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)
 
 
  =
 
  1   2
1     -1
2   -1  
 
 
A-λI=
 
  1   2
1     -1
2   -1  
 

  =  -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
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