6. Matrices and Linear Equations

by M. Bourne

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

If we let

, and

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 use CA^-1 because matrix multiplication is not commutative.

Example - solving a system using the Inverse Matrix

Solve the system using matrices.

Always check your solutions!

Answer

Now let's see how LiveMath can do this for us.

LIVEMath

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 LiveMath (or similar) to find inverses larger than 2×2.

Example - 3×3 System of Equations

Solve the system using matrix methods.

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

Answer

Now let's see how LiveMath can do this for us.

LIVEMath

Example - Electronics application of 3×3 System of Equations

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

Answer

Exercise 1

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

Answer

Exercise 2

Recall this problem from before? Now we know how to solve it, using inverse matrices on a computer.

The circuit equations, using Kirschoff's Law:

-26 = 72I₁ - 17I₃ - 35I₄

34 = 122I₂ - 35I₃ - 87I₇

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

5 = 105I₄ - 35I₁ - 43I₅

-27 = 105I₅ - 28I₃ - 43I₄ - 34I₆

24 = 141I₆ - 35I₃ - 34I₅ - 72I₇

-4 = 233I₇ - 87I₂ - 34I₃ - 72I₆