It's Professor Dave, let's look at inverse matrices. We know about inverse functions, and the concept of an inverse can apply to a matrix as well. With a function, we know how to swap x and y, and then solve for y to find the inverse function.
So how does this work for matrices? Certainly it's a different algorithm, so let's learn about this now. Although a matrix is different from a function, when denoting the inverse of a matrix, we will use similar notation as we did for functions, where the inverse of matrix A will be A with a negative one in superscript, just like a function F and its inverse.
We must realize that this is not an exponent. A inverse does not mean the reciprocal of A, or one over A, because there is no such thing as matrix division. But an inverse matrix is similar to the reciprocal of a number in another way.
We can recall that a number times its reciprocal equals one. Similarly, a matrix times its inverse, or the inverse times the original matrix, will yield an identity matrix, a square matrix with all zeros except ones across its main diagonal, which is kind of like the number one in matrix form. So in that sense...
there is a correlation. Now we know how to perform matrix multiplication, so how exactly can we structure an inverse matrix so as to get an identity matrix when multiplying by the original matrix? Let's start with a simple two by two matrix with entries A, B, C, and D.
The inverse of this matrix will be one over the determinant of this matrix times this new matrix, with entries D, negative B, negative C, A. So A and D have swapped positions, B and C have gone from positive to negative, and the resulting matrix is being divided by the determinant of the original matrix. To see why this is the case, let's multiply this inverse in this general form by the original matrix. First, let's just generate the two by two matrix that is the product of these two, without worrying about this determinant. For the first entry, we get DA plus negative BC, or keeping everything in alphabetical order and simplifying the sign, AD minus BC. For the second entry, we get BD minus BD.
Then moving down a row, we get negative AC plus AC. And for the last entry, negative BC plus AD. or AD minus BC. The upper right and lower left terms clearly become zero, leaving us with this. Now we can divide by the determinant like we originally said, meaning each entry in the matrix will be divided by AD minus BC.
AD minus BC divided by AD minus BC is 1, and the zeros don't change, so we are left with this two by two identity matrix, just like we expected. So that's how you get the inverse of a 2x2 matrix. You swap A and D, change the sign on B and C, and divide by the determinant of the original matrix. Let's try a concrete example to make sure we have this.
How about a matrix with entries four, three, three, two. To get the inverse, we find one over the determinant. and the determinant is 8 minus 9. So here we have 1 over negative 1 and after that comes the 2 by 2 matrix with these two entries swapped and the signs of these two inverted.
That gives us 2, negative 3, negative 3, 4. Incorporating this negative 1 out here is easy, that just gives us negative 2, 3, 3, negative 4. And that is the inverse of this original matrix. Just to make sure this worked, let's multiply them together and see what we get. We do get the identity matrix, so we know our answer was correct.
So what do we need inverse matrices for? Well as we said earlier, there is no such thing as matrix division, so we can use this technique to solve equations with matrices where division would be necessary from an algebraic standpoint. If we have matrices X, A, and B, where X times A equals B, and we want to solve for X, we can't divide both sides by matrix A.
Instead, we can multiply both sides by A. A inverse. On the left side of the equation, A times A inverse will give us the identity matrix, which we can essentially treat like the number one and just get rid of it. And on the right side of the equation, we have a product of two matrices that we can compute, which will be equal to X. Of course this assumes that the dimensions of these matrices are such that the product can be calculated, otherwise this will not work.
Don't forget that matrix multiplication is not commutative, so wherever we place A inverse on one side, we have to place it the same way on the other side, so in this case as the rightmost entry on each side. If the original equation was A times X equals B, we would have to place A inverse all the way on the left, because A inverse times A should also equal the identity matrix, in which case the right side must be A inverse. A inverse times B and not B times A inverse. In addition, when taking the inverse of a product of matrices like the quantity B times A inverse, we get A inverse times B inverse in that order. We should also note that not every matrix will have an inverse matrix.
If a matrix has a determinant that is equal to zero, its inverse will involve dividing by zero, which is undefined, and that matrix will therefore not have an inverse. Such a matrix is called a singular matrix. Lastly, you may be wondering how we can find the inverse of any matrix other than a two by two matrix.
Let's look at a three by three matrix and see how this would work. To find the inverse of this, we will first find something called the matrix of minors. This will be a new 3x3 matrix, and each entry will be a particular determinant. For the first entry, we locate the corresponding entry in the original matrix, and we block out its row and column. There is a 2x2 array of numbers left, and we find its determinant.
That determinant will become this first entry in the matrix of minors. Then for the second entry, again we go to the corresponding entry in the original matrix. in the original matrix, block out its row and column, and find the determinant of what is left. We continue doing this for every single entry until we have completed our matrix of minors.
Unfortunately, that was just step one. Now for step two, we use our matrix of minors to generate a matrix of cofactors. This isn't too hard.
we just have to apply a checkerboard of plus and minuses to get the right sign on each term. The corner entries and the central entry will remain as they are, and the other four entries will have their signs inverted. Then step three is to find the adjugate, or the adjoint. This involves reflecting all the entries across the diagonal. The diagonal stays as it is, but the other terms are reflected, so these two swap, these two corners swap, and these two swap.
Finally, the last step involves dividing by the determinant of the whole original three by three matrix, which won't seem so hard after doing that whole business with the matrix of minors. We just have this first entry times the determinant of this two by two matrix, minus the second entry times the determinant of this two by two matrix, plus the third entry times the determinant of this two by two matrix. Whatever we get, we divide the last matrix we had by this value, meaning we divide every entry in the matrix by that value, and that's what we are left with for the inverse of the original three by three matrix. So to summarize the four steps we just took, first we found the matrix of minors, going one entry at a time.
Then we applied the checkerboard pattern of signs to find the matrix of cofactors. Then we transposed all entries across the diagonal to get the adjugate. And lastly, we divided by the determinant of the original matrix to get the inverse matrix.
This is essentially the same algorithm as we used for a two by two matrix, just many more steps to achieve the same result. So as we can see, even with just a three by three matrix, finding the inverse becomes rather laborious. If we were to extend this to a four by four matrix, it wouldn't really be any harder, we would just end up with an absurd amount of steps to follow. For this reason, inverses of larger matrices are usually found using some kind of matrix calculator to avoid doing pages of arithmetic.
So that's what we need to know about inverse matrices. Now that we have covered the basics about matrices, their operations, determinants, and inverses, we have many of the important algorithms out of the way, and we are ready to examine some of the more abstract ideas in linear algebra. But before we move forward, let's check comprehension. Thanks for watching, guys. Subscribe to my channel for more tutorials, support me on patreon so I can keep making content, and as always feel free to email me professordaveexplains at gmail.com