrix}$$\;\;$(adding x(C3) to C2))

$\displaystyle\hspace{2.6 in}=\begin{vmatrix}x+2&-4x-7\\2&x^2-2x-6\end{vmatrix}=x^3-2x+2$

score 0 · Answer 2 · edited Apr 13 '17 at 12:20

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.

And note that $E$ is the product of elementary matrices, which is invert-able. — 9301293, Aug 27 '15 at 19:11

calculating the characteristic polynomial

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