Engineering Fundamentals: An Introduction to Engineering, 4th ed.c

18.5 Matrix Algebra 615

Let us summarize the results of the operations performed during steps 1 through 3.

These operations eliminatedx 1 from Equations (18.18b) and (18.18c).

(18.24a)

(18.24b)

(18.24c)

Step 4: To eliminatex 2 from Equation (18.24c), first we divide Equation (18.24b) by , the

coefficient ofx 2.

(18.25)

Then, we multiply Equation (18.25) by : the coefficient ofx 2 in Equation (18.24c), and

subtract that equation from Equation (18.24c). These operations lead to

(18.26)

Dividing both sides of Equation (18.26) by 18, we get

Summarizing the results of the previous steps, we have

(18.27)

(18.28)

(18.29)

Step 5: Now we can use back substitution to compute the values ofx 2 andx 3. We substitute

forx 3 in Equation (18.28) and solve forx 2.

Next, we substitute forx 3 andx 2 in Equation (18.27) and solve forx 1.

Inverse of a Matrix

In the previous sections, we discussed matrix addition, subtraction and multiplication, but you

may have noticed that we did not say anything about matrix division. That is because such an

operation is not defined formally. Instead, we define an inverse of a matrix in such a way that

when it is multiplied by the original matrix, the identity matrix is obtained.

x 1 

1

2

152 

1

2

142 

13

2

S x 1  2

x 2  5142  25 S x 2  5

x 3  4

x 2  5 x 3  25

x 1 

1

2

x 2 

1

2

x 3 

13

2

x 3  4

a

7

2

x 2 

35

2

x 3 

175

2

b

18 x 3  72



7

2

x 2 

1

2

x 3 

31

2

 7

2

x 2  5 x 3  25

1

2



7

2

x 2 

1

2

x 3 

31

2

1

2

x 2 

5

2

x 3 

25

2

x 1 

1

2

x 2 

1

2

x 3 

13

2

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