Compute the characteristic equation 3x3 matrix

The easy and quick way to compute the characteristic equation of 3x3 matrix is to use the formulae $$x^3-tr(A)x^2+(A_{11}+A_{22}+A_{33})x-det(A)=0$$ For given matrix $$tr(A)=4, A_{11}(cofa_{11})=3, A_{22}(cofa_{22})=1, A_{33}(cofa_{33})=1, det(A)=2$$ so the char equation will be $x^3-4x^2+5x-2=0$

Quick way: Note that sum of each column of $A$ is $2$, hence $\lambda_1=2$ is an eigenvalue of $A$.Let $\lambda_1$ and $\lambda_2$ be other eigenvalues,using trace and determinant $\lambda_1 +\lambda_2=2$ and $\lambda_1 \lambda_2=1$,hence $\lambda_1=\lambda_2=1$ Hence Characteristic polynomial $p(x)$ is $$p(x)=(x-1)^2(x-2)$$

The characteristic equation is used to find the eigenvalues of a square matrix A.

First: Know that an eigenvector of some square matrix A is a non-zero vector x such that Ax = λx.

Second: Through standard mathematical operations we can go from this: Ax = λx, to this: (A - λI)x = 0

The solutions to the equation det(A - λI) = 0 will yield your eigenvalues. The previously mentioned equation is the characteristic equation.

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Given some square matrix A, the characteristic equation is det(A - λI) = 0.

Example:

$$ A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$$$ λI = \begin{bmatrix} λ & 0 \\ 0 & λ \\ \end{bmatrix}$$

$$ A - λI = \begin{bmatrix} a-λ & b \\ c & d-λ \\ \end{bmatrix}$$

$$det\begin{bmatrix} a-λ & b \\ c & d-λ \\ \end{bmatrix} = 0$$ Characteristic Equation: $$(a-λ)(d-λ) - (c)(b) = 0$$

Source Cited: Lay, David C., et al. Linear Algebra and Its Applications, Fifth Edition. Pearson, 2016.