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Problem: diagonal [[8,-6,2],[-6,7,-4],[2,-4,3]] [ Calculator, Method and examples ]

Solution:
Your problem -> diagonal [[8,-6,2],[-6,7,-4],[2,-4,3]]

A can be diagonalized if there exists an invertible matrix P and diagonal matrix D such that A=PDP^-1

Here A =
 8 -6 2 -6 7 -4 2 -4 3

Find eigenvalues of the matrix A

|A-lamdaI|=0

 (8-lamda) -6 2 -6 (7-lamda) -4 2 -4 (3-lamda)
= 0

:.(8-lamda)((7-lamda) × (3-lamda) - (-4) × (-4))-(-6)((-6) × (3-lamda) - (-4) × 2)+2((-6) × (-4) - (7-lamda) × 2)=0

:.(8-lamda)((21-10lamda+lamda^2)-16)+6((-18+6lamda)-(-8))+2(24-(14-2lamda))=0

:.(8-lamda)(5-10lamda+lamda^2)+6(-10+6lamda)+2(10+2lamda)=0

:. (40-85lamda+18lamda^2-lamda^3)+(-60+36lamda)+(20+4lamda)=0

:.(-lamda^3+18lamda^2-45lamda)=0

:.-lamda(lamda-3)(lamda-15)=0

:.lamda=0 or(lamda-3)=0 or(lamda-15)=0

:. The eigenvalues of the matrix A are given by lamda=0,3,15,

1. Eigenvectors for lamda=0

1. Eigenvectors for lamda=0

A-lamdaI =
 8 -6 2 -6 7 -4 2 -4 3
- 0
 1 0 0 0 1 0 0 0 1

=
 8 -6 2 -6 7 -4 2 -4 3

Now, reduce this matrix
R_1 larr R_1-:8

=
 1 -3/4 1/4 -6 7 -4 2 -4 3

R_2 larr R_2+6xx R_1

=
 1 -3/4 1/4 0 5/2 -5/2 2 -4 3

R_3 larr R_3-2xx R_1

=
 1 -3/4 1/4 0 5/2 -5/2 0 -5/2 5/2

R_2 larr R_2-:5/2

=
 1 -3/4 1/4 0 1 -1 0 -5/2 5/2

R_1 larr R_1+(3/4)xx R_2

=
 1 0 -1/2 0 1 -1 0 -5/2 5/2

R_3 larr R_3+(5/2)xx R_2

=
 1 0 -1/2 0 1 -1 0 0 0

The system associated with the eigenvalue lamda=0

(A-0I)
 x_1 x_2 x_3
=
 1 0 -1/2 0 1 -1 0 0 0

 x_1 x_2 x_3
=
 0 0 0

=>x_1-(1/2)x_3=0,x_2-x_3=0

=>x_1=(1/2)x_3,x_2=x_3

:. eigenvectors corresponding to the eigenvalue lamda=0 is

v=
 (1/2)x_3 x_3 x_3

Let x_3=1

v_1=
 1/2 1 1
v_1=
 1/2 1 1

2. Eigenvectors for lamda=3

2. Eigenvectors for lamda=3

A-lamdaI =
 8 -6 2 -6 7 -4 2 -4 3
- 3
 1 0 0 0 1 0 0 0 1

=
 8 -6 2 -6 7 -4 2 -4 3
-
 3 0 0 0 3 0 0 0 3

=
 5 -6 2 -6 4 -4 2 -4 0

Now, reduce this matrix
interchanging rows R_1 harr R_2

=
 -6 4 -4 5 -6 2 2 -4 0

R_1 larr R_1-:-6

=
 1 -2/3 2/3 5 -6 2 2 -4 0

R_2 larr R_2-5xx R_1

=
 1 -2/3 2/3 0 -8/3 -4/3 2 -4 0

R_3 larr R_3-2xx R_1

=
 1 -2/3 2/3 0 -8/3 -4/3 0 -8/3 -4/3

R_2 larr R_2-:-8/3

=
 1 -2/3 2/3 0 1 1/2 0 -8/3 -4/3

R_1 larr R_1+(2/3)xx R_2

=
 1 0 1 0 1 1/2 0 -8/3 -4/3

R_3 larr R_3+(8/3)xx R_2

=
 1 0 1 0 1 1/2 0 0 0

The system associated with the eigenvalue lamda=3

(A-3I)
 x_1 x_2 x_3
=
 1 0 1 0 1 1/2 0 0 0

 x_1 x_2 x_3
=
 0 0 0

=>x_1+x_3=0,x_2+(1/2)x_3=0

=>x_1=-x_3,x_2=-(1/2)x_3

:. eigenvectors corresponding to the eigenvalue lamda=3 is

v=
 -x_3 -(1/2)x_3 x_3

Let x_3=1

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