# 6. Matrices and Linear Equations

by M. Bourne

We wish to solve the system of simultaneous linear equations using matrices:

a1x + b1y = c1
a2x + b2y = c2

If we let

A=((a_1,b_1),(a_2,b_2)), \ X=((x),(y))\  and \ C=((c_1),(c_2))

then AX=C. (We first saw this in Multiplication of Matrices).

If we now multiply each side of

AX = C

on the left by

A-1, we have:

A-1AX = A-1C.

However, we know that A-1A = I, the Identity matrix. So we obtain

IX = A-1C.

But IX = X, so the solution to the system of equations is given by:

X = A-1C

See the box at the top of Inverse of a Matrix for more explanation about why this works.

Note: We cannot reverse the order of multiplication and use CA-1 because matrix multiplication is not commutative.

Continues below

### Example - solving a system using the Inverse Matrix

Solve the system using matrices.

x + 5y = 4

2x + 5y = −2

We have:

A=((-1,5),(2,5)),  \ X=((x),(y))\  and \ C=((4),(-2))

To solve the system, we need the inverse of A, which we write as A-1.

((5,5),(2,-1))

Change signs of the other 2 elements:

((5,-5),(-2,-1))

Now we find the determinant of A:

|A| = -5 - 10 = -15

So

A^-1 =-1/15((5,-5),(-2,-1))  = ((-1/3,1/3),(2/15,1/15))  = ((-0.333,0.333),(0.133,0.067))

So the solution to the system is given by:

X=A^-1C =((-0.333,0.333),(0.133,0.067))((4),(-2)) =((-2),(0.4))

This answer means that we have found the solution x = -2 and y = 0.4.

Is the solution correct?

We check it in the original set of equations:

{:(-x+5y,=4),(2x+5y,=-2):}

Substituting x = -2 and y = 0.4, we get:

−(−2) + 5×(0.4) = 2 + 2 = 4 [Checks OK]

2×(−2) + 5×(0.4)  = −4 + 2  = −2 [Checks OK]

So the solution to the original system of equations is

x = -2,\ \ y = 0.4.

## Solving 3×3 Systems of Equations

We can extend the above method to systems of any size. We cannot use the same method for finding inverses of matrices bigger than 2×2.

We will use a Computer Algebra System to find inverses larger than 2×2.

### Example - 3×3 System of Equations

Solve the system using matrix methods.

{: (x+2y-z=6),(3x+5y-z=2),(-2x-y-2z=4) :}

Did I mention? It's a good idea to always check your solutions.

A=((1,2,-1),(3,5,-1),(-2,-1,-2)), X=((x),(y),(z)), and C=((6),(2),(4))

Using Scientific Notebook, we find the inverse of A to be:

A^-1=((5.5,-2.5,-1.5),(-4,2,1),(-3.5,1.5,0.5))

(We could have used Gauss-Jordan Elimination if we need to show all steps.)

So the solution to the system of equations is:

X=A^-1C

=((5.5,-2.5,-1.5),(-4,2,1),(-3.5,1.5,0.5))((6),(2),(4))

=((22),(-16),(-16))

Check:

22 + 2(-16) - (-16) = 6 [Checks OK]

3(22) + 5(-16) - (-16) = 2 [Checks OK]

-2(22) - (16) - 2(-16) = 4 [Checks OK]

So the solution is x = 22, y = -16 and z = -16.

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### Example - Electronics application of 3×3 System of Equations

Find the electric currents shown by solving the matrix equation (obtained using Kirchhoff's Law) arising from this circuit:

((I_1+I_2+I_3),(-2I_1+3I_2),(-3I_2+6I_3))=((0),(24),(0))

(You can explore what the solution for this example really means in this 3D interactive systems of equations applet.)

We can write this as:

((1,1,1),(-2,3,0),(0,-3,6))((I_1),(I_2),(I_3))=((0),(24),(0))

So we have:

((I_1),(I_2),(I_3))=((1,1,1),(-2,3,0),(0,-3,6))^-1((0),(24),(0))

Using a computer algebra system to perform the inverse and multiply by the constant matrix, we get:

I_1= -6\ "A"

I_2= 4\ "A"

I_3= 2\ "A"

We observe that I1 is negative, as expected from the circuit diagram.

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### Exercise 1

The following equations are found in a particular electrical circuit. Find the currents using matrix methods.

{: (I_A+I_B+I_C=0),(2I_A-5I_B=6),(5I_B-I_C=-3) :}

(This example is also included in the 3D interactive systems of equations applet.)

We need to form the matrices:

A=((1,1,1),(2,-5,0),(0,5,-1)), \ X=((I_A),(I_B),(I_C))\  and \ C=((0),(6),(-3))

Using Scientific Notebook (or any Computer Algebra System), we find:

A^-1  = ((0.294,0.353,0.294),(0.118,-0.059,0.118),(0.588,-0.294,-0.412))

and so (continuing to use Scientific Notebook, with rounding to 3 decimal places):

((I_A),(I_B),(I_C))=A^-1C

=((0.294,0.353,0.294),(0.118,-0.059,0.118),(0.588,-0.294,-0.412))((0),(6),(-3))

=((1.236),(-0.708),(-0.528))

Therefore

I_A= 1.236\ "A",

I_B= -0.708\ "A" and

I_C= -0.528\ "A"

### Exercise 2

Recall this problem from before? If we know the simultaneous equations involved, we will be able to solve the system using inverse matrices on a computer.

The circuit equations, using Kirchhoff's Law:

−26 = 72I1 − 17I3 − 35I4

34 = 122I2 − 35I3 − 87I7

−4 = 233I7 − 87I2 − 34I3 − 72I6

−13 = 149I3 − 17I1 − 35I2 − 28I5 − 35I6 − 34I7

−27 = 105I5 − 28I3 − 43I4 − 34I6

24 = 141I6 − 35I3 − 34I5 − 72I7

5 = 105I4 − 35I1 − 43I5

What are the individual currents, I1 to I7?

### Phone users

NOTE: If you're on a phone, you can scroll any wide matrices on this page to the right or left to see the whole expression.

#### Solving currents in a Circuit (7 × 7 system)

We solve this using a computer as follows. We just write the coefficient matrix on the left, find the inverse (raise the matrix to the power -1) and multiply the result by the constant matrix.

You can use Matlab, Mathcad or similar math software to do this. Wolfram|Alpha is a free alternative.

X=[(72,0,-17,-35,0,0,0), (0,122,-35,0,0,0,-87), (0,-87,-34,0,0,-72,233), (-17,-35,149,0,-28,-35,-34), (0,0,-28,-43,105,-34,0), (0,0,-35,0,-34,141,-72), (-35,0,0,105,-43,0,0)]^-1 [(-26),(34),(-4),(-13),(-27),(24),(5)]

=[(-0.46801),(0.42932),(5.193xx10^-3),(-0.22243),(-0.27848),(0.21115),(0.20914)]

The answer means that the currents in this circuit are (to 4 decimal places):

I_1 = -0.4680\ "A"

I_2= 0.4293\ "A"

I_3= 0.0005\ "A"

I_4= -0.2224\ "A"

I_5= -0.2785\ "A"

I_6= 0.2112 \ "A"

I_7= 0.2091 \ "A"

Easy to understand math videos:
MathTutorDVD.com

### Exercise 3

We want 10 L of gasoline containing 2% additive. We have drums of the following:

We need to use 4 times as much pure gasoline as 5% additive gasoline. How much of each is needed?

Let

x = no. of litres of pure gasoline

y = no. of litres of 5% gasoline

z = no. of litres of 6% gasoline

From the first sentence, we have:

x + y + z = 10

The second sentence gives us:

We get NO additive from the pure gasoline.

We get (5% of y) L of additive from the second drum.

We get (6% of z) L of additive from the third drum.

We NEED 2% of 10 L of additive = 0.2 L = 200 mL.

So

0.05y + 0.06z = 0.2

Multiplying through by 100 gives us:

5y + 6z = 20

The second last sentence gives us:

x = 4y

We can write this as:

x - 4y = 0

This gives us the set of simultaneous equations:

x + y + z = 10

5y + 6z = 20

x − 4y = 0

So

A=((1,1,1),(0,5,6),(1,-4,0)), \ C=((10),(20),(0))

Using Scientific Notebook for the inverse:

((1,1,1),(0,5,6),(1,-4,0))^-1 =((0.96,-0.16,0.04),(0.24,-0.04,-0.24),(-0.2,0.2,0.2))

Multiplying the inverse by matrix C:

((0.96,-0.16,0.04),(0.24,-0.04,-0.24),(-0.2,0.2,0.2))((10),(20),(0)) =((6.4),(1.6),(2))

So we have 6.4 L of pure gasoline, 1.6 L of 5% additive and 2 L of 6% additive.

Is it correct?

6.4 + 1.6 + 2 = 10 L [Checks OK]

5% xx 1.6 + 6% xx 2 = 200 mL [Checks OK]

4 × 1.6 = 6.4 [Checks OK]

### Exercise 4

This statics problem was presented earlier in Section 3: Matrices.

From the diagram, we obtain the following equations (these equations come from statics theory):

Vertical forces:

F1 sin 69.3° − F2 sin 71.1° − F3 sin 56.6° + 926 = 0

Horizontal forces:

F1 cos 69.3° − F2 cos 71.1° + F3 cos 56.6° = 0

Moments:

7.80 F1 sin 69.3° − 1.50 F2 sin 71.1° − 5.20 F3 sin 56.6° = 0

Using matrices, find the forces F1, F2 and F3.

We write the first equation so that the constant term is on the right hand side:

F1 sin 69.3° − F2 sin 71.1° − F3 sin 56.6° = −926

In matrix form, we write the equations as:

((sin 69.3°,-sin 71.1°,-sin 56.6°),(cos 69.3°,-cos 71.1°,cos 56.6°),(7.80 sin 69.3°,-1.50 sin 71.1°,-5.20 sin 56.6°))((F_1),(F_2),(F_3))

=((-926),(0),(0))

So the solution for the system is:

((F_1),(F_2),(F_3))=((sin 69.3°,-sin 71.1°,-sin 56.6°),(cos 69.3°,-cos 71.1°,cos 56.6°),(7.80 sin 69.3°,-1.50 sin 71.1°,-5.20 sin 56.6°))^-1((-926),(0),(0))

=((425.5),(1079.9),(362.2))

So

F_1= 425.5\ "N"

F_2= 1079.9\ "N"

F_3= 362.2\ "N"

This is very easy and quick in Scientific Notebook, Matlab or any other computer algebra system!