# Linear Algebra

**Exercise 1** Solve the system using matrix method:

x_{1} + x_{2} = 1

x_{1} - x_{2} = 3

-x_{1} + 2x_{2} = -2

**Exercise 2** Solve the system using matrix method

-x_{1} + x_{2} - x_{3} + 3x_{4} = 0

3x_{1} + x_{2} - x_{3} - x_{4} = 0

2x_{1} - x_{2} - x_{3} - x_{4} = 0

**Exercise 3** In the downtown section of a certain city, two sets of one way

streets intersect as shown below. The average hourly volume traffic entering and

leaving this section during rush hour is given in the diagram. Determine the

amount of traffic between each of the four intersections.

**Rank,** Uniqueness and Consistency As the exercises suggest, there are

two fundamental questions that we may want to address when we are solving

for a system of linear equations

1) Is the system consistent? (Does a solution exist?)

2) If a solution exists, is it unique? (Is there one and only one

solution?)

We can determine whether one infinite, or no solutions exist if we know (1)

the number of equations m, (2) the number of unknowns n, and (3) the rank of

the matrix representing the linear system.

**Rank: **number of nonzero rows in its row echelon form.

**Example 4** Rank = 3

**Example 5** Rank = 2

**Example 6** Rank = 4

Let A be the coefficient matrix and
be the augmented matrix.

Then,

1. rank A ≤ rank Â

Augmenting A with b can never result in more zero rows than originally

in A itself. Suppose row i in A is all zeros and that bi is non-zero. Aug-

menting A with b will yield a non-zero row i in Â.

2. rank A ≤ rows A

By definition of a rank.3

3. rank A ≤ columns A

Suppose there are more rows than columns (otherwise the previous rule

applies). Each column can contain at most one pivot. By pivoting, all

other entries in a column below the pivot are zeroed. Hence, there will

only be as many non-zero rows as pivots, which will equal the number of

columns.

**Existence of solutions:**

1. Exactly one solution

rank A = rank Â = rows A = columns A

Necessary condition for a system to have a unique solution: that there be

exactly as many equations as unknowns.

2. Infinite solutions

rank A = rank; and columns A > rank A

If a system has a solution and has more unknowns than equations, then

it has infinitely many solutions.

3. No solution

rank A < rank Â

Then there is a zero-row i in A's reduced echelon that corresponds to a

non-zero row i in Â's reduced echelon. Row i of the translates to the

equation:

Where . Hence, the system has no solution.

**Exercise 7 **

x_{1} + 2x_{2} + x_{3} = 1

2x_{1} + 4x_{2} + 2x_{3} = 3

**Exercise 8 **

x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 2

x_{1} + x_{2} + x_{3} + 2x_{4} + 2x_{5} = 3

x_{1} + x_{2} + x_{3} + 2x_{4} + 3x_{5} = 2

## 2 Linear Algebra: Module 2

**2.1 Module Topics:**

Linear Independence

Matrix Algebra

Inverse

Determinants

Cramers Rule

Eigenvalues & Eigenvectors

**2.1.1 Linear Independence**

Let v_{1}, v_{2}, …, v_{n} be a set of n vectors each of which is of order m. Then the
set

of vectors is linearly dependent if there exist scalars a_{1}, a_{2}, …, a_{n} at least
one of

which is not 0 such that

a_{1}v_{1} + a_{2}v_{2} + a_{3}v_{3} + … + a_{n}v_{n} = 0

Or

**Example 9 **Is there linear dependence?

**Hint: **We can try to augment these vectors into a matrix and solve for a

reduced row echelon form and see if we can find some relationship between the

rows. If we can determine some functional linear relationship between these

vectors, we can say that there is linear dependence.

**2.1.2 Matrix Algebra**

**Matrix: **A matrix is an array of mn real numbers arranged in m rows by n

column.

Matrix addition (or subtraction): Let A and B be two m
× n matrices.

Then

Note: A and B must be the same size!

**Example 10**

**Example 11
**

**Example 12
**

**Scalar Multiplication:** Given the scalar s, the scalar multiplication of sA

is

**Matrix multiplication: **If A is an m
× n matrix and B is a k × n matrix,

then their product C = AB is an m × n matrix where

So

If and then, BA =?

Note: the number of columns of the first matrix must equal the number of

rows of the second matrix. The size of the matrices (including the resulting

product) must be:

(m × k)(k × n) = (m × n)

Using matrix multiplication, we can now write the linear equation.

Let A be a m × n coefficient matrix

x be a m × 1 vector array of unknowns

b be a m × 1 vector array of constants

Then,