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Engineering LibreTexts

4.25: Untitled Page 80

  • Page ID
    18213
  • Chapter 4

    (4‐134)

    We begin by eliminating the first term in the second equation. This is accomplished by multiplying the first equation by 0.045 and subtracting the result from the second equation to obtain

    (4‐135)

    Directing our attention to the third equation, we multiply the first equation by 0.091 and subtract the result from the third equation to obtain (4‐136)

    The second equation can be conditioned by dividing by 0.024 so that the equation set takes the form

    (4‐137)

    We now multiply the second equation by 0 . 810 and subtract the result from the third equation to obtain

    Multicomponent systems

    142

    (4‐138)

    and division of the third equation by 30 . 763 leads to (4‐139)

    Having worked our way forward through this set of three equations in order to determine

    , we can work our way backward through the set to determine and

    . The results are the same as we obtained earlier in Sec. 4.8 and we list the result again as

    (4‐140)

    The procedure represented by Eqs. 4‐133 through 4‐139 is extremely convenient for automated computation and large systems of equations can be quickly solved using a variety of software. Using systems such as MATLAB or Mathematica, problems of this type become quite simple.

    4.9.2 Determination of the inverse of a square matrix

    The Gaussian elimination procedure described in the previous section is closely related to the determination of the inverse of a square matrix. In order to illustrate how the inverse matrix can be calculated, we consider the coefficient matrix associated with Eq. 4‐134 which we express as

     1

    1

    1 

    A

    0 . 045 0 . 069 0 955

    .

    (4‐141)

    0 . 091 0 901

    .

    0 041

    .

    We can express Eq. 4‐134 in the compact form represented by Eq. 4‐119 to obtain A u =

    b

    (4‐142)

    143