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Given the matrices E size 12{E} {} , F size 12{F} {} , G size 12{G} {} and H size 12{H} {} , below

E = 1 2 4 2 3 1 size 12{E= left [ matrix { 1 {} # 2 {} ##4 {} # 2 {} ## 3 {} # 1{}} right ]} {} F = 2 1 3 2 size 12{F= left [ matrix { 2 {} # - 1 {} ##3 {} # 2{} } right ]} {} G = 4 1 size 12{G= left [ matrix { 4 {} # 1{}} right ]} {} H = 3 1 size 12{H= left [ matrix { - 3 {} ##- 1 } right ]} {}

Find, if possible.

  1. EF size 12{ ital "EF"} {}
  2. FE size 12{ ital "FE"} {}
  3. FH size 12{ ital "FH"} {}
  4. GH size 12{ ital "GH"} {}
  1. To find EF size 12{ ital "EF"} {} , we multiply the first row 1 2 size 12{ left [ matrix { 1 {} # 2{}} right ]} {} of E size 12{E} {} with the columns 2 3 size 12{ left [ matrix { 2 {} ##3 } right ]} {} and 1 2 size 12{ left [ matrix { - 1 {} ##2 } right ]} {} of the matrix F size 12{F} {} , and then repeat the process by multiplying the other two rows of E size 12{E} {} with these columns of F size 12{F} {} . The result is as follows:

    EF = 1 2 4 2 3 1 2 1 3 2 = 1 2 + 2 3 1 1 + 2 2 4 2 + 2 3 4 1 + 2 2 3 2 + 1 3 3 1 + 1 2 = 8 3 14 0 9 1 size 12{ matrix { ital "EF" {} # ={} {} # left [ matrix {1 {} # 2 {} ## 4 {} # 2 {} ##3 {} # 1{} } right ]left [ matrix { 2 {} # - 1 {} ##3 {} # 2{} } right ]{} ## {} # ={} {} # left [ matrix {1 cdot 2+2 cdot 3 {} # 1 cdot - 1+2 cdot 2 {} ## 4 cdot 2+2 cdot 3 {} # 4 cdot - 1+2 cdot 2 {} ##3 cdot 2+1 cdot 3 {} # 3 cdot - 1+1 cdot 2{} } right ]= left [ matrix { 8 {} # 3 {} ##"14" {} # 0 {} ## 9 {} # - 1{}} right ]{} } } {}
  2. The product FE size 12{ ital "FE"} {} is not possible because the matrix F size 12{F} {} has two entries in each row, while the matrix E size 12{E} {} has three entries in each column. In other words, the matrix F size 12{F} {} has two columns, while the matrix E size 12{E} {} has three rows.

  3. FH = 2 1 3 2 3 1 = 2 3 + 1 1 3 3 + 2 1 = 5 11 size 12{ ital "FH"= left [ matrix { 2 {} # - 1 {} ##3 {} # 2{} } right ]left [ matrix { - 3 {} ##- 1 } right ]= left [ matrix { 2 cdot - 3+ - 1 cdot - 1 {} ##3 cdot - 3+2 cdot - 1 } right ]= left [ matrix { - 5 {} ##- "11" } right ]} {}

  4. GH = 4 1 3 1 = 4 3 + 1 1 = 13 size 12{ ital "GH"= left [ matrix { 4 {} # 1{}} right ] left [ matrix {- 3 {} ## - 1} right ]= left [4 cdot - 3+1 cdot - 1 right ]= left [ - "13" right ]} {}

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We summarize matrix multiplication as follows:

In order for product AB size 12{ ital "AB"} {} to exist, the number of columns of A size 12{A} {} , must equal the number of rows of B size 12{B} {} . If matrix A size 12{A} {} is of dimension m × n size 12{m times n} {} and B size 12{B} {} of dimension n × p size 12{n times p} {} , the product will have the dimension m × p size 12{m times p} {} . Furthermore, matrix multiplication is not commutative.

Given the matrices R size 12{R} {} , S size 12{S} {} , and T size 12{T} {} below.

R = 1 0 2 2 1 5 2 3 1 size 12{R= left [ matrix { 1 {} # 0 {} # 2 {} ##2 {} # 1 {} # 5 {} ## 2 {} # 3 {} # 1{}} right ]} {} S = 0 1 2 3 1 0 4 2 1 size 12{S= left [ matrix { 0 {} # - 1 {} # 2 {} ##3 {} # 1 {} # 0 {} ## 4 {} # 2 {} # 1{}} right ]} {} T = 2 3 0 3 2 2 1 1 0 size 12{T= left [ matrix { - 2 {} # 3 {} # 0 {} ##- 3 {} # 2 {} # 2 {} ## - 1 {} # 1 {} # 0{}} right ]} {}

Find 2 RS 3 ST size 12{2 ital "RS" - 3 ital "ST"} {} .

We multiply the matrices R size 12{R} {} and S size 12{S} {} .

RS = 8 3 4 23 9 9 13 3 5 size 12{ ital "RS"= left [ matrix { 8 {} # 3 {} # 4 {} ##"23" {} # 9 {} # 9 {} ## "13" {} # 3 {} # 5{}} right ]} {}
2 RS = 2 8 3 4 23 9 9 13 3 5 = 16 6 8 46 18 18 26 6 10 size 12{2 ital "RS"=2 left [ matrix { 8 {} # 3 {} # 4 {} ##"23" {} # 9 {} # 9 {} ## "13" {} # 3 {} # 5{}} right ]= left [ matrix {"16" {} # 6 {} # 8 {} ## "46" {} # "18" {} # "18" {} ##"26" {} # 6 {} # "10"{} } right ]} {}
ST = 1 0 2 9 11 2 15 17 4 size 12{ ital "ST"= left [ matrix { 1 {} # 0 {} # - 2 {} ##- 9 {} # "11" {} # 2 {} ## - "15" {} # "17" {} # 4{}} right ]} {}
3 ST = 3 1 0 2 9 11 2 15 17 4 = 3 0 6 27 33 6 45 51 12 size 12{3 ital "ST"=3 left [ matrix { 1 {} # 0 {} # - 2 {} ##- 9 {} # "11" {} # 2 {} ## - "15" {} # "17" {} # 4{}} right ]= left [ matrix {3 {} # 0 {} # - 6 {} ## - "27" {} # "33" {} # 6 {} ##- "45" {} # "51" {} # "12"{} } right ]} {}
2 RS 3 ST = 16 6 8 46 18 18 26 6 10 3 0 6 27 33 6 45 51 12 = 13 6 14 73 15 12 71 45 2 size 12{2 ital "RS" - 3 ital "ST"= left [ matrix { "16" {} # 6 {} # 8 {} ##"46" {} # "18" {} # "18" {} ## "26" {} # 6 {} # "10"{}} right ] - left [ matrix {3 {} # 0 {} # - 6 {} ## - "27" {} # "33" {} # 6 {} ##- "45" {} # "51" {} # "12"{} } right ]= left [ matrix { "13" {} # 6 {} # "14" {} ##"73" {} # - "15" {} # "12" {} ## "71" {} # - "45" {} # - 2{}} right ]} {}
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In this chapter, we will be using matrices to solve linear systems. In [link] , we will be asked to express linear systems as the matrix equation AX = B size 12{ ital "AX"=B} {} , where A size 12{A} {} , X size 12{X} {} , and B size 12{B} {} are matrices. The matrix A size 12{A} {} is called the coefficient matrix .

Verify that the system of two linear equations with two unknowns:

ax + by = h size 12{ ital "ax"+ ital "by"=h} {}
cx + dy = k size 12{ ital "cx"+ ital "dy"=k} {}

can be written as AX = B size 12{ ital "AX"=B} {} , where

A = a b c d size 12{A= left [ matrix { a {} # b {} ##c {} # d{} } right ]} {} X = x y size 12{X= left [ matrix { x {} ##y } right ]} {} and   B = h k size 12{B= left [ matrix { h {} ##k } right ]} {}

If we multiply the matrices A size 12{A} {} and X size 12{X} {} , we get

AX = a b c d x y = ax + by cx + dy size 12{ ital "AX"= left [ matrix { a {} # b {} ##c {} # d{} } right ]left [ matrix { x {} ##y } right ]= left [ matrix { ital "ax"+ ital "by" {} ##ital "cx"+ ital "dy" } right ]} {}

If AX = B size 12{ ital "AX"=B} {} then

ax + by cx + dy = h k size 12{ left [ matrix { ital "ax"+ ital "by" {} ##ital "cx"+ ital "dy" } right ]= left [ matrix { h {} ##k } right ]} {}

If two matrices are equal, then their corresponding entries are equal. Therefore, it follows that

ax + by = h size 12{ ital "ax"+ ital "by"=h} {}
cx + dy = k size 12{ ital "cx"+ ital "dy"=k} {}
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Express the following system as AX = B size 12{ ital "AX"=B} {} .

2x + 3y 4z = 5 size 12{2x+"3y"–4z=5} {}
3x + 4y 5z = 6 size 12{3x+4y - 5z=6} {}
5x 6z = 7 size 12{5x - 6z=7} {}

The above system of equations can be expressed in the form AX = B size 12{ ital "AX"=B} {} as shown below.

2 3 4 3 4 5 5 0 6 x y z = 5 6 7 size 12{ left [ matrix { 2 {} # 3 {} # - 4 {} ##3 {} # 4 {} # - 5 {} ## 5 {} # 0 {} # - 6{}} right ] left [ matrix {x {} ## y {} ##z } right ]= left [ matrix { 5 {} ##6 {} ## 7} right ]} {}
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Systems of linear equations; gauss-jordan method

In this section, we learn to solve systems of linear equations using a process called the Gauss-Jordan method. The process begins by first expressing the system as a matrix, and then reducing it to an equivalent system by simple row operations. The process is continued until the solution is obvious from the matrix. The matrix that represents the system is called the augmented matrix , and the arithmetic manipulation that is used to move from a system to a reduced equivalent system is called a row operation .

Write the following system as an augmented matrix.

2x + 3y 4z = 5 size 12{2x+3y - 4z=5} {}
3x + 4y 5z = 6 size 12{3x+4y - 5z= - 6} {}
4x + 5y 6z = 7 size 12{4x+5y - 6z=7} {}

We express the above information in matrix form. Since a system is entirely determined by its coefficient matrix and by its matrix of constant terms, the augmented matrix will include only the coefficient matrix and the constant matrix. So the augmented matrix we get is as follows:

2 3 4 5 3 4 5 6 4 5 6 7 size 12{ left [ matrix { 2 {} # 3 {} # - 4 {} # \lline {} # 5 {} ##3 {} # 4 {} # - 5 {} # \lline {} # - 6 {} ## 4 {} # 5 {} # - 6 {} # \lline {} # 7{}} right ]} {}
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In the [link] , we expressed the system of equations as AX = B size 12{ ital "AX"=B} {} , where A size 12{A} {} represented the coefficient matrix, and B size 12{B} {} the matrix of constant terms. As an augmented matrix, we write the matrix as A B size 12{ left [A \lline B right ]} {} . It is clear that all of the information is maintained in this matrix form, and only the letters x size 12{x} {} , y size 12{y} {} and z size 12{z} {} are missing. A student may choose to write x size 12{x} {} , y size 12{y} {} and z size 12{z} {} on top of the first three columns to help ease the transition.

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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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