Finite Test #1

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A Brief Introduction to
Matrices
A matrix (plural is matrices) is a rectangular array of numbers arranged in rows (horizontal) and
columns (vertical)
An m x n (read m by n) matrix is a matrix with m rows and n columns. The numbers m and n are
the dimensions of the matrix.
 a11
A  a 21
a31
a12
a 22
a32
a13 
a 23  where a23 represents the entry in the second row and the third column.
a33 
6
7
1. Consider the matrices : A  
 2

0
1 12
3 5
6 1 12
6 1 12 
 B 
C 



4 7
7 3 5 
7 3 5

0 1
a) Write the dimension of A.
_______________
b) List the elements in the 3rd row of A.
_______________
c) Which element is indicated by c32?
_______________
d) Why are A and B unequal?
e) Why are B and C unequal?
2. A matrix X has dimension m x n.
a) How many elements are there in X?
______________
b) How many columns are there in X?
______________
c) If you remove one row from X, what are the
new dimensions of X?
______________
If A and B are two matrices with the same dimensions, then we define their sum A+B to the be
matrix obtained by adding their corresponding elements.
Scalar multiplication is the operation of multiplying a matrix by a scalar. The result is obtained
by multiplying each element of the matrix by the scalar.
2 4 
2 5 0
  7 3 5 
1 2 3
B 
C  4 8 and D = 
3. Given A  
.


4 5 6
1 4 3
 8 4 2

6 9
Determine, if possible a) A + B
b)
1
C
2
c) A - B
d) 2B + 3D
e) 2B + 3A - C
4. Solve for x.
0 3   0 4  x 
x


1 1  x  x 
Matrix multiplication is based on row by column multiplication. To multiply a row by a column,
multiply the first element in the row by the first element in the column; then multiply the second
element in the row by the second element in the column; then the third and so on. Finally add the
products.
 2
 
e.g. 9  2 4 3 = (9)(2)  (2)(3)  (4)(5)  32
 
5
Notice that the row and the column must have the same number of elements in order to be
multiplied.
In general, a product AB exists only when the number of columns of A equals the number of
rows of B. In this case, the elements of AB are obtained by multiplying each row of A by each
column of B. Also, when the product exists, AB has the same number of rows as A and the same
number of columns as B.
This can be seen as: Amxn x Bnxp = ABmxp
2 4 
2 5 0
  7 3 5 
1 2 3
B 
C  4 8 and D = 
5. Given A  
. Find


4 5 6
1 4 3
 8 4 2

6 9
a) AC
b) BC
c) CD
Matrices that have a main diagonal (top left to bottom right) whose elements are all 1’s and the
rest of the matrix elements are all 0’s, are called identity matrices. An identity matrix is also
always a square matrix (same number of rows as columns). Any square matrix A remains
unchanged when multiplied by an identity matrix. AI = IA = A
1 2
 and I =
3 4
Evaluate AI and IA, for A = 
1 0
0 1  .


Matrices whose product results in I, are called inverse matrices.
6. Evaluate the following:
4 1  2  1


7 2    7 4 
a) 
5 2  3  2


7 3    7 5 
b) 
Find the inverse of a matrix (if it exists) by created an augmented matrix of the form [A|I] and
row-reduce the augmented matrix into the form [I|A-1]. This can be completed algebraically and
on the TI-83+.
Algebraically - the goal is to form the identity matrix on the left side using the following
Elementary Row Operations:
1) interchange the rows of the matrix.
2) add/subtract multiples of one row to (or from) another row.
3) multiply any row by a non-zero real number.
Using the TI-83+ - make the technology do the work!
clear your work space
row by column
2nd Matrix Edit – to enter in the matrix.
- Choose a name and enter the information
2nd Quit
2nd Matrix Math RREF(
scroll down
until you find
rref(
Then choose the matrix name that you edited earlier
highlite the number of the matrix
you wish to use
Close the bracket, enter and Interpret your results
note the identity
matrix shown
7. Find the inverse of the following matrices algebraically:
 2  1

 5 3 
a) 
 2 2
c) 

 3 4
5 2

2 1 
b) 
14 8 3


d) 8 5 2


 3 2 1 
Cryptography is a branch of mathematics that involves the encoding of a message to that the
message can be safely passed on only to the person who has the decoding of information. By
multiplying a matrix on either the left or the right by an invertible matrix, information can be
scrambled so that no one else can interpret. To decode, multiply on the same side by the inverse
of the encode matrix, called the decode matrix.
4
8. A message was encoded, by multiplying on the left, using the encode matrix E = 
 7
 20
The coded message EM = 
32
4
.
3
a) Find the matrix that will decode EM.
b) Find the original message ( assume A = 1, B = 2, C = 3, … )
1
.
2 
Solving a system of equations can be completed algebraically using elimination or by matrix
methods, as well as by using the row-reduced echelon form (RREF- when the first non-zero entry
in any row is a 1 and when a column contains the leading 1 of any row, then all other entries of
that column are zero) matrix operation built into the TI-83+. A system yielding a particular
solution is said to be independent. Some systems will result in an infinite number of solutions, as
represented by a parametric equation, and are referred to as dependent. Other systems have no
solutions and are called inconsistent.
3x  5 y  14
can be expressed as
 x  2y  1
The system of equations 
3  5 14
1 2 1  , where the left side


is referred to as the matrix of coefficients and combined with the constants forms an augmented
matrix. The goal is to change the matrix of coefficients into the identity matrix, yielding the
solution to the system.
9. Solve each of the following systems of equations by reducing the corresponding augmented
matrix to row-reduced echelon form. In each case, state whether the system is independent,
dependent or inconsistent. If the system is dependent, state the number of parameters needed to
express the solution and find two particular solutions.
a) 4x – 2y – z = 12, 3x – y = 9, x + y + z = 5
b) 2x + y + z = 2, -x +10y + 13z = -19, 3x - 2y - 3z = 9
c) x + y + z = 2, x + 9y +3z = 4, 2x – 2y + z = 3
d) 2x – y = 5, x – 2y + z = 0, 3y – 2z = 4
e) 4x – 3y = 2, 3x – y + 2z = 9, 2y – z = -9
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