Diagonalization of Matrix 3
In this page diagonalization of matrix 3 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 3 :
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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= (-2-λ)[ -λ(1-λ) - 12 ] - 2[-2 λ - 6] - 3 [-4-(-1)(1-λ) ]
= (-2-λ)[ -λ + λ² - 12 ] + 4 λ + 12 - 3 [-4+1-λ ]
= (-2-λ)[ λ² -λ - 12 ] + 4 λ + 12 - 3 [-3-λ ]
= (-2-λ) [λ² -λ - 12 ] + 4 λ + 12 + 9 + 3 λ
= -2λ² + 2λ + 24 - λ³ + λ² + 12 λ + 4 λ + 12 + 9 + 3 λ
= - λ³ - λ² + 2λ + 12 λ + 4 λ + 3 λ + 24 + 12 + 9
= - λ³ - λ² + 21λ + 45
= λ³ + λ² - 21λ - 45
To find roots let |A-λI| = 0
λ³ + λ² - 21λ - 45 = 0 diagonalization of matrix 3
For solving this equation first let us do synthetic division.
By using synthetic division we have found one value of λ that is λ = -3.
Now we have to solve λ² - 2 λ - 15 to get another two values. For that let us factorize
λ² - 2 λ - 15 = 0
λ² + 3 λ - 5 λ - 15 = 0
λ (λ + 3) - 5 (λ + 3) = 0
(λ - 5) (λ + 3) = 0
λ - 5 = 0
λ = 5
λ + 3 = 0
λ = - 3
Therefore the characteristic roots (or) Eigen values are x = -3,-3,5
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.
1x + 2y - 3z = 0 ------ (1)
2x + 4y - 6z = 0 ------ (2)
-1x - 2y + 3z = 0 ------ (3)
By solving (1) and (3) we get the eigen vector
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The eigen vector x = |
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Substitute λ = 5 in the matrix A - λI Diagonalization of Matrix3
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From this matrix we are going to form three linear equations using variables x,y and z.
-7x + 2y - 3z = 0 ------ (4)
2x - 4y - 6z = 0 ------ (5)
-1x - 2y - 5z = 0 ------ (6)
By solving (4) and (5) we get the eigen vector Diagonalization of Matrix3
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The eigen vector z = |
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Let P = |
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Eigen vectors of x and y are linearly dependent. So we cannot find diagonal matrix. diagonalization of matrix 3
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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 4 : Diagonalize the following matrix
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Question 5 : Diagonalize the following matrix diagonalization of matrix 3 diagonalization of matrix 3
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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
