Unit 1: Linear Algebra and Partial Differentiation Matrices
Unit 1: Linear Algebra and Partial Differentiation Matrices
Unit 1: Linear Algebra and Partial Differentiation Matrices
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Unit
MATRICES
1
Introduction
A matrix (plural: matrices), The term matrix was introduced by the 19th-century
English mathematician James Sylvester, but it was his friend the mathematician Arthur
Cayley who developed the algebraic aspect of matrices in two papers in the 1850s. Cayley
first applied them to the study of systems of linear equations, where they are still very useful.
They are also important because, as Cayley recognized, certain sets of matrices form
algebraic systems in which many of the ordinary laws of arithmetic (e.g., the associative and
distributive laws) are valid but in which other laws (e.g., the commutative law) are not valid.
Historically, it was not the matrix but a certain number associated with a square array of
numbers called the determinant that was first recognized. Only gradually did the idea of the
matrix as an algebraic entity emerge. Applications of matrices are found in most scientific
fields.
Applications in Engineering
1 − 3 a 22 matrix; an
2 7.2 example of a square 14.3 a 11 matrix;
matrix.
termed a scalar
8 2.5 A 32
6 − 3 1 a 23 rectangular 0 − 1.1
2 0 9 rectangular
matrix
− 5 − 3 matrix
A 31 matrix; an
9 .1
− 8 0 1.4 − 1.3
a 14 matrix; an
example of a row
2 .8
example of a
column matrix
matrix or row vector. − 0.7 or column
vector.
TYPES OF MATRICES
1) Row matrix or vector: Any number of columns but only one row is called row matrix.
[1 1 6], [0 3 5 2]
2) Column matrix or vector: Only one column and any number of rows is called column
matrix.
1
1
[4] , [ ]
−3
2
3) Zero matrix or null matrix :A matrix containing all zero elements is called zero matrix.
0 0
0 0
[0 0] , [ ]
0 0
0 0
4) Rectangular matrix: Contains more than one element and number of rows is not equal to
the number of column(𝑚 ≠ 𝑛)
3 1 2
1) A = 2 0 5
1 2 3
Solution: The largest square sub matrix possible is of order 3 and that is [ A] itself. Since
det( A) = −23 0, the rank of [ A] = 3.
3 1 2
2) A = 2 0 5
5 1 7
Solution: The largest square sub matrix of [ A] is of order 3 and that is [ A] itself. Since
det( A) = 0 , the rank of [ A] is less than 3. The next largest square sub matrix would be a 2
2 matrix. One of the square sub matrices of [ A] is
B =
3 1
2 0
and det( B) = −2 0 . Hence the rank of [ A] is 2. There is no need to look at other 2 2 sub
matrices to establish that the rank of [ A] is 2.
Properties
1. Rank of a null matrix is zero
2. For a nonzero matrix A,𝜌(𝐴) ≥ 1
3.Non-singular matrix of order n has rank n
4.Rank of singular matrix of order n is always less than n
1 2 −2 3 1 3 4 3
3. 𝐴=[ 2 5 −4 6] 8. 𝐴 = [3 9 12 9]
−1 −3 2 −2
1 3 4 1
2 4 −1 6
4 2 −1 2 3 𝑘 𝑘
𝐴 = [1 −1 2 1] Find K if [𝑘 3 3] has ranks ‘3’.
4. 2 2 −2 0 9. 𝑘 𝑘 3
1 2 3 𝑘 𝑘 2
𝐴 = [4 5 6] Find K if 𝐴 = [2 1 1 ] has rank ‘2’.
5. 7 8 9 10. 0 0 −5
1 0 0 0
𝐴~ [0 −5 2 0]
0 0 −2 0
Step V: Operate 𝐶3 → 𝐶3 − 2𝐶2
1 0 0 0
𝐴~ [0 −5 0 0]
0 0 10 0
𝑅2
Step VI: Operate 𝑅2 → , 𝑅3 → 𝑅3/(10)
−5
1 0 0 0
𝐴~ [0 1 0 0] = [𝐼3: 0]
0 0 1 0
∴ 𝜌(𝐴) = 3
Exercise 1.2
Reduce the matrix A to its normal form and hence find its matrix:-
2 3 −1 −1
2 1 −3−6
1. 𝐴 = [1 −1 −2 −4]
6. 𝐴 = [3 −3 1 2 ]
3 1 3 −2
6 3 0 −7 1 1 1 2
2 3 4 5 2 −2 −2 0 2
𝐴=[3 4 5 6 ] 3 −1 −1 4 1
2. 4 5 6 7 7. 𝐴 = 0 −1 0 1 2
4 10 11 12 1 1 3 2 −2
[2 2 4 6 −3]
2 −1 31 3 2 1 3 2
𝐴 = [1 −4 −21]
3. 8. 𝐴 = [1 0 2 1 4]
5 2 43 2 3 3 4 6
1 2 4 1 8
0 1 2−2 1 1 −1 1
𝐴 = [4 0 26] 𝐴 = [1 −1 2 −1]
4. 2 1 31 9. 3 1 0 1
2 −3 4 4 3 2 5 7 12
𝐴 = [1 1 1 2] 𝐴 = [1 1 2 3 5]
5. 3 −2 3 6 10. 3 3 6 9 15
𝑖. 𝑒. 𝐴𝑋 = 𝐵
Where𝐴 is called the coefficient matrix, 𝐵 is called the constant matrix and 𝑋 is called the
variable matrix or solution vector.
System of equations
𝐴𝑋 = 𝐵
𝐴𝑋 = 0 𝐴𝑋 = 𝐵
Augmented Matrix:
𝐴𝑋 = 𝐵 is a systems of equations then the augmented matrix is defined
Reduce augmented matrix [𝐴: 𝐵] to echelon form by using elementary row operations
and find rank of matrix A and augmented matrix[𝐴: 𝐵].
i) If 𝜌(𝐴) = 𝜌(𝐴: 𝐵) = number of unknowns, then system is consistence and has unique
solution.
ii) If 𝜌(𝐴) = 𝜌(𝐴: 𝐵) < number of unknowns, then system is consistence and has
infinitely many solutions.
iii) If 𝜌(𝐴) ≠ 𝜌(𝐴: 𝐵) then system is inconsistence i.e. it has no solution.
No Solutions
Unique Solution if Infinite Solutions if
Rank (A) =number of unknowns Rank (A) < number of unknowns
Step IV:-
𝜌(𝐴) = 2 , 𝜌(𝐴: 𝐵) = 3
System consistent iff𝜌(𝐴: 𝐵) = 2
𝑖. 𝑒. 12 − 2𝜆 = 0
12=2 𝜆
𝜆=6
𝜌(𝐴) = 𝜌(𝐴: 𝐵) = 2
n=3
It has infinite no. of solutions
Step V:-
1 1 1 𝑥1 6
𝑥
[0 −1 1] [ 2 ] = [−15]
0 0 0 𝑥3 0
𝑥1 + 𝑥2 + 𝑥3 = 6, (1)
−𝑥2 + 𝑥3 = −15, (2)
𝑙𝑒𝑡 𝑥3 = 𝑡, 𝑥2 = 15 + 𝑡
𝑥1 = 6 − 𝑥2 − 𝑥3
𝑥1 = 6 − 15 − 𝑡
𝑥1 = −9 − 2𝑡
If 𝜆 ≠ 6 then system is inconsistent
𝜌(𝐴) ≠ 𝜌(𝐴: 𝐵)and it has no solutions.
Example 1.7: For what value of 𝜆 and 𝜇 the system
𝑥 + 𝑦 + 𝑧 = 6,
𝑥 + 2𝑦 + 3𝑧 = 10
𝑥 + 2𝑦 + 𝜆𝑧 = 𝜇
has i) No Solution ii) Unique Solution iii) Infinitely many solution.
Solution:
Step I: write in matrix form
𝐴𝑋 = 𝐵
1 1 1 𝑥 6
[1 2 3] [𝑦] = [10]
1 2 𝜆 𝑧 𝜇
Step II:-
1 1 1 : 6
𝑪 = (𝑨 ∶ 𝑩) = [1 2 3 : 10],
1 2 𝜆 : 𝜇
Step III:- Operating 𝑅2 → 𝑅2 − 𝑅1 , 𝑅3 → 𝑅3 − 𝑅1 ``
1 1 1 : 6
𝑪~ [0 1 2 : 4 ],
0 0 𝜆 − 3 : 𝜇 − 10
If𝜆 = 3 and 𝜇 ≠ 10
𝜌(𝐴) = 2 , 𝜌(𝐶) = 3
𝜌(𝐴) ≠ 𝜌(𝐶)
System is inconsistent and has no solution
Step V:-1) For Unique Solution
If𝜆 ≠ 3 and 𝜇 can take any value
𝜌(𝐴) = 3 , 𝜌(𝐶) = 3
𝜌(𝐴) = 𝜌(𝐶) = 3 = 𝑟,
System is consistent, number of unknowns = n = 3, n = r
System has unique solution
Step VI:-1) For Infinite Solution
If𝜆 = 3 and 𝜇 = 10
𝜌(𝐴) = 2 , 𝜌(𝐶) = 2
𝜌(𝐴) = 𝜌(𝐶) = 3 = 𝑟, System is consistent
n > r, System have infinite solution
Exercise 1.3
1) Test the consistency and solve if consistent
12𝑥 − 𝑦 − 𝑧 = 2, 2. 2𝑥 + 3𝑦 − 4𝑧 = −2,
𝑥 + 2𝑦 + 𝑧 = 2,
𝑥 − 𝑦 + 3𝑧 = 4,
4𝑥 − 7𝑦 − 5𝑧 = 2
3𝑥 + 2𝑦 − 𝑧 = −5
5. 4𝑥 − 2𝑦 +3𝑥
6𝑧−=2𝑦
8, + 𝑧 = 4 6. 𝑥 + 𝑦 + 𝑧 = 6,
𝑥 + 𝑦 − 3𝑧 = −1, 𝑥 − 𝑦 + 2𝑧 = 5,
15𝑥 − 3𝑦 + 9𝑧 = 21 3𝑥 + 𝑦 + 𝑧 = 8
7. 2𝑥 + 𝑧 = 4, 8. 3𝑥 + 3𝑦 +2𝑥2𝑧−=2𝑦1,+ 3𝑧 = 7
𝑥 − 2𝑦 + 2𝑧 = 7 𝑥 + 2𝑦 = 4,
3𝑥 − 2𝑦 = 1 10𝑦 + 3𝑧 = −2
9. 3𝑥 + 𝑦 + 2𝑧 = 3, 10. 2𝑥1 + 𝑥2 2𝑥
− 𝑥−3 3𝑦 −4𝑧==8,5
+ 3𝑥
2𝑥 − 3𝑦 − 𝑧 = −3 𝑥1 + 𝑥2 + 𝑥3 − 𝑥4 + 2 = 0,
𝑥 + 2𝑦 + 𝑧 = 4 3𝑥1 + 2𝑥2 − 𝑥3 = 6
4𝑥2 + 3𝑥3 + 2𝑥4 = −8
Homogeneous System
Consider a homogeneous linear system 𝐴𝑋 = 0, the augmented matrix is [𝐴 ∶ 0]
𝜌(𝐴) = 𝜌(𝐴: 0)
∴ Every homogeneous system is always consistent and has unique or infinite solution
Homogeneous System
[A][X]= [0]
Example1.9: For different values of k discuss the nature of solution of the following
equation
𝑥 + 2𝑦 − 𝑧 = 0
3𝑥 + (𝑘 + 7)𝑦 − 3𝑧 = 0
2𝑥 + 4𝑦 + (𝑘 − 3)𝑧 = 0
Solution: Step 1: Write in matrix form
𝐴𝑋 = 0
1 2 −1 𝑥 0
[3 𝑘+7 −3 ] [𝑦] = [0]
2 4 𝑘−3 𝑧 0
1 2 −1
Step 2:-|𝐴| = |3 𝑘 + 7 −3 |
2 4 𝑘−3
= [(𝑘 + 7)(𝑘 + 3) + 12] − 2[3𝑘 − 9 + 6] − [12 − 2𝑘 − 14] = 𝑘 2 − 1
Step 3:-1) For Unique Solution
If|𝐴| ≠ 0 , 𝑘 2 − 1 ≠ 0 i.e. 𝑘 ≠ ±1
System is consistent and has unique solution 𝑥 = 0, 𝑦 = 0, 𝑧 = 0
Step 4:-1) For Infinite Solution
If|𝐴| = 0, 𝑘 2 − 1 = 0 i.e. 𝑘 = ±1
System is consistent and has nontrivial or infinite solution
Step 5:-1)Solution Fork=1
1 2 −1 : 0
𝑪 = (𝑨 ∶ 𝟎) = [3 8 −3 : 0],
2 4 −2 : 0
Step 6:- Operating 𝑅2 → 𝑅2 − 3𝑅1 , 𝑅3 → 𝑅3 − 2𝑅1
1 2 −1 : 0
𝑪~ [0 2 0 : 0],
0 0 0 : 0
Step 7:- again write in matrix form AX = 0
1 2 −1 𝑥 0
[0 2 0 ] [𝑦] = [0]
0 0 0 𝑧 0
𝑥 + 2𝑦 − 𝑧 = 0 (1)
2y = 0 (2)
4 −1 1
Step 2:-|𝐴| = |1 2 −1| = 47 ≠ 0
3 1 5
System has trivial solution 𝑥1 = 0, 𝑥2 = 0, 𝑥3 = 0
Exercise 1.4
1) Solve the following system of equations
1 𝑥 + 2𝑦 + 3𝑧 = 0 2 𝑥 + 𝑦 + 2𝑧 = 0
2𝑥 + 3𝑦 + 𝑧 =0 𝑥 + 2𝑦 + 3𝑧 =0
4𝑥 + 5𝑦 + 4𝑧 = 0 𝑥 + 3𝑦 + 4𝑧 = 0
𝑥 + 𝑦 − 2𝑧 = 0 3𝑥 + 4𝑦 + 7𝑧 = 0
3 𝑥 + 𝑦 + 3𝑧 = 0 4 𝑥 + 2𝑦 + 3𝑧 = 0
𝑥 − 𝑦 + 𝑧 =0 2𝑥 + 3𝑦 + 𝑧 =0
−𝑥 + 2𝑦 = 0 4𝑥 + 5𝑦 + 4𝑧 = 0
Sanjivani College of Engineering, Kopargaon (An Autonomous Institute) 1.17
Linear Algebra and Partial Differentiation Matrices
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𝑥−𝑦+𝑧 =0
5 2𝑥 − 𝑦 + 3𝑧 = 0 6 𝑥 + 3𝑦 + 𝑧 = 0
3𝑥 + 2𝑦 + 𝑧 =0 2𝑥 − 2𝑦 − 6𝑧 =0
𝑥 − 4𝑦 + 5𝑧 = 0 3𝑥 + 𝑦 − 5𝑧 = 0
6) Find the value of 𝑘 such that the system of equations has non-trivial solution, Find the
solution
2𝑥 + 3𝑦 − 2𝑧 = 0
3𝑥 − 𝑦 + 3𝑧 =0
7𝑥 + 𝑘𝑦 − 𝑧 = 0
Syllabus content: - Gauss Jordan Elimination Method
• This method is used for nonhomogeneous system of equations.
• Reduce the augumented matrix by using only row transformation into identity
matrix.
• The solution is obtained for the desired variables.
𝑥+𝑦+𝑧 = 5
2𝑥 + 3𝑦 + 5𝑧 = 8
4𝑥 + 5𝑧 = 2
Solution:
Step I: AX=B
1 1 1 𝑥 5
[2 3 5] [𝑦] = [8]
4 0 5 𝑧 2
Step II:-
1 1 1 ∶ 5
(𝐴: 𝐵) = [2 3 5 ∶ 8],
4 0 5 ∶ 2
Step III:- Operate 𝑅2 → 𝑅2 − 2𝑅1 , 𝑅3 → 𝑅3 − 4𝑅1
1 1 1 ∶ 5
(𝐴: 𝐵) = [0 1 3 : −2 ],
0 −4 1 ∶ −18
Operate 𝑅3 → 𝑅3 + 4𝑅2
1 1 1 ∶ 5
(𝐴: 𝐵) = [0 1 3 : −2 ],
0 0 13 ∶ −26
Operate 𝑅3 → 𝑅3 /13
1 1 1 ∶ 5
(𝐴: 𝐵) = [0 1 3 : −2],
0 0 1 ∶ −2
Operate 𝑅2 → 𝑅2 − 3𝑅3 , 𝑅1 → 𝑅1 − 𝑅3 ,
1 1 0 ∶ 7
(𝐴: 𝐵) = [0 1 0 : 4]
0 0 1 ∶ −2
Operate 𝑅1 → 𝑅1 − 𝑅2
1 0 0 ∶ 3
(𝐴: 𝐵) = [0 1 0 : 4]
0 0 1 ∶ −2
1 0 0 𝑥 3
Step IV:- [0 1 0] [𝑦] = [ 4 ]
0 0 1 𝑧 −2
𝑥 = 3, 𝑦 = 4, 𝑧 = −2
Exercise 1.5
Solve the following system of equations by using Gauss Jordan elimination method
1) 𝑥 + 2𝑦 − 3𝑧 = 2, 6𝑥 + 3𝑦 − 9𝑧 = 6, 7𝑥 + 14𝑦 − 21𝑧 = 13
2) 4𝑦 + 𝑧 = 2, 2𝑥 + 6𝑦 − 2𝑧 = 3, 4𝑥 + 8𝑦 − 5𝑧 = 4
3) 𝑥 + 𝑦 + 2𝑧 = 1, 2𝑥 − 𝑦 + 𝑡 = −2, 𝑥 − 𝑦 − 𝑧 − 2𝑡 = 4, 2𝑥 − 𝑦 + 2𝑧 − 𝑡 = 0
4) 2𝑥 + 4𝑦 + 6𝑧 = 22, 3𝑥 + 8𝑦 + 5𝑧 = 27, −𝑥 + 𝑦 + 2𝑧 = 2
5) 𝑥 − 2𝑦 + 𝑧 = 5, 2𝑥 − 5𝑦 + 4𝑧 = −3, 𝑥 − 4𝑦 + 6𝑧 = 10
Solution:
StepI:
To verify A is orthogonal matrix check that 𝐴𝐴𝑇 = 𝐼
Step II:
cos 𝜃 0 sin 𝜃 cos 𝜃 0 − sin 𝜃 1 0 0
𝑇
𝐴𝐴 = [ 0 1 0 ][ 0 1 0 ] = [0 1 0] = 𝐼
− sin 𝜃 0 cos 𝜃 sin 𝜃 0 cos 𝜃 0 0 1
∴ A is orthogonal matrix
0 − sin 𝜃 cos 𝜃
−1
𝐴 1 0 ]=[ 0
0 cos 𝜃 sin 𝜃
0 2𝑏 𝑐
Example1.14: Determine the values of a,b,c when [𝑎 𝑏 −𝑐] is orthogonal
𝑎 −𝑏 𝑐
Solution:
StepI:
Given that A is orthogonal matrix ∴ 𝐴𝐴𝑇 = 𝐼
Step II:
0 2𝑏 𝑐 0 𝑎 𝑎 1 0 0
[𝑎 𝑏 −𝑐] [2𝑏 𝑏 −𝑏] = [0 1 0]
𝑎 −𝑏 𝑐 𝑐 −𝑐 𝑐 0 0 1
4𝑏 2 + 𝑐 2 = 1, 2𝑏 2 − 𝑐 2 = 0, 𝑎2 + 𝑏 2 + 𝑐 2 = 1, 𝑎2 − 𝑏 2 − 𝑐 2 = 0
1 1 1
𝑎=± ,𝑏 = ± ,𝑐 = ±
√2 √6 √3
Exercise 1.5
cos 𝜃 sin 𝜃
1) Show that [ ] is orthogonal and find 𝐴−1
−sin 𝜃 cos 𝜃
1/3 2/3 𝑎
2) If 𝐴 = [1/3 1/3 𝑏 ] is orthogonal and find a, b, c .
2/3 −2/3 𝑐
2 1 −2
3) Is 𝐴 = [−2 2 1 ] is an orthogonal matrix? If not can it be converted into an
1 2 −2
orthogonal matrix.