Mat🍚s
- A set of mn element arranged in a rectangular array of m rows and n columns is known as matrix of order m by n, which is written as m x n.
_ _
A = | a11 a12 a13 |
| a21 a22 a23 |
| a31 a32 a33 |
- -
- It can also be expressed in the form
A = [aij]mxn Where, aij is the element in the ith row and jth column. - Applications of matrices in mathematics:
- Computer graphics
- Graph theory
- Statistics, etc.
Types of Matrices
Row matrix
- Only one row and any no. of columns.
- Example: [2 -1 6] and [-4 7]
- Note: Row matrix → Row vector
Column matrix
- Only one column and any no. of rows.
Example:
_ _ _ _
| 2 | | 4 |
| 4 | | 5 |
| 6 | - -
- -
- Note: Column matrix → Column vector
Square matrix
- No. of rows = no. of columns.
Example:
_ _ _ _
| 1 -3 | | 1 2 3 |
| 2 2 | | 4 5 6 |
- - 2x2 | 7 8 9 |
- - 3x3
Diagonal matrix
- A square matrix all of whose non-diagonal elements are zero
Example:
_ _ _ _
| 1 0 | | 1 0 0 |
| 0 2 | | 0 5 0 |
- - | 0 0 9 |
- -
NULL matrix
- Also known as zero matrix.
- All the elements are 0.
- It is denoted by 0.
Example:
_ _
| 0 0 |
| 0 0 |
- -
Unit / Identity matrix
- A diagonal matrix all of whose diagonal elements are unity. It is denoted by I.
Example:
_ _ _ _
| 1 0 | | 1 0 0 |
| 0 1 | | 0 1 0 |
- - | 0 0 1 |
- -
Scalar matrix:
- A diagonal matrix, all of whose diagonal elements are equal.
Example:
_ _ _ _
| 4 0 | | 2 0 0 |
| 0 4 | | 0 2 0 |
- - | 0 0 2 |
- -
Upper triangular matrix
- A square matrix in which all the elements below the diagonal are zero.
Example:
_ _ _ _
| 4 3 | | 2 3 3 |
| 0 4 | | 0 2 5 |
- - | 0 0 2 |
- -
Lower triangular matrix
- A square matrix in which all the elements above the diagonal are zero.
Example:
_ _ _ _
| 4 0 | | 2 0 0 |
| 4 5 | | 1 2 0 |
- - | 3 4 2 |
- -
Some Matrix and Determinant concept
Trace of a matrix
- The sum of all the diagonal elements of a square matrix.
Example:
_ _
A = | 4 0 |
| 4 5 |
- -
trace of A = 4 + 5 = 9
Transpose of a matrix:
- Interchanging rows and columns.
- Denoted by AT
Example:
_ _
A = | 4 0 |
| 4 5 |
- -
_ _
Transpose of A = | 4 4 |
| 0 5 |
- -
Determinant of a matrix
_ _
A = | 4 0 |
| 4 5 |
- -
Det(A) = | 4 0 |
| 4 5 |
= 5*4 - 4*0
= 20
_ _
Now if A = | 1 2 3 |
| 4 5 6 |
| 7 8 9 |
- -
Note: first element (+), second element (-) and so on
Expand it in any row or column.
Det of (A) = | 1 2 3 |
| 4 5 6 |
| 7 8 9 |
= 1(5*9 - 8*6) - 2(4*9 - 7*6) + 3(4*8 - 7*5)
= (45 - 48) - 2(36 - 42) + 3(32 - 35)
= -3 - 2(-6) + 3(-3)
= -3 + 12 - 9
= 0
Singular and Non singular matrix
- If a square matrix A has det(A) = 0
- If det(A) ≠ 0 then it is non-singular matrix
Minor of an element of a matrix
_ _
if X = | x11 x12 x13 |
| x21 x22 x23 |
| x31 x32 x33 |
- -
minor of element x11 = | x22 x23 |
| x32 x33 |
minor of element x32 = | x11 x13 |
| x21 x23 |
Co-factor of an element of a matrix
_ _
if X = | x11 x12 x13 |
| x21 x22 x23 |
| x31 x32 x33 |
- -
co-factor of element x11 = (-1) ^ 1+1 | x22 x23 |
| x32 x33 |
or (-1) ^ 1+1 * minor of x11
co-factor of element x32 = (-1) ^ 3+2 * minor of x32
Adjoint of a matrix
- Adj(A) = transpose of co-factor matrix
_ _
A = | 3 4 5 |
| 2 3 4 |
| 2 5 6 |
- -
_ _
First we make co-factor matrix = | -2 -4 4 |
| 1 8 -7 |
| 1 -2 1 |
- -
_ _
Now adjoint is transpose of co-factor matrix = | -2 1 1 |
| -4 8 -2 |
| 4 -7 1 |
- -
Special types of matrices
Symmetric matrix: A = AT
- i.e., aij = aji for all i and j
Example:
_ _ _ _
| 1 4 | | 1 -i 2i |
| 4 -2 | | -i -3 7 |
- - | 2i 7 5 |
- -
- Note: Diagonal elements must be real
Skew-symmetric matrix: A = -AT
- i.e., aij = -aji for all i and j
Example:
_ _ _ _
| 0 5 | | 0 -5 2i |
| -5 0 | | 5 0 -i |
- - | -2i i 0 |
- -
- Note: Diagonal elements must be all zero.
Algebra of matrices
Transpose of a matrix : A' or AT
- aij = aji
- Imp points:
- (A')' = A
- If A is symmetric matrix then A = A'
- If A is Skew-symmetric matrix then A = -A'
Equality of matrices
- Order same → corresponding elements are also equal
_ _ _ _
| a b | | 1 2 |
| c d | = | 3 4 |
- - - -
then a = 1, b = 2, c = 3 and d = 4
Addition / Subtraction of matrices:
- Both matrices should be of same order.
_ _ _ _
A = | 2 3 4 | , B = | 1 2 3 |
| 5 6 7 | | 9 8 6 |
- - - -
_ _ _ _
A+B = | 3 5 7 | , A-B = | 1 1 1 |
| 14 14 13 | | -4 -2 1 |
- - - -
Scalar multiplication
_ _
A = | 3 7 5 |
| 4 2 1 |
- -
_ _
2A = | 6 14 10 |
| 8 4 2 |
- -
Matrix multiplication
- Note: Two matrix can only be multiplied when total number of columns of 1st matrix = total number of rows of 2nd matrix.
A mxn B nxp
n is common for both
and mxp is dimension of multiplied matrix.
_ _ _ _
A = | 0 1 | B = | 2 3 |
| 2 3 | 2x2 | 4 5 | 2x2
- - - -
A and B can be multiplied
_ _
C = | 1 4 |
| 3 4 |
| 5 7 | 3x2
- -
A & C can't be multiplied
_ _
D = | 1 3 5 |
| 3 6 8 | 2x3
- -
A & D can be multiplied
_ _ _ _
A*B = | 0*2 + 1*4 0*3 + 1*5 | = | 4 5 |
| 2*2 + 3*4 2*3 + 3*5 | | 16 21 |
- - - -
Rank of a matrix by reducing to Normal Form
- Now some rules to make normal form.
- Both Row and column operation are allowed.
Questions
Find the rank of following matrices:
_ _
A = | 1 2 3 4 |
| 2 1 4 3 |
| 3 0 5 -10 |
- -
_ _
A = | 1 2 3 2 |
| 2 3 5 1 |
| 1 3 4 5 |
- -
_ _
A = | 1 2 -1 3 |
| 3 4 0 -1 |
| -1 0 -2 7 |
- -
_ _
A = | 1 2 -1 3 |
| 4 1 2 1 |
| 3 -1 1 2 |
| 1 2 0 1 |
- -
_ _
A = | 2 -2 3 |
| 3 -1 2 |
| 1 2 -1 |
- -
Inverse of a matrix
(using elementary transformation)
- Let A be any non-singular matrix then A = I.A
- Applying suitable elementary row transformations to A on he LHS and I on the RHS, A reduces to I and
I reduces to any matrix (say B)
I = B.A
A-1 = B - When finding the inverse of a matrix using elementary row operations, the restriction is that you can only perform row operations. Similarly, when using elementary column operations, you can only perform column operations.
Question ↓
Find the inverse of a matrix by elementary transformation.
_ _
A = | 2 3 4 |
| 4 3 1 |
| 1 2 4 |
- -
Solution of Non-Homogeneous linear equations
Consider a system of non-homogeneous system of linear equations in 'n' unknowns (x1, x2, x3, ... , xn) is represented by
a11 x1 + a12 x2 + ... + a1n xn = b1
a21 x1 + a22 x2 + ... + a2n xn = b2
. . . .
. . . .
. . . .
an1 x1 + am2 x2 + ... + ann xn = bn
In matrix form, we can write this system of equations as
AX = B, where
_ _
A = | a11 a12 a13 ... ... ... a1n |
| a21 a22 a23 ... ... ... a2n |
| ... ... ... ... ... ... ... |
| ... ... ... ... ... ... ... |
| an1 an2 an3 ... ... ... ann | nxn
- -
_ _
X = | x1 |
| x2 |
| . |
| . |
| xn | nx1
- -
_ _
B = | b1 |
| b2 |
| . |
| . |
| bn | nx1
- -
Conditions for consistency or inconsistency:
- If |A| ≠ 0, then the system of equations is consistent and has unique solution X = A-1B.
- If |A| = 0 and Adj(A) B = 0, then the system is consistent and has infinitely many solutions.
- If |A| = 0 and Adj(A) B = ≠ 0, then the system is inconsistent.
Questions:
- Solve the following system of equations, using matrix method:
x + 2y + z = 7, x + 3z = 11, 2x - 3y = 1 - Solve the following system of equations, using matrix method:
x + y + z = 6, x - y + z = 2, 2x + y - z = 1
Solution of homogeneous system of linear equations
- Let AX = 0 is a homogeneous system of linear equations in n variables , then
Case 1: When |A| ≠ 0 i.e. the coefficient matrix A is non-singular, then the system is consistent and has unique solution i.e. zero solution.
Case 2: When |A| = 0 i.e. the coefficient matrix A is singular, then the system is consistent and has infinitely many solutions.
Example: Solve the following system of equations by matrix method
2x - y + z = 0, 3x + 2y - z = 0, x + 4y + 3z = 0.
Solution: Here we can write the matrix form of the system of equations as
AX = 0, where
_ _ _ _ _ _
| 2 1 1 | | x | | 0 |
A = | 3 2 -1 |, X = | y | and 0 = | 0 |
| 1 4 3 | | z | | 0 |
- - - - - -
|A| = 2(6 + 4) - 1(9 + 1) + 1(12 - 2) = 20 - 10 + 10 = 20 ≠ 0
Thus |A| ≠, so the system of equations is consistent and has zero solution i.e.
x = 0, y = 0, z = 0
Applications of Determinants to solve a system of linear equations:
Consider a system of linear equations
a11 x1 + a12 x2 + ... + a1n xn = b1
a21 x1 + a22 x2 + ... + a2n xn = b2
. . . .
. . . .
. . . .
an1 x1 + am2 x2 + ... + ann xn = bn
In matrix form, we can write this system of equations as
AX = B, where
_ _
A = | a11 a12 a13 ... ... ... a1n |
| a21 a22 a23 ... ... ... a2n |
| ... ... ... ... ... ... ... |
| ... ... ... ... ... ... ... |
| an1 an2 an3 ... ... ... ann | nxn
- -
_ _
X = | x1 |
| x2 |
| . |
| . |
| xn | nx1
- -
_ _
B = | b1 |
| b2 |
| . |
| . |
| bn | nx1
- -
A system of linear equations may have a unique solution or many solutions or no solutions at all.
If it has a solution (whether unique or many), the system is said to be consistent.
If it has no solution, it is called an inconsistent system.
If B = 0 (i.e. b1 = b2 = b3 ... ... ... = bn = 0) the system is said to be homogenous other wise it is called a non-homogeneous system of equations.
Solution of a non-homogeneous system of linear equations (ramer's rule)
_ _
| a11 a12 a13 ... ... ... a1n |
▵ = | a21 a22 a23 ... ... ... a2n |
| ... ... ... ... ... ... ... |
| ... ... ... ... ... ... ... |
| an1 an2 an3 ... ... ... ann |
- -
_ _
| b1 a12 a13 ... ... ... a1n |
▵1 = | b2 a22 a23 ... ... ... a2n |
| ... ... ... ... ... ... ... |
| ... ... ... ... ... ... ... |
| bn an2 an3 ... ... ... ann |
- -
_ _
| a11 b1 a13 ... ... ... a1n |
▵2 = | a21 b2 a23 ... ... ... a2n |
| ... ... ... ... ... ... ... |
| ... ... ... ... ... ... ... |
| an1 bn an3 ... ... ... ann |
- -
_ _
| a11 a12 b1 ... ... ... a1n |
▵3 = | a21 a22 b2 ... ... ... a2n |
| ... ... ... ... ... ... ... |
| ... ... ... ... ... ... ... |
| an1 an2 bn ... ... ... ann |
- -
_ _
| a11 a12 a13 ... ... ... b1 |
▵n = | a21 a22 a23 ... ... ... b2 |
| ... ... ... ... ... ... ... |
| ... ... ... ... ... ... ... |
| an1 an2 an3 ... ... ... bn |
- -
x1 = ▵1/▵, x2 = ▵2/▵, x3 = ▵3/▵, ... ... ... and xn = ▵n/▵,
The determinants ▵1, ▵2, ▵3, ... ... ▵n are obtained by replacing 1st, 2nd, 3rd, ..., nth column in ▵ by
the column of the right hand side of the given equations i.e. B = - -
| b1 |
| b2 |
| . |
| . |
| bn |
- -
Conditions of consistency:
- Case 1 : If ▵ = 0 and one of ▵1, ▵2, ▵3, ..., ... ▵n is non-zero, the system of equations is inconsistent.
- Case 2: If ▵ = 0 and ▵1 = ▵2 = ▵3 = ... = ▵n = 0, then the system of equations may or may not be consistent. However, it is consistent, and then it has infinitely many solutions.
- Case 3: If ▵ ≠ -, then the given system of equations is consitent and has a unique
solution given by
x1 = ▵1/▵, x2 ▵2/▵1, x3 = ▵3/▵, ... ... ... and xn = ▵n/▵1.
x1 = ▵1/▵
Questions:
- Solve by Cramer's rule 5x - 7y + z = 11, 6x - 8y - z = 15 and 3x + 2y - 6z = 7.
Eigen values
- A.K.A proper values, latent values, characteristic values and spectral values.
- Let A = [aij]nxn be a square matrix of order n over the field F. An element λ in F is called an Eigen values of A if |A - λI| = 0 where, I is the unit matrix of order n.
Procedure:
_ _
let A = | 2 2 1 |
| 1 3 1 |
| 1 2 2 |
- -
characteristic matrix
_ _ _ _ _ _
[A - λI] = | 2 2 1 | | 1 0 0 | | 2-λ 2 1 |
| 1 3 1 | - λ | 0 1 0 | = | 1 3-λ 1 |
| 1 2 2 | | 0 0 1 | | 1 2 2-λ |
- - - - - -
characteristic polynomial
|A - λI| = | 2-λ 2 1 |
| 1 3-λ 1 |
| 1 2 2-λ |
characteristic equation
|A - λI| = 0
-λ^3 + 7λ^2 - 11λ + 5 = 0
λ^3 - 7λ^2 + 11λ - 5 = 0
characteristics values / Eigen values
λ = 1, 1, 5
Important points on Eigen values: (shortcuts)
- The characteristics equation of the matrix A of order 2 can be obtained from
λ2 - S1λ + S2 = 0
Where S1 = sum of principal diagonal elements (trace)
S2 = determinant A
_ _ Example | 3 2 | | -1 0 | - - S1 = 3 + 0 S2 = | 3 2 | = 0 - (-1*2) = 2 | -1 0 | characteristic equation = λ2 - 3λ + 2 = 0 - The characteristics equation of the matrix A of order 3 can be obtained from
λ3 - S1λ2 + S2λ - S3 = 0
Where S1 = sum of principal diagonal elements (trace)
S2 = sum of minors of principal diagonal elements
S3 = |A|
_ _ Example | 0 1 0 | | 0 0 1 | | 4 -17 8 | - - S1 = 0 + 0 + 8 = 8 S2 = | 0 1 | + | 0 0 | + | 0 1 | = 17 + 0 + 0 = 17 | -17 8 | | 4 8 | | 0 0 | S3 = -4 characteristic equation = λ^3 - 8λ^2 + 17λ + 4 = 0 - The sum of Eigen values of a matrix is the sum of its principal diagonal elements (or trace of
the matrix)
_ _ | 0 1 0 | | 0 0 1 | | 4 -17 8 | 3x3 - - as the order is 3 so there will be 3 Eigen values λ1, λ2, λ3
Questions
Find the Eigen values of matrix
_ _
A = | 3 2 |
| -1 0 |
- -
Find the Eigen values of matrix
_ _
A = | 0 1 0 |
| 0 0 1 |
| 4 -17 8 |
- -
Eigen Vector
- Definition: Let λ be an Eigen value of n square matrix A, then a non-zero matrix X of the
order
nx1 (i.e. column matrix) such that (A - λI)X = 0
X is called an Eigen vector of A corresponding to that Eigen value λ
Procedure of finding Eigen vectors:
If λ is an Eigen value of A then the corresponding Eigen vector of A will be given by a non-zero vector
_ _
X = | x1 |
| x2 |
| . |
| . |
| xn |
- -
Satifying the equation
(A - λI)X = 0
= AX = λX (short cut trick as λI = λ)
Properties of Eigen vectors
- The given vector X of a matrix A is not unique but ratio will be same
Example _ _ _ _ X1 = | 1 | X2 = | 2 | | 2 | | 4 | - - - -
Questions
(Non-repeated Eigen values)
1- Find the Eigen values and Eigen vectors of the matrix
_ _
| 3 1 4 |
A = | 0 2 6 |
| 0 0 5 |
- -
(Repeated Eigen values)
2- Find the Eigen values and Eigen vectors of the matrix
_ _
| -2 2 -3 |
A = | 2 1 -6 |
| -1 -2 0 |
- -
Normalized form of vector
_ _
To find the normalized form of | a | , we will divide each element by √(a^2 + b^2 + c^2)
| b |
| c |
- -
Example: _ _ _ _
Normalized form of | 1 | is | 1/3 | as √(1^2 + 2^2 + 2^2) = √9 = 3
| 2 | | 2/3 |
| 2 | | 2/2 |
- - - -
Cayley Hamilton theorem
- Every square matrix satisfies its own characteristic equation.
- Let A be a sqaure matrix and λ3 - 2λ2 + 3λ - 4 = 0 be its
characteristic equation then accorting to Cayley Hamilton theorem,
A3 - 2A2 + 3A - 4I = 0 - To find A-1 we use Cayley Hamilton theorem
Multiply A-1 on both side we get
A2 - 2A + 3I - 4A-1 = 0
4A-1 = A2 - 2A + 3I
A-1 = 1/4(A2 - 2A + 3I)
Questions
_ _
Verify CHT and find inverse of A = | 1 4 | and find its inverse.
| 2 3 |
- -
Find the characteristic equation of matrix
_ _
| 2 -1 1 |
A = | -1 2 -1 | and verify that it is staified by A and hence find A^-1.
| 1 -1 2 |
- -
Find the characteristic equation of matrix
_ _
| 2 1 1 |
A = | 0 1 0 |
| 1 1 2 |
- -
and hence find the matrix represented by A^8 - 5A^7 +7A^6 - 3A^5 + A^4 -5A^3 + 8A^2 - 2A + I
_ _
Using CHT find A^8 if A = | 1 2 |
| 2 -1 |
- -