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.

Example - solving a system using the Inverse Matrix

Solve the system using matrices.

x + 5y = 4

2x + 5y = −2

Always check your solutions!

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.

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:

circuit diagram


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

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) :}`

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.

circuit diagram - finding 7 currents

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?

Exercise 3

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

Gasoline without additive

Gasoline with 5% additive

Gasoline with 6% additive

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

Always check your solutions!

Exercise 4

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

statics forces diagram

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.

Didn't find what you are looking for on this page? Try search:

Online Algebra Solver

This algebra solver can solve a wide range of math problems. (Please be patient while it loads.)

Ready for a break?

 

Play a math game.

(Well, not really a math game, but each game was made using math...)

The IntMath Newsletter

Sign up for the free IntMath Newsletter. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!

Given name: * required

Family name:

email: * required

See the Interactive Mathematics spam guarantee.

Share IntMath!

Short URL for this Page

Save typing! You can use this URL to reach this page:

intmath.com/matlineq

Algebra Lessons on DVD

 

Easy to understand algebra lessons on DVD. See samples before you commit.

More info: Algebra videos

Loading...
Loading...