Question

Is matrix A consistent, motivate your answer in terms of eigenvalues. A = 2 1 0 0 2 0 2 3 1

Answer 1

If $A$ is an $n × n$ matrix, then the sum of the $n$ eigenvalues of $A$ is the trace of $A$ and the product of the $n$ eigenvalues is the determinant of $A.$

$$A=\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0\\ 2 &3 & 1 \end{pmatrix}$$

$$A-\lambda I=\begin{pmatrix} 2-\lambda & 1 & 0 \\ 0 & 2-\lambda & 0\\ 2 &3 & 1-\lambda \end{pmatrix}$$

$$\det (A-\lambda I)=\begin{vmatrix} 2-\lambda & 1 & 0 \\ 0 & 2-\lambda & 0\\ 2 &3 & 1-\lambda \end{vmatrix}$$

$$=(2-\lambda)\begin{vmatrix} 2-\lambda & 0 \\ 3 & 1-\lambda \end{vmatrix}-1\begin{vmatrix} 0 & 0 \\ 2 & 1-\lambda \end{vmatrix}+0\begin{vmatrix} 0 & 2-\lambda \\ 2 & 3 \end{vmatrix}$$

$$=(2-\lambda)(2-\lambda)(1-\lambda)$$

Characteristic equation

$$\det (A-\lambda I)=0$$

$$(2-\lambda)(2-\lambda)(1-\lambda)=0$$

$$\lambda_1=1, \lambda_2=2, \lambda_3=2$$

These are the eigenvalues.

Hence

$$\det A=\lambda_1\lambda_2\lambda_3=1(2)(2)=4\not=0$$

=> matrix $A$ consistent.