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What you just witnessed is no coincidence. This is the method that is often employed in finding the inverse of a matrix.

We list the steps, as follows:

The method for finding the inverse of a matrix

  1. Write the augmented matrix A I n size 12{ left [A \lline I rSub { size 8{n} } right ]} {} .
  2. Write the augmented matrix in step 1 in reduced row echelon form.
  3. If the reduced row echelon form in 2 is I n B size 12{ left [ matrix { I rSub { size 8{n} } {} # \lline {} # B{}} right ]} {} , then B size 12{B} {} is the inverse of A size 12{A} {} .
  4. If the left side of the row reduced echelon is not an identity matrix, the inverse does not exist.
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Given the matrix A size 12{A} {} below, find its inverse.

A = 1 1 1 2 3 0 0 2 1 size 12{A= left [ matrix { 1 {} # - 1 {} # 1 {} ##2 {} # 3 {} # 0 {} ## 0 {} # - 2 {} # 1{}} right ]} {}

We write the augmented matrix as follows.

1 1 1 1 0 0 2 3 0 0 1 0 0 2 1 0 0 1 size 12{ left [ matrix { 1 {} # - 1 {} # 1 {} # \lline {} # 1 {} # 0 {} # 0 {} ##2 {} # 3 {} # 0 {} # \lline {} # 0 {} # 1 {} # 0 {} ## 0 {} # - 2 {} # 1 {} # \lline {} # 0 {} # 0 {} # 1{}} right ]} {}

We reduce this matrix using the Gauss-Jordan method.

Multiplying the first row by –2 and adding it to the second row, we get

1 1 1 1 0 0 0 5 2 2 1 0 0 2 1 0 0 1 size 12{ left [ matrix { 1 {} # - 1 {} # 1 {} # \lline {} # 1 {} # 0 {} # 0 {} ##0 {} # 5 {} # - 2 {} # \lline {} # - 2 {} # 1 {} # 0 {} ## 0 {} # - 2 {} # 1 {} # \lline {} # 0 {} # 0 {} # 1{}} right ]} {}

If we swap the second and third rows, we get

1 1 1 1 0 0 0 2 1 0 0 1 0 5 2 2 1 0 size 12{ left [ matrix { 1 {} # - 1 {} # 1 {} # \lline {} # 1 {} # 0 {} # 0 {} ##0 {} # - 2 {} # 1 {} # \lline {} # 0 {} # 0 {} # 1 {} ## 0 {} # 5 {} # - 2 {} # \lline {} # - 2 {} # 1 {} # 0{}} right ]} {}

Divide the second row by –2. The result is

1 1 1 1 0 0 0 1 1 / 2 0 0 1 / 2 0 5 2 2 1 0 size 12{ left [ matrix { 1 {} # - 1 {} # 1 {} # \lline {} # 1 {} # 0 {} # 0 {} ##0 {} # 1 {} # - 1/2 {} # \lline {} # 0 {} # 0 {} # - 1/2 {} ## 0 {} # 5 {} # - 2 {} # \lline {} # - 2 {} # 1 {} # 0{}} right ]} {}

Let us do two operations here. 1) Add the second row to first, 2) Add -5 times the second row to the third. And we get

1 0 1 / 2 1 0 1 / 2 0 1 1 / 2 0 0 1 / 2 0 0 1 / 2 2 1 5 / 2 size 12{ left [ matrix { 1 {} # 0 {} # 1/2 {} # \lline {} # 1 {} # 0 {} # - 1/2 {} ##0 {} # 1 {} # - 1/2 {} # \lline {} # 0 {} # 0 {} # - 1/2 {} ## 0 {} # 0 {} # 1/2 {} # \lline {} # - 2 {} # 1 {} # 5/2{}} right ]} {}

Multiplication of the third row by 2 results in

1 0 1 / 2 1 0 1 / 2 0 1 1 / 2 0 0 1 / 2 0 0 1 4 2 5 size 12{ left [ matrix { 1 {} # 0 {} # 1/2 {} # \lline {} # 1 {} # 0 {} # - 1/2 {} ##0 {} # 1 {} # - 1/2 {} # \lline {} # 0 {} # 0 {} # - 1/2 {} ## 0 {} # 0 {} # 1 {} # \lline {} # - 4 {} # 2 {} # 5{}} right ]} {}

Multiply the third row by 1 / 2 size 12{1/2} {} and add it to the second. Also, multiply the third row by 1 / 2 size 12{ - 1/2} {} and add it to the first. We get

1 0 0 3 1 3 0 1 0 2 1 2 0 0 1 4 2 5 size 12{ left [ matrix { 1 {} # 0 {} # 0 {} # \lline {} # 3 {} # - 1 {} # - 3 {} ##0 {} # 1 {} # 0 {} # \lline {} # - 2 {} # 1 {} # 2 {} ## 0 {} # 0 {} # 1 {} # \lline {} # - 4 {} # 2 {} # 5{}} right ]} {}

Therefore, the inverse of matrix A size 12{A} {} is

3 1 3 2 1 2 4 2 5 size 12{ left [ matrix { 3 {} # - 1 {} # - 3 {} ##- 2 {} # 1 {} # 2 {} ## - 4 {} # 2 {} # 5{}} right ]} {}

One should verify the result by multiplying the two matrices to see if the product does, indeed, equal the identity matrix.

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Now that we know how to find the inverse of a matrix, we will use inverses to solve systems of equations. The method is analogous to solving a simple equation like the one below.

2 3 x = 4 size 12{ { {2} over {3} } x=4} {}

Solve the following equation .

2 3 x = 4 size 12{ { {2} over {3} } x=4} {}

To solve the above equation, we multiply both sides of the equation by the multiplicative inverse of 2 3 size 12{ { {2} over {3} } } {} which happens to be 3 2 size 12{ { {3} over {2} } } {} . We get

3 2 2 3 x = 4 3 2 size 12{ { {3} over {2} } cdot { {2} over {3} } x=4 cdot { {3} over {2} } } {}
x = 6 size 12{x=6} {}

We use the above example as an analogy to show how linear systems of the form AX = B size 12{ ital "AX"=B} {} are solved.

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To solve a linear system, we first write the system in the matrix equation AX = B size 12{ ital "AX"=B} {} , where A size 12{A} {} is the coefficient matrix, X size 12{X} {} the matrix of variables, and B size 12{B} {} the matrix of constant terms. We then multiply both sides of this equation by the multiplicative inverse of the matrix A size 12{A} {} .

Consider the following example.

Solve the following system

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

To solve the above equation, first we express the system as

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

where A size 12{A} {} is the coefficient matrix, and B size 12{B} {} is the matrix of constant terms. We get

3 1 5 2 x y = 3 4 size 12{ left [ matrix { 3 {} # 1 {} ##5 {} # 2{} } right ]left [ matrix { x {} ##y } right ]= left [ matrix { 3 {} ##4 } right ]} {}

To solve this system, we multiply both sides of the matrix equation AX = B size 12{ ital "AX"=B} {} by A 1 size 12{A rSup { size 8{ - 1} } } {} . Since the matrix A size 12{A} {} is the same matrix A size 12{A} {} whose inverse we found in [link] ,

A 1 = 2 1 5 3 size 12{A rSup { size 8{ - 1} } = left [ matrix { 2 {} # - 1 {} ##- 5 {} # 3{} } right ]} {}

Multiplying both sides by A 1 size 12{A rSup { size 8{ - 1} } } {} , we get

2 1 5 3 3 1 5 2 x y = 2 1 5 3 3 4 size 12{ left [ matrix { 2 {} # - 1 {} ##- 5 {} # 3{} } right ]left [ matrix { 3 {} # 1 {} ##5 {} # 2{} } right ]left [ matrix { x {} ##y } right ]= left [ matrix { 2 {} # - 1 {} ##- 5 {} # 3{} } right ]left [ matrix { 3 {} ##4 } right ]} {}
1 0 0 1 x y = 2 3 size 12{ left [ matrix { 1 {} # 0 {} ##0 {} # 1{} } right ]left [ matrix { x {} ##y } right ]= left [ matrix { 2 {} ##- 3 } right ]} {}
x y = 2 3 size 12{ left [ matrix { x {} ##y } right ]= left [ matrix { 2 {} ##- 3 } right ]} {}

Therefore, x = 2 size 12{x=2} {} , and y = 3 size 12{y= - 3} {} .

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Solve the following system

x y + z = 6 size 12{x - y+z=6} {}
2x + 3y = 1 size 12{2x+3y=1} {}
2y + z = 5 size 12{ - 2y+z=5} {}

To solve the above equation, we write the system in the matrix form AX = B size 12{ ital "AX"=B} {} as follows:

1 1 1 2 3 0 0 2 1 x y z = 6 1 5 size 12{ left [ matrix { 1 {} # - 1 {} # 1 {} ##2 {} # 3 {} # 0 {} ## 0 {} # - 2 {} # 1{}} right ] left [ matrix {x {} ## y {} ##z } right ]= left [ matrix { 6 {} ##1 {} ## 5} right ]} {}

To solve this system, we need inverse of A size 12{A} {} . From [link] , we have

A 1 = 3 1 3 2 1 2 4 2 5 size 12{A rSup { size 8{ - 1} } = left [ matrix { 3 {} # - 1 {} # - 3 {} ##- 2 {} # 1 {} # 2 {} ## - 4 {} # 2 {} # 5{}} right ]} {}

We multiply both sides of the matrix equation AX = B size 12{ ital "AX"=B} {} , by A 1 size 12{A rSup { size 8{ - 1} } } {} , we get

3 1 3 2 1 2 4 2 5 1 1 3 2 3 0 0 2 1 x y z = 3 1 3 2 1 2 4 2 5 6 1 5 size 12{ left [ matrix { 3 {} # - 1 {} # - 3 {} ##- 2 {} # 1 {} # 2 {} ## - 4 {} # 2 {} # 5{}} right ] left [ matrix {1 {} # - 1 {} # - 3 {} ## 2 {} # 3 {} # 0 {} ##0 {} # - 2 {} # 1{} } right ]left [ matrix { x {} ##y {} ## z} right ]= left [ matrix {3 {} # - 1 {} # - 3 {} ## - 2 {} # 1 {} # 2 {} ##- 4 {} # 2 {} # 5{} } right ]left [ matrix { 6 {} ##1 {} ## 5} right ]} {}

After multiplying the matrices, we get

1 0 0 0 1 0 0 0 1 x y z = 2 1 3 size 12{ left [ matrix { 1 {} # 0 {} # 0 {} ##0 {} # 1 {} # 0 {} ## 0 {} # 0 {} # 1{}} right ] left [ matrix {x {} ## y {} ##z } right ]= left [ matrix { 2 {} ##- 1 {} ## 3} right ]} {}

x y z = 2 1 3 size 12{ left [ matrix { x {} ##y {} ## z} right ]= left [ matrix {2 {} ## - 1 {} ##3 } right ]} {}

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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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