Diagonalization of Matrix 4
In this page diagonalization of matrix 4 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 4 :
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.diagonalization of matrix4
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λI = |
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| A-λI= |
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- |
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| = |
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= (4-λ)[(10-λ)(-13- λ)+120]+
20[-2(-13-λ)-24]-10[60-6(10-λ)]
= (4-λ)[-130-10 λ+13λ+λ²+120]+20[26+2λ-24]-10[60-60+6λ]
= (4-λ)[-10+3λ+λ²]+20[2+2λ]-10[6λ]
= (4-λ)[λ²+3λ-10]+20[2+2λ]-10[6λ]
= 4λ²+12λ-40-λ³-3λ²+10λ+40λ+40-60λ
= -λ³ + 1λ² + 2λ
To find roots let |A-λI| = 0
-λ³ + 1λ² + 2λ = 0
For solving this equation -λ from all the terms
-λ (λ² - 1λ - 2) = 0
-λ = 0 (or) λ² - 1 λ - 2 = 0
λ = 0 (λ+1) (λ-2) = 0
λ + 1 = 0 λ - 2 = 0
λ = - 1 λ = 2
Therefore the characteristic roots (or) Eigen values are x = 0,-1,2
Substitute λ = 0 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.
4x - 20y - 10z = 0 ------ (1)
-2x + 10y + 4z = 0 ------ (2)
6x - 30y - 13z = 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 = -1 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 - 20y - 10z = 0 ------ (4)
-2x + 10y + 4z = 0 ------ (5)
6x - 30y - 12z = 0 ------ (6)
By solving (4) and (5) we get the eigen vector diagonalization of matrix 4
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The eigen vector y = |
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Substitute λ = 2 in the matrix A - λI diagonalization of matrix 4
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From this matrix we are going to form three linear equations using variables x,y and z.
2x - 20y - 10z = 0 ------ (7)
-2x + 8y + 4z = 0 ------ (8)
6x - 30y - 15z = 0 ------ (9)
By solving (7) and (8) we get the eigen vector
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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 4 diagonalization of matrix 4 |
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Questions |
Solution |
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Question 1 : Diagonalize the following matrix
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Question 2 : Diagonalize the following matrix
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Question 3 : Diagonalize the following matrix
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Question 5 : Diagonalize the following matrix
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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
