Solving Matrix A: Characteristic Equation and Eigenvectors

[wefweff]

good evening everyone!
Decided to solve the problems from last year's exams. I came across this example. Honestly, I didn't understand it. Who can help a young student? :)
Find characteristic equation of the matrix A in the form of the polynomial of degree of 3 (you do not need to find eigenvalues) and associated eigenvectors of the matrix. Eigenvalues of the matrix: -2, -2, 1.
А= 7 0 -3
-9 -2 3
18 0 -8

 

[HOI]

$\lambda$ is an eigenvalue of matrix A if there exist some non-zero vector, v, such that $Av= \lambda v$. That is the same as $Av-\lambda v= 0$ or $(A- \lambda I)v= 0$. v= 0 is obviously a solution. In order that there be another solution $A- \lambda$ must not have an inverse. That requires that the determinant or $A- \lambda$ be 0.

Here $A= \begin{bmatrix}7 & 0 & -3 \\ -9 & -2 & 3 \\ 10 & 0 & -8 \end{bmatrix}$ so $A- \lambda I=\begin{bmatrix}7- \lambda & 0 & 3 \\ -9 & -2- \lambda & 3 \\ 18 & 0 & -8- \lambda \end{bmatrix}$.

The determinant is $\left|\begin{array}{ccc}7-\lambda & 0 & -3 \\ -9 & -2-\lambda & 3 \\ 18 & 0 & -8-\lambda \end{array}\right|$ so the characteristic equation is $|A- \lambda I|=\left|\begin{array}{ccc}7-\lambda & 0 & -3 \\ -9 & -2-\lambda & 3 \\ 18 & 0 & -8-\lambda \end{array}\right|= 0$. Since this is a 3 by 3 matrix, that will be a cubic equation.

 

[HOI]

To calculate $\left|\begin{array}{ccc} 7- \lambda & 0 & -3 \\ -9 & -2- \lambda & 3 \\ 18 & 0 & -8- \lambda \end{array}\right|$ expand on the middle column: $(-2- \lambda)\left|\begin{array}{cc} 7- \lambda & -3 \\ 18 & -8- \lambda \end{array}\right|= (-2- \lambda)((7- \lambda)(-8- \lambda)+ 54)= (-2- \lambda)(-56+ \lambda+ \lambda^2+ 54)= (-2- \lambda)(\lambda^2+ \lambda- 2)= (-2- \lambda)(-2- \lambda)(1- \lambda)$.