| by Kenneth Chase | 15 comments

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


Mohmmed Adil

Jan 1, 2019, 6:18 pm Reply


Amit Das

Jan 1, 2019, 6:39 pm Reply



Jan 1, 2019, 8:30 pm Reply

Matrices exam tomorrow ! thanks

Do You Even Game

Jan 1, 2019, 9:11 pm Reply

You release this after my math IA was due, damn


Jan 1, 2019, 11:41 pm Reply

well covered & explained, thankyou.

Narendra Nirala

Jan 1, 2019, 3:49 am Reply


surya k

Jan 1, 2019, 6:44 am Reply

Very helpful video man 🙂


Jan 1, 2019, 2:18 pm Reply

Professor, how's the merch sale going on?

Mohmmed Adil

Jan 1, 2019, 5:02 pm Reply

So what's the next plan Professor?
Differential Equations or multi variable calculus?

Jack Cornison

Jan 1, 2019, 11:56 am Reply

Thank 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 am Reply

Thanks 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 am Reply

excuse me, i love you.

veerraju annamdevula

Feb 2, 2019, 7:17 pm Reply

Sir 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 am Reply

Best explanation of Inverse Matrices on youtube, thank you sir!

arif hossain

Sep 9, 2019, 4:25 pm Reply

3×3 matrix A multiply A inverse =Identity matrix????

Leave a Reply