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An introduction to matrices, including their definition, operations such as addition, scalar multiplication, and matrix multiplication, special matrices like square, transpose, symmetric, diagonal, and identity matrices, determinants, and inverses.
What you will learn
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Definition. A m × n matrix is a rectangular array of numbers having m rows and n columns.
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.
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
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
Choose row i = 1.
det (A 11 ) = − 2
det (A 11 ) = − 6
det (A 11 ) = − 2
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:
Example. Let A =
. Find A−^1.
det(A) = 1(0 − (−3)(1)) − (−1)(0 − (−2)(1)) + 1(0 − (−2)(−2)) = 3 + 2 − 4 = 1 6 = 0
Therefore the inverse exists.
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 →
Steps to find eigenvalues and eigenvectors:
det(λI − A) = 0.
(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
(λ − 4)(λ − 3) − 2 = 0 λ^2 − 7 λ + 12 − 2 = 0 λ^2 − 7 λ + 10 = 0
λ^2 − 7 λ + 10 = 0 (λ − 5)(λ − 2) = 0
So we have two eigenvalues λ 1 = 2, λ 2 = 5.
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