# From the characteristic matrix to solutions for the characteristic equation (Polynomial)

There's something about the determinant which I don't get, so if someone could explain step-by-step how this is done, it would be much appreciated:

It's about finding eigenvalues via. the characteristic polynomial

• Example $$K_A(\lambda)=A-\lambda E=\left( \begin{matrix}4 && -2 \\ 3 && -1\end{matrix} \right)- \left(\begin{matrix}\lambda && 0 \\ 0 && \lambda \end{matrix} \right)=\left( \begin{matrix} 4-\lambda && -2 \\ 3 && -1-\lambda \end{matrix} \right)$$

We then calculate: $K_A(\lambda)=det\big[\left( \begin{matrix}4-\lambda && -2 \\ 3 && -1-\lambda\end{matrix} \right) \big]$

I don't get how you deduce that $det\big[\left( \begin{matrix}4-\lambda && -2 \\ 3 && -1-\lambda\end{matrix} \right) \big]=(4-\lambda)\cdot(-1-\lambda)-(-2)\cdot 3=\lambda^2-3\lambda+2$

$det\big[\left( \begin{matrix}4-\lambda && -2 \\ 3 && -1-\lambda\end{matrix} \right) \big]=(4-\lambda)\color {red}-(-1-\lambda)-(-2)\cdot 3=\lambda^2-3\lambda+2$
$det\big[\left( \begin{matrix}4-\lambda && -2 \\ 3 && -1-\lambda\end{matrix} \right) \big]=(4-\lambda)\cdot(-1-\lambda)-(-2)\cdot 3=\lambda^2-3\lambda+2$
• Great! I totally forgot about the $$det\big[ \begin{matrix} a && b \\ c && d \end{matrix} \big] = ad - bc$$ Easy to see that's what happening here. Thank you! Nov 15, 2017 at 9:44