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Precalculus
Course: Precalculus > Unit 7
Lesson 11: Properties of matrix multiplication- Defined matrix operations
- Matrix multiplication dimensions
- Intro to identity matrix
- Intro to identity matrices
- Dimensions of identity matrix
- Is matrix multiplication commutative?
- Associative property of matrix multiplication
- Zero matrix & matrix multiplication
- Properties of matrix multiplication
- Using properties of matrix operations
- Using identity & zero matrices
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Using identity & zero matrices
Sal solves a problem where he has to determine whether unknown matrices are zero or identity to make an equation true. Created by Sal Khan.
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- So there's an answer to a question here that gives a good reason behind why identity matrices exist, so now I'm wondering about zero matrices. If you multiply a matrix by any zero matrix, you get another zero matrix. Always a bunch of zeros, though potentially a different number of zeros.
Why would you ever need a bunch of zeros?(5 votes)- For the same reason, more or less. Imagine you have an equation such as (A+B)(C+D). None of these are zero matrices, but when you add A and B, you get a zero matrix. Now you know that no matter what C+D is, as long as the multiplication is defined, the answer is a zero matrix without ever having to do the math.
There might be other applications I'm not seeing, but I'm just starting to learn about matrices, so you'll have to forgive the oversight.(15 votes)
- Isn't what Sal did basically guess and check? Is there a faster way to determine the solution rather than exhaust all the possibilities?(6 votes)
- I wouldn't call it faster, but you could write equations to solve for scalar values of A, B, and C:
A+B+C=2
3A-5B+C=4
4A+B+3C=7
-2A+3B+2C=0
and then solve for A, B, and C, with answers of 1 meaning identity and 0 meaning, well, zero. To solve this, though, you'd probably need to solve another matrix. In this case, it's easier to do it Sal's way.(1 vote)
- what if the leading diagonal is from left to right ? does properties of identity matrix still holds.and why the leading diagonal is always from right to left in a diagonal matrix. why it is not other way around.(2 votes)
- Unless certain elements are equal, no.
http://www.wolframalpha.com/input/?i=%7B%7B0,1%7D,%7B1,0%7D%7D*%7B%7Ba,b%7D,%7Bc,d%7D%7D%3D%7B%7B1,0%7D,%7B0,1%7D%7D*%7B%7Ba,b%7D,%7Bc,d%7D%7D(2 votes)
- my teacher said that identity matrix is used to replace division in matrix, because there are no dividing in matrix, is that right?(2 votes)
- Not quite. The identity matrix is used to prove that your inverse matrix (which is the matrix equivalent of division, also providing the matrix is invertible), will be the result when multiplied to your original matrix.
http://www.wolframalpha.com/input/?i=%7B%7B1,2,3%7D,%7B1,0,1%7D,%7B3,2,1%7D%7D*inverse(%7B%7B1,2,3%7D,%7B1,0,1%7D,%7B3,2,1%7D%7D)(1 vote)
- How do you divide two matrices?
If it's the opposite of multiplication, I guess it should be the same as multiplying by power -1.
But, how do you find the matrix^-1?(1 vote)- We cannot divide matrices, but as you suggested we can multiply by the inverse of a matrix. Much like how you cannot divide by 0, there are certain matrices we cannot find an inverse for. The actual process for finding an inverse involves augmenting the original matrix with the identity matrix and using row operations to transform the matrix on the left into the identity matrix which leaves the right matrix as the inverse. There are later lessons in Khan academy that give far more detailed explanations on this topic though.(2 votes)
- Is matrix multipcation commutativ or no(1 vote)
- In general, no, it isn't. Only with identity and zero matrices (and possibly some other coincidental special cases).(1 vote)
- Is this solution unique? Is matrix algebra similar to ordinary algebra in the sense that you would need 3 equations in this case to arrive at a unique solution?(1 vote)
- I don't know if you have a video specifically on the question i have, but this is my question: What is the identity element of the operation a o b = a + b - ab, if a and b are real numbers(1 vote)
- You're defining an operation on the reals a•b=a+b-ab (I assume those symbols on the right-hand side mean standard addition, subtraction, and multiplication.)
The identity is the element i such that a•i=i•a=a for every real a. If we take i=0, we get
a•0=a+0-a*0=a
0•a=0+a-0*a=a
So the identity is 0.(1 vote)
- In, I didn't understand that matrix A and C should be identity matrix. I belive that matrix A and C should stay in their own form. 3:11(1 vote)
- Are the "glues" a part of the problem setting? In other case this is not by any means proofed to be the only solution. What is the idea of this video at all?(1 vote)
Video transcript
So we have a matrix equation set up right over here, we have matrix A times this plus matrix B times this plus matrix C times this is equal to this two by two matrix. And the way it's set up, it's a little bit of a puzzle. And I'll give you one clue as to what type of matrices A, B, and C could be. They're each either going to be an Identity Matrix, or a Zero Matrix. So given that clue, that each of these are either an identity matrix or a zero matrix, can you pause this video, and essentially solve this puzzle? Which of these are identity matrices, and which of these are zero matrices? I'm assuming you've given a go at it. So let's go entry by entry. So if you look at this first entry right over here, how can we get a two
for this top left entry? Well, let's see. When we multiply, if any
of these are identities, and essentially you just
get the value of the matrix, if any of these are zero, then essentially that
product doesn't get added. It's one way of thinking about it, is that matrix won't matter anymore, because it's going to
be multiplied by zero. So if A was a zero matrix and B and C were identity matrices, you would add one plus one to get to two. Just like oh, maybe that's the case. But it could be the other way around. It could be that A is identity matrix, B is a zero matrix, and C is an identity matrix, and you add one plus one over there to get two. Or you could say that
maybe C is the zero matrix, and B is the identity matrix, and you add one plus one here. So really, all this is telling us is two of these matrix, two of A, B and C are going
to be identity matrices, and one is going to be a zero matrix, but we don't know which
one is which just yet. Any of them can be the zero matrix, at least based on looking
at that first entry. Now let's look at, I don't know, let's look at this entry right over here. How can we add up to four? Well, let's see. Three, if this was an identity matrix and this was an identity matrix, then you're going to, then essentially you'll just be left with this matrix plus this matrix. This is going to be the zero. So three plus negative five. That is not equal to four. So both of these. You can't have A and
B be identity matrices and C be the zero matrix. So let's think about
other combinations here. So what about B and C being
the identity matrices, and A being the zero matrix? In that situation, so A is the zero matrix,
that's not going to matter, and we're essentially, when you multiply B times this you're just
going to get this matrix, and C times this, you're
going to get that matrix, if B and C are identity matrices. And so you have negative five plus one would be negative four. So that doesn't work either. So our last scenario is going to be so we can essentially rule out B as an identity matrix. Because when we had B be one of the identity matrices, and we picked the other two options, we still couldn't get to four here. So B is going to be a zero matrix. And let's see if that works out. A is identity, then A would be an identity and C would be an idenitty matrix as well. So let's see if that actually makes sense. So if this right over
here is identity matrix, and that over there is an identity matrix, this whole thing simplifies to one three four comma negative two plus one one three two, and it does indeed equal one one plus is two, three plus one is four, four plus three is seven, negative two plus two is indeed zero.