## Characteristic Equation of Matrix

In this page characteristic equation of matrix we are going to see how to find characteristic equation of any matrix with detailed example.

Definition :

Let A be any square matrix of order n x n and I be a unit matrix of same order. Then |A-λI| is called characteristic polynomial of matrix.

Then the equation |A-λI| = 0 is called characteristic roots of matrix.  The roots of this equation is called characteristic roots of matrix.

Another name of characteristic roots:

characteristic roots are also known as latent roots or eigenvalues of a matrix.

Example :

Determine the characteristic roots of the matrix

 0 1 2 1 0 -1 2 -1 0

Solution :

Let A =

 0 1 2 1 0 -1 2 -1 0

The order of A is 3 x 3. So the unit matrix I =

 1 0 0 0 1 0 0 0 1

Now we have to multiply λ with unit matrix I.

λI =

 λ 0 0 0 λ 0 0 0 λ

A-λI=

 0 1 2 1 0 -1 2 -1 0

-

 λ 0 0 0 λ 0 0 0 λ

=

 (0-λ) (1-0) (2-0) (1-0) (0-λ) (-1-0) (2-0) (-1-0) (0-λ)

=

 -λ 1 2 1 -λ -1 2 -1 -λ

A-λI=

 -λ 1 2 1 -λ -1 2 -1 -λ

=  -λ( λ ² - 1) - 1 (-λ - (-2) ) + 2 (-1 - (-2 λ) )

=  -λ( λ ² - 1) - 1 (-λ + 2) ) + 2 (-1 +2 λ)

=  -λ³ + λ  + λ - 2 - 2 + 4 λ

=  -λ³ + 2λ  - 2 - 2 + 4 λ

=  -λ³ + 6λ - 4

To find roots let |A-λI| = 0

-λ³ + 6λ - 4 = 0

λ³ - 6λ + 4 = 0

For solving this equation first let us do synthetic division.  characteristic equation of matrix

By using synthetic division we have found one value of λ that is λ = 2.

Now we have to solve λ² + 2 λ - 2 to get another two values. For that we have to use quadratic formula (-b ± √b² -4ac)/2a

a = 1 b = 2 and c = -2

x = [-2 ± √2²-4(1)(-2)]/2(1)

x = [-2 ± √4+8]/2(1)

x = [-2 ± √12]/2

x = [-2 ± 2√3]/2

x = 2[-1 ± √3]/2

x = -1 ± √3

Therefore the characteristic roots are x = 1, -1 ± √3.

Characteristic Equation of Matrix to Rank method