Diagonalization of Matrix 5
In this page diagonalization of matrix 5 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 5 :
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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= (11-λ)[(-2-λ)(-6-λ)-20]+4[7(-6-λ)+50]-7[-28-10(-2-λ)] = (11-λ)[(2+λ)(6+λ)-20]+4[-42-7λ+50]-7[-28+20+10λ] = (11-λ)[12+2λ+6λ+λ²-20]+4[8-7λ]-7[-8+10λ] = (11-λ)[λ²+8λ-8]+32-28λ+56-70λ = 11λ²+8λ-88-λ³-8λ²+88λ-98λ+88 = - λ³+3λ²-2λ = -λ³+3λ²-2λ = -λ(λ²-3λ²+2) |
To find roots let |A-λI| = 0
-λ(λ²-3λ²+2) = 0
λ = 0
Now we have to solve λ²-3λ²+2 to get another two values. For that let us factorize
λ²-3λ²+2 = 0
λ²-1λ-2λ+2 = 0
λ(λ-1)-2(λ-1) = 0
(λ-1)(λ-2) = 0
λ - 1 = 0
λ = 1
λ - 2 = 0
λ = 2
Therefore the characteristic roots (or) Eigen values are x = 0,1,2
Substitute λ = 0 in the matrix A - λI diagonalization of matrix 5
| A-λI= |
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From this matrix we are going to form three linear equations using variables x,y and z.
11x - 4y - 7z = 0 ------ (1)
7x - 2y - 5z = 0 ------ (2)
10x - 4y - 6z = 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.
10x - 4y - 7z = 0 ------ (4)
7x - 3y - 5z = 0 ------ (5)
10x - 4y - 7z = 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 = 2 in the matrix A - λI diagonalization of matrix 5
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From this matrix we are going to form three linear equations using variables x,y and z.
9x - 4y - 7z = 0 ------ (7)
7x - 4y - 5z = 0 ------ (8)
10x - 4y - 8z = 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 5 |
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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 4 : Diagonalize the following matrix diagonalization of matrix 5
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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
