linear equations in one variable

consider ax = b where a and b are real numbers

there are then 3 cases:

1. a ≠ 0: unique solution → x = b/a

2. a = 0: if b = 0 then any x E R solutions (infinitely many solutions)

3. a = 0: if b ≠ 0 then the equation 0x = b has no solution

solving systems of linear equations with 2 unknowns

method 1

1. change one equation to be one variable = to another equation in terms of the other variable

2. plug into the other equation

3. solve for the variable

4. use the variable you’ve solved for to solve for the other variable

e.g.

1) x - y = 3 → x = 3+y plug into 2) for x

2) x + y = 1

plug in:

(3+y) + y = 1

3 + 2y = 1

2y = -2

y = -1

x = 3+y = 3-1 = 2

therefore: x =2, y = -1

method 2 (substraction)

1. add/subtract the equations should there be a variable that can be cancelled - need variables that cancel each other out

e.g.

1) x - y = 3

2) x + y = 1

add 1) and 2)

x - y = 3

x + y = 1

2x + 0y = 4

2x = 4

x = 1

plug into equation:

x - y = 3

2 - y = 3

y = -1

therefore: x =2, y = -1

row operations

1. multiply rows by scalars

2. swap rows

3. add and subtract rows or multiples of rows

classifying matrices: m x n - rows x columns

sum of 2 matrices

• a and b have to be the same size

multiplication of a matrix by a scalar

• multiply all entries by the scalar

product of 2 matrices

must have the same inner value (e.g. 2 × 1 • 1 × 3 is defined, 1 × 3 • 2 × 1 is not defined)

matrix multiplication isn’t communitive

let a, b, c be matrices and suppose the operations are defined (rows and columns “agree”) then:

1. a + b = b + a

2. a + (b + c) = (a + b) + c

3. a(bc) = (ab)c

4. a (b + c) = ab + ac

5. (b+c)a = ba + ca

a different way to look at systems

x + y = 1

2x - 3y = 5

can be written as

inverses

formula for the inverse of a 2 × 2 matrix

• multiplying a matrix by its inverse gives you the identity

1. let a and b be nxn matrices and suppose ab = Inxn, then ba = Inxn

2. if a and b are nxn matrices such that ab = Inxn

   then b = a⁻¹ and b⁻¹ = a

   (a⁻¹)⁻¹ = a

3. let a and b be invertible nxn matrices then ab is invertible then

   (ab)⁻¹ = b⁻¹a⁻¹

4. let a be an invertible nxn matrix and z ≠ 0 be a real number. then za is invertible and (za)⁻¹ = 1/z a⁻¹

5. x Inxn = Inxn x = x

linear optimization (graphing to maximize)

maximize the equation

subject to (st)) constraints (inequalities)

1. set the constraints = (change inequality to =)

2. find x and y intercepts

3. draw the lines of the constraints

4. use a point not on the line to test if statement is true or not while considering inequalities (for the constraints)- if true, shade on the side of that point → use the origin most of the time for simple math

5. then calculate the points of intersection → from here we can get every corner point of the shaded/allowed region

6. test corner points by plugging them into the original equation and see what’s the highest/lowest output for max/min

simplex tableau

1. change the maximize equation to have z in it by making other variables negative and setting to zero (e.g. if x1 + 5x2 + 2x3 = z then change to 0 = -x1 - 5x2 -2x3 + z)

2. change the constraints by making the inequalities into equal signs and adding slack variables (s) - add a different slack variable to the different constraints; s1, s2…

3. put this into a table, put the equation with z in the bottom and the =s in like an augmented matrix

4. choose column with most negative entry in the bottom

5. take ratios of the end column over the selected column, choose the smaller number as your pivot

6. make the pivot 1 and clear out the other rows in that column to be zero - do this using regular row operations

7. keep doing this until you have no negative entries in the bottom

8. read the table

standard maximum form

a linear program is said to be in standard max form if

1. the objective function is to be maximized

2. x1…xn ≥ 0

3. the constraints of the form a1x1 +…+anxn ≤ b where b≥0

   to every problem in standard max form, an initial simplex tableau is associated. solving a linear problem in 2 variables can be done graphically

   • if the number of variables is ≥3, a graphical solution may be difficult

   • simplex tableau provides a way of solving linear programs in standard max form

standard minimum form

a linear program is in standard minimum form if

1. the linear objective function is to be minimized (w = c1,…,cn E R)

2. all variables are non negative (y1,…, yn ≥ 0)

3. constraints are of the form a1y1 +a2y2+…+anyn ≥ b with b≥0

a transpose

let a be an mxn matrix A^t (A transpose) is the matrix obtained by turning the I-th row of A into the I-th column of A^t → A^t is an nom matrix

(A^t)^t = A if a is an nxn matrix, and A = A^t then A is called symmetric

changing from min to max (the dual problem)

1. without considering slack variables, write data in an augmented matrix

2. take the transpose of such matrix

3. rewrite question given new info and use different variables → this is the dual of the original problem

rules for invertibility

• no row of zeros on middle

• must be square

• if 2×2, must not have ad-bc = 0