## Inverse Matrices and Their Properties

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 zeroes 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 one,

and the zeroes 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 two

by two 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 eight minus nine, so here we have one over negative one, and after

that comes the two by two matrix with these two entries swapped, and the signs of these

two inverted. That gives us two, negative three, negative

three, four. Incorporating this negative one out here is

easy, that just gives us negative two, three, three, negative four. 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 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

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 three by three 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 two by two 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, 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.

## Mohmmed Adil

Jan 1, 2019, 6:18 pmFinally!

## Amit Das

Jan 1, 2019, 6:39 pmFirst

## Julien TCHILINGUIRIAN

Jan 1, 2019, 8:30 pmMatrices exam tomorrow ! thanks

## Do You Even Game

Jan 1, 2019, 9:11 pmYou release this after my math IA was due, damn

## Bruce

Jan 1, 2019, 11:41 pmwell covered & explained, thankyou.

## Narendra Nirala

Jan 1, 2019, 3:49 amWell

## surya k

Jan 1, 2019, 6:44 amVery helpful video man ðŸ™‚

## H1N1

Jan 1, 2019, 2:18 pmProfessor, how's the merch sale going on?

## Mohmmed Adil

Jan 1, 2019, 5:02 pmSo what's the next plan Professor?

Differential Equations or multi variable calculus?

## Jack Cornison

Jan 1, 2019, 11:56 amThank you so much for everything! I now know physics & mathematics in a very high level, and I find it very interesting, I'm 15 so what we learn at school is a mere bagatelle…

## cham fast

Jan 1, 2019, 7:55 amThanks professor it's really halpern ful. ..

Sir my math portin of physical chemistry is so week Sir give some advice. ….

## Molly Pete

Jan 1, 2019, 12:53 amexcuse me, i love you.

## veerraju annamdevula

Feb 2, 2019, 7:17 pmSir is angle AOB same as BOA.I feel that the sign of the angle changes bcuz when we measure an angle we measure it from initial to final ray right!then while measuring AOB angle if we move in clockwise then obviously while measuring BOA that must be antivlockwise .Am I correct sir??

## Dexter Nierva

May 5, 2019, 1:21 amBest explanation of Inverse Matrices on youtube, thank you sir!

## arif hossain

Sep 9, 2019, 4:25 pm3Ã—3 matrix A multiply A inverse =Identity matrix????