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# Matrices

Module by: Scott Starks. E-mail the author

## Matrices

### 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 v1 and v2. 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

10 V v 1 1 Ω 10 V v 1 1 Ω size 12{ { {"10"V - v rSub { size 8{1} } } over {1 %OMEGA } } } {}
(1)

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

v 2 v 1 1 Ω v 2 v 1 1 Ω size 12{ { {v rSub { size 8{2} } - v rSub { size 8{1} } } over {1 %OMEGA } } } {}
(2)

The current that travels downward from node 1 is

v 1 2 Ω v 1 2 Ω size 12{ { {v rSub { size 8{1} } } over {2 %OMEGA } } } {}
(3)

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

10 V v 1 1 Ω + v 2 v 1 1 Ω = v 1 2 Ω 10 V v 1 1 Ω + v 2 v 1 1 Ω = v 1 2 Ω size 12{ { {"10"V - v rSub { size 8{1} } } over {1 %OMEGA } } + { {v rSub { size 8{2} } - v rSub { size 8{1} } } over {1 %OMEGA } } = { {v rSub { size 8{1} } } over {2 %OMEGA } } } {}
(4)

We can combine and rearrange these terms into the equation

5 v 1 2 v 2 = 20 V 5 v 1 2 v 2 = 20 V size 12{5v rSub { size 8{1} } - 2v rSub { size 8{2} } ="20"V} {}
(5)

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

v 1 v 2 1 Ω v 1 v 2 1 Ω size 12{ { {v rSub { size 8{1} } - v rSub { size 8{2} } } over {1 %OMEGA } } } {}
(6)

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

v 2 4 Ω v 2 4 Ω size 12{ { {v rSub { size 8{2} } } over {4 %OMEGA } } } {}
(7)

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

v 1 v 2 1 Ω + 2 A = v 2 4 Ω v 1 v 2 1 Ω + 2 A = v 2 4 Ω size 12{ { {v rSub { size 8{1} } - v rSub { size 8{2} } } over {1 %OMEGA } } +2A= { {v rSub { size 8{2} } } over {4 %OMEGA } } } {}
(8)

This equation can be rearranged as

4 v 1 + 5 v 2 = 8 V 4 v 1 + 5 v 2 = 8 V size 12{ - 4v rSub { size 8{1} } +5v rSub { size 8{2} } = - 8V} {}
(9)

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

5 v 1 2 v 2 = 20 V 5 v 1 2 v 2 = 20 V size 12{5v rSub { size 8{1} } - 2v rSub { size 8{2} } ="20"V} {}
(10)

and

4 v 1 + 5 v 2 = 8 V 4 v 1 + 5 v 2 = 8 V size 12{ - 4v rSub { size 8{1} } +5v rSub { size 8{2} } = - 8V} {}
(11)

These equations may be expressed in matrix-vector form as

5 2 4 5 v 1 v 2 = 20 8 5 2 4 5 v 1 v 2 = 20 8 size 12{ left [ matrix { 5 {} # - 2 {} ## - 4 {} # 5{} } right ] left [ matrix { v rSub { size 8{1} } {} ## v rSub { size 8{2} } } right ]= left [ matrix { "20" {} ## - 8 } right ]} {}
(12)

or

A v 1 v 2 = 20 8 A v 1 v 2 = 20 8 size 12{A left [ matrix { v rSub { size 8{1} } {} ## v rSub { size 8{2} } } right ]= left [ matrix { "20" {} ## - 8 } right ]} {}
(13)
A = 5 2 4 5 A = 5 2 4 5 size 12{A= left [ matrix { 5 {} # - 2 {} ## - 4 {} # 5{} } right ]} {}
(14)

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

a 1,1 = 5 a 1,1 = 5 size 12{a rSub { size 8{1,1} } =5} {}
(15)
a 1,2 = 2 a 1,2 = 2 size 12{a rSub { size 8{1,2} } = - 2} {}
(16)
a 2,1 = 4 a 2,1 = 4 size 12{a rSub { size 8{2,1} } = - 4} {}
(17)
a 2,2 = 5 a 2,2 = 5 size 12{a rSub { size 8{2,2} } =5} {}
(18)

The inverse can be found making use of the following formula

A 1 = 1 det A a 2,2 a 1,2 a 2,1 a 1,1 A 1 = 1 det A a 2,2 a 1,2 a 2,1 a 1,1 size 12{A rSup { size 8{ - 1} } = { {1} over {"det"A} } left [ matrix { a rSub { size 8{2,2} } {} # - a rSub { size 8{1,2} } {} ## - a rSub { size 8{2,1} } {} # a rSub { size 8{1,1} } {} } right ]} {}
(19)

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

det A = ( 5 ) ( 5 ) ( 4 ) ( 2 ) = 25 8 = 17 det A = ( 5 ) ( 5 ) ( 4 ) ( 2 ) = 25 8 = 17 size 12{"det"A= $$5$$ $$5$$ - $$- 4$$ $$- 2$$ ="25" - 8="17"} {}
(20)

We can incorporate this information to express the inverse matrix as

A 1 = 1 17 5 2 4 5 A 1 = 1 17 5 2 4 5 size 12{A rSup { size 8{ - 1} } = { {1} over {"17"} }  left [ matrix { 5 {} # 2 {} ## 4 {} # 5{} } right ]} {}
(21)

which can be written as

A 1 = 5 17 2 17 4 17 5 17 A 1 = 5 17 2 17 4 17 5 17 size 12{A rSup { size 8{ - 1} } = left [ matrix { { {5} over {"17"} } {} # { {2} over {"17"} } {} ## { {4} over {"17"} } {} # { {5} over {"17"} } {} } right ]} {}
(22)

We can apply A-1 to solve for the unknowns

A 1 A v 1 v 2 = A 1 20 8 A 1 A v 1 v 2 = A 1 20 8 size 12{A rSup { size 8{ - 1} } A left [ matrix { v rSub { size 8{1} } {} ## v rSub { size 8{2} } } right ]=A rSup { size 8{ - 1} }  left [ matrix { "20" {} ## - 8 } right ]} {}
(23)

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

v 1 v 2 = A 1 20 8 = 1 17 5 2 4 5 20 8 = 1 17 84 40 = 4 . 94 2 . 35 v 1 v 2 = A 1 20 8 = 1 17 5 2 4 5 20 8 = 1 17 84 40 = 4 . 94 2 . 35 size 12{ left [ matrix { v rSub { size 8{1} } {} ## v rSub { size 8{2} } } right ]=A rSup { size 8{ - 1} } left [ matrix { "20" {} ## - 8 } right ]= { {1} over {"17"} } left [ matrix { 5 {} # 2 {} ## 4 {} # 5{} } right ] left [ matrix { "20" {} ## - 8 } right ]= { {1} over {"17"} }  left [ matrix { "84" {} ## "40" } right ]= left [ matrix { 4 "." "94" {} ## 2 "." "35" } right ]} {}
(24)

So the node voltages are given as

v 1 = 4 . 94 V v 1 = 4 . 94 V size 12{v rSub { size 8{1} } =4 "." "94"V} {}
(25)

and

v 2 = 2 . 35 V v 2 = 2 . 35 V size 12{v rSub { size 8{2} } =2 "." "35"V} {}
(26)

### 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 I1, I2 and I3. 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 (I1, I2 and I3), we must have three independent, linear equations.

I 1 + I 2 + I 3 = 0 I 1 + I 2 + I 3 = 0 size 12{I rSub { size 8{1} } +I rSub { size 8{2} } +I rSub { size 8{3} } =0} {}
(27)
2 I 1 + 3 I 2 = 24 2 I 1 + 3 I 2 = 24 size 12{ - 2I rSub { size 8{1} } +3I rSub { size 8{2} } ="24"} {}
(28)
3 I 2 + 6 I 3 = 0 3 I 2 + 6 I 3 = 0 size 12{ - 3I rSub { size 8{2} } +6I rSub { size 8{3} } =0} {}
(29)

Let us define the matrix A=111230036A=111230036 size 12{A= left [ matrix { 1 {} # 1 {} # 1 {} ## - 2 {} # 3 {} # 0 {} ## 0 {} # - 3 {} # 6{} } 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.

( 1 1 1 2 3 0 0 3 6 0 24 0 ) ( 1 1 1 2 3 0 0 3 6 0 24 0 ) size 12{$$matrix { 1 {} # 1 {} # 1 {} ## - 2 {} # 3 {} # 0 {} ## 0 {} # - 3 {} # 6{} } ~ matrix { 0 {} ## "24" {} ## 0 }$$} {}
(30)

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

( 1 1 1 0 5 2 0 3 6 0 24 0 ) ( 1 1 1 0 5 2 0 3 6 0 24 0 ) size 12{$$matrix { 1 {} # 1 {} # 1 {} ## 0 {} # 5 {} # 2 {} ## 0 {} # - 3 {} # 6{} } ~ matrix { 0 {} ## "24" {} ## 0 }$$} {}
(31)

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

( 1 1 1 0 1 2 / 5 0 3 6 0 24 / 5 0 ) ( 1 1 1 0 1 2 / 5 0 3 6 0 24 / 5 0 ) size 12{$$matrix { 1 {} # 1 {} # 1 {} ## 0 {} # 1 {} # 2/5 {} ## 0 {} # - 3 {} # 6{} } ~ matrix { 0 {} ## "24"/5 {} ## 0 }$$} {}
(32)

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

1 1 1 0 1 2 / 5 0 0 36 / 5 0 24 / 5 72 / 5 1 1 1 0 1 2 / 5 0 0 36 / 5 0 24 / 5 72 / 5 size 12{ left ( matrix { 1 {} # 1 {} # 1 {} ## 0 {} # 1 {} # 2/5 {} ## 0 {} # 0 {} # "36"/5{} } ~ matrix { 0 {} ## "24"/5 {} ## "72"/5 } right )} {}
(33)

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

( 1 1 1 0 1 2 / 5 0 0 1 0 24 / 5 2 ) ( 1 1 1 0 1 2 / 5 0 0 1 0 24 / 5 2 ) size 12{$$matrix { 1 {} # 1 {} # 1 {} ## 0 {} # 1 {} # 2/5 {} ## 0 {} # 0 {} # 1{} } ~ matrix { 0 {} ## "24"/5 {} ## 2 }$$} {}
(34)

Interpretation of the third row tells us that the value for the third unknown (I3) is 2 A. We can use the coefficients from the second row along with the value for I3 to solve for I2.

I 2 + 2 5 ( I 3 ) = 24 5 I 2 + 2 5 ( I 3 ) = 24 5 size 12{I rSub { size 8{2} } + { {2} over {5} } $$I rSub { size 8{3} }$$ = { {"24"} over {5} } } {}
(35)

which yields the result

I 2 = 4 A . I 2 = 4 A . size 12{I rSub { size 8{2} } =4A "." } {}
(36)

Lastly, we may use the coefficients of the first row along with the previously determined values for I2 and I3 to produce the result for I1.

I 1 + I 2 + I 3 = 0 I 1 + I 2 + I 3 = 0 size 12{I rSub { size 8{1} } +I rSub { size 8{2} } +I rSub { size 8{3} } =0} {}

Insertion of the previously found unknowns yields

I 1 + 4 + 2 = 0 I 1 + 4 + 2 = 0 size 12{I rSub { size 8{1} } +4+2=0} {}
(37)

So we find the value for I1 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=213062101842A=213062101842 size 12{A= left [ matrix { 2 {} # 1 {} # 3 {} ## 0 {} # 6 {} # 2 {} ## 1 {} # 0 {} # 1{} } ~ matrix { 8 {} ## 4 {} ## 2 } 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 i1+i2+i2=0i1+i2+i2=0 size 12{i rSub { size 8{1} } +i rSub { size 8{2} } +i rSub { size 8{2} } =0} {}, i124+2i2=0i124+2i2=0 size 12{ - i rSub { size 8{1} } - "24"+2i rSub { size 8{2} } =0} {}, and 2i2+4i3=02i2+4i3=0 size 12{ - 2i rSub { size 8{2} } +4i rSub { size 8{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=2iaib6V=2iaib size 12{6V=2i rSub { size 8{a} } - i rSub { size 8{b} } } {} and 2ibia+9V=02ibia+9V=0 size 12{2`i rSub { size 8{b} } - i rSub { size 8{a} } +9V=0} {}. Use the matrix inversion method to find the two mesh current.

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