# 4.9: Matrix Algebra

In Sec. 4.7 we examined a distillation process with the objective of determining molar flow rates, and our analysis led to a set of two equations and two unknowns given by

Species $$A$$: $0.6 \dot{M}_{2} + 0.1 \dot{M}_{3} = 360 \text{ mol/hr} \label{103a}$

Species $$B$$: $0.4 \dot{M}_{2} + 0.9 \dot{M}_{3} = 840 \text{ mol/hr} \label{103b}$

Some thought was necessary in order to set up the macroscopic mole balances for the process illustrated in Figure $$4.5.2$$; however, the algebraic effort required to solve the governing macroscopic balances was trivial. In Sec. 4.8, we considered the system illustrated in Figure $$4.8.1$$ and the analysis led to the following set of three equations and three unknowns:

species $$A$$: $(\omega_{A} )_{2} \dot{m}_{2} + (\omega_{A} )_{3} \dot{m}_{3} + (\omega_{A} )_{4} \dot{m}_{4} = (\omega_{A} )_{1} \dot{m}_{1} \label{104a}$

species $$B$$: $(\omega_{B} )_{2} \dot{m}_{2} + (\omega_{B} )_{3} \dot{m}_{3} + (\omega_{B} )_{4} \dot{m}_{4} = (\omega_{B} )_{1} \dot{m}_{1} \label{104b}$

Total: $\dot{m}_{2} +\dot{m}_{3} +\dot{m}_{4} = \dot{m}_{1} \label{104c}$

The algebraic effort required to solve these three equations for $$\dot{m}_{2}$$, $$\dot{m}_{3}$$ and $$\dot{m}_{4}$$ was considerable as one can see from the solution given by Equation $$(4.8.17)$$. It should not be difficult to imagine that solving sets of four or five equations can become exceedingly difficult to do by hand; however, computer routines are available that can be used to solve virtually any set of equations have the form given by Eqs. \ref{104a} - \ref{104c}.

In dealing with sets of many equations, it is convenient to use the language of matrix algebra. For example, in matrix notation we would express Eqs. \ref{104a} - \ref{104c} according to

$\begin{bmatrix} {(\omega_{A} )_{2} } & {(\omega_{A} )_{3} } & {(\omega_{A} )_{4} } \\ {(\omega_{B} )_{2} } & {(\omega_{B} )_{3} } & {(\omega_{B} )_{4} } \\ {1} & {1} & {1} \end{bmatrix}\begin{bmatrix} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{bmatrix} = \begin{bmatrix} {(\omega_{A} )_{1} \dot{m}_{1} } \\ {(\omega_{B} )_{1} \dot{m}_{1} } \\ {\dot{m}_{1} } \end{bmatrix} \label{105}$

In this representation of Eqs. \ref{104a} - \ref{104c}, the $$3\times 3$$ matrix of mass fractions multiplies the $$3\times 1$$ column matrix of mass flow rates to produce a $$3\times 1$$ column matrix that is equal to the right hand side of Equation \ref{105}. In working with matrices, it is generally convenient to make use of a nomenclature in which subscripts are used to identify the row and column in which an element is located. We used this type of nomenclature in Chapter 2 where the $$m\times n$$ matrix $$\mathbf{A}$$ was represented by

$\mathbf{A} = \begin{bmatrix} {a_{11} } & {a_{12} } & {......} & {a_{1n} } \\ {a_{21} } & {a_{22} } & {......} & {a_{2n} } \\ {....} & {....} & {......} & {....} \\ {a_{m1} } & {a_{m2} } & {......} & {a_{mn} } \end{bmatrix} \label{106}$

Here the first subscript identifies the row in which an element is located while the second subscript identifies the column. In Chapter 2 we discussed matrix addition and subtraction, and here we wish to discuss matrix multiplication and the matrix operation that is analogous to division. Matrix multiplication between $$\mathbf{A}$$ and $$\mathbf{B}$$ is defined only if the number of columns of $$\mathbf{A}$$ (in this case $$n$$) is equal to the number of rows of $$\mathbf{B}$$. Given an $$m\times n$$ matrix $$\mathbf{A}$$ and an $$n\times p$$ matrix, $$\mathbf{B}$$, the product between $$\mathbf{A}$$ and $$\mathbf{B}$$ is illustrated by the following equation:

$\label{107}$

Here we see that the elements of the $$i^{th}$$ row in matrix $$\mathbf{A}$$ multiply the elements of the $$j^{th}$$ column in matrix $$\mathbf{B}$$ to produce the element in the $$i^{th}$$ row and the $$j^{th}$$ column of the matrix $$\mathbf{C}$$. For example, the specific element $$c_{11}$$ is given by

$c_{11} = a_{11} b_{11} + a_{12} b_{21} + ....... + a_{1n} b_{n1} \label{108}$

while the general element $$c_{ij}$$ is given by

$c_{ij} = a_{i1} b_{1j} + a_{i2} b_{2j} + ....... + a_{in} b_{nj} \label{109}$

In Equation \ref{107} we see that an $$m\times n$$ matrix can multiply an $$n\times p$$ to produce an $$m\times p$$, and we see that the matrix multiplication represented by $$\mathbf{AB}$$ is only defined when the number of columns in $$\mathbf{A}$$ is equal to the number of rows in $$\mathbf{B}$$. The matrix multiplication illustrated in Equation \ref{105} conforms to this rule since there are three columns in the matrix of mass fractions and three rows in the column matrix of mass flow rates. The configuration illustrated in Equation \ref{105} is extremely common since it is associated with the solution of $$n$$ equations for $$n$$ unknowns. Our generic representation for this type of matrix equation is given by

$\mathbf{Au} = \mathbf{ b} \label{110}$

in which $$\mathbf{A}$$ is a square matrix, $$u$$ is a column matrix of unknowns, and $$b$$ is a column matrix of knowns. These matrices are represented explicitly by

$\mathbf{ A} = \begin{bmatrix} {a_{11} } & {a_{12} } & {......} & {a_{1n} } \\ {a_{21} } & {a_{22} } & {......} & {a_{2n} } \\ {....} & {....} & {......} & {....} \\ {a_{n1} } & {a_{n2} } & {......} & {a_{nn} } \end{bmatrix} , \quad \mathbf{ u} = \begin{bmatrix} {u_{1} } \\ {.} \\ {.} \\ {u_{n} } \end{bmatrix} , \quad \mathbf{ b} = \begin{bmatrix} {b_{1} } \\ {.} \\ {.} \\ {b_{n} } \end{bmatrix} \label{111}$

Sometimes the coefficients in $$\mathbf{A}$$ depend on the unknowns, $$\mathbf{u}$$, and the matrix equation may be bi-linear as indicated in Eqs. $$(4.7.5 - 4.7.7)$$.

The transpose of a matrix is defined in the same manner as the transpose of an array that was discussed in Sec. 2.5, thus the transpose of the matrix $$\mathbf{A}$$ is constructed by interchanging the rows and columns to obtain

$\mathbf{ A} = \begin{bmatrix} {a_{11} } & {a_{12} } & {......} & {a_{1n} } \\ {a_{21} } & {a_{22} } & {......} & {a_{2n} } \\ {.} & {.} & {......} & {.} \\ {a_{m1} } & {a_{m2} } & {......} & {a_{mn} } \end{bmatrix} , \quad \mathbf{ A}^{ T} = \begin{bmatrix} {a_{11} } & {a_{21} } & {...} & {a_{m1} } \\ {a_{12} } & {a_{22} } & {...} & {a_{m2} } \\ {.} & {.} & {...} & {.} \\ {.} & {.} & {...} & {.} \\ {a_{1n} } & {a_{2n} } & {...} & {a_{mn} } \end{bmatrix} \label{112}$

Here it is important to note that $$\mathbf{A}$$ is an $$m\times n$$ matrix while $$\mathbf{ A}^{ T}$$ is an $$n\times m$$ matrix.

## Inverse of a square matrix

In order to solve Equation \ref{110}, one cannot "divide” by $$\mathbf{A}$$ to determine the unknown, $$\mathbf{u}$$, since matrix division is not defined. There is, however, a related operation involving the inverse of a matrix. The inverse of a matrix, $$\mathbf{A}$$, is another matrix, $$\mathbf{ A}^{-1}$$, such that the product of $$\mathbf{A}$$ and $$\mathbf{A}^{-1}$$ is given by

$\mathbf{ A} \mathbf{ A}^{-1} = { I} \label{113}$

in which $$\mathbf{I}$$ is the identity matrix. Identity matrices have ones in the diagonal elements and zeros in the off-diagonal elements as illustrated by the following $$4\times 4$$ matrix:

$\mathbf{ I} = \begin{bmatrix} {1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {1} \end{bmatrix} \label{114}$

For the inverse of a matrix to exist, the matrix must be a square matrix, i.e., the number of rows must be equal to the number of columns. In addition, the determinant of the matrix must be different from zero. Thus for Equation \ref{113} to be valid we require that the determinant of $$\mathbf{A}$$ be different from zero, i.e.,

$\left|\mathbf{A}\right| \neq 0 \label{115}$

This type of requirement plays an important role in the derivation of the pivot theorem (see Appendix C1) that forms the basis for one of the key developments in Chapter 6.

As an example of the use of the inverse of a matrix, we consider the following set of four equations containing four unknowns:

${a_{11} u_{1} + a_{12} u_{2} + a_{13} u_{3} + a_{14} u_{4} = b_{1} } \\ {a_{21} u_{1} + a_{22} u_{2} + a_{23} u_{3} + a_{24} u_{4} = b_{2} } \\ {a_{31} u_{1} + a_{32} u_{2} + a_{33} u_{3} + a_{34} u_{4} = b_{3} } \\ {a_{41} u_{1} + a_{42} u_{2} + a_{43} u_{3} + a_{44} u_{4} = b_{4} } \label{116}$

In compact notation, we would represent these equations according to

$\mathbf{ Au} = \mathbf{ b} \label{117}$

We suppose that $$\mathbf{A}$$ is invertible and that the inverse, along with $$\mathbf{u}$$ and $$\mathbf{b}$$, are given by

$\mathbf{ A}^{-1} = \begin{bmatrix} {\bar{a}_{11} } & {\bar{a}_{12} } & {\bar{a}_{13} } & {\bar{a}_{14} } \\ {\bar{a}_{21} } & {\bar{a}_{22} } & {\bar{a}_{23} } & {\bar{a}_{24} } \\ {\bar{a}_{31} } & {\bar{a}_{32} } & {\bar{a}_{33} } & {\bar{a}_{34} } \\ {\bar{a}_{41} } & {\bar{a}_{42} } & {\bar{a}_{43} } & {\bar{a}_{44} } \end{bmatrix} , \quad \mathbf{ u} = \begin{bmatrix} {u_{1} } \\ {u_{2} } \\ {u_{3} } \\ {u_{n} } \end{bmatrix} , \quad \mathbf{ b} = \begin{bmatrix} {b_{1} } \\ {b_{2} } \\ {b_{3} } \\ {b_{n} } \label{118} \end{bmatrix}$

We can multiply Equation \ref{117} by $$\mathbf{ A}^{-1}$$ to obtain

$\mathbf{ A}^{-1} \mathbf{ Au} = \mathbf{ Iu} = \mathbf{ u} = \mathbf{ A}^{-1} \mathbf{ b} \label{119}$

and the details are given by

$\begin{bmatrix} {1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {1} \end{bmatrix}\begin{bmatrix} {u_{1} } \\ {u_{2} } \\ {u_{3} } \\ {u_{n} } \end{bmatrix} = \begin{bmatrix} {\bar{a}_{11} } & {\bar{a}_{12} } & {\bar{a}_{13} } & {\bar{a}_{14} } \\ {\bar{a}_{21} } & {\bar{a}_{22} } & {\bar{a}_{23} } & {\bar{a}_{24} } \\ {\bar{a}_{31} } & {\bar{a}_{32} } & {\bar{a}_{33} } & {\bar{a}_{34} } \\ {\bar{a}_{41} } & {\bar{a}_{42} } & {\bar{a}_{43} } & {\bar{a}_{44} } \end{bmatrix}\begin{bmatrix} {b_{1} } \\ {b_{2} } \\ {b_{3} } \\ {b_{n} } \label{120} \end{bmatrix}$

Carrying out the matrix multiplication on the left hand side leads to

$\begin{bmatrix} {u_{1} } \\ {u_{2} } \\ {u_{3} } \\ {u_{n} } \end{bmatrix} = \begin{bmatrix} {\bar{a}_{11} } & {\bar{a}_{12} } & {\bar{a}_{13} } & {\bar{a}_{14} } \\ {\bar{a}_{21} } & {\bar{a}_{22} } & {\bar{a}_{23} } & {\bar{a}_{24} } \\ {\bar{a}_{31} } & {\bar{a}_{32} } & {\bar{a}_{33} } & {\bar{a}_{34} } \\ {\bar{a}_{41} } & {\bar{a}_{42} } & {\bar{a}_{43} } & {\bar{a}_{44} } \end{bmatrix}\begin{bmatrix} {b_{1} } \\ {b_{2} } \\ {b_{3} } \\ {b_{n} } \label{121} \end{bmatrix}$

while the more complex multiplication on the right hand side provides

$\begin{bmatrix} {u_{1} } \\ {u_{2} } \\ {u_{3} } \\ {u_{n} } \end{bmatrix} = \begin{bmatrix} {\bar{a}_{11} b_{1} } & {\bar{a}_{12} b_{2} } & {\bar{a}_{13} b_{3} } & {\bar{a}_{14} b_{4} } \\ {\bar{a}_{21} b_{1} } & {\bar{a}_{22} b_{2} } & {\bar{a}_{23} b_{3} } & {\bar{a}_{24} b_{4} } \\ {\bar{a}_{31} b_{1} } & {\bar{a}_{32} b_{2} } & {\bar{a}_{33} b_{3} } & {\bar{a}_{34} b_{4} } \\ {\bar{a}_{41} b_{1} } & {\bar{a}_{42} b_{2} } & {\bar{a}_{43} b_{3} } & {\bar{a}_{44} b_{4} } \end{bmatrix} \label{122}$

Each element of the column matrix on the left hand side is equal to the corresponding element on the right hand side and this leads to the solution for the unknowns given by

$u_{1}=\bar{a}_{11} b_{1}+\bar{a}_{12} b_{2}+\bar{a}_{13} b_{3}+\bar{a}_{14} b_{4} \\ u_{2}=\bar{a}_{21} b_{1}+\bar{a}_{22} b_{2}+\bar{a}_{23} b_{3}+\bar{a}_{24} b_{4} \\ u_{3}=\bar{a}_{31} b_{1}+\bar{a}_{32} b_{2}+\bar{a}_{33} b_{3}+\bar{a}_{34} b_{4} \\ u_{4}=\bar{a}_{41} b_{1}+\bar{a}_{42} b_{2}+\bar{a}_{43} b_{3}+\bar{a}_{44} b_{4} \label{123}$

It should be clear that there are two main problems associated with the use of Eqs. \ref{116} to obtain the solution for the unknowns given by Equation \ref{123}. The first problem is the correct interpretation of a physical process to arrive at the original set of equations, and the second problem is the determination of the inverse of the matrix $$\mathbf{A}$$.

There are a variety of methods for developing the inverse for a matrix; however, sets of equations are usually solved numerically without calculating the inverse of a matrix. One of the classic methods is known as Gaussian elimination and we can illustrate this technique with the problem that was studied in Sec. 4.8. We can make use of the data provided in Figure $$4.7.2$$ to express Eqs. $$(9.8.8 - 9.8.10)$$ in the form

Species $$A$$: $0.045 \dot{m}_{2} + 0.069 \dot{m}_{3} + 0.955 \dot{m}_{4} = 500{ \ lb}_{ m} /{ hr} \label{124a}$

Species $$B$$: $0.091 \dot{m}_{2} + 0.901 \dot{m}_{3} + 0.041 \dot{m}_{4} = 300 { \ lb}_{ m} /{ hr} \label{124b}$

Total: $\dot{m}_{2} +\dot{m}_{3} +\dot{m}_{4} = { 1000 \ lb}_{ m} /{ hr} \label{124c}$

and our objective is to determine the three mass flow rates, $$\dot{m}_{2}$$, $$\dot{m}_{3}$$ and $$\dot{m}_{4}$$. The solution is obtained by making use of the following three rules which are referred to as elementary row operations:

I. Any equation in the set can be modified by multiplying or dividing by a non-zero scalar without affecting the solution.

II. Any equation can be added or subtracted from the set without affecting the solution.

III. Any two equations can be interchanged without affecting the solution.

For this particular problem, it is convenient to arrange the three equations in the form

$\dot{m}_{2} + \dot{m}_{3} + \dot{m}_{4} = { 1000 \ lb}_{ m} /{ hr} \\ {0.045 \dot{m}_{2} + 0.069 \dot{m}_{3} + 0.955 \dot{m}_{4} = 500{ \ lb}_{ m} /{ hr}} \\ {0.091 \dot{m}_{2} + 0.901 \dot{m}_{3} + 0.041 \dot{m}_{4} = 300 { \ lb}_{ m} /{ hr}} \label{125}$

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

${\dot{m}_{2} +\dot{m}_{3} +\dot{m}_{4} = { 1000 \ lb}_{ m} /{ hr}} \\ {0 + 0.024 \dot{m}_{3} + 0.910 \dot{m}_{4} = 455{ \ lb}_{ m} /{ hr}} \\ {0.091 \dot{m}_{2} + 0.901 \dot{m}_{3} + 0.041 \dot{m}_{4} = 300 { \ lb}_{ m} /{ hr}} \label{126}$

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

${\dot{m}_{2} +\dot{m}_{3} +\dot{m}_{4} = { 1000 \ lb}_{ m} /{ hr}} \\ {0 + 0.024 \dot{m}_{3} + 0.910 \dot{m}_{4} = 455{ \ lb}_{ m} /{ hr}} \\ {0 + 0.810 \dot{m}_{3} - 0.050 \dot{m}_{4} = 209 { \ lb}_{ m} /{ hr}} \label{127}$

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

${\dot{m}_{2} +\dot{m}_{3} +\dot{m}_{4} = { 1000 \ lb}_{ m} /{ hr}} \\ {0+\dot{m}_{3} +37.917 \dot{m}_{4} = 18958{ \ lb}_{ m} /{ hr}} \\ {0 + 0.810 \dot{m}_{3} - 0.050 \dot{m}_{4} = 209 { \ lb}_{ m} /{ hr}} \label{128}$

We now multiply the second equation by $$0.810$$ and subtract the result from the third equation to obtain

$\dot{m}_{2} +\dot{m}_{3} +\dot{m}_{4} = 1000 \ lb_{ m} /{ hr} \\ {0+\dot{m}_{3} +37.917 \dot{m}_{4} = 18958{ \ lb}_{ m} /{ hr}} \\ {0+0 - 30.763 \dot{m}_{4} = -15147 { \ lb}_{ m} /{ hr}} \label{129}$

and division of the third equation by $$-30.763$$ leads to

${\dot{m}_{2} +\dot{m}_{3} +\dot{m}_{4} = { 1000 \ lb}_{ m} /{ hr}} \\ {0+\dot{m}_{3} +37.917 \dot{m}_{4} = 18958.3{ \ lb}_{ m} /{ hr}} \\ {0+0+ \dot{m}_{4} = 492.39 { \ lb}_{ m} /{ hr}} \label{130}$

Having worked our way forward through this set of three equations in order to determine $$\dot{m}_{4}$$, we can work our way backward through the set to determine $$\dot{m}_{3}$$ and $$\dot{m}_{2}$$. The results are the same as we obtained earlier in Sec. 4.8 and we list the result again as

$\dot{m}_{2} = { 219 \ lb}_{ m} /{ hr} , \quad \dot{m}_{3} = { 288 \ lb}_{ m} /{ hr} , \quad \dot{m}_{4} = { 492 \ lb}_{ m} /{ hr} \label{131}$

The procedure represented by Eqs. \ref{124a} - \ref{124c} through \ref{130} 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.

## 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 Equation \ref{125} which we express as

${ A} = \begin{bmatrix} {1} & {1} & {1} \\ {0.045} & {0.069} & {0.955} \\ {0.091} & {0.901} & {0.041} \end{bmatrix} \label{132}$

We can express Equation \ref{125} in the compact form represented by Equation \ref{110}

$\mathbf{ Au} = \mathbf{b} \label{133}$

in which $$\mathbf{u}$$ is the column matrix of unknown mass flow rates and $$\mathbf{b}$$ is the column matrix of known mass flow rates.

$\mathbf{ u} = \begin{bmatrix} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{bmatrix} , \quad \mathbf{ b} = \begin{bmatrix} 1000 \ lb_{m} /hr \\ 500 \ lb_{m} /hr \\ 300 \ lb_{m} /hr \end{bmatrix} = \begin{bmatrix} {b_{2} } \\ {b_{3} } \\ {b_{4} } \end{bmatrix} \label{134}$

We can also make use of the unit matrix

$\mathbf{ I} = \begin{bmatrix} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{bmatrix} \label{135}$

to express Equation \ref{133} in the form

$\mathbf{ Au} = \mathbf{Ib} \label{136}$

Now we wish to repeat the Gaussian elimination used in Sec. 4.9.1, but in this case we will make use of Equation \ref{136} rather than Equation \ref{133} in order to retain the terms associated with $$\mathbf{Ib}$$. Multiplying the first of Eqs. \ref{125} by -0.045 and adding the result to the second equation leads to

$\dot{m}_{2} + \dot{m}_{3} +\dot{m}_{4} = b_{2} +0+ 0 = 1000 \ lb_{ m} /{ hr} \\ 0 + 0.024 \dot{m}_{3} +0.910 \dot{m}_{4} = -0.045 b_{2} + b_{3} + {0} = 455 \ lb_{ m} /{ hr} \\ 0.091 \dot{m}_{2} + 0.901 \dot{m}_{3} + 0.041 \dot{m}_{4} = {0+} { 0+ } {b_{4} } = 300 \ lb_{ m} /{ hr} \label{137}$

Note that this set of equations is identical to Eqs. \ref{126} except that we have included the terms associated with $$\mathbf{Ib}$$. We now proceed with the Gaussian elimination represented by Eqs. \ref{127} through \ref{130} to arrive at

${\dot{m}_{2} + \dot{m}_{3} +\dot{m}_{4} = b_{2} + 0 + 0 = 1000 { \ lb}_{ m} /{ hr}} \\ {0 + \dot{m}_{3} +37.917 \dot{m}_{4} = {-1.875 b_{2} +} {41.667b_{3} +} { 0} \\ = 18958.3{ \ lb}_{ m} /{ hr}} \\ {0 + 0 + \dot{m}_{4} = {-0.0464 b_{2} +} {1.0971 b_{3} -} {0.03251 b_{4} } = 492.39 { \ lb}_{ m} /{ hr}} \label{138}$

This result is identical to Equation \ref{130} except for the fact that we have retained the terms in the second column matrix that are associated with $$\mathbf{Ib}$$ in Equation \ref{136}. In Sec. 4.9.1, we used Equation \ref{130} to carry out a backward elimination in order to solve for the mass flow rates given by Equation \ref{131}, and here we want to present that backward elimination explicitly in order to demonstrate that it leads to the inverse of the coefficient matrix, $$\mathbf{A}$$. This procedure is known as the Gauss-Jordan algorithm and it consists of elementary row operations that reduce the left hand side of Equation \ref{138} to a diagonal form.

We begin to construct a diagonal form by multiplying the third equation by -37.917 and adding it to the second equation so that our set of equations take the form

${\dot{m}_{2} +\dot{m}_{3} +\dot{m}_{4} = b_{2} + 0 + 0 = 1000 { \ lb}_{ m} /{ hr}} \\ {0 + \dot{m}_{3} + 0 = {-0.1152 b_{2} + } {0.0677b_{3} +} {1.2326 b_{4} } = 288.42{ \ lb}_{ m} /{ hr}} \\ {0 + 0 +\dot{m}_{4} = {-0.0464 b_{2} +} {1.0971 b_{3} -} {0.03251 b_{4} } = 492.39 { \ lb}_{ m} /{ hr}} \label{139}$

We now multiply the third equation by $$-1$$ and add the result to the first equation to obtain

${\dot{m}_{2} +\dot{m}_{3} +0 = 1.0464 b_{2} - 1.09711 b_{3} + 0.03251 b_{4} = 507.61 { \ lb}_{ m} /{ hr}} \\ {0 + \dot{m}_{3} + 0 = {-0.1152 b_{2} +} {0.0677b_{3} +} {1.2326 b_{4} } = 288.42{ \ lb}_{ m} /{ hr}} \\ {0 + 0 +\dot{m}_{4} = {-0.0464 b_{2} +} {1.0971 b_{3} -} {0.03251 b_{4} } = 492.39 { \ lb}_{ m} /{ hr}} \label{140}$

and finally we multiply the second equation by $$-1$$ and add the result to the first equation to obtain the desired diagonal form

${\dot{m}_{2} + 0 +0 = 1.1616 b_{2} - 1.16484 b_{3} - 1.20005 b_{4} = 219.187{ \ lb}_{ m} /{ hr}} \\ {0 + \dot{m}_{3} +0 = {-0.1152 b_{2} +} {0.0677b_{3} +} {1.2326 b_{4} } = 288.42{ \ lb}_{ m} /{ hr}} \\ {0 + 0 + \dot{m}_{4} = {-0.0464 b_{2} +} {1.0971 b_{3} -} {0.03251 b_{4} } = 492.39{ \ lb}_{ m} /{ hr}}\label{141}$

Here we see that the unknown mass flow rates are determined by

$\dot{m}_{2} = { 219 \ lb}_{ m} /{ hr} , \quad \dot{m}_{3} = { 288 \ lb}_{ m} /{ hr} , \quad \dot{m}_{4} = { 492 \ lb}_{ m} /{ hr} \label{142}$

where only three significant figures have been listed since it is unlikely that the input data are accurate to more than 1%. The coefficient matrix contained in the central term of Eqs. \ref{141} is the inverse of $$\mathbf{A}$$ and we express this result as

$\mathbf{ A}^{-1} = \begin{bmatrix} {1.1616} & {-1.16484} & {-1.20005} \\ {-0.1152} & {0.0677} & {1.2326} \\ {-0.0464} & {1.0971} & {-0.0325} \end{bmatrix} \label{143}$

The solution procedure that led from Equation \ref{133} to the final answer can be expressed in compact form according to,

$\mathbf{ Au} = \mathbf{b} , \quad \Rightarrow \quad \mathbf{A}^{-1} \mathbf{Au} = \mathbf{A}^{-1} \mathbf{b} , \quad \Rightarrow \quad \mathbf{Iu} = \mathbf{A}^{-1} \mathbf{b} , \quad \Rightarrow \quad \mathbf{u} = \mathbf{A}^{-1} \mathbf{b} \label{144}$

and in terms of the details of the inverse matrix the last of these four equations can be expressed as

$\begin{bmatrix} {\dot{m}_{2} } \\ {\dot{m}_{3} } \\ {\dot{m}_{4} } \end{bmatrix} = \begin{bmatrix} {1.1616} & {-1.16484} & {-1.20005} \\ {-0.1152} & {0.0677} & {1.2326} \\ {-0.0464} & {1.0971} & {-0.0325} \end{bmatrix} \begin{bmatrix} 1000 \ lb_{m} /hr \\ 500 \ lb_{m} /hr \\ 300 \ lb_{m} /hr \end{bmatrix} \label{145}$

Given that computer routines are available to carry out the Gauss-Jordan algorithm, it should be clear that the solution of a set of linear mass balance equations is a routine matter.