Chapter 16, Problem 2E

For the following system of equations, (a) write the set of equations in matrix form, (b) Use ∆_Y to calculate V₂ only.

To determine

The set of equations in matrix form.

Answer to Problem 2E

The set of equations in matrix form is $\left[\begin{array}{ccc}100& -45& 30\\ 75& 0& 80\\ 48& 200& 42\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\\ V_3\end{array}\right]=\left[\begin{array}{c}0.2\\ -0.1\\ 0.5\end{array}\right]$.

Explanation of Solution

Given data:

The set of equations are,

$\begin{array}{l}100V_1-45V_2+30V_3=0.2\\ 75V_1+80V_3=-0.1\\ 48V_1+200V_2+42V_3=0.5\end{array}$

Calculation:

The expression for matrix representation of system is given by,

$\begin{array}{l}\left[Y\right]\left[V\right]=\left[B\right]\\ \left[\begin{array}{ccc}100& -45& 30\\ 75& 0& 80\\ 48& 200& 42\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\\ V_3\end{array}\right]=\left[\begin{array}{c}0.2\\ -0.1\\ 0.5\end{array}\right]\end{array}$

Here,

$\begin{array}{l}\left[Y\right]=\left[\begin{array}{ccc}100& -45& 30\\ 75& 0& 80\\ 48& 200& 42\end{array}\right]\\ \left[V\right]=\left[\begin{array}{c}{V}_1\\ V_2\\ V_3\end{array}\right]\\ \left[B\right]=\left[\begin{array}{c}0.2\\ -0.1\\ 0.5\end{array}\right]\end{array}$

Conclusion:

Therefore, the set of equations in matrix form is $\left[\begin{array}{ccc}100& -45& 30\\ 75& 0& 80\\ 48& 200& 42\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\\ V_3\end{array}\right]=\left[\begin{array}{c}0.2\\ -0.1\\ 0.5\end{array}\right]$.

To determine

The value of voltage $V_2$.

Answer to Problem 2E

The value of voltage $V_2$ is $2.551095\times {10}^{-3}\text{V}$.

Explanation of Solution

The set of equations in matrix form is given by,

$\left[Y\right]=\left[\begin{array}{ccc}100& -45& 30\\ 75& 0& 80\\ 48& 200& 42\end{array}\right]$

The expression for the ${\Delta }_Y$ is given by,

$\begin{array}{c}{\Delta }_Y=\left|\begin{array}{ccc}100& -45& 30\\ 75& 0& 80\\ 48& 200& 42\end{array}\right|\\ =100\left(0\times 42-200\times 80\right)+45\left(75\times 42-80\times 48\right)+30\left(75\times 200-0\right)\\ =-1600000-31050+450000\\ =-1181050\end{array}$

The expression for voltage using determinant is given by,

$\left[V\right]={\left[Y\right]}^{-1}\left[B\right]$        (1)

The expression of ${\left[Y\right]}^{-1}$ is given by,

${\left[Y\right]}^{-1}=\frac{\text{adj}Y}{{\Delta }_Y}$        (2)

Here,

$\text{adj}Y$ is the adjoint of the matrix.

The transpose of the matrix $Y$ is given by,

$Y^{\text{T}}=\left[\begin{array}{ccc}100& 75& 48\\ -45& 0& 200\\ 30& 80& 42\end{array}\right]$

The adjacent of the matrix is determined by the co-factors of the transpose matrix which is given by,

$\text{adj}Y=\left[\begin{array}{ccc}{\text{A}}_{11}& {\text{A}}_{12}& {\text{A}}_{13}\\ {\text{B}}_{21}& {\text{B}}_{22}& {\text{B}}_{23}\\ {\text{C}}_{31}& {\text{C}}_{32}& {\text{C}}_{33}\end{array}\right]$        (3)

Here,

${\text{A}}_{\text{11}}$, ${\text{A}}_{\text{12}}$, ${\text{A}}_{\text{13}}$, ${\text{B}}_{\text{21}}$, ${\text{B}}_{\text{22}}$, ${\text{B}}_{\text{23}}$, ${\text{C}}_{\text{31}}$, ${\text{C}}_{\text{32}}$, and ${\text{C}}_{\text{33}}$ are the co-factors of the matrix.

The expression for ${\text{A}}_{\text{11}}$ is given by,

$\begin{array}{c}{\text{A}}_{\text{11}}=\left(0\times 42-80\times 200\right)\\ =-16000\end{array}$

The expression for ${\text{A}}_{\text{12}}$ is given by,

$\begin{array}{c}{\text{A}}_{\text{12}}=-\left(\left(-45\times 42\right)-\left(200\times 30\right)\right)\\ =7890\end{array}$

The expression for ${\text{A}}_{\text{13}}$ is given by,

$\begin{array}{c}{\text{A}}_{\text{13}}=\left(-45\times 80-0\times 30\right)\\ =-3600\end{array}$

The expression for ${\text{B}}_{\text{21}}$ is given by,

$\begin{array}{c}{\text{B}}_{\text{21}}=-\left(75\times 42-80\times 48\right)\\ =690\end{array}$

The expression for ${\text{B}}_{\text{22}}$ is given by,

$\begin{array}{c}{\text{B}}_{\text{22}}=\left(100\times 42-30\times 48\right)\\ =2760\end{array}$

The expression for ${\text{B}}_{\text{23}}$ is given by,

$\begin{array}{c}{\text{B}}_{\text{23}}=-\left(100\times 80-75\times 30\right)\\ =-5750\end{array}$

The expression for ${\text{C}}_{\text{31}}$ is given by,

$\begin{array}{c}{\text{C}}_{\text{31}}=\left(75\times 200-0\times 48\right)\\ =15000\end{array}$

The expression for ${\text{C}}_{\text{32}}$ is given by,

$\begin{array}{c}{\text{C}}_{\text{32}}=-\left(100\times 200+45\times 48\right)\\ =-22160\end{array}$

The expression for ${\text{C}}_{\text{33}}$ is given by,

$\begin{array}{c}{\text{C}}_{\text{33}}=\left(100\times 0+45\times 75\right)\\ =3375\end{array}$

Substitute $-16000$ for ${\text{A}}_{\text{11}}$, $7890$ for ${\text{A}}_{\text{12}}$, $-3600$ for ${\text{A}}_{\text{13}}$, $690$ for ${\text{B}}_{\text{21}}$, $2760$ for ${\text{B}}_{\text{22}}$, $-5750$ for ${\text{B}}_{\text{23}}$, $15000$ for ${\text{C}}_{\text{31}}$, $-22160$ for ${\text{C}}_{\text{32}}$, and $3375$ for ${\text{C}}_{\text{33}}$ in equation (3).

$\text{adj}Y=\left[\begin{array}{ccc}-16000& 7890& -3600\\ 690& 2760& -5750\\ 15000& -22160& 3375\end{array}\right]$

Substitute $\left[\begin{array}{ccc}-16000& 7890& -3600\\ 690& 2760& -5750\\ 15000& -22160& 3375\end{array}\right]$ for $\text{adj}Y$ and $-1181050$ for ${\Delta }_Y$ in equation (2).

$\begin{array}{c}{\left[Y\right]}^{-1}=\frac{\left[\begin{array}{ccc}-16000& 7890& -3600\\ 690& 2760& -5750\\ 15000& -22160& 3375\end{array}\right]}{-1181050}\\ =\left[\begin{array}{ccc}0.01354& -6.68049\times {10}^{-3}& 3.048\times {10}^{-3}\\ -5.8422\times {10}^{-4}& -2.3369\times {10}^{-3}& 4.8685\times {10}^{-3}\\ -0.0127& 0.01876& -2.8576\times {10}^{-3}\end{array}\right]\end{array}$

Substitute $\left[\begin{array}{ccc}0.01354& -6.68049\times {10}^{-3}& 3.048\times {10}^{-3}\\ -5.8422\times {10}^{-4}& -2.3369\times {10}^{-3}& 4.8685\times {10}^{-3}\\ -0.0127& 0.01876& -2.8576\times {10}^{-3}\end{array}\right]$ for ${\left[Y\right]}^{-1}$, $\left[\begin{array}{c}{V}_1\\ V_2\\ V_3\end{array}\right]$ for $\left[V\right]$, and $\left[\begin{array}{c}0.2\\ -0.1\\ 0.5\end{array}\right]$ for $\left[B\right]$ in equation (1).

$\begin{array}{c}\left[\begin{array}{c}{V}_1\\ V_2\\ V_3\end{array}\right]=\left[\begin{array}{ccc}0.01354& -6.68049\times {10}^{-3}& 3.048\times {10}^{-3}\\ -5.8422\times {10}^{-4}& -2.3369\times {10}^{-3}& 4.8685\times {10}^{-3}\\ -0.0127& 0.01876& -2.8576\times {10}^{-3}\end{array}\right]\left[\begin{array}{c}0.2\\ -0.1\\ 0.5\end{array}\right]\\ =\left[\begin{array}{c}\left(0.01354\times 0.2\right)+\left(-6.68049\times {10}^{-3}\times 0.1\right)+\left(3.048\times {10}^{-3}\times 0.5\right)\\ \left(-5.8422\times {10}^{-4}\times 0.2\right)+\left(-2.3369\times {10}^{-3}\times 0.1\right)+\left(4.8685\times {10}^{-3}\times 0.5\right)\\ \left(-0.0127\times 0.2\right)-\left(0.01876\times 0.1\right)-\left(-2.8576\times {10}^{-3}\times 0.5\right)\end{array}\right]\\ =\left[\begin{array}{c}4.89204\times {10}^{-3}\\ 2.551095\times {10}^{-3}\\ -5.84388\times {10}^{-3}\end{array}\right]\end{array}$

Thus, the value of voltage $V_2$ is $2.551095\times {10}^{-3}\text{V}$.

Conclusion:

Therefore, the value of voltage $V_2$ is $2.551095\times {10}^{-3}\text{V}$.