Tutorial 7: Engineering Applications of Matrices and Vectors

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Tutorial PDF

Class Recording

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Q1: An electrical engineer supervises the production of three types of electrical components. Three kinds of materials are; metal, plastic and rubber – are required for production. The amounts needed to produce each component are

If totals of 2.12, 0.0434 and 0.164 kg of metal, plastic and rubber respectively are available each day, how many components can be produced per day?

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Solution

\[\begin{aligned} \begin{bmatrix} 15 & 17 & 19\\ 0.25 & 0.33 & 0.42\\ 1 & 1.2 & 1.6 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} & =\begin{bmatrix} 2120\\ 43.4\\ 164 \end{bmatrix} \end{aligned}\] \[D=\begin{vmatrix} 15 & 17 & 19\\ 0.25 & 0.33 & 0.42\\ 1 & 1.2 & 1.6 \end{vmatrix} =0.13\] \[D_{x_{1}} =\begin{vmatrix} 2120 & 17 & 19\\ 43.4 & 0.33 & 0.42\\ 164 & 1.2 & 1.6 \end{vmatrix} =2.6\] \[D_{x_{2}} =\begin{vmatrix} 15 & 2120 & 19\\ 0.25 & 43.4 & 0.42\\ 1 & 164 & 1.6 \end{vmatrix} =5.2\] \[D_{x_{3}} =\begin{vmatrix} 15 & 17 & 2120\\ 0.25 & 0.33 & 43.4\\ 1 & 1.2 & 164 \end{vmatrix} =7.8\] \[\begin{aligned} x_{1} & =\frac{D_{x_{1}}}{D}\\ & =\frac{2.6}{0.13}\\ & =\mathbf{20} \end{aligned}\] \[\begin{aligned} x_{2} & =\frac{D_{x_{2}}}{D}\\ & =\frac{5.2}{0.13}\\ & =\mathbf{40} \end{aligned}\] \[\begin{aligned} x_{3} & =\frac{D_{x_{3}}}{D}\\ & =\frac{7.8}{0.13}\\ & =\mathbf{60} \end{aligned}\]

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Q2: A civil engineer involved in construction requires 4800, 5800 and 5700 m\(^3\) of sand, fine gravel and coarse gravel respectively for building a project. There are three pits from which these materials can be obtained. The composition of these pits is

How many cubic meters must be hauled from each pit in order to meet the engineer’s needs?

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Solution

\[\begin{aligned} \begin{bmatrix} 0.52 & 0.30 & 0.18\\ 0.20 & 0.50 & 0.30\\ 0.25 & 0.20 & 0.55 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix} & =\begin{bmatrix} 4800\\ 5800\\ 5700 \end{bmatrix} \end{aligned}\] \[D=\begin{vmatrix} 0.52 & 0.30 & 0.18\\ 0.20 & 0.50 & 0.30\\ 0.25 & 0.20 & 0.55 \end{vmatrix} =0.086\] \[D_{x_{1}} =\begin{vmatrix} 4800 & 0.30 & 0.18\\ 5800 & 0.50 & 0.30\\ 5700 & 0.20 & 0.55 \end{vmatrix} =344.5\] \[D_{x_{2}} =\begin{vmatrix} 0.52 & 4800 & 0.18\\ 0.20 & 5800 & 0.30\\ 0.25 & 5700 & 0.55 \end{vmatrix} =613.3\] \[D_{x_{3}} =\begin{vmatrix} 0.52 & 0.30 & 4800\\ 0.20 & 0.50 & 5800\\ 0.25 & 0.20 & 5700 \end{vmatrix} =444\] \[\begin{aligned} x_{1} & =\frac{D_{x_{1}}}{D}\\ & =\frac{344.5}{0.086}\\ & =\mathbf{4005.82} \end{aligned}\] \[\begin{aligned} x_{2} & =\frac{D_{x_{2}}}{D}\\ & =\frac{613.3}{0.086}\\ & =\mathbf{7131.40} \end{aligned}\] \[\begin{aligned} x_{3} & =\frac{D_{x_{3}}}{D}\\ & =\frac{444}{0.086}\\ & =\mathbf{5162.79} \end{aligned}\]

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Q3: By referring to the schematic below, find V1, V2, V3 and V4. Solve the problem using Cramer’s rule.

Solution

Node equations

Node Equation #1: \(V_{1} =1.2 \qquad \color{red} {\scriptsize{\triangleleft [\text{Since Node 1 directly connectted to source}]}}\)

Node Equation #2: \(\frac{V_{2} -1.2}{R_{11}} +\frac{V_{2} -V_{3}}{R_{13}} +\frac{V_{2} -V_{4}}{R_{12}} =0\)

Node Equation #3: \(\frac{V_{3} -V_{2}}{R_{13}} +\frac{V_{3} -8.7}{R_{14}} =0\)

Node Equation #4: \(\frac{V_{4} -V_{2}}{R_{12}} +\frac{V_{4} -0}{R_{15}} =0\)

Solving the node equations:

Substituting all values to Node Equation #2:

\[\begin{aligned} \frac{V_{2} -1.2}{1000} +\frac{V_{2} -V_{3}}{1000} +\frac{V_{2} -V_{4}}{10000} & =0\\ 10V_{2} -12+10V_{2} -10V_{3} +V_{2} -V_{4} & =0\\ 21V_{2} -10V_{3} -V_{4} & =12 \end{aligned} \tag{1}\]

Substituting all values to Node Equation #3:

\[\begin{aligned} \frac{V_{3} -V_{2}}{1000} +\frac{V_{3} -8.7}{1000} & =0\\ V_{3} -V_{2} +V_{3} -8.7 & =0\\ -V_{2} +2V_{3} & =8.7 \end{aligned} \tag{2}\]

Substituting all values to Node Equation #4:

\[\begin{aligned} \frac{V_{4} -V_{2}}{10000} +\frac{V_{4} -0}{10000} & =0\\ V_{4} -V_{2} +V_{4} & =0\\ -V_{2} +2V_{4} & =0 \end{aligned} \tag{3}\]

Putting Equation (1) - (3) into matrix and solve for \(V_{2}\), \(V_{3}\), \(V_4\):

\[\begin{bmatrix} 21 & -10 & -1\\ -1 & 2 & 0\\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} V_{2}\\ V_{3}\\ V_{4} \end{bmatrix} =\begin{bmatrix} 12\\ 8.7\\ 0 \end{bmatrix}\] \[D=\begin{vmatrix} 21 & -10 & -1\\ -1 & 2 & 0\\ -1 & 0 & 2 \end{vmatrix} =62\] \[D_{V_{2}} =\begin{vmatrix} 12 & -10 & -1\\ 8.7 & 2 & 0\\ 0 & 0 & 2 \end{vmatrix} =222\] \[D_{V_{3}} =\begin{vmatrix} 21 & 12 & -1\\ -1 & 8.7 & 0\\ -1 & 0 & 2 \end{vmatrix} =380.7\] \[D_{V_{4}} =\begin{vmatrix} 21 & -10 & 12\\ -1 & 2 & 8.7\\ -1 & 0 & 0 \end{vmatrix} =111\] \[\begin{aligned} V_{2} & =\frac{D_{V_{2}}}{D}\\ & =\frac{222}{62}\\ & =\mathbf{3.58\ V} \end{aligned}\] \[\begin{aligned} V_{3} & =\frac{D_{V_{3}}}{D}\\ & =\frac{380.7}{62}\\ & =\mathbf{6.14\ V} \end{aligned}\] \[\begin{aligned} V_{4} & =\frac{D_{V_{4}}}{D}\\ & =\frac{111}{62}\\ & =\mathbf{1.79\ V} \end{aligned}\] \[\therefore \ V_{1} =\mathbf{1.2\ V} ;\ \ V_{2} =\mathbf{3.58\ V} ;\ \ V_{3} =\mathbf{6.14\ V} ;\ \ V_{4} =\mathbf{1.79\ V}\]

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Q4: By referring to the schematic below, find V2, and V3. Solve the problem using Cramer’s rule.

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Solution

Node equations

Node Equation #1: \(V_{1} =8.7 \qquad \color{red} {\scriptsize{\triangleleft [\text{Since Node 1 directly connectted to source}]}}\)

Node Equation #2: \(\frac{V_{2} -V_{1}}{R_{1}} +\frac{V_{2} -0}{R_{2}} +\frac{V_{2} -V_{3}}{R_{3}} =0\)

Node Equation #3: \(\frac{V_{3} -V_{2}}{R_{3}} +\frac{V_{3} -0}{R_{4}} =0\)

Solving the node equations:

Substituting all values to Node Equation #2:

\[\begin{aligned} \frac{V_{2} -8.7}{1000} +\frac{V_{2} -0}{10000} +\frac{V_{2} -V_{3}}{1000} & =0\\ 10V_{2} -87+V_{2} +10V_{2} -10V_{3} & =0\\ 21V_{2} -10V_{3} & =87 \end{aligned} \tag{1}\]

Substituting all values to Node Equation #3:

\[\begin{aligned} \frac{V_{3} -V_{2}}{1000} +\frac{V_{3} -0}{10000} & =0\\ 10V_{3} -10V_{2} +V_{3} & =0\\ -10V_{2} +11V_{3} & =0 \end{aligned} \tag{2}\]

Putting Equation (1) and (2) into matrix and solve for \(V_2\), \(V_3\):

\[\begin{bmatrix} 21 & -10\\ -10 & 11 \end{bmatrix}\begin{bmatrix} V_{2}\\ V_{3} \end{bmatrix} =\begin{bmatrix} 87\\ 0 \end{bmatrix}\] \[D=\begin{vmatrix} 21 & -10\\ -10 & 11 \end{vmatrix} =131\] \[D_{V_{2}} =\begin{vmatrix} 87 & -10\\ 0 & 11 \end{vmatrix} =957\] \[D_{V_{3}} =\begin{vmatrix} 21 & -87\\ -10 & 0 \end{vmatrix} =870\] \[\begin{aligned} V_{2} & =\frac{D_{V_{2}}}{D}\\ & =\frac{957}{131}\\ & =\mathbf{7.31\ V} \end{aligned}\] \[\begin{aligned} V_{3} & =\frac{D_{V_{3}}}{D}\\ & =\frac{870}{131}\\ & =\mathbf{6.64\ V} \end{aligned}\] \[\therefore \ V_{1} =\mathbf{8.7\ V} ;\ \ V_{2} =\mathbf{7.31\ V} ;\ \ V_{3} =\mathbf{6.64\ V}\]

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