# 6. Matrices and Linear Equations

by M. Bourne

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

a_{1}x+b_{1}y=c_{1}

a_{2}x+b_{2}y=c_{2}

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^{-1}AX=A^{-1}C.

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

IX=A^{-1}C.

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

X=A^{-1}C

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= 42

x+ 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:

`((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 = 72

I_{1}- 17I_{3}- 35I_{4}34 = 122

I_{2}- 35I_{3}- 87I_{7}-4 = 233

I_{7}- 87I_{2}- 34I_{3}- 72I_{6}-13 = 149

I_{3}- 17I_{1}- 35I_{2}- 28I_{5}- 35I_{6}- 34I_{7}-27 = 105

I_{5}- 28I_{3}- 43I_{4}- 34I_{6}24 = 141

I_{6}- 35I_{3}- 34I_{5}- 72I_{7}5 = 105

I_{4}- 35I_{1}- 43I_{5}

What are the individual currents, *I*_{1} to *I*_{7}?

### 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.

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

**Vertical forces: **

F_{1} sin 69.3° − F_{2} sin 71.1° − F_{3} sin 56.6° + 926 = 0

**Horizontal forces: **

F_{1} cos 69.3° − F_{2} cos 71.1° + F_{3} cos 56.6° = 0

**Moments:**

7.80 F_{1} sin 69.3° − 1.50 F_{2} sin 71.1° − 5.20 F_{3} sin
56.6° = 0

Using matrices, find the forces F_{1}, F_{2} and F_{3}.

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