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7 Matrices
7.1 Introduction
Definition. A m × n matrix is a rectangular array of numbers having m rows and n columns.
A =
a 11 a 12... a 1 n a 21 a 22... a 2 n a 31 a 32... a 3 n
.... .... am 1 am 2... amn
The element aij represents the entry in the ith row and jth column. We sometimes denote A by (aij )m×n.
7.2 Matrix Operations
Addition
We can only add two matrices of the same dimension i.e. same number of rows and columns. We then add element-wise.
Example. (^) ( 2 1 1 3
Scalar Multiplication
If c is a real number and A = (aij )m×n is a matrix then cA = (caij )m×n.
Example.
5
Matrix Multiplication
Given a matrix A = (aij )m×n and a matrix B = (bij )r×s, we can only multiply them if n = r. In such a case the multiplication is defined to be the matrix
C = (cij )m×s as follows:
cij =
∑^ n
k=
aikbkj.
We may view the cij th element as the dot product of the ith row of the matrix A and jth column of the matrix B.
Example. 1. ( 2 1 1 3
7.4 Determinants
Definition. The determinant of a 2 × 2 matrix is defined as follows: ∣ ∣ ∣ ∣
a b c d
∣ =^ ad^ −^ bc.
Definition. The determinant of a 3 × 3 matrix is defined as follows:
∣ ∣ ∣ ∣ ∣ ∣ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33
= a 11
a 22 a 23 a 32 a 33
∣ −^ a^12
a 21 a 23 a 31 a 33
∣ +^ a^13
a 21 a 22 a 31 a 33
= a 11 (a 22 a 33 − a 32 a 23 ) − a 12 (a 21 a 33 − a 31 a 23 ) + a 13 (a 21 a 32 − a 31 a 22 )
Definition. The definition of determinants can be generalized for a square matrix of any size. Let A = (aij ) be a square n × n matrix. Fix a row i. Then the determinant is defined to be,
det (A) =
∑^ n
j=
(−1)i+j^ aij det (Aij ),
where Aij is the minor matrix obtained by deleting row i and column j. It is called the ij Cof actor of A. T
Example. Find the determinant of
A =
Choose row i = 1.
A 11 =
det (A 11 ) = − 2
A 12 =
det (A 11 ) = − 6
A 13 =
det (A 11 ) = − 2
A 14 =
det (A 11 ) = − 2
det (A) =
∑^ n
j=
(−1)1+j^ a 1 j det (A 1 j )
= (−1)^2 (1)det (A 11 ) + (−1)^3 (1)det (A 12 ) + (−1)^4 (0)det (A 13 ) + (−1)^5 (−1)det (A 14 ) = (1)(1)(−2) + (−1)(1)(−6) + (1)(0)(−2) + (−1)(−1)(−2) = 2
We can find the inverse in another way: Write as (A|I 2 ) ( 1 1 1 0 1 2 0 1
Then use matrix row operations to get it into the form (I 2 |B). ( 1 0 a b 0 1 c d
You will find that B = A−^1. ( 1 1 1 0 1 2 0 1
R 2 =R 2 −R 1 →
R 1 =R 1 −R 2 →
Observe that the matrix obtained ( a b c d
is the inverse of A.
Inverse of a 3 × 3 matrix
The method is the same as in the 2 × 2 case:
- Check determinant A is non-zero.
- Rewrite as (A|I 3 ).
- Use row operations to put in the form (I 3 |A−^1 ).
Example. Let A =
. Find A−^1.
- Check that det(A) 6 = 0.
det(A) = 1(0 − (−3)(1)) − (−1)(0 − (−2)(1)) + 1(0 − (−2)(−2)) = 3 + 2 − 4 = 1 6 = 0
Therefore the inverse exists.
- Rewrite as (A|I 3 ). (^)
- Use row operations to put in the form (I 3 |A−^1 ).
R^3 =R →^3 +2R^1
R 2 = − 21 R 2 →
R 3 =R 3 +5R 2 →
R 3 =− 2 R 3 →
R 2 =R 2 + 12 R 3 →
R 1 =R 1 −R 3 →
R 1 =R 1 +R 2 →
A−^1 =
Steps to find eigenvalues and eigenvectors:
- Form the characteristic equation
det(λI − A) = 0.
- To find all the eigenvalues of A, solve the characteristic equation.
- For each eigenvalue λ, to find the corresponding set of eigenvectors, solve the linear system of equations
(λI − A)~x = 0
Example. 1. A =
. Find the eigenvalues and eigenvectors of A.
Solution: • We will find the characteristic equation of A. We will first find the matrix λI − A.
λI − A = λ
λ 0 0 λ
λ − 4 2 1 λ − 3
The determinant of this matrix gives us the characteristic polyno- mial.
det(λI − A) = 0 ∣ ∣ ∣ ∣
λ − 4 2 1 λ − 3
∣ =^0
(λ − 4)(λ − 3) − 2 = 0 λ^2 − 7 λ + 12 − 2 = 0 λ^2 − 7 λ + 10 = 0
- To find all the eigenvalues of A, solve the characteristic equation.
λ^2 − 7 λ + 10 = 0 (λ − 5)(λ − 2) = 0
So we have two eigenvalues λ 1 = 2, λ 2 = 5.
- For each eigenvalue λ, to find the corresponding set of eigenvectors, we simply solve the linear system of equations given by, (λI − A)~x = 0.
case(i) λ 1 = 2.
(2I − A)~x = 0 ( 2 − 4 − 2 − 1 2 − 3
x 1 x 2
x 1 x 2
− 2 x 1 − 2 x 2 −x 1 − x 2
We get two equations: 2 x 1 + 2x 2 = 0 x 1 + x 2 = 0 Both equations give the relation x 1 = −x 2. Therefore the set of eigenvectors corresponding to λ = 2 is given by:
{
x 1 x 2
|x 1 = −x 2 }
−x 2 x 2
{ x 2
|x 2 is a real number.}
An eigenvector corresponding to λ 1 = 2 is
case(i) λ 2 = 5.
(5I − A)~x = 0 ( 5 − 4 − 2 − 1 5 − 3
x 1 x 2
x 1 x 2
x 1 − 2 x 2 −x 1 + 2x 2
