# 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

## 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:

((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.

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:

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

### 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.

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