Diagonalization of Matrix 2
In this page diagonalization of matrix 2 we are going to see how to diagonalize a matrix.
Definition :
A square matrix of order n is diagonalizable if it is having linearly independent eigen values.
We can say that the given matrix is diagonalizable if it is alike to the diagonal matrix. Then there exists a non singular matrix P such that P⁻¹ AP = D where D is a diagonal matrix.
Question 2 :
Diagonalize the following matrix
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Let A = |
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The order of A is 3 x 3. So the unit matrix I = |
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Now we have to multiply λ with unit matrix I.
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λI = |
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| A-λI= |
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- |
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| = |
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| A-λI= |
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= (1-λ)[ (5-λ)(1-λ) - 1 ] - 1[1 - λ - 3] + 3 [1 - 3 (5-λ) ] = (1-λ)[ 5 - 5 λ - λ + λ² - 1 ] - 1[ -λ - 2] + 3 [ 1 - 15 +3 λ ] = (1-λ)[ λ² - 6 λ + 4 ] + 1 λ + 2 + 3 [ - 14 +3 λ ] = λ² - 6 λ + 4 - λ³ + 6 λ² - 4 λ + λ + 2 - 42 + 9 λ = - λ³ + λ² + 6 λ² - 6 λ - 4 λ + λ + 9 λ + 4 + 2 - 42 = - λ³ + 7 λ² - 10 λ + 10 λ + 6 - 42 = - λ³ + 7 λ² - 36 = λ³ - 7 λ² + 36 |
To find roots let |A-λI| = 0
λ³ - 7 λ² + 36 = 0
For solving this equation first let us do synthetic division.diagonalization of matrix 2
By using synthetic division we have found one value of λ that is λ = -2.
Now we have to solve λ² - 10 λ + 24 to get another two values. For that let us factorize
λ² - 9 λ + 18 = 0
λ² - 3 λ - 6 λ + 18 = 0
λ (λ - 3) - 6 (λ - 3) = 0
(λ - 6) (λ - 3) = 0
λ - 6 = 0
λ = 6
λ - 3 = 0
λ = 3
Therefore the characteristic roots (or) Eigen values are x = -2,3,6
Substitute λ = -2 in the matrix A - λI
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From this matrix we are going to form three linear equations using variables x,y and z.
3x + 1y + 3z = 0 ------ (1)
1x + 7y + 1z = 0 ------ (2)
3x + 1y + 3z = 0 ------ (3)
By solving (1) and (2) we get the eigen vector
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The eigen vector x = |
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Substitute λ = 3 in the matrix A - λI
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From this matrix we are going to form three linear equations using variables x,y and z.
-2x + 1y + 3z = 0 ------ (4)
1x + 2y + 1z = 0 ------ (5)
3x + 1y - 2z = 0 ------ (6)
By solving (4) and (5) we get the eigen vector
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The eigen vector y = |
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Substitute λ = 6 in the matrix A - λI
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From this matrix we are going to form three linear equations using variables x,y and z.
-5x + 1y + 3z = 0 ------ (7)
1x - 1y + 1z = 0 ------ (8)
3x + 1y - 5z = 0 ------ (9)
By solving (7) and (8) we get the eigen vector diagonalization of matrix 2
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The eigen vector z = |
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Let P = |
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The column of P are linearly independent eigen vectors of A . Therefore the diagonal matrix = diagonalization of matrix 2 |
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Questions |
Solution |
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Question 1 : Diagonalize the following matrix
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Question 3 : Diagonalize the following matrix
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Question 4 : Diagonalize the following matrix diagonalization of matrix 2
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Question 5 : Diagonalize the following matrix diagonalization of matrix 2
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- Types of matrices
- Equality of matrices
- Operation on matrices
- Algebraic properties of matrices
- Multiplication properties
- Minor of a matrix
- Practice questions of minor of matrix
- Adjoint of matrix
- Determinant of matrix
- Inverse of a matrix
- Practice questions of inverse of matrix
- Inversion method
- Inversion method worksheets
- Cramer's Rule for 3 equations
- Cramer's Rule for 2 equations
- Solving 2 equations using Cramer's method
- Rank Method in Matrix
- Rank of a matrix
- Solving using rank method
- Linear dependence of vectors
- Linear dependence of vectors in rank method
- Characteristic Equation of matrix
- Characteristic vector of matrix
- Diagonalization of matrix
- Gauss elimination method
- Matrix calculator
