The principle of this method is A×X=B. For normal algebra we find X by dividing each side by A: X=B/A.
Another way is to write X=B×(1/A) or X=(1/A)×B. The expression 1/A is called the reciprocal of A or the inverse of A and can be written A-1.
In the world of matrices, we effectively employ the same method as X=B×(1/A) but when X, A and B are matrices then 1/A is written in different ways: A-1 or A' or A*. The most usual representation is A-1.
The definition of an inverse is found in the following equation AA-1=A-1A=I where I is the identity matrix, consisting of diagonal 1s in the matrix elements and all other elements are zero. So we have AX=B, so we multiply both sides by A-1 ro solve the equation:
A-1AX=IX=A-1B, which is the same as X=A-1B.
So what is matrix X? If we represent the elements vertically, X is:
⎡x⎤
⎢y⎥
⎣z⎦
So matrix X effectively combines all the unknown variables, which we're trying to find.
Now we need A-1. A is represented by a 3×3 matrix made up of the coefficients of the system of equations. The matrix size is square and the number of rows and columns has to be 3 because there are 3 variables.
⎡1 1 1⎤
⎢1 2 3⎥
⎣1 4 -9⎦
The question is incomplete because x+4y-9z=? so I'll use the letter a to represent the missing constant and later we'll make some substitutions and arrive at solutions in terms of a. Matrix B is:
⎡3⎤
⎢4⎥
⎣a⎦
Before moving on to find the matrix inverse, let's show why the matrices restate the system of equations:
⎡1 1 1⎤⎡x⎤ ⎡3⎤
⎢1 2 3⎥⎢y⎥=⎢4⎥
⎣1 4 -9⎦⎣z⎦ ⎣a⎦
Using normal matrix multiplication we arrive at the three given equations.
Finding the inverse requires a degree of concentration so here's what I find easiest for me. But please note that there are other methods of finding the inverse matrix and you may find them easier.
We need to find 10 determinants. The first one is |A|:
1(2×(-9)-3×4)-1(1×(-9)-3×1)+1(1×4-2×1)=1(-18-12)-1(-9-3)+1(4-2)=-30+12+2=-16. The reciprocal of this is applied to the matrix we calculate next, so we have -1/16 as the multiplier. The constant will be applied to every element in the matrix. We have 9 more determinants to calculate and they're all 2×2 matrices. Let's apply some colour coding to A:
⎡1 1 1⎤
⎢1 2 3⎥
⎣1 4 -9⎦
Now we create a matrix "guide" by interchanging rows and columns:
⎡1 1 1⎤
⎢1 2 4⎥
⎣1 3 -9⎦
These 9 elements are going to be replaced by 9 determinants taken from the original A. Let's start with the first row of the guide. We take the first red 1 and find the 2×2 determinant by referring to the original A:
|2 3|
|4 -9| =-18-12=-30.
Back to the guide, and the next element in the row is blue 1 so, going back to A the determinant is:
|1 1|
|4 -9|=-9-4=-13.
The last element in the first row of the guide is green 1. From A we get the determinant:
|1 1|
|2 3|=3-2=1.
These 3 determinants form the first row of what will become A-1.
We now move to the second row of the guide to get the second row of what will become A-1, then on to the third row.
It's easy to make mistakes so be careful!
The matrix we get should be:
⎡-30 -13 1⎤
⎢-12 -10 2⎥
⎣ 2 3 1⎦
Now we apply the sign rule matrix:
⎡+ - +⎤
⎢- + -⎥
⎣+ - +⎦
Where there's a minus we change the sign of the element:
⎡-30 13 1⎤
⎢ 12 -10 -2⎥
⎣ 2 -3 1⎦
Finally we apply the multiplier -1/16:
⎡15/8 -13/16 -1/16⎤
⎢-3/4 5/8 1/8⎥
⎣-1/8 3/16 -1/16⎦
This matrix is applied to matrix B:
⎡45/8-13/4-a/16⎤ ⎡19/8-a/16⎤
⎢ -9/4+5/2+a/8 ⎥= ⎢ 1/4+a/8 ⎥
⎣ -3/8+3/4-a/16 ⎦ ⎣ 3/8-a/16 ⎦
These are x, y and z values.
If a=0, then x=19/8, y=1/4, z=3/8. The third equation would be x+4y-9z=0.
If a=6, then x=2, y=1, z=0.
My guess is that equation 3 should have been x+4y-9z=6 making x=2, y=1, z=0 (which satisfy the system of equations), because the variables are integers whereas if a=0, the variables are fractions, which nevertheless satisfy the system of equations.
I hope this helps.