# 0.12 Matrices

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## Solution of an electric circuit with 2 unknowns by matrix inversion

Let us apply our knowledge of matrices to assist us in the analysis of an electric circuit. We consider the circuit shown below.

In this example, we wish to solve for the two node voltages v 1 and v 2 . Since there are two unknowns in this problem, we must first establish two independent equations that reflect the operation of the circuit.

Kirchoff’s Current Law tells us that the sum of the currents that enter a node must equal the sum of the currents that leave a node. Let us focus first on node 1. The current that enters node 1 from the left can be stated mathematically as

$\frac{\text{10}V-{v}_{1}}{1\Omega }$

The current that enters node 1 from the right can be stated as

$\frac{{v}_{2}-{v}_{1}}{1\Omega }$

The current that travels downward from node 1 is

$\frac{{v}_{1}}{2\Omega }$

We can arrange the expressions for each of the currents in terms of an equation via Kirchoff’s Current Law

$\frac{\text{10}V-{v}_{1}}{1\Omega }+\frac{{v}_{2}-{v}_{1}}{1\Omega }=\frac{{v}_{1}}{2\Omega }$

We can combine and rearrange these terms into the equation

$5{v}_{1}-2{v}_{2}=\text{20}V$

Now let us turn our attention to node 2. The current entering node 2 from the left is given by the expression

$\frac{{v}_{1}-{v}_{2}}{1\Omega }$

The current entering node 2 from the right is 2 A. The current leaving node 2 in a downward direction is

$\frac{{v}_{2}}{4\Omega }$

We proceed to combine these currents via Kirchoff’s Current Law

$\frac{{v}_{1}-{v}_{2}}{1\Omega }+2A=\frac{{v}_{2}}{4\Omega }$

This equation can be rearranged as

$-4{v}_{1}+5{v}_{2}=-8V$

So the pair of equations that we will use to solve for the two unknowns are

$5{v}_{1}-2{v}_{2}=\text{20}V$

and

$-4{v}_{1}+5{v}_{2}=-8V$

These equations may be expressed in matrix-vector form as

$\left[\begin{array}{cc}5& -2\\ -4& 5\end{array}\right]\left[\begin{array}{c}{v}_{1}\\ {v}_{2}\end{array}\right]=\left[\begin{array}{c}\text{20}\\ -8\end{array}\right]$

or

$A\left[\begin{array}{c}{v}_{1}\\ {v}_{2}\end{array}\right]=\left[\begin{array}{c}\text{20}\\ -8\end{array}\right]$
$A=\left[\begin{array}{cc}5& -2\\ -4& 5\end{array}\right]$

Let us find the inverse of the matrix A. The coefficients of this matrix are given by

${a}_{1,1}=5$
${a}_{1,2}=-2$
${a}_{2,1}=-4$
${a}_{2,2}=5$

The inverse can be found making use of the following formula

${A}^{-1}=\frac{1}{\text{det}A}\left[\begin{array}{cc}{a}_{2,2}& -{a}_{1,2}\\ -{a}_{2,1}& {a}_{1,1}\end{array}\right]$

It should be noted that this formula works only with (2 x 2) matrices. For matrices of higher rank, other methods need to be applied.

For this example, the determinant of A is found as

$\text{det}A=\left(5\right)\left(5\right)-\left(-4\right)\left(-2\right)=\text{25}-8=\text{17}$

We can incorporate this information to express the inverse matrix as

${A}^{-1}=\frac{1}{\text{17}}\left[\begin{array}{cc}5& 2\\ 4& 5\end{array}\right]$

which can be written as

${A}^{-1}=\left[\begin{array}{cc}\frac{5}{\text{17}}& \frac{2}{\text{17}}\\ \frac{4}{\text{17}}& \frac{5}{\text{17}}\end{array}\right]$

We can apply A -1 to solve for the unknowns

${A}^{-1}A\left[\begin{array}{c}{v}_{1}\\ {v}_{2}\end{array}\right]={A}^{-1}\left[\begin{array}{c}\text{20}\\ -8\end{array}\right]$

Recognizing that A -1 A = I , we find that

$\left[\begin{array}{c}{v}_{1}\\ {v}_{2}\end{array}\right]={A}^{-1}\left[\begin{array}{c}\text{20}\\ -8\end{array}\right]=\frac{1}{\text{17}}\left[\begin{array}{cc}5& 2\\ 4& 5\end{array}\right]\left[\begin{array}{c}\text{20}\\ -8\end{array}\right]=\frac{1}{\text{17}}\left[\begin{array}{c}\text{84}\\ \text{40}\end{array}\right]=\left[\begin{array}{c}4\text{.}\text{94}\\ 2\text{.}\text{35}\end{array}\right]$

So the node voltages are given as

${v}_{1}=4\text{.}\text{94}V$

and

${v}_{2}=2\text{.}\text{35}V$

## Solution of an electric circuit with 3 unknowns by gaussian elimination

Let us consider the electric circuit that is shown below.

Suppose that we are interested in determining the value of the three unknown currents I 1 , I 2 and I 3 . In order to do so, we rely upon Ohm’s Law and Kirchoff’s Laws to develop a system of three independent, linear equations. We should note that because we have three unknowns ( I 1 , I 2 and I 3 ), we must have three independent, linear equations.

${I}_{1}+{I}_{2}+{I}_{3}=0$
$-2{I}_{1}+3{I}_{2}=\text{24}$
$-3{I}_{2}+6{I}_{3}=0$

Let us define the matrix $A=\left[\begin{array}{ccc}1& 1& 1\\ -2& 3& 0\\ 0& -3& 6\end{array}\right]$

In order to find the unknowns, we must first find the inverse of the matrix A. This can be accomplished using elimination. To start the process, we adjoin the vector [0 24 0] T to the matrix A.

$\left(\begin{array}{ccc}1& 1& 1\\ -2& 3& 0\\ 0& -3& 6\end{array}\begin{array}{c}0\\ \text{24}\\ 0\end{array}\right)$

Next, we wish to force the left-most constant of row 2 to take on a value of 0. We can do so by multiplying each value in the first row by (-2) and subtracting the result from the corresponding value in row 2. This process yields

$\left(\begin{array}{ccc}1& 1& 1\\ 0& 5& 2\\ 0& -3& 6\end{array}\begin{array}{c}0\\ \text{24}\\ 0\end{array}\right)$

Now, we divide each term in row 2 by (5) to yield

$\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 2/5\\ 0& -3& 6\end{array}\begin{array}{c}0\\ \text{24}/5\\ 0\end{array}\right)$

Next, we turn our attention to eliminating the (-3) term in row 3. We can do so by multiplying each term of row 2 by (-3) and subtracting the results from the corresponding terms in row 3. This produces the matrix

$\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 2/5\\ 0& 0& \text{36}/5\end{array}\begin{array}{c}0\\ \text{24}/5\\ \text{72}/5\end{array}\right)$

We can then divide the terms of row 3 by (36/5) to produce

$\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 2/5\\ 0& 0& 1\end{array}\begin{array}{c}0\\ \text{24}/5\\ 2\end{array}\right)$

Interpretation of the third row tells us that the value for the third unknown ( I 3 ) is 2 A. We can use the coefficients from the second row along with the value for I 3 to solve for I 2 .

${I}_{2}+\frac{2}{5}\left({I}_{3}\right)=\frac{\text{24}}{5}$

which yields the result

${I}_{2}=4A\text{.}$

Lastly, we may use the coefficients of the first row along with the previously determined values for I 2 and I 3 to produce the result for I 1 .

${I}_{1}+{I}_{2}+{I}_{3}=0$

Insertion of the previously found unknowns yields

${I}_{1}+4+2=0$

So we find the value for I 1 to be -6 A.

## Exercises

1. Company A has more cash than Company B. If Company A lends $20 million to Company B, then the two companies would have the same amount of cash. If instead Company B gave Company A$22 million, then Company A would have twice as much cash as Company B. Use the matrix inversion method to find how much cash each company has.
2. A computer manufacturer sells two types of units. One unit is primarily marketed to the professional community and sells for $1,700. Another unit is marketed to students and sells for$900. In a typical month, the manufacturer sells 2,000 units. This accounts for \$1,380,000 in sales. Use the matrix inversion method to find how many units of each type are sold.
3. A ship can travel 300 miles upstream in 80 hours. Under the same conditions, the same ship can travel 275 miles downstream in 65 hours. Use the matrix inversion method to find the speed of the current along with the speed of the ship.
4. The matrix $A=\left[\begin{array}{ccc}2& 1& 3\\ 0& 6& 2\\ 1& 0& 1\end{array}\begin{array}{c}8\\ 4\\ 2\end{array}\right]$ represents a linear system with three unknowns. Use Gaussian elimination to solve for the three unknowns.
5. A system of 3 independent linear equations that govern the operation of the circuit below are ${i}_{1}+{i}_{2}+{i}_{2}=0$ , $-{i}_{1}-\text{24}+2{i}_{2}=0$ , and $-2{i}_{2}+4{i}_{3}=0$ . Use Gaussian elimination to solve for the three currents.
6. Suppose that the value of each resistor in the figure below is 1 Ω. The mesh equations that govern the circuit are $6V=2{i}_{a}-{i}_{b}$ and $2{i}_{b}-{i}_{a}+9V=0$ . Use the matrix inversion method to find the two mesh current.

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