Transcript PowerPoint
159 Lecture 19
Elementary Matrix Theory
Matrix Definition
A matrix is a
rectangular array
of elements
(usually numbers)
written in rows and
columns.
Example 1: Some
matrices:
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Matrix Definition
Example 1 (cont.):
Matrix A is a 3 x 2
matrix of integers.
A has 3 rows and 2
columns.
Matrix B is a 2 x 2
matrix of rational
numbers.
Matrix C is a 1 x 4
matrix of real numbers.
We also call C a row
vector.
A matrix consisting of a
single column is often
called a column
vector.
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Matrix Definition
Notation:
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Arithmetic with Matrices
Matrices of the same size (i.e. same
number of rows and same number of
columns), with elements from the
same set, can be added or
subtracted!
The way to do this is to add or
subtract corresponding entries!
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Arithmetic with Matrices
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Arithmetic with Matrices
Example 2: For matrices A and B
given below, find A+B and A-B.
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Arithmetic with Matrices
Example 2 (cont):
Solution:
Note that A+B and A-B are the same
size as A and B, namely 2 x 3.
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Arithmetic with Matrices
Matrices can also be multiplied. For AB to make sense,
the number of columns in A must equal the number of
rows in B.
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Arithmetic with Matrices
Example 3: For matrices A and B
given below, find AB and BA.
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Arithmetic with Matrices
Example 3 (cont.):
A x B is a 3 x 2 matrix. To get the row i, column j entry of this matrix,
multiply corresponding entries of row i of A with column j of B and add.
Since B has 2 columns and A has 3 rows, we cannot find the product
BA (# columns of 1st matrix must equal # rows of 2cd matrix).
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Arithmetic with Matrices
Another useful operation with
matrices is scalar multiplication,
i.e. multiplying a matrix by a number.
For scalar k and matrix A, kA=Ak is
the matrix formed by multiplying
every entry of A by k.
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Arithmetic with Matrices
Example 4:
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Identities and Inverses
Recall that for any real number a,
a+0 = 0+a = a and (a)(1) = (1)(a) = a.
We call 0 the additive identity and 1 the
multiplicative identity for the set of real
numbers.
For any real number a, there exists a real
number –a, such that
a+(-a) = -a+a = 0.
Also, for any non-zero real number a, there
exists a real number a-1 = 1/a, such that
(a-1)(a) = (a)(a-1) = 1.
We all –a and a-1 the additive inverse and
multiplicative inverse of a, respectively.
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Identities and Inverses
For matrices, we also have an additive identity
and multiplicative identity!
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Identities and Inverses
A+0 = 0+A = A and AI = IA = A holds.
(HW-check!)
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Identities and Inverses
Clearly, A+(-A) = -A + A = 0 follows! Note also that B-A
= B+(-A) holds for any m x n matrices A and B.
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Identities and Inverses
Example 5:
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Identities and Inverses
Example 5 (cont):
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Identities and Inverses
Example 5 (cont.)
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Identities and Inverses
Example 5 (cont.)
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Identities and Inverses
Example 5 (cont):
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Identities and Inverses
For multiplicative inverses, more work
is needed.
For example, here is one way to find
the matrix A-1, given matrix A, in the
2 x 2 case!
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Identities and Inverses
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Identities and Inverses
From the first matrix equation, we see that e, f, g, and h
must satisfy the system of equations:
ae + bg = 1
af + bh = 0
ce + dg = 0
cf + dh = 1.
It follows that if e, f, g, and h satisfy this system, then the
second matrix equation above also holds!
Solving the system of equations, we find that ad-bc 0
must hold and
e = d/(ad-bc),
f = -b/(ad-bc),
g = -c/(ad-bc),
h = a/(ad-bc).
Thus, we have the following result for 2 x 2 matrices:
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Identities and Inverses
In this case, we say A is invertible.
If ad-bc = 0, A-1 does not exist and
we say A is not invertible.
We call the quantity ad-bc the
determinant of matrix A.
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Identities and Inverses
Example 6: For matrices A and B
below, find A-1 and B-1, if possible.
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Identities and Inverses
Example 6 (cont.)
Solution: For matrix A, ad-bc = (1)(4)-(2)(3)= 4-6 =
-2 0, so A is invertible. For matrix B, ad-bc = (3)(2)(1)(6) = 6-6 = 0, so B is not invertible.
HW-Check that AA-1 = A-1A = I!!
Note: For any n x n matrix, A-1 exists, provided the
determinant of A is non-zero.
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Linear Systems of Equations
One use of matrices is to solve
systems of linear equations.
Example 7: Solve the system
x + 2y = 1
3x + 4y = -1
Solution: This system can be written
in matrix form AX=b with:
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Linear Systems of Equations
Example 7 (cont.)
Since we know from Example 6 that
A-1 exists, we can multiply both sides
of AX = b by A-1 on the left to get:
A-1AX = A-1b => X = A-1b.
Thus, we get in this case:
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Linear Systems of Equations
Example 7 (cont.):
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References
Elementary Linear Algebra (4th ed) by
Howard Anton.
Cryptological Mathematics by Robert
Edward Lewand (section on
matrices).
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