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x + y + 2z = 0 size 12{x+y+2z=0} {}
x + 2y + z = 0 size 12{x+2y+z=0} {}
2x + 3y + 3z = 0 size 12{2x+3y+3z=0} {}
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Find three solutions to the following system of equations.

x + 2y + z = 12 size 12{x+2y+z="12"} {}
y = 3 size 12{y=3} {}

(5, 3, 1), (4, 3, 2) (3, 3, 3)

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For what values of k size 12{k} {} the following system of equations have a). No solution? b). Infinitely many solutions?

x + 2y = 5 size 12{x+2y=5} {}
2x + 4y = k size 12{2x+4y=k} {}
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x + 3y z = 5 size 12{x+3y - z=5} {}

( 5 3s + t size 12{5 - 3s+t} {} , s size 12{s} {} , t size 12{t} {} )

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Why is it not possible for a linear system to have exactly two solutions? Explain geometrically.

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Inverse matrices

In the next two problems, verify that the given matrices are inverses of each other.

7 3 2 1 1 3 2 7 size 12{ left [ matrix { 7 {} # 3 {} ##2 {} # 1{} } right ]left [ matrix { 1 {} # - 3 {} ##- 2 {} # 7{} } right ]} {}
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1 1 0 1 0 1 2 3 4 3 4 1 2 4 1 3 5 1 size 12{ left [ matrix { 1 {} # - 1 {} # 0 {} ##1 {} # 0 {} # - 1 {} ## 2 {} # 3 {} # - 4{}} right ] left [ matrix {3 {} # - 4 {} # 1 {} ## 2 {} # - 4 {} # 1 {} ##3 {} # - 5 {} # 1{} } right ]} {}
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In the following problems, find the inverse of each matrix by the row-reduction method.

3 5 1 2 size 12{ left [ matrix { 3 {} # - 5 {} ##- 1 {} # 2{} } right ]} {}
2 5 1 3 size 12{ left [ matrix { 2 {} # 5 {} ##1 {} # 3{} } right ]} {}
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1 0 2 0 1 4 0 0 1 size 12{ left [ matrix { 1 {} # 0 {} # 2 {} ##0 {} # 1 {} # 4 {} ## 0 {} # 0 {} # 1{}} right ]} {}
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1 1 1 1 0 1 2 1 1 size 12{ left [ matrix { 1 {} # 1 {} # - 1 {} ##1 {} # 0 {} # 1 {} ## 2 {} # 1 {} # 1{}} right ]} {}
1 2 –1 –1 –3 2 –1 –1 1 size 12{ left [ matrix { 1 {} # 2 {} # –1 {} ##–1 {} # –3 {} # 2 {} ## –1 {} # –1 {} # 1{}} right ]} {}
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1 1 1 3 1 0 1 1 2 size 12{ left [ matrix { 1 {} # 1 {} # 1 {} ##3 {} # 1 {} # 0 {} ## 1 {} # 1 {} # 2{}} right ]} {}
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In the following problems, first express the system as AX = B, and then solve using matrix inverses found in the preceding four problems.

3x 5y = 2 size 12{3x - 5y=2} {}
x + 2y = 0 size 12{ - x+2y=0} {}

(4, 2)

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x + 2z = 8 y + 4z = 8 z = 3 size 12{ matrix { x {} # +{} {} # {} # {} # 2z {} # ={} {} # 8 {} ##{} # {} # y {} # +{} {} # 4z {} # ={} {} # 8 {} ## {} # {} # {} # {} # z {} # ={} {} # 3{}} } {}
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x + y z = 2 x + z = 7 2x + y + z = 13 size 12{ matrix { x {} # +{} {} # y {} # - {} {} # z {} # ={} {} # 2 {} ##x {} # +{} {} # {} # {} # z {} # ={} {} # 7 {} ## 2x {} # +{} {} # y {} # +{} {} # z {} # ={} {} # "13"{}} } {}

(3, 3, 4)

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x + y + z = 2 3x + y = 7 x + y + 2z = 3 size 12{ matrix { x {} # +{} {} # y {} # +{} {} # z {} # ={} {} # 2 {} ##3x {} # +{} {} # y {} # {} # {} # ={} {} # 7 {} ## x {} # +{} {} # y {} # +{} {} # 2z {} # ={} {} # 3{}} } {}
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Why is it necessary that a matrix be a square matrix for its inverse to exist? Explain by relating the matrix to a system of equations.

If a matrix M size 12{M} {} has an inverse, then the system of linear equations that has M size 12{M} {} as its coefficient matrix has a unique solution. If a system of linear equations has a unique solution, then the number of equations must be the same as the number of variables. Therefore, the matrix that represents its coefficient matrix must be a square matrix.

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Suppose we are solving a system AX = B size 12{ ital "AX"=B} {} by the matrix inverse method, but discover A size 12{A} {} has no inverse. How else can we solve this system? What can be said about the solutions of this system?

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Application of matrices in cryptography

In the following problems, the letters A to Z correspond to the numbers 1 to 26, as shown below, and a space is represented by the number 27.

A B C D E F G H I J K L M
1 2 3 4 5 6 7 8 9 10 11 12 13
N O P Q R S T U V W X Y Z
14 15 16 17 18 19 20 21 22 23 24 25 26

In the next two problems, use the matrix A, given below, to encode the given messages.

A = 3 2 1 1 size 12{A= left [ matrix { 3 {} # 2 {} ##1 {} # 1{} } right ]} {}

In the two problems following, decode the messages that were encoded using matrix A.

Make sure to consider the spaces between words, but ignore all punctuation. Add a final space if necessary.

Encode the message: WATCH OUT!

71 24 66 23 78 35 87 36 114 47 size 12{ left [ matrix { "71" {} ##"24" } right ]left [ matrix { "66" {} ##"23" } right ]left [ matrix { "78" {} ##"35" } right ]left [ matrix { "87" {} ##"36" } right ]left [ matrix { "114" {} ##"47" } right ]} {}
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Encode the message: HELP IS ON THE WAY.

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Decode the following message:

64 23 102 41 82 32 97 35 71 28 69 32

RETURN HOME

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Decode the following message:

105 40 117 48 39 19 69 32 72 27 37 15 114 47

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In the next two problems, use the matrix B, given below, to encode the given messages.

B = 1 0 0 2 1 2 1 0 1 size 12{B= left [ matrix { 1 {} # 0 {} # 0 {} ##2 {} # 1 {} # 2 {} ## 1 {} # 0 {} # - 1{}} right ]} {}

In the two problems following, decode the messages that were encoded using matrix B size 12{B} {} .

Make sure to consider the spaces between words, but ignore all punctuation. Add a final space(s) if necessary.

Encode the message using matrix B size 12{B} {} :

LUCK IS ON YOUR SIDE.

12 51 9 11 67 2 19 95 14 14 105 11 15 87 3 27 91 18 4 67 23 size 12{ left [ matrix { "12" {} ##"51" {} ## 9} right ] left [ matrix {"11" {} ## "67" {} ##2 } right ]left [ matrix { "19" {} ##"95" {} ## "14"} right ] left [ matrix {"14" {} ## "105" {} ##- "11" } right ]left [ matrix { "15" {} ##"87" {} ## - 3} right ] left [ matrix {"27" {} ## "91" {} ##"18" } right ]left [ matrix { 4 {} ##"67" {} ## - "23"} right ]} {}
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Encode the message using matrix B size 12{B} {} :

MAY THE FORCE BE WITH YOU.

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Decode the following message that was encoded using matrix B:

8 23 7 4 47 –2 15 102 –12 20 58 15 27 80 18 12 74 –7

HEAD FOR THE HILLS

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Decode the following message that was encoded using matrix B:

12 69 –3 11 53 9 5 46 –10 18 95 –9 25 107 4 27 76 22 1 72 –26

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Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
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