Chapter 10, Problem 8P

The following system of equations is designed to determine concentrations $\left(\text{the }c\text{'}s{\text{ in g/m}}^3\right)$ in a series of coupled reactors as a function of the amount of mass input to each reactor (the right-hand sides in g/day),

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ -3c_1+18c_2-6c_3=1200\\ -4c_1-c_2+12c_3=2350\end{array}$

(a) Determine the matrix inverse.

(b) Use the inverse to determine the solution.

(c) Determine how much the rate of mass input to reactor 3 must be increased to induce a $10{\text{ g/m}}^3$ rise in the concentration of reactor 1.

(d) How much will the concentration in reactor 3 be reduced if the rate of mass input to reactors 1 and 2 is reduced by 500 and 250 g/day, respectively?

To determine

To calculate: The inverse for the given system:

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ -3c_1+18c_2-6c_3=1200\\ -4c_1-c_2+12c_3=2350\end{array}$

Answer to Problem 8P

Solution:

The inverse matrix is, $A^{-1}=\left[\begin{array}{ccc}0.072538& 0.012781& 0.012435\\ 0.020738& 0.060794& 0.032124\\ 0.025906& 0.009326& 0.090155\end{array}\right]$.

Explanation of Solution

Given:

The system of equations,

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ -3c_1+18c_2-6c_3=1200\\ -4c_1-c_2+12c_3=2350\end{array}$

Formula used:

(1) The forward substitution equations for L can be expressed as,

$\left[L\right]\left[D\right]=\left[B\right]$

(2) The backward substitution equation for U can be expressed as,

$\left[U\right]\left[X\right]=\left[D\right]$

(3) $f_{21}=\frac{a_{21}}{a_{11}}$, $f_{31}=\frac{a_{31}}{a_{11}}$ and $f_{32}=\frac{{a^{\prime }}_{32}}{{a^{\prime }}_{22}}$

Calculation:

Consider the system of equations,

$15c_1-3c_2-c_3=3800$ …… (1)

$-3c_1+18c_2-6c_3=1200$ …… (2)

$-4c_1-c_2+12c_3=2350$ …… (3)

The coefficient $a_{21}$ is eliminated by multiplying equation (1) by $f_{21}=\frac{a_{21}}{a_{11}}$,

$f_{21}=\frac{-3}{15}$

And subtracting the result from equation (2).

Thus, multiply equation (1) by $\left(\frac{-3}{15}\right)$,

$\begin{array}{c}15\left(\frac{-3}{15}\right)c_1-3\left(\frac{-3}{15}\right)c_2-\left(\frac{-3}{15}\right)c_3=3800\left(\frac{-3}{15}\right)\\ -3c_1+\frac{3}{5}{c}_2+\frac{1}{5}{c}_3=-760\end{array}$

Now subtract this equation from equation (2),

$\begin{array}{c}-3c_1+18c_2-6c_3+3c_1-\frac{3}{5}{c}_2-\frac{1}{5}{c}_3=1200+760\\ \frac{87}{5}{c}_2-\frac{31}{5}{c}_3=1960\\ 17.4c_2-6.2c_3=1960\end{array}$ …… (4)

The coefficient $a_{31}$ is eliminated by multiplying equation (1) by $f_{31}=\frac{a_{31}}{a_{11}}$,

$f_{31}=\frac{-4}{15}$

And subtracting the result from equation (3).

Thus, multiply equation (1) by $\left(\frac{-4}{15}\right)$,

$\begin{array}{c}15\left(\frac{-4}{15}\right)c_1-3\left(\frac{-4}{15}\right)c_2-\left(\frac{-4}{15}\right)c_3=3800\left(\frac{-4}{15}\right)\\ -4c_1+\frac{4}{5}{c}_2+\frac{4}{15}{c}_3=\frac{-3040}{3}\end{array}$

Now subtract this equation from equation (3),

$\begin{array}{c}-4c_1-c_2+12c_3+4c_1-\frac{4}{5}{c}_2-\frac{4}{15}{c}_3=2350+\frac{3040}{3}\\ -\frac{9}{5}{c}_2+\frac{176}{15}{c}_3=3363.333\\ -1.8c_2+11.73333c_3=3363.333\end{array}$ …… (5)

Now the set of equations is,

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ 17.4c_2-6.2c_3=1960\\ -1.8c_2+11.73333c_3=3363.333\end{array}$

The factors $f_{21}$ and $f_{31}$ can be stored in $a_{21}$ and $a_{31}$.

$\left[\begin{array}{ccc}15& -3& -1\\ -0.2& 17.4& -6.2\\ -0.26667& -1.8& 11.73333\end{array}\right]$

The coefficient $a_{32}$ is eliminated by multiplying equation (4) by $f_{32}=\frac{{a^{\prime }}_{32}}{{a^{\prime }}_{22}}$,

$f_{32}=\frac{-1.8}{17.4}$

And subtracting the result from equation (5).

Thus, multiply equation (4) by $\left(\frac{-1.8}{17.4}\right)$,

$\begin{array}{c}17.4\left(\frac{-1.8}{17.4}\right)c_2-6.2\left(\frac{-1.8}{17.4}\right)c_3=1960\left(\frac{-1.8}{17.4}\right)\\ -1.8c_2+0.641379c_3=-202.7586\end{array}$

Now, subtract this equation from equation (5),

$\begin{array}{c}-1.8c_2+11.73333c_3+1.8c_2-0.641379c_3=3363.333+202.7586\\ 11.091951c_3=3566.0916\end{array}$ …… (6)

The factor $f_{32}$ can be stored in $a_{32}$.

Thus, the matrix obtained is:

$\left[\begin{array}{ccc}15& -3& -1\\ -0.2& 17.4& -6.2\\ -0.26667& -0.103448& 11.091951\end{array}\right]$

Therefore, the LU decomposition is

$\left[L\right]=\left[\begin{array}{ccc}1& 0& 0\\ -0.2& 1& 0\\ -0.26667& -0.103448& 1\end{array}\right]\left[U\right]=\left[\begin{array}{ccc}15& -3& -1\\ 0& 17.4& -6.2\\ 0& 0& 11.091951\end{array}\right]$

Now, to find the inverse of the given system.

The first column of the inverse matrix can be determined by performing the forward substitution solution with a unit vector (with 1 in the first row) of right-hand-side vector.

The forward substitution equations for L can be expressed as,

$\left[L\right]\left[D\right]=\left[B\right]$

Where,

$\left[B\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$

Determine [D] by substituting L and B as shown below,

$\left[\begin{array}{ccc}1& 0& 0\\ -0.2& 1& 0\\ -0.26667& -0.103448& 1\end{array}\right]\left[\begin{array}{c}d1\\ d2\\ d3\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$

Solve for $d_1$,

$d_1=1$

Solve for $d_2$,

$\begin{array}{c}-0.2d_1+d_2=0\\ d_2=0.2d_1\\ d_2=0.2\left(1\right)\\ d_2=0.2\end{array}$

Solve for $d_3$,

$\begin{array}{c}-0.26667d_1-0.103448d_2+d_3=0\\ d_3=0.26667d_1+0.103448d_2\\ d_3=0.26667\left(1\right)+0.103448\left(0.2\right)\\ d_3=0.287359\end{array}$

Hence, the values obtained are $d_1=1,d_2=0.2,\text{and}{d}_3=0.287359$.

Solve with forward substitution of ${\left[D\right]}^T=\left[\begin{array}{ccc}1& 0.2& 0.287359\end{array}\right]$,

This vector can be used as right-hand side vector of equation,

$\begin{array}{c}\left[U\right]\left[X\right]=\left[D\right]\\ \left[\begin{array}{ccc}15& -3& -1\\ 0& 17.4& -6.2\\ 0& 0& 11.091951\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}1\\ 0.2\\ 0.287359\end{array}\right]\end{array}$

Solve the above matrix by back substitution, which gives the first column of the inverse matrix as:

$\left[X\right]=\left[\begin{array}{c}0.072538\\ 0.020738\\ 0.025906\end{array}\right]$

Similarly, the second column of the inverse matrix can be determined by performing the forward substitution solution with a unit vector (with 1 in the second row) of right-hand-side vector.

The forward substitution equations for L can be expressed as,

$\left[L\right]\left[D\right]=\left[B\right]$

Where,

$\left[B\right]=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$

Determine[D] by substituting L and B as shown below,

$\left[\begin{array}{ccc}1& 0& 0\\ -0.2& 1& 0\\ -0.26667& -0.103448& 1\end{array}\right]\left[\begin{array}{c}d1\\ d2\\ d3\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$

Solve for $d_1$,

$d_1=0$

Solve for $d_2$,

$\begin{array}{c}-0.2d_1+d_2=1\\ d_2=1+0.2d_1\\ d_2=1+0.2\left(0\right)\\ d_2=1\end{array}$

Solve for $d_3$,

$\begin{array}{c}-0.26667d_1-0.103448d_2+d_3=0\\ d_3=0.26667d_1+0.103448d_2\\ d_3=0.26667\left(0\right)+0.103448\left(1\right)\\ d_3=0.103448\end{array}$

Hence, the values obtained are $d_1=0,d_2=1,\text{and}{d}_3=0.103448$.

Solve with forward substitution of ${\left[D\right]}^T=\left[\begin{array}{ccc}0& 1& 0.103448\end{array}\right]$,

This vector can be used as right-hand side vector of equation,

$\begin{array}{c}\left[U\right]\left[X\right]=\left[D\right]\\ \left[\begin{array}{ccc}15& -3& -1\\ 0& 17.4& -6.2\\ 0& 0& 11.091951\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ 0.103448\end{array}\right]\end{array}$

Solve the above matrix by back substitution, which gives the second column of the inverse matrix as:

$\left[X\right]=\left[\begin{array}{c}0.012781\\ 0.060794\\ 0.009326\end{array}\right]$

Similarly, the third column of the inverse matrix can be determined by performing the forward substitution solution with a unit vector (with 1 in the third row) of right-hand-side vector.

The forward substitution equations for L can be expressed as,

$\left[L\right]\left[D\right]=\left[B\right]$

Where,

$\left[B\right]=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

Determine [D] by substituting L and B as shown below,

$\left[\begin{array}{ccc}1& 0& 0\\ -0.2& 1& 0\\ -0.26667& -0.103448& 1\end{array}\right]\left[\begin{array}{c}d1\\ d2\\ d3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

Solve for $d_1$,

$d_1=0$

Solve for $d_2$,

$\begin{array}{c}-0.2d_1+d_2=0\\ d_2=0.2d_1\\ d_2=0.2\left(0\right)\\ d_2=0\end{array}$

Solve for $d_3$,

$\begin{array}{c}-0.26667d_1-0.103448d_2+d_3=1\\ d_3=1+0.26667d_1+0.103448d_2\\ d_3=1+0.26667\left(0\right)+0.103448\left(0.2\right)\\ d_3=1\end{array}$

Hence, the values obtained are $d_1=0,d_2=0\text{and}{d}_3=1$.

Solve with forward substitution of $D^T=\left[\begin{array}{ccc}0& 0& 1\end{array}\right]$,

This vector can be used as right-hand side vector of equation,

$\begin{array}{c}\left[U\right]\left[X\right]=\left[D\right]\\ \left[\begin{array}{ccc}15& -3& -1\\ 0& 17.4& -6.2\\ 0& 0& 11.091951\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]\end{array}$

Solve the above matrix by back substitution, which gives the third column of the inverse matrix as:

$\left[X\right]=\left[\begin{array}{c}0.012435\\ 0.032124\\ 0.090155\end{array}\right]$

Thus, the inverse matrix is:

$A^{-1}=\left[\begin{array}{ccc}0.072538& 0.012781& 0.012435\\ 0.020738& 0.060794& 0.032124\\ 0.025906& 0.009326& 0.090155\end{array}\right]$

To determine

To calculate: The solution of the given system using inverse:

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ -3c_1+18c_2-6c_3=1200\\ -4c_1-c_2+12c_3=2350\end{array}$

Answer to Problem 8P

Solution:

The solution of the given system is $c_1=320.207,c_2=227.202\text{and}{c}_3=321.503$.

Explanation of Solution

Given:

The system of equations,

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ -3c_1+18c_2-6c_3=1200\\ -4c_1-c_2+12c_3=2350\end{array}$

Where,

$A^{-1}=\left[\begin{array}{ccc}0.072538& 0.012781& 0.012435\\ 0.020738& 0.060794& 0.032124\\ 0.025906& 0.009326& 0.090155\end{array}\right]$

Formula used:

If $\left[A\right],\left[X\right]\text{and}\left[B\right]$ are matrices such that $\left[A\right]\left[X\right]=\left[B\right]$. Then the solution vector $\left[X\right]$ is given as:

$\left[X\right]={\left[A\right]}^{-1}\left[B\right]$

Calculation:

Consider the given system of equations:

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ -3c_1+18c_2-6c_3=1200\\ -4c_1-c_2+12c_3=2350\end{array}$

Where,

$\left[A\right]=\begin{array}{ccc}15& -3& -1\\ -3& 18& -6\\ -4& -1& 12\end{array}$

$\left[C\right]=\left[\begin{array}{l}{c}_1\\ c_2\\ c_3\end{array}\right]$

$\left[B\right]=\left[\begin{array}{l}3800\\ 1200\\ 2350\end{array}\right]$

The inverse of the given system is:

$A^{-1}=\left[\begin{array}{ccc}0.072538& 0.012781& 0.012435\\ 0.020738& 0.060794& 0.032124\\ 0.025906& 0.009326& 0.090155\end{array}\right]$

Thus, the solution vector [C] is given by:

$\begin{array}{c}\left[C\right]={\left[A\right]}^{-1}\left[B\right]\\ =\left[\begin{array}{ccc}0.072538& 0.012781& 0.012435\\ 0.020738& 0.060794& 0.032124\\ 0.025906& 0.009326& 0.090155\end{array}\right]\left[\begin{array}{l}3800\\ 1200\\ 2350\end{array}\right]\\ =\left[\begin{array}{l}320.207\\ 227.202\\ 321.503\end{array}\right]\end{array}$

Thus, the solution of the given system is $c_1=320.207,c_2=227.202\text{and}{c}_3=321.503$.

To determine

To calculate: The rate of mass input to reactor 3 that is to be increased to induce a $10{\text{g/m}}^{\text{3}}$ rise in the concentration of reactor 1, where the following system of equations is designed to determine concentrations $\left(\text{the }c{\text{'s in g/m}}^{\text{3}}\right)$ in a series of coupled reactors as a function of the amount of mass input to each reactor $\left(\text{the right-hand sides in g/day}\right)$.

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ -3c_1+18c_2-6c_3=1200\\ -4c_1-c_2+12c_3=2350\end{array}$

Answer to Problem 8P

Solution:

The rate of mass input must be increased to $804.1817\text{g/day}$.

Explanation of Solution

Given:

The system of equations,

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ -3c_1+18c_2-6c_3=1200\\ -4c_1-c_2+12c_3=2350\end{array}$

And the inverse of the given system is:

$A^{-1}=\left[\begin{array}{ccc}0.072538& 0.012781& 0.012435\\ 0.020738& 0.060794& 0.032124\\ 0.025906& 0.009326& 0.090155\end{array}\right]$

Formula used:

$\Delta W_3=\frac{\Delta c_1}{a^{-1}{}_{13}}$

Calculation:

Consider the given system of equations:

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ -3c_1+18c_2-6c_3=1200\\ -4c_1-c_2+12c_3=2350\end{array}$

Where,

$\left[A\right]=\begin{array}{ccc}15& -3& -1\\ -3& 18& -6\\ -4& -1& 12\end{array}$

Let rate of mass input to reactor 3 be $\Delta W_3$.

Rise in concentration of reactor 1 be $\Delta c_1=10{\text{gm/m}}^{\text{3}}$.

The inverse of the given system is:

$A^{-1}=\left[\begin{array}{ccc}0.072538& 0.012781& 0.012435\\ 0.020738& 0.060794& 0.032124\\ 0.025906& 0.009326& 0.090155\end{array}\right]$

So, $a_{13}{}^{-1}=0.012435$.

Thus,

$\begin{array}{c}\Delta W_3=\frac{\Delta c_1}{a^{-1}{}_{13}}\\ =\frac{10}{0.012435}\\ =804.1817\end{array}$

Hence, the rate of mass input must be increased to $804.1817\text{g/day}$.

To determine

To calculate: The reduced concentration in reactor 3 if the rate of mass input to reactors 1 and 2 is reduced by 500 and 250 g/day, where the following system of equations is designed to determine concentrations $\left(\text{the }c{\text{'s in g/m}}^{\text{3}}\right)$ in a series of coupled reactors as a function of the amount of mass input to each reactor $\left(\text{the right-hand sides in g/day}\right)$.

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ -3c_1+18c_2-6c_3=1200\\ -4c_1-c_2+12c_3=2350\end{array}$

Answer to Problem 8P

Solution:

The reduced concentration in reactor 3 is $15.285{\text{gm/m}}^3$.

Explanation of Solution

Given:

The system of equations,

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ -3c_1+18c_2-6c_3=1200\\ -4c_1-c_2+12c_3=2350\end{array}$

And the inverse of the given system is:

$A^{-1}=\left[\begin{array}{ccc}0.072538& 0.012781& 0.012435\\ 0.020738& 0.060794& 0.032124\\ 0.025906& 0.009326& 0.090155\end{array}\right]$

Formula used:

$\Delta c_3=a_{31}{}^{-1}\Delta W_1+a_{32}{}^{-1}\Delta W_2$

Calculation:

Consider the given system of equations:

$\begin{array}{c}15c_1-3c_2-c_3=3800\\ -3c_1+18c_2-6c_3=1200\\ -4c_1-c_2+12c_3=2350\end{array}$

Where,

$\left[A\right]=\begin{array}{ccc}15& -3& -1\\ -3& 18& -6\\ -4& -1& 12\end{array}$

Let, the reduced concentration of reactor 3 is $\Delta c_3$.

The reduced rate of mass input to reactor 1 is $\Delta W_1=-500\text{g/day}$.

The reduced rate of mass input to reactor 2 is $\Delta W_2=-250\text{g/day}$.

The inverse of the given system is:

$A^{-1}=\left[\begin{array}{ccc}0.072538& 0.012781& 0.012435\\ 0.020738& 0.060794& 0.032124\\ 0.025906& 0.009326& 0.090155\end{array}\right]$

So, $a_{31}{}^{-1}=0.025906\text{ and}{a}_{32}{}^{-1}=0.009326$.

Thus,

$\begin{array}{c}\Delta c_3=a_{31}{}^{-1}\Delta W_1+a_{32}{}^{-1}\Delta W_2\\ =0.025906\left(-500\right)+0.009326\left(-250\right)\\ =-15.285\end{array}$

Hence, the reduced concentration in reactor 3 is $15.285{\text{gm/m}}^3$.