# Verify that a matrix is a solution of its own characteristic polynomial??

I have a $2\times 2$ matrix $A=\begin{pmatrix} 17 & 19\\ 9 & -7\end{pmatrix}$ and I need to verify that it is a solution of its own characteristic polynomial. How do I do that? I know to get the polynomial equation is $\det(A-\lambda I)=0$ but I tried factoring out the polynomial equation but $\lambda$ is not a real number? The equation I got is $\lambda^2 -10\lambda +38=0$?

• A matrix isn't a real number either, so that shouldn't be a problem. But I don't think your characteristic polynomial is quite right either. – Sam Cassidy Apr 6 '18 at 13:45

## 2 Answers

You have $A=\left(\begin{smallmatrix}17&19\\9&-7\end{smallmatrix}\right)$ and its characteristic polynomial is $x^2-10x-290$. On the other hand $A^2=\left(\begin{smallmatrix}460 & 190 \\90 & 220\end{smallmatrix}\right)$ and $10A=\left(\begin{smallmatrix}170&190\\90&-70\end{smallmatrix}\right)$. So…

The characteristic polynomial of matrix $A$ is defined as $p(\lambda)=det(A-\lambda I)$, therefore to verify whether $A$ is a solution of the charcteristic polynomial just substitute $A$ directly into the polynomial to see if the result is a zero matrix that is, $p(A) = A^2 - 10A + 32 I_2 = 0$