259 Lecture 14

Elementary Matrix Theory

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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 = 0ce + dg = 0cf + 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 ande = 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 = 13x + 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).