Matrix Multiplication

Lesson Features »

Lesson Priority: Normal

Algebra Two $\longrightarrow$
Matrices $\longrightarrow$
  • First, understand when two matrices can and cannot be multiplied together
  • Learn the pattern-based approach for systematically multiplying two matrices
  • Understand why the associative property of multiplication FAILS for matrices, and thus multiplication order matters
  • For square matrices, learn how the identity matrix and the zero matrix act just like the numbers 1 and 0 do for number multiplication, respectively
Lesson Description

Multiplying matrices is a bit tedious, but its a staple item on the menu of things your teacher will serve up on a test, when you study matrices. It's also occasionally useful to know in the future, having a few important applications in probability and statistics. This lesson covers the basics and then some, for everything there is to know about multiplying two matrices together.

Practice Problems

Practice problems and worksheet coming soon!


Multiplication Meets Misery

Multiplying matrices will give you new appreciation for multiplying good ol' integers together. Regrettably, they are quite a bit more complex to multiply than numbers, for a few reasons.First, it is difficult to picture or interpret what matrix multiplication means, unlike integers. For example, when we were in grammar school and we started memorizing facts like $4 \times 3$, we could at least interpret the result of $12$ by thinking about having $4$ sets of $3$ or vice versa. Visually, we might interpret $4 \times 3$ as something like this:In any case, multiplying two numbers had a physical interpretation to accompany the rote memorization facts that we were asked to know. With matrix multiplication, while there is reason and purpose to doing it, it lacks the intrinsic interpretation that integer multiplication has.Second, order matters. That's right - it sounds incredulous, after spending so long with numbers and the well-known and convenient Commutative Property of Multiplication, but it's true. $5 \times 6$ may equal $6 \times 5$, but matrix $A$ times matrix $B$ is NOT equal to matrix $B$ times matrix $A$ (technically this CAN happen but it is unlikely and uncommon).Third, if we have to do it by hand, there are many little calculations involved, and we need to make zero mistakes. As we learn the process, we'll see why each number in the resulting matrix is calculated using several numbers from each of the two parent matrices, which is why it's easy to slip. Of course, if you follow my method and do a little practice, you won't fall victim to this inherent pitfall.

Before We Multiply

A very important prerequisite for multiplying matrices is ensuring that you can even multiply them in the first place. Unlike integers, you cannot just pick any two and multiply them together. Matrices actually have to be compatible before they can be multiplied.Recall that each matrix has what we called a dimension, given by the number of rows and columns the matrix has. Specifically, a matrix with $m$ rows and $n$ columns is said to have dimension $m$ by $n$.Two matrices can only be multiplied if they have a match on certain dimension numbers.
Define: Matrix CompatibilityTwo matrices, $A$ and $B$ can only be multiplied ($A$ times $B$, order matters) if $A$ has the same number of columns as the number of rows in $B$. The resulting matrix will have $A$'s number of rows, and $B$'s number of columns.
Because multiplication of square matrices is common, it is helpful to point out a stronger statement applicable only to square matrices:
Define: Square Matrix CompatibilitySquare matrix $A$ and square matrix $B$ can only be multiplied if they are the same dimension, and their resulting matrix will have the same dimension as well.
The definition of matrix multiplication is not easy to use or even remember, symbolically. The word definition serves us fine for the need to formally define matrix multiplication eligibility, but you will do much, much better to learn and understand this visually with a small diagram.Let matrix $A$ have dimension $m \times n$, and let matrix $B$ have dimension $n \times p$. Then matrix $A$ and matrix $B$ can be multiplied together, according to the definition ($A$'s number of columns match $B$'s number of rows). Furthermore, the answer you get will be a matrix with dimension $m \times p$. The visual diagram is this:
You Should Know
When you're asked to multiply matrices, you usually don't have to tediously check if they are compatible. Most of the time you'll be asked to multiply square matrices, which makes it super obvious that they can or can't multiply together (they must be the same square size in that case). If you are not given square matrices, you will usually be given compatible ones to multiply.
Teachers don't usually give you incompatible matrices, but you'll want to chart out these $m \times n$ and $n \times p$ dimensions every time you multiply non-square matrices, as you will want to know the resulting matrix's dimensions in advance of actually performing the multiplication. As we practice the first few examples, it will be clearer why we really want to know the dimensions of the result in advance.
Pro Tip
The advice here is to write down the dimensions of each matrix you are about to multiply. The payoff is that knowing what dimensions to expect of your answer makes the multiplication process easier.
Examples 1-3Determine if matrix multiplication is possible, and if so, state the dimensions of the resulting matrix.
Example 1$$ \left[ \begin{array}{cccc} 5 & 6 & -9 & 4 \\ 7 & -3 & 4 & -2 \\ 8 & -1 & 1 & 1 \end{array} \right] \left[\begin{array}{cc} 4 & -2 \\ 0 & 6 \\ -3 & 10 \\ 7 & 7 \end{array} \right] $$$\blacktriangleright$ The first matrix has four columns and the second matrix has four rows - therefore multiplication is possible. Write down these dimensions and draw the brackets to better see this, and also to quickly see the dimensions of the result. We can see that the result will be a $3 \times 2$ matrix.
Example 2$$ \left[ \begin{array}{ccc} 9 & -2 & 6 \\ 1 & 2 & -6 \end{array} \right] \left[ \begin{array}{ccc} 5 & 8 & 9 \\ -7 & 12 & -3 \end{array} \right] $$$\blacktriangleright$ Let's line up the dimensions once again using the bracket method. Since the dimensions do not match in the way they would need to, matrix multiplication between these two matrices is not possible.
Example 3$$ \left[ \begin{array}{ccc} 7 & -5 & 3 \\ 4 & 1 & 3 \\ 0 & 5 & -2 \end{array} \right] \left[ \begin{array}{ccc} 4 & 3 & -1 \\ 8 & 17 & -20 \\ 10 & 0 & 100 \end{array} \right] $$$\blacktriangleright$ These two matrices are compatible. Specifically, the number of columns of the first matrix match the number of rows in the second matrix. More clearly, we can see the dimensions match using the bracket method: Note also that as stated before, because we have two square $3 \times 3$ matrices multiplied together, we should expect them to be compatible, and for the result to be a $3 \times 3$ matrix.
Hopefully now you feel ready to check whether or not matrix multiplication is possible. Now we'll turn our attention to a more interesting matrix multiplication idea - how to actually multiply two matrices.

Matrix Multiplication How-To

Matrix multiplication is a matter of computing each entry in the resulting matrix one-at-a-time. Because this is the case, it will quickly become clear why it is beneficial for us to map out and know what the dimensions of the resulting matrix will be using the Bracket Method, before we even multiply.Each entry in the resulting matrix is obtained by taking the dot product of the first matrix's row with the second matrix's column. In fact, this is exactly the reason why these dimensions must match! If you aren't familiar with "dot" products, it means to take the two sets of numbers and multiply each corresponding number from each set together, and add up all the results. An example or two will make this process clearer.In this illustrative example, we will multiply a $2 \times 4$ matrix with a $4 \times 2$ matrix, the product of which yields a $2 \times 2$ result.
Example 4Multiply matrix $A$ with matrix $B$ (i.e. find $AB$).$$A = \left[ \begin{array}{cccc} 1 & 3 & -2 & -2\\ 2 & -3 & 1 & -4 \end{array} \right] $$$$B = \left[ \begin{array}{cc} 4 & -1 \\ 0 & -2 \\ 2 & -3 \\ 3 & 1 \end{array} \right] $$$\blacktriangleright$ As stated, this operation will result in a $2 \times 2$ matrix. Let's set up a blank matrix to fill in:$$ \left[ \begin{array}{cc} \boxed{\,\,\,} & \boxed{\,\,\,} \\ \boxed{\,\,\,} & \boxed{\,\,\,} \end{array} \right] $$The best way to go about this is to fill in the resulting matrix one number at a time. You can go in any order you like but it's best to work systemically, especially if you're working on a bigger matrix.Use the following definition to guide your work.
Define: Matrix MultiplicationWhen multiplying two matrices $A$ and $B$, with $A$ on the left and $B$ on the right, the entry in the resulting matrix that goes in the $i$-th row and $j$-th column is obtained by taking the dot product of the $i$-th row from $A$ with the $j$-th column from $B$.
I promise this will look more digestible as we work through real examples.Returning to Example 4, let's fill in the top-left entry in the resulting matrix, which is row 1 column 1. We will dot product row 1 from the left matrix ($A$) with column 1 from the right matrix ($B$). As we said above, taking the dot product means we are going to add together the product of each corresponding number in the two lists of numbers - the first list being matrix $A$ row 1 and the second list being matrix $B$ column 1.$$(1)(4) + (3)(0) + (-2)(2) + (-2)(3)$$$$=4 + 0 - 4 -6=\boxed{-6}$$Again, the first number in each product came from the entries in matrix $A$ row 1, and the second number in each product came from matrix $B$ column 1, since we are calculating the answer in row 1 column 1 of the resulting matrix.Now we'll calculate the entry in row 2 column 1 of the resulting matrix. To do this, we will dot product row 2 from matrix $A$ with column 1 from matrix $B$.$$(2)(4) + (-3)(0) + (1)(2) + (-4)(3)$$$$8 - 0 + 2 - 12 = \boxed{-2}$$Therefore we have, so far:$$ \left[ \begin{array}{cc} -6 & \boxed{\,\,\,} \\ -2 & \boxed{\,\,\,} \end{array} \right] $$Let's finish off this example by performing the last two calculations. To obtain the entry that goes in row 1 column 2, we need to dot product row 1 from matrix $A$ with column 2 from matrix $B$:$$(1)(-1) + (3)(-2) + (-2)(-3) + (-2)(1)$$$$=-1 -6 + 6 -2 = \boxed{-3}$$Finally, we'll get the entry in row 2 column 2 by taking the dot product of matrix $A$ row 2 with matrix $B$ column 2.$$(2)(-1) + (-3)(-2) + (1)(-3) + (-4)(1)$$$$= -2 + 6 -3 -4= \boxed{-3}$$With all four calculations complete, we have our final answer.$$ \left[ \begin{array}{cccc} 1 & 3 & -2 & -2\\ 2 & -3 & 1 & -4 \end{array} \right] \left[ \begin{array}{cc} 4 & -1 \\ 0 & -2 \\ 2 & -3 \\ 3 & 1 \end{array} \right] $$$$= \left[ \begin{array}{cc} -6 & -3 \\ -2 & -3 \end{array} \right] $$
Pro Tip
Once again, the general procedure is to ask yourself which entry you are populating and get the corresponding row from the first matrix to multiply with the corresponding column of the second matrix. E.g. if you are working on a large matrix multiplication problem $EF$ where $E$ and $F$ are matrices that are compatible to multiply together, and you wanted to calculate the entry in the 5th row, 8th column of the resulting matrix, you would take the dot product of the 5th row of $E$ with the 8th column of $F$.
You can populate the resulting matrix in any order you want, but you cannot rearrange the order of the numbers you work with when taking the dot products in each calculation. If row 1 of matrix $A$ is$$[1 \,\, 2 \,\, 3 \,\, 4]$$then when you go to use it, you must use the numbers in the order $1$, $2$, $3$, $4$.
Let's take another look at the process with a fresh example.
Example 5Multiply matrices $C$ and $D$ together (i.e. find $CD$).$$ C= \left[ \begin{array}{ccc} 2 & -2 & -1 \\ -3 & 0 & 4 \\ 1 & 5 & 3 \end{array} \right] $$$$ D= \left[ \begin{array}{ccc} 0 & 9 & 4 \\ -3 & 1 & -4 \\ 2 & -2 & 5 \end{array} \right] $$$\blacktriangleright$ We can see that we are being asked to multiply a $3 \times 3$ matrix with another $3 \times 3$ matrix, thus telling us first that we are able to carry out the multiplication, and second that the result we should expect will also be a $3 \times 3$ matrix.With a $3 \times 3$ answer, we have nine calculations to perform. We will handle this systematically:Row 1 Column 1$$(2)(0) + (-2)(-3) + (-1)(2) = \boxed{4}$$Row 1 Column 2$$(2)(9) + (-2)(1) + (-1)(-2) = \boxed{18}$$Row 1 Column 3$$(2)(4) + (-2)(-4) + (-1)(5) = \boxed{11}$$Row 2 Column 1$$(-3)(0) + (0)(-3) + (4)(2) = \boxed{8}$$Row 2 Column 2$$(-3)(9) + (0)(1) + (4)(-2) = \boxed{-35}$$Row 2 Column 3$$(-3)(4) + (0)(-4) + (4)(5) = \boxed{8}$$Row 3 Column 1$$(1)(0) + (5)(-3) + (3)(2) = \boxed{-9}$$Row 3 Column 2$$(1)(9) + (5)(1) + (3)(-2) = \boxed{8}$$Row 3 Column 3$$(1)(4) + (5)(-4) + (3)(5) = \boxed{-1}$$$$ \therefore CD = \left[ \begin{array}{ccc} 4 & 18 & 11 \\ 8 & -35 & 8 \\ -9 & 8 & -1 \end{array} \right] $$Make sure you can follow along with each calculation and where the numbers are coming from. For at least 3-4 of these calculations, make sure you can reproduce the results. If you're flawless, you can probably be sure you could do the rest just fine.
You Should Know
Matrix multiplication is tedious, mistake-prone, and, truth be told, a little complicated. You may experience episodes of anger - if so please direct them at your teacher. The TI-84 and similar Casio calculators will multiply matrices for you instantly. So unless you're planning on majoring in Math or taking Actuarial exams, I personally don't see why we can't give in to the 21st century and just rely on computers for this particular task. If you can use a calculator, you should.

Properties of Matrix Multiplication

There are three major properties of matrix multiplication that we need to understand, regardless of whether or not our teacher will let us use the aide of a calculator to get it done.Commutative Property FailureWe are not strangers to the fact that $3 \times 7$ and $7 \times 3$ are both equal to $21$. In early Algebra, we formally defined this multiplication property for all numbers and variables, and we called it the Commutative Property of Multiplication. Formally, we say that $a \cdot b$ is equal to $b \cdot a$ for any two real numbers $a$ and $b$. Oddly enough, this property fails for matrix multiplication. While it is possible to find two matrices $A$ and $B$ such that $AB = BA$, it is not guaranteed to be true, and indeed well over 90% of the matrix multiplication you do will be with matrices such that $AB$ is not equal to $BA$.
Theorem: Matrix Multiplication ResultsLet $A$ and $B$ be matrices with dimensions $m \times n$ and $n \times m$, respectively. It follows that
  • $AB$ will never equal $BA$ unless $A$ and $B$ are square matrices of the same dimension
  • Even if $A$ and $B$ are square matrices of the same dimension, $AB$ and $BA$ will only be the same in very specific circumstances. One such common circumstance is that $A$ and $B$ are inverses of one another. The other ones are more coincidental and not worth outlining here and now. The takeaway here is that there is no guarantee that $AB$ and $BA$ will give the same result.
We will learn more about Matrix Inverses in the near future.So what does the failure of the Commutative Property of Multiplication mean for us? Simply that we cannot expect $AB$ and $BA$ to give the same answer, and if we are asked to compute both, we must do each separately.Multiplication by the Zero MatrixThe next property we are expected to know is that multiplication of a matrix by a zero matrix will yield a resulting matrix that is also a zero matrix. Recall from the recent lesson on defining what matrices are » that a zero matrix is one where all of its entries are populated with $0$. This property is a parallel to multiplying by $0$ in the realm of working with real numbers.
Define: Multiplying with the Zero MatrixJust as we have$$a \cdot 0 = 0$$for all real numbers $a$, it is similarly true that$$A \boldsymbol{0} = \boldsymbol{0}$$for all matrices $A$ and $\boldsymbol{0}$ such that the dimensions of $A$ and $\boldsymbol{0}$ permit the two matrices to be multiplied.
The reason that this happens is the fact that all dot products that you could try and calculate in the matrix multiplication process will all come out to be zero, since the columns of the second matrix will all be zero, coming from the zero matrix.Multiplication by the Identity MatrixThe last property that we are often expected to know, understand, and leverage, is the fact that multiplication by the "identity" matrix will behave like multiplication by the number $1$ did for regular numeric multiplication.Recall that an identity matrix is defined to be any $n \times n$ square matrix with entries of $1$ in its diagonal starting in row $1$ column $1$ and ending in row $n$ column $n$, with zeroes in all other entries. The identity matrix will behave a lot like the number $1$ does in Algebra multiplication, if it is being used with another square matrix.
Define: Multiplying with the Identity MatrixLet $A$ be an $n \times n$ square matrix, and let $I$ be an $n \times n$ identity matrix. It follows that$$A I = A$$and$$I A = A$$
This behavior echoes the fact that$$a \cdot 1 = a$$for any real number $a$, and is the reason why the identity matrix is named the way it is (the number $1$ in Algebra is sometimes referred to as "the multiplicative identity").
You Should Know
If you multiply a non-square matrix by the Identity matrix, this concept doesn't work. As the definition states, this multiplicative identity behavior that we can parallel to multiplying by the number $1$ in number-based Algebra is specific to square matrices.

Put It To The Test

Here are a few examples on how you can expect this material to show up on a quiz or test. We'll do three straight-up multiplication exercises followed by some concept-based questions.Examples 6-8Multiply each given matrix product.
Example 6$$ \left[ \begin{array}{cc} 5 & -2 \\ 1 & 3 \end{array} \right] \left[ \begin{array}{cc} -3 & -1 \\ 4 & 6 \end{array} \right] $$
Show solution
$\blacktriangleright$ Multiplying a $2 \times 2$ matrix with another $2 \times 2$ matrix will get us a resulting $2 \times 2$ matrix answer. Remember to work methodically and let the entry that you're calculating guide your choice of which rows and columns to use in each dot product calculation.Row 1, Column 1$$(5)(-3) + (-2)(4) = \boxed{-23}$$Row 1, Column 2$$(5)(-1) + (-2)(6) = \boxed{-17}$$Row 2, Column 1$$(1)(-3) + (3)(4) = \boxed{9}$$Row 2, Column 2$$(1)(-1) + (3)(6) = \boxed{17}$$Putting the pieces in the right place, we have$$ \left[ \begin{array}{cc} 5 & -2 \\ 1 & 3 \end{array} \right] \left[ \begin{array}{cc} -3 & -1 \\ 4 & 6 \end{array} \right]$$$$= \left[ \begin{array}{cc} -23 & -17 \\ 9 & 17 \end{array} \right] $$
Example 7$$ \left[ \begin{array}{cccc} 5 & 0 & -9 & 4 \\ -6 & 0 & -2 & 1 \\ 4 & -5 & 1 & -4 \end{array} \right] \left[ \begin{array}{c} 3 \\ -2 \\ 5 \\ 1 \end{array} \right] $$
Show solution
$\blacktriangleright$ This time we are looking at the product of a $3 \times 4$ matrix and a $4 \times 1$ matrix. While there are a lot of numbers in play here, note that, using the bracket rule, we know that our answer will be a matrix of dimension $3 \times 1$. This means there are only three calculations we need to do. Once again we'll follow the row / column dot product process, but since our answer only has one column, we'll always be using the single entire column from the second matrix in each calculation.Row 1, Column 1$$(5)(3) + (0)(-2) + (-9)(5) + (4)(1)$$$$=\boxed{-26}$$Row 2, Column 1$$(-6)(3) + (0)(-2) + (-2)(5) + (1)(1)$$$$=\boxed{-27}$$Row 3, Column 1$$(4)(3) + (-5)(-2) + (1)(5) + (-4)(1)$$$$=\boxed{23}$$Therefore, the answer for this matrix product is$$ \left[ \begin{array}{c} -26 \\ -27 \\ 23 \end{array} \right] $$
Calculate entries one at a time, and do it carefully and systematically. This process begs for mistakes - don't oblige!
Example 8$$ \left[ \begin{array}{cccc} 4 & 1 & 5 & 8 \\ 0 & 3 & 6 & 1 \\ 3 & 5 & 0 & 9 \\ 2 & 4 & 6 & 1 \end{array} \right] \left[ \begin{array}{cccc} 1 & 5 & 1 & 0 \\ 0 & 3 & 6 & 1 \\ 3 & 5 & 7 & 2 \\ 2 & 0 & 6 & 1 \end{array} \right] $$
Show solution
$\blacktriangleright$ Multiplying $4 \times 4$ matrices may just qualify as cruel and unusual punishment. With this many calculations (16) the process becomes a bit mind numbing and difficult. Fortunately, the process is the same, and furthermore, we usually will not be asked to work with matrices this large when doing calculations "by hand" - though an occasional mean-spirited teacher will give you one on a quiz or test to snub out some would-be A's.Here, we will provide three sample calculations of the sixteen required. I recommend you try and replicate them before seeing the result below, just to confirm that you know what you're doing. Realistically this is a calculator problem - so it's also a good chance for you to grab your TI-84 or similar device, and see if you can figure out how to make it do matrix multiplication for you.Row 2, Column 1$$(0)(1) + (3)(0) + (6)(3) + (1)(2)$$$$=\boxed{20}$$Row 3, Column 4$$(3)(0) + (5)(1) + (0)(2) + (9)(1)$$$$=\boxed{14}$$Row 4, Column 2$$(2)(5) + (4)(3) + (6)(5) + (1)(0)$$$$=\boxed{52}$$Via calculator, or, if you're bored and brave, your own brute force, you should obtain the answer$$ \left[ \begin{array}{cccc} 23 & 40 & 89 & 15 \\ 20 & 39 & 66 & 16 \\ 21 & 30 & 87 & 14 \\ 22 & 52 & 74 & 17 \end{array} \right] $$
Teachers virtually never give us $4 \times 4$ or larger matrices to work with. They are just too large to multiply by hand in a reasonable amount of time. Even in forthcoming matrix topics, we'll rarely be asked to use them - other processes we will learn in addition to matrix multiplication are also too slow and cumbersome by hand, when the matrices are so large.
Examples 9-10Without performing any specific calculations, determine the result of the following matrix products.
Example 9$$ \left[ \begin{array}{ccc} 5 & 6 & -3 \\ 10 & -1 & 2 \\ -4 & -8 & 0 \end{array} \right] \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] $$
Show solution
$\blacktriangleright$ Hopefully the somewhat odd instructions were a hint here that there isn't any need to actually go through the process of nine calculations of dot products. Questions like this are meant to test whether or not we remember the identity property of matrix multiplication for square matrices multiplied by the identity matrix. According to the rules, we know that any square matrix $A$ multiplied by $I$ is itself. Therefore$$ \left[ \begin{array}{ccc} 5 & 6 & -3 \\ 10 & -1 & 2 \\ -4 & -8 & 0 \end{array} \right] \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] $$$$= \left[ \begin{array}{ccc} 5 & 6 & -3 \\ 10 & -1 & 2 \\ -4 & -8 & 0 \end{array} \right] $$It's worth mentioning that teachers will occasionally give you something silly like an $8 \times 8$ matrix multiplied with the identity matrix and expect that you just simply write a sentence along the lines of "This matrix multiplied by $I$ will be itself."
Example 10$$ \left[ \begin{array}{cc} 4 & 3 \\ 6 & -1 \\ 0 & 7 \\ -3 & -3 \end{array} \right] \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] $$
Show solution
$\blacktriangleright$ Similar to Example 9, we are not expected to do any tedious calculation, but rather recognize a multiplication property that we are expected to be familiar with. Multiplying any matrix by a zero matrix will yield a resulting zero matrix. So the only real work to do here is make sure you give a zero matrix that has the correct dimensions. Since we are multiplying a $4 \times 2$ matrix by a $2 \times 3$ matrix, our result will be a $4 \times 3$ matrix. It follows that$$ \left[ \begin{array}{cc} 4 & 3 \\ 6 & -1 \\ 0 & 7 \\ -3 & -3 \end{array} \right] \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] $$$$= \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] $$Note that some teachers will allow you to write a zero matrix answer in a shorthand notation where you put a zero inside of matrix brackets with its dimensions on the side, such as$$\left[ 0 \right]_{4 \times 3}$$However, if you are unsure if that is allowed, you should write it properly.
Example 11Let $A$ and $B$ be two square matrices compatible for multiplication. If you know the result $AB$, what can you say about the result $BA$?


Show solution
$\blacktriangleright$ Answer: Nothing meaningful. There is no guarantee the answer is the same as $AB$ (in fact it's not likely) and furthermore, there is no pattern or other implicit information that we can garner from $AB$ to tell us anything about specific entries of the matrix we get from multiplying $BA$. The Commutative Property of Multiplication does not work for matrices!
Lesson Takeaways
  • Understand specifically when two matrices can and can't be multiplied together
  • Use the method of dot products to obtain the answer for the product of two matrices, one entry at a time
  • Be lazy and use a calculator, if you can - some teachers don't mind, and even if you have to do it long-hand, it's nice to check your work
  • Be familiar with the three properties of matrix multiplication - that order matters (Commutative Property fails), and products involving the zero or identity matrices

Lesson Metrics

At the top of the lesson, you can see the lesson title, as well as the DNA of Math path to see how it fits into the big picture. You can also see the description of what will be covered in the lesson, the specific objectives that the lesson will cover, and links to the section's practice problems (if available).

Key Lesson Sections

Headlines - Every lesson is subdivided into mini sections that help you go from "no clue" to "pro, dude". If you keep getting lost skimming the lesson, start from scratch, read through, and you'll be set straight super fast.

Examples - there is no better way to learn than by doing. Some examples are instructional, while others are elective (such examples have their solutions hidden).

Perils and Pitfalls - common mistakes to avoid.

Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!

Put It To The Test - Shows you examples of the most common ways in which the concept is tested. Just knowing the concept is a great first step, but understanding the variation in how a concept can be tested will help you maximize your grades!

Lesson Takeaways - A laundry list of things you should be able to do at lesson end. Make sure you have a lock on the whole list!

Special Notes

Definitions and Theorems: These are important rules that govern how a particular topic works. Some of the more important ones will be used again in future lessons, implicitly or explicitly.

Pro-Tip: Knowing these will make your life easier.

Remember! - Remember notes need to be in your head at the peril of losing points on tests.

You Should Know - Somewhat elective information that may give you a broader understanding.

Warning! - Something you should be careful about.

Return to Lesson