Matrix Operations and Properties in Mathematics

Econ 213

Todays Class Inverse of a Matrix Determinant of a Matrix Matrix Solution of a system of equations Use of Cramer s Rule

Multiplicative Identity The multiplicative identity for real numbers is 1. The property is written as: a x 1 = 1 x a = a In terms of matrices, we need a matrix that can be multiplied by a matrix (A) and give a product which is the same matrix (A) This matrix exists and it is called the identity matrix. It is named I and it comes in different sizes It is a square matrix with all 1 s on the main diagonal and all other elements are 0

Identity Matrices 1 0 0 1 0 = 2I = 0 1 0 3I 0 1 0 0 1 1 0 0 0 0 1 0 0 = I 0 0 1 0 0 0 0 1

Given the matrix A below multiply AI 2 5 = A 4 0 2 5 1 0 4 0 0 1

The identity Matrix for Multiplication Let A be a square matrix with n rows and n columns. Let I be a square matrix with the same dimensions with 1 s on the main diagonal and 0 s elsewhere Then AI = IA = A

The multiplicative Inverse For every non-zero real number a, there is a real number 1/a such that a(1/a)=1 In terms of matrices, the product of a square matrix and its inverse is I + + 3 1 1 1 ) 1 ( 3 ( 1 ) 2 ( 3 ) 1 ) 3 ( 1 1 0 = = + + 2 1 2 3 ) 1 ( 2 ( 1 ) 2 ( 2 ) 1 ) 3 ( 1 0 1

The inverse of a Matrix Let A be a square matrix with n rows and n columns In terms of matrices, the product of a square matrix and its inverse is I + + 3 1 1 1 ) 1 ( 3 ( 1 ) 2 ( 3 ) 1 ) 3 ( 1 1 0 = = + + 2 1 2 3 ) 1 ( 2 ( 1 ) 2 ( 2 ) 1 ) 3 ( 1 0 1

The inverse of a Matrix Let A be a square matrix with n rows and n columns. If there is an n x n matrix B such that AB = I and BA = I , then A and B are inverses of one another. The inverse of a matrix A is denoted by A-1.

Inverse of a Matrix To show that matrices are inverses of one another, show that the multiplication of the matrices is commutative and the results is the identity matrix. Example: Show that A and B are inverses of each other. 5 3 2 3 5 3 = = A and B 3 2

2 3 5 3 = AB 3 5 3 2 + + ) 5 ( 2 ( 3 ) 3 ( 2 ) 3 ) 2 ( 3 = + + ) 5 ( 3 ) 3 ( 5 ( 3 ) 3 ) 2 ( 5 1 0 = 0 1

5 3 2 3 = BA 3 2 3 5 + + ) 2 ( 5 ( ) 3 )( 3 ) 3 ( 5 ( ) 5 )( 3 = + + ) 2 ( 3 ) 3 ( 2 ) 3 ( 3 ) 5 ( 2 1 0 = 0 1

Finding the Inverse of a Matrix: Method 1 Use the equation AB = I 1 2 a b = = Let A and B 3 5 c d Write and Solve the equation 1 2 1 0 a b = 3 5 0 1 c d

1 2 1 0 a b = 3 5 0 1 c d + + 2 2 1 0 a c b d = + + 3 5 3 5 0 1 a c b d + a = + b = 2 1 2 0 a c b d + = + = d 3 = 5 0 c 3 = 5 and 1 c d a b = = 5 3 2 1 and

So the inverse of A is 5 2 3 1 We can check this my multiplying A x A-1 + + 1 2 5 2 ( 1 ) 5 ) 3 ( 2 ) 2 ( 1 ( 2 ) 1 = 0 ( 3 + + 3 5 3 1 ) 5 ) 3 ( 5 ) 2 ( 3 ( 5 ) 1 1 = 0 1

Method 2: Determinants Each matrix can be assigned a real number called the determinant of the matrix. It is denoted by the symbol d c a b = If A a b means the determinant of A c d

The determinant of a 2 x 2 matrix is found as follows: ad d c a b = cb 7 8 = G Find the determinant of the matrix 6 7 7 8 = ) 8 ( 6 = = ) 7 ( 7 49 48 1 6 7

1 1 Find the determinant of the matrix = H 2 2 1 1 ) 2 ( 1 = ) 1 ( 2 = 0 2 2 If the determinant of a matrix is 0, the matrix does not have an inverse. The matrix is then said to be invertible

Determinants can be used to find the inverse of a matrix. a b = ( ) , 0 If A and det A then c d d b 1 = 1 A c a ( ) det A

is called the adjoint of the original d b c a Matrix It is found by found by switching the entries on the main diagonal and changing the signs of the entries on the other diagonal.

Find the multiplicative inverse of: 1 2 1 2 = A ) 4 ( 1 = ) 2 ( 3 = 2 3 4 3 4 3 2 1 4 2 1 = = 1 A 1 3 1 2 2 2

We can check to see if we are correct by multiplying. Remember that AA-1 = I 3 2 1 1 2 ) 2 1 3 4 2 2 + + ( 1 / 3 ( 2 ) 2 ) 1 ( 1 ( 2 / 1 ) 2 = ( 3 + + ) 2 / 3 ( 4 ) 2 ) 1 ( 3 ( 4 / 1 ) 2 1 0 = 0 1

Find the inverse using determinants 2 1 1 3 0 3 1 1 4 8 2 4

Determinant of a 3 x 3 Matrix One way to find the determinant of a 3 x 3 is the formula below: a b c e f d f d e = + d e f a b c h i g i g h g h i

Find the determinant using the formula 2 0 5 3 1 5 0 2 4 2 0 5 1 5 3 5 3 1 = + 3 1 5 2 0 5 2 4 0 4 0 2 0 2 4 ) 4 ( 1 2 ) = ) 5 ) 4 ( 3 ) 5 + ) 2 ( 3 ) 1 ( 0 ( 2 0 ( 0 5 = + ) 6 ( 2 14 ) ( 0 12 ( 5 = 28 30 = 2

Find the determinant 2 1 1 1 2 3 4 1 2 2 1 1 2 3 1 3 1 2 = ) 1 + 1 2 3 2 ( 1 1 2 4 2 4 1 4 1 2 ) 2 ( 1 1 ) 1 ( 1 1 = ) 2 ( 2 + ) 3 ( 4 + ) 2 2 ) 3 ( 1 ( 4 = ( 2 ) 7 ( 1 + ) 9 ( 1 + 10 ) = + 14 10 9 = 13

Systems of Equations Matrices can be used to find the solutions of systems of equations. First, the system of equations must be put in the matrix form Consider a system of equations with two unknowns, ?1 and ?2. a?1+ b?2 = e c?1 + d?2 = f

This can be represented in matrix form as follows: Ax=B where the Matrix A is called the matrix of coefficients written as a b 1 x A= x= C= e f c d 2 x

Cramers Rule (2 x 2) Cramer s rule is a method that uses determinants to solve a system of equations. This method was named after the Swiss mathematician Gabriel Cramer (1704-1752)

+ + = = + + + = = ce bf ce ax dx by ey c f + = = aex bdx bey bey bey ce bf a x dx e b y e c e bf Lets eliminate y b = = aex bdx x(ae bd) bf c f a d b e b e ce ae bf bd = x = x Look familiar?

If you apply the same process but eliminate x a d a d c f b e af ae cd bd = y = y So, what does Cramer s Rule say?

ax + by = e is (x,y) In general the solution to the system cx + dy = f b e f d x= b a c d b a where and = 0 c d e a If we let A be the coefficient matrix of the linear system, notice this is just det A. c f y= b a c d

Cramers Rule Given a system + + = = ax dx by ey c f Replace solutions in y column to solve for y Replace solutions in x column to solve for x a d a d c f b e c f a d b e b e = = y x Denominators are coefficient deterimanants

Examples Solve using Cramer s Rule + = = 5 6x 7y x 9 1. y + = 10 5x 2x 4b b 1 2. =

Cramer Rule can be use to solve a 3 x 3 system. + + + + + = = = ax by dx gx + cz fz z i j a d g b e h c f i Let A be the coefficient matrix of this linear system: ey hy k = A l If det A is not 0, then the system has exactly one solution. The solution is: a b c b e h j j b e h c f i a d g j c f i k l k l k l = = = z x y det A det A det A

Lets solve this system equations by Cramer s rule 2x 3y + z = 5 x + 2y + z = -1 x 3y + 2z = 1 Need to find the determinants of 2 3 1 2 5 1 5 3 1 2 3 5 1 2 1 1 1 1 1 2 1 1 2 1 1 3 2 1 1 2 1 3 2 1 3 1

Find the determinant We will use this for the denominators in the all the fractions. 2 3 1 2 1 1 1 1 2 = + ) 3 + 1 2 1 2 ( 1 3 2 1 2 1 3 1 3 2 2 ( 3 + ) 1 + 4 ( 2 ( 3 )) ( 3 ) 2 ) 1 ( 3 + + ) 5 = + = ) 7 ( 2 ( 14 3 5 12

Solve for x Replace the x column with the answers. 5 3 1 2 1 1 1 1 2 = + ) 3 + 1 2 1 5 ( 1 3 2 1 2 1 3 1 3 2 + ) 1 + 4 ( 5 ( 3 )) ( 3 2 3 ( ) 2 + ) 3 + = + = ) 7 ( 5 ( 3 1 35 9 1 27 27= 9 So = x 12 4

Solve for y Replace the y column with the answers. 2 5 1 1 1 1 1 1 1 = + + 1 1 1 2 5 1 1 2 1 2 1 1 1 1 2 ) 1 2 ( 5 ) 1 + ( 2 2 1 ( ( 1 )) ) 3 ) 1 ( 5 + = + = ( 2 2 6 5 2 9 So 9 3 = = y 12 4

Solve for z Replace the z column with the answers. 2 3 5 2 1 1 1 1 2 = ) 3 + 1 2 1 2 ( 5 3 1 1 1 1 3 1 3 1 1 ( 3 + + 2 ( 2 ) 3 ( 1 )) ( 5 3 ) 2 ) 1 ) 2 ( 3 + + ) 5 = + = ( 2 ( 5 2 6 25 21 So 21 7 = = z 12 4

Trial Question 1 1 4 2 Given find the following: 2 3 2 = 3 5 4 A a) The determinant of A b) The inverse of A c) Form a new matrix B by interchanging row 1 and row 2 of A and find the determinant and inverse of B.

Trial Questions 2 The equilibrium conditions for two related markets (pork and beef) are given by 18?? - ??=87 -2??+36??=98 Find the equilibrium price for each market using matrices.

Trial Question 3 Use Cramer s rule to solve the unknowns in the system of linear equations given below A) 2?1+ 4?2- 3?3=12 3?1- 5?2+2?3=13 -?1 +3?2+ 2?3= 17 B) 11?1- ?2-?3=31 -?1+6?2-2?3= 26 -?1-2?2+ 7?3 = 24

Trial Question 4 Given Y= C + ?0 where C= ?0+ bY. Use matrix inversion to find the equilibrium levels of Y and C.