When working in the real numbers, the equation ax=b could be solved for x by dividing both sides of the equation by a to get x=b/a, as long as a wasn't zero. It would therefore seem logical that when working with matrices, one could take the matrix equation AX=B and divide both sides by A to get X=B/A.
However, that won't work because ...
There is NO matrix division!
Ok, you say. Subtraction was defined in terms of addition and division was defined in terms of multiplication. So, instead of dividing, I'll just multiply by the inverse. This is the way that it has to be done.
So, what is the inverse of a matrix?
Well, in real numbers, the inverse of any real number a was the number a^{-1}, such that a times a^{-1} equaled 1. We knew that for a real number, the inverse of the number was the reciprocal of the number, as long as the number wasn't zero.
The inverse of a square matrix A, denoted by A^{-1}, is the matrix so that the product of A and A^{-1} is the Identity matrix. The identity matrix that results will be the same size as the matrix A. Wow, there's a lot of similarities there between real numbers and matrices. That's good, right - you don't want it to be something completely different.
A(A^{-1}) = I or A^{-1}(A) = I
There are a couple of exceptions, though. First of all, A^{-1} does not mean 1/A. Remember, "There is no Matrix Division!" Secondly, A^{-1} does not mean take the reciprocal of every element in the matrix A.
A square matrix that has an inverse is called invertible or non-singular. A matrix that does not have an inverse is called singular.
A matrix does not have to have an inverse, but if it does, the inverse is unique.
The inverse of a matrix A will satisfy the equation A(A^{-1}) = I.
The inverse of a 2×2 matrix can be found by ...
Let's go with an original matrix of
7 | -2 | ||
3 | 5 |
Step 1, switch the elements on the main diagonal would involve switching the 5 and 7.
5 | -2 | ||
3 | 7 |
Step 2, take the opposite of the other two elements, but leave them where they are.
5 | 2 | ||
-3 | 7 |
Step 3, find the determinant and divide every element by that. The determinant is the product of the elements on the main diagonal minus the product of the elements off the main diagonal. That means the determinant of this matrix is 7(5) - (-3)(2) = 35 + 6 = 41. We divide every element by 41.
The inverse of the original matrix is ...
5/41 | 2/41 | ||
-3/41 | 7/41 |
Now, you're saying, wait a minute - you said there was no matrix division. There is no division by a matrix. You may multiply or divide a matrix by a scalar (real number) and the determinant is a scalar.
Now that you know how to find the identity matrix by hand, let's talk practicality. The calculator will do it for you.
Whenever you need to access a matrix that you have created, just hit the Matrix key and choose the appropriate matrix. I would suggest that you start using Matrix 1, Matrix 2, etc, instead of Matrix, arrow down, enter. It will go faster, and you will be doing a lot with these matrices.
Enter the expression [A]^{-1} by going Matrix 1, and then hitting the x^{-1} key. It will not work if you try to raise the matrix to the -1 power as in [A]^(-1).
You may have to use the right or left arrow keys to scroll through the entire matrix to write it down. Please give exact answers whenever possible.
One way of giving exact answers is to have the calculator convert the decimals to fractions for you. After all, fractions really are your friends (and I seriously mean that here). You can have the calculator do a decimal to fraction conversion by hitting Math, Enter, Enter.
Also, if you get an answer like 1.2E-12, chances are really good that number is zero and it is because of inaccuracies in the calculator that you are getting that response. Convert the number to zero.
I am so glad you asked that.
One of the major uses of inverses is to solve a system of linear equations. You can write a system in matrix form as AX = B.
Now, pre-multiply both sides by the inverse of A. Make sure you meet these two conditions.
Matrix Multiplication is NOT Commutative!
A^{-1}(AX) = A^{-1}(B) ... pre-multiply both sides by A^{-1}
(A^{-1}A) X = A^{-1} B ... use the associative property to regroup factors
I X = A^{-1} B ... when you multiply inverses together, they become the identity matrix
X = A^{-1} B ... the identity matrix is like multiplying by 1.
If AX = B, then X = A^{-1} B
So what you're asking in your normal cynical way is "You've just solved another equation, what does that have to do with anything?"
Consider the system of linear equations
3x + 2y - 5z = 12 x - 3y + 2z = -13 5x - y + 4z = 10
Write the coefficients in an A matrix.
x | y | z | ||
---|---|---|---|---|
3 | 2 | -5 | ||
1 | -3 | 2 | ||
5 | -1 | 4 |
Write the variables in an X matrix.
x | ||
y | ||
z |
Write the constants in a B matrix.
12 | ||
-13 | ||
10 |
This step isn't really needed, but I wanted to show you that this thing really does work.
AX will be a (3×3) × (3×1) = 3×1 matrix. The B matrix is also a 3×1 matrix, so at least the dimensions work out right.
Here's A times X.
3 | 2 | -5 | x | 3x + 2y - 5z | ||||||||
1 | -3 | 2 | y | = | 1x - 3y + 2z | |||||||
5 | -1 | 4 | z | 5x - 1y + 4z |
Notice that turns out to be the left side of the system of equations. The B is the right hand side, so we have achieved equality. Woohoo! You can write a system of linear equations as AX = B.
So, if you can write a system of linear equations as AX=B where A is the coefficient matrix, X is the variable matrix, and B is the right hand side, you can find the solution to the system by X = A^{-1} B.
Place the coefficient matrix into [A] on the calculator and the right hand side into [B].
If you asked the calculator to find the inverse of the coefficient matrix, it would give you this for A^{-1}
5/44 | 3/88 | 1/8 | ||
-3/44 | -37/88 | 1/8 | ||
-7/44 | -13/88 | 1/8 |
You could do that, and then multiply that by B, but it would be easier just to put the whole expression into the calculator and get the answer directly. Even what is shown below is more work than is necessary.
X = A^{-1} B = ...
x | 5/44 | 3/88 | 1/8 | 12 | 191/88 | |||||||||||
y | = | -3/44 | -37/88 | 1/8 | -13 | = | 519/88 | |||||||||
z | -7/44 | -13/88 | 1/8 | 10 | 111/88 |
There you go, x = 191/88, y = 519/88, and z = 111/88. That would be a real pain to solve by hand.
The main reason is because it doesn't always work.