The MatLab system simply performs mathematical operations on matrices and vectors. Consider first the simple operations of addition and multiplication of matrices and vectors. Let two vectors be given

a = ; % row vector
b = ; % column vector

then the multiplication of these two vectors can be written as

c = a*b; %c=1+2+3+4+5=16
d = b*a; %d - matrix of 5x5 elements

In accordance with operations on vectors, multiplying a row vector by a column vector gives a number, and multiplying a column vector by a row vector gives a two-dimensional matrix, which is the result of the calculations in the above example, i.e.

Addition and subtraction of two vectors is written as

a1 = ;
a2 = ;
c = a1+a2; % c = ;
c = a2-a1; % c = ;

Note that addition and subtraction operations can be performed between two column vectors or two row vectors. Otherwise, MatLab will issue an error message, because vectors of different types cannot be added. This is the case with all illegal arithmetic operations: if they cannot be calculated, the MatLab system will report an error and the program will end on the corresponding line.

Similarly, operations of multiplication and addition between matrices are performed:

A = ;
B = ones(3);
C=A+B; % addition of two matrices of the same size
D=A+5; % addition of a matrix and a number
E=A*B; % multiplication of matrix A by B
F=B*A; % multiplication of matrix B by A
G=5*A; % multiplication of a matrix by a number

The operations of calculating the inverse matrix, as well as transposing matrices and vectors, are written as follows:

a = ; % row vector
b = a'; % column vector formed by
% transpose of the row vector a.
A = ; % matrix 3x3 elements
B = a*A; %b= - row vector
C=A*b; % C = - column vector
D = a*A*a'; % D = 45 – number, sum of matrix A
E = A'; % E is the transposed matrix A
F = inv(A); % F - inverse matrix A
G = A^-1; % G - inverse matrix A

From the above example, it can be seen that the operation of transposing matrices and vectors is denoted by the symbol ‘ (apostrophe), which is placed after the name of the vector or matrix. The calculation of the inverse matrix can be done by calling the inv() function or by raising the matrix to the -1 power. The result in both cases will be the same, and two calculation methods are made for ease of use when implementing various algorithms.

If in the course of calculations it is required to multiply, divide or raise elements of a vector or matrix element by element, then the following operators are used for this:

.* - element-wise multiplication;
./ and .\ - element-wise divisions;
.^ - element-wise exponentiation.

Consider the operation of these operators in the following example.

a = ; % row vector
b = ; % row vector
c = a.*b; %c=
A = ones(3); % 3x3 matrix consisting of ones
B = ; % matrix 3x3
C = A.*B; % matrix 3x3, consisting of
D = A./B; % matrix 3x3, consisting of
E = A.\B; % matrix 3x3, consisting of
F = A.^2; % squaring of matrix A elements

To conclude this section, consider a few functions that are useful when working with vectors and matrices.

To find the maximum value of a vector element, the standard function max() is used, which returns the found maximum value of the element and its position (index):

a = ;
= max(a); % v = 6, i = 2;

v = max(a); %v = 6;

The above example shows two different ways to call the max() function. In the first case, both the maximum value of the element and its index in the vector are determined, and in the second, only the maximum value of the element is determined.

In the case of matrices, this function determines the maximum values ​​in the columns, as shown in the example below:

A = ;
= max(A); % V=, I=
V = max(A); %V=

The full syntax of the max() function can be found by typing the command in the MatLab command window

help<название функции>

The min() function works in a similar way, which determines the minimum value of a vector or matrix element and its index.

Another useful function for working with matrices and vectors is the sum() function, which calculates the sum of the values ​​of the elements of a vector or columns of a matrix:

a = ;
s = sum(a); %s = 3+5+4+2+1=15
A = ;
S1 = sum(A); %S1=
S2 = sum(sum(A)); % S2=39

When calculating the sum S2, the sum of the values ​​of the elements of the matrix A is first calculated by columns, and then by rows. As a result, the variable S2 contains the sum of the values ​​of all elements of the matrix A.

To sort the values ​​of the elements of a vector or matrix in ascending or descending order, use the sort() function as follows:

a = ;

b1 = sort(a); %b1=
b2 = sort(a, 'descend'); %b2=
b3 = sort(a, 'ascend'); %b3=

for matrices

A = ;
B1 = sort(A); %B1=
B2 = sort(A, 'descend'); %B2=

In many practical problems, it is often required to find a specific element in a vector or matrix. This can be done using the standard find() function, which takes as an argument a condition according to which the required elements are found, for example:

a = ;
b1 = find(a == 2); %b1 = 4 - element index 2
b2 = find(a ~= 2); % b2 = - indices without 2
b3 = find(a > 3); %b3=

In the above example, the symbol ‘==’ means checking for equality, and the symbol ‘~=’ performs a check for inequality of the values ​​of the elements of the vector a. More details about these operators will be described in the section on conditional operators.

Another useful function for working with vectors and matrices is the mean() function for calculating the arithmetic mean, which works like this:

a = ;
m = mean(a); %m = 3
A = ;
M1 = mean(A); %M1=
M2 = mean(mean(A)); % M2 = 4.333

So, in the previous lesson, we analyzed the rules for adding and subtracting matrices. These are such simple operations that most students understand them literally right off the bat.

However, you rejoice early. The freebie is over - let's move on to multiplication. I’ll warn you right away: multiplying two matrices is not at all multiplying the numbers in cells with the same coordinates, as you might think. Everything is much more fun here. And you have to start with preliminary definitions.

Consistent matrices

One of the most important characteristics of a matrix is ​​its size. We have already talked about this a hundred times: $A=\left[ m\times n \right]$ means that the matrix has exactly $m$ rows and $n$ columns. We have already discussed how not to confuse rows with columns. Now something else is important.

Definition. Matrices of the form $A=\left[ m\times n \right]$ and $B=\left[ n\times k \right]$, in which the number of columns in the first matrix is ​​the same as the number of rows in the second, are called consistent.

Once again: the number of columns in the first matrix is ​​equal to the number of rows in the second! From this we get two conclusions at once:

1. We care about the order of the matrices. For example, the matrices $A=\left[ 3\times 2 \right]$ and $B=\left[ 2\times 5 \right]$ are consistent (2 columns in the first matrix and 2 rows in the second), but vice versa — the matrices $B=\left[ 2\times 5 \right]$ and $A=\left[ 3\times 2 \right]$ are no longer consistent (5 columns in the first matrix are, as it were, not 3 rows in the second ).
2. Consistency is easy to check if you write out all the dimensions one after another. Using the example from the previous paragraph: "3 2 2 5" - the same numbers are in the middle, so the matrices are consistent. But “2 5 3 2” is not agreed, because there are different numbers in the middle.

Besides, the captain seems to hint that square matrices of the same size $\left[ n\times n \right]$ are always consistent.

In mathematics, when the order of enumeration of objects is important (for example, in the definition discussed above, the order of matrices is important), one often speaks of ordered pairs. We met them at school: I think it's a no brainer that the coordinates $\left(1;0 \right)$ and $\left(0;1 \right)$ define different points on the plane.

So: coordinates are also ordered pairs, which are made up of numbers. But nothing prevents you from making such a pair of matrices. Then it will be possible to say: "An ordered pair of matrices $\left(A;B \right)$ is consistent if the number of columns in the first matrix is ​​the same as the number of rows in the second."

Well, so what?

Definition of multiplication

Consider two consistent matrices: $A=\left[ m\times n \right]$ and $B=\left[ n\times k \right]$. And we define for them the operation of multiplication.

Definition. The product of two consistent matrices $A=\left[ m\times n \right]$ and $B=\left[ n\times k \right]$ is the new matrix $C=\left[ m\times k \right] $, the elements of which are calculated according to the formula:

\[\begin(align) & ((c)_(i;j))=((a)_(i;1))\cdot ((b)_(1;j))+((a)_ (i;2))\cdot ((b)_(2;j))+\ldots +((a)_(i;n))\cdot ((b)_(n;j))= \\ & =\sum\limits_(t=1)^(n)(((a)_(i;t))\cdot ((b)_(t;j))) \end(align)\]

Such a product is denoted in the standard way: $C=A\cdot B$.

For those who see this definition for the first time, two questions immediately arise:

1. What kind of wild game is this?
2. Why is it so difficult?

Well, first things first. Let's start with the first question. What do all these indexes mean? And how not to make mistakes when working with real matrices?

First of all, we note that the long line for calculating $((c)_(i;j))$ (specially put a semicolon between the indices so as not to get confused, but you don’t need to put them in general - I myself got tired of typing the formula in the definition) really boils down to a simple rule:

1. Take the $i$-th row in the first matrix;
2. Take the $j$-th column in the second matrix;
3. We get two sequences of numbers. We multiply the elements of these sequences with the same numbers, and then add the resulting products.

This process is easy to understand from the picture:

Scheme for multiplying two matrices

Once again: we fix the row $i$ in the first matrix, the column $j$ in the second matrix, multiply the elements with the same numbers, and then add the resulting products - we get $((c)_(ij))$. And so for all $1\le i\le m$ and $1\le j\le k$. Those. there will be $m\times k$ such "perversions" in total.

In fact, we have already met with matrix multiplication in the school curriculum, only in a greatly truncated form. Let vectors be given:

\[\begin(align) & \vec(a)=\left(((x)_(a));((y)_(a));((z)_(a)) \right); \\ & \overrightarrow(b)=\left(((x)_(b));((y)_(b));((z)_(b)) \right). \\ \end(align)\]

Then their scalar product will be exactly the sum of pairwise products:

\[\overrightarrow(a)\times \overrightarrow(b)=((x)_(a))\cdot ((x)_(b))+((y)_(a))\cdot ((y )_(b))+((z)_(a))\cdot ((z)_(b))\]

In fact, in those distant years, when the trees were greener and the sky brighter, we simply multiplied the row vector $\overrightarrow(a)$ by the column vector $\overrightarrow(b)$.

Nothing has changed today. It's just that now there are more of these row and column vectors.

But enough theory! Let's look at real examples. And let's start with the simplest case - square matrices.

Multiplication of square matrices

Task 1. Perform the multiplication:

\[\left[ \begin(array)(*(35)(r)) 1 & 2 \\ -3 & 4 \\\end(array) \right]\cdot \left[ \begin(array)(* (35)(r)) -2 & 4 \\ 3 & 1 \\\end(array) \right]\]

Solution. So, we have two matrices: $A=\left[ 2\times 2 \right]$ and $B=\left[ 2\times 2 \right]$. It is clear that they are consistent (square matrices of the same size are always consistent). So we do the multiplication:

\[\begin(align) & \left[ \begin(array)(*(35)(r)) 1 & 2 \\ -3 & 4 \\\end(array) \right]\cdot \left[ \ begin(array)(*(35)(r)) -2 & 4 \\ 3 & 1 \\\end(array) \right]=\left[ \begin(array)(*(35)(r)) 1\cdot \left(-2 \right)+2\cdot 3 & 1\cdot 4+2\cdot 1 \\ -3\cdot \left(-2 \right)+4\cdot 3 & -3\cdot 4+4\cdot 1 \\\end(array) \right]= \\ & =\left[ \begin(array)(*(35)(r)) 4 & 6 \\ 18 & -8 \\\ end(array)\right]. \end(align)\]

That's all!

Answer: $\left[ \begin(array)(*(35)(r))4 & 6 \\ 18 & -8 \\\end(array) \right]$.

Task 2. Perform the multiplication:

\[\left[ \begin(matrix) 1 & 3 \\ 2 & 6 \\\end(matrix) \right]\cdot \left[ \begin(array)(*(35)(r))9 & 6 \\ -3 & -2 \\\end(array) \right]\]

Solution. Again, consistent matrices, so we perform the following actions:\[\]

\[\begin(align) & \left[ \begin(matrix) 1 & 3 \\ 2 & 6 \\\end(matrix) \right]\cdot \left[ \begin(array)(*(35)( r)) 9 & 6 \\ -3 & -2 \\\end(array) \right]=\left[ \begin(array)(*(35)(r)) 1\cdot 9+3\cdot \ left(-3 \right) & 1\cdot 6+3\cdot \left(-2 \right) \\ 2\cdot 9+6\cdot \left(-3 \right) & 2\cdot 6+6\ cdot \left(-2 \right) \\\end(array) \right]= \\ & =\left[ \begin(matrix) 0 & 0 \\ 0 & 0 \\\end(matrix) \right] . \end(align)\]

As you can see, the result is a matrix filled with zeros

Answer: $\left[ \begin(matrix) 0 & 0 \\ 0 & 0 \\\end(matrix) \right]$.

From the above examples, it is obvious that matrix multiplication is not such a complicated operation. At least for 2 by 2 square matrices.

In the process of calculations, we compiled an intermediate matrix, where we directly painted what numbers are included in a particular cell. This is exactly what should be done when solving real problems.

Basic properties of the matrix product

In a nutshell. Matrix multiplication:

1. Non-commutative: $A\cdot B\ne B\cdot A$ in general. There are, of course, special matrices for which the equality $A\cdot B=B\cdot A$ (for example, if $B=E$ is the identity matrix), but in the vast majority of cases this does not work;
2. Associative: $\left(A\cdot B \right)\cdot C=A\cdot \left(B\cdot C \right)$. There are no options here: adjacent matrices can be multiplied without worrying about what is to the left and to the right of these two matrices.
3. Distributively: $A\cdot \left(B+C \right)=A\cdot B+A\cdot C$ and $\left(A+B \right)\cdot C=A\cdot C+B\cdot C $

And now - all the same, but in more detail.

Matrix multiplication is a lot like classical number multiplication. But there are differences, the most important of which is that matrix multiplication is, generally speaking, non-commutative.

Consider again the matrices from Problem 1. We already know their direct product:

\[\left[ \begin(array)(*(35)(r)) 1 & 2 \\ -3 & 4 \\\end(array) \right]\cdot \left[ \begin(array)(* (35)(r)) -2 & 4 \\ 3 & 1 \\\end(array) \right]=\left[ \begin(array)(*(35)(r))4 & 6 \\ 18 & -8 \\\end(array) \right]\]

But if we swap the matrices, we get a completely different result:

\[\left[ \begin(array)(*(35)(r)) -2 & 4 \\ 3 & 1 \\\end(array) \right]\cdot \left[ \begin(array)(* (35)(r)) 1 & 2 \\ -3 & 4 \\\end(array) \right]=\left[ \begin(matrix) -14 & 4 \\ 0 & 10 \\\end(matrix )\right]\]

It turns out that $A\cdot B\ne B\cdot A$. Also, the multiplication operation is only defined for the consistent matrices $A=\left[ m\times n \right]$ and $B=\left[ n\times k \right]$, but no one guaranteed that they would remain consistent, if they are swapped. For example, the matrices $\left[ 2\times 3 \right]$ and $\left[ 3\times 5 \right]$ are quite consistent in this order, but the same matrices $\left[ 3\times 5 \right] $ and $\left[ 2\times 3 \right]$ written in reverse order no longer match. Sadness :(

Among square matrices of a given size $n$, there will always be those that give the same result both when multiplied in direct and in reverse order. How to describe all such matrices (and how many of them in general) is a topic for a separate lesson. Today we will not talk about it. :)

However, matrix multiplication is associative:

\[\left(A\cdot B \right)\cdot C=A\cdot \left(B\cdot C \right)\]

Therefore, when you need to multiply several matrices in a row at once, it is not at all necessary to do it ahead of time: it is quite possible that some adjacent matrices, when multiplied, give an interesting result. For example, a zero matrix, as in Problem 2 discussed above.

In real problems, most often one has to multiply square matrices of size $\left[ n\times n \right]$. The set of all such matrices is denoted by $((M)^(n))$ (i.e., the entries $A=\left[ n\times n \right]$ and \ mean the same thing), and it will definitely contain matrix $E$, which is called the identity matrix.

Definition. The identity matrix of size $n$ is a matrix $E$ such that for any square matrix $A=\left[ n\times n \right]$ the equality holds:

Such a matrix always looks the same: there are units on its main diagonal, and zeros in all other cells.

\[\begin(align) & A\cdot \left(B+C \right)=A\cdot B+A\cdot C; \\ & \left(A+B \right)\cdot C=A\cdot C+B\cdot C. \\ \end(align)\]

In other words, if you need to multiply one matrix by the sum of two others, then you can multiply it by each of these "other two", and then add the results. In practice, you usually have to perform the inverse operation: we notice the same matrix, take it out of the bracket, perform addition, and thereby simplify our life. :)

Note that to describe distributivity, we had to write two formulas: where the sum is in the second factor and where the sum is in the first. This is precisely due to the fact that matrix multiplication is non-commutative (and in general, in non-commutative algebra, there are a lot of all sorts of jokes that do not even come to mind when working with ordinary numbers). And if, for example, you need to write down this property during the exam, then be sure to write both formulas, otherwise the teacher may get a little angry.

Okay, these were all fairy tales about square matrices. What about rectangles?

The case of rectangular matrices

But nothing - everything is the same as with square ones.

Task 3. Perform the multiplication:

\[\left[ \begin(matrix) \begin(matrix) 5 \\ 2 \\ 3 \\\end(matrix) & \begin(matrix) 4 \\ 5 \\ 1 \\\end(matrix) \ \\end(matrix) \right]\cdot \left[ \begin(array)(*(35)(r)) -2 & 5 \\ 3 & 4 \\\end(array) \right]\]

Solution. We have two matrices: $A=\left[ 3\times 2 \right]$ and $B=\left[ 2\times 2 \right]$. Let's write the numbers indicating the sizes in a row:

As you can see, the central two numbers are the same. This means that the matrices are consistent, and they can be multiplied. And at the output we get the matrix $C=\left[ 3\times 2 \right]$:

\[\begin(align) & \left[ \begin(matrix) \begin(matrix) 5 \\ 2 \\ 3 \\\end(matrix) & \begin(matrix) 4 \\ 5 \\ 1 \\ \end(matrix) \\\end(matrix) \right]\cdot \left[ \begin(array)(*(35)(r)) -2 & 5 \\ 3 & 4 \\\end(array) \right]=\left[ \begin(array)(*(35)(r)) 5\cdot \left(-2 \right)+4\cdot 3 & 5\cdot 5+4\cdot 4 \\ 2 \cdot \left(-2 \right)+5\cdot 3 & 2\cdot 5+5\cdot 4 \\ 3\cdot \left(-2 \right)+1\cdot 3 & 3\cdot 5+1 \cdot 4 \\\end(array) \right]= \\ & =\left[ \begin(array)(*(35)(r)) 2 & 41 \\ 11 & 30 \\ -3 & 19 \ \\end(array)\right]. \end(align)\]

Everything is clear: the final matrix has 3 rows and 2 columns. Quite $=\left[ 3\times 2 \right]$.

Answer: $\left[ \begin(array)(*(35)(r)) \begin(array)(*(35)(r)) 2 \\ 11 \\ -3 \\\end(array) & \begin(matrix) 41 \\ 30 \\ 19 \\\end(matrix) \\\end(array) \right]$.

Now consider one of the best training tasks for those who are just starting to work with matrices. In it, you need not just to multiply some two tablets, but first to determine: is such a multiplication permissible?

Problem 4. Find all possible pairwise products of matrices:

\\]; $B=\left[ \begin(matrix) \begin(matrix) 0 \\ 2 \\ 0 \\ 4 \\\end(matrix) & \begin(matrix) 1 \\ 0 \\ 3 \\ 0 \ \\end(matrix) \\\end(matrix) \right]$; $C=\left[ \begin(matrix)0 & 1 \\ 1 & 0 \\\end(matrix) \right]$.

Solution. First, let's write down the dimensions of the matrices:

\;\ B=\left[ 4\times 2 \right];\ C=\left[ 2\times 2 \right]\]

We get that the matrix $A$ can be matched only with the matrix $B$, since the number of columns in $A$ is 4, and only $B$ has this number of rows. Therefore, we can find the product:

\\cdot \left[ \begin(array)(*(35)(r)) 0 & 1 \\ 2 & 0 \\ 0 & 3 \\ 4 & 0 \\\end(array) \right]=\ left[ \begin(array)(*(35)(r))-10 & 7 \\ 10 & 7 \\\end(array) \right]\]

I suggest that the reader perform the intermediate steps on their own. I will only note that it is better to determine the size of the resulting matrix in advance, even before any calculations:

\\cdot \left[ 4\times 2 \right]=\left[ 2\times 2 \right]\]

In other words, we simply remove the "transitional" coefficients that ensured the consistency of the matrices.

What other options are possible? It is certainly possible to find $B\cdot A$, since $B=\left[ 4\times 2 \right]$, $A=\left[ 2\times 4 \right]$, so the ordered pair $\left(B ;A \right)$ is consistent, and the dimension of the product will be:

\\cdot \left[ 2\times 4 \right]=\left[ 4\times 4 \right]\]

In short, the output will be a matrix $\left[ 4\times 4 \right]$, whose coefficients are easy to calculate:

\\cdot \left[ \begin(array)(*(35)(r)) 1 & -1 & 2 & -2 \\ 1 & 1 & 2 & 2 \\\end(array) \right]=\ left[ \begin(array)(*(35)(r))1 & 1 & 2 & 2 \\ 2 & -2 & 4 & -4 \\ 3 & 3 & 6 & 6 \\ 4 & -4 & 8 & -8 \\\end(array) \right]\]

Obviously, you can also match $C\cdot A$ and $B\cdot C$, and that's it. Therefore, we simply write the resulting products:

It was easy.:)

Answer: $AB=\left[ \begin(array)(*(35)(r)) -10 & 7 \\ 10 & 7 \\\end(array) \right]$; $BA=\left[ \begin(array)(*(35)(r)) 1 & 1 & 2 & 2 \\ 2 & -2 & 4 & -4 \\ 3 & 3 & 6 & 6 \\ 4 & -4 & 8 & -8 \\\end(array) \right]$; $CA=\left[ \begin(array)(*(35)(r)) 1 & 1 & 2 & 2 \\ 1 & -1 & 2 & -2 \\\end(array) \right]$; $BC=\left[ \begin(array)(*(35)(r))1 & 0 \\ 0 & 2 \\ 3 & 0 \\ 0 & 4 \\\end(array) \right]$.

In general, I highly recommend doing this task yourself. And another similar task that is in homework. These seemingly simple thoughts will help you work out all the key steps in matrix multiplication.

But the story doesn't end there. Let's move on to special cases of multiplication. :)

Row vectors and column vectors

One of the most common matrix operations is multiplication by a matrix that has one row or one column.

Definition. A column vector is a $\left[ m\times 1 \right]$ matrix, i.e. consisting of several rows and only one column.

A row vector is a matrix of size $\left[ 1\times n \right]$, i.e. consisting of one row and several columns.

In fact, we have already met with these objects. For example, an ordinary three-dimensional vector from stereometry $\overrightarrow(a)=\left(x;y;z \right)$ is nothing but a row vector. From a theoretical point of view, there is almost no difference between rows and columns. You need to be careful only when coordinating with the surrounding multiplier matrices.

Task 5. Multiply:

\[\left[ \begin(array)(*(35)(r)) 2 & -1 & 3 \\ 4 & 2 & 0 \\ -1 & 1 & 1 \\\end(array) \right] \cdot \left[ \begin(array)(*(35)(r)) 1 \\ 2 \\ -1 \\\end(array) \right]\]

Solution. We have a product of consistent matrices: $\left[ 3\times 3 \right]\cdot \left[ 3\times 1 \right]=\left[ 3\times 1 \right]$. Find this piece:

\[\left[ \begin(array)(*(35)(r)) 2 & -1 & 3 \\ 4 & 2 & 0 \\ -1 & 1 & 1 \\\end(array) \right] \cdot \left[ \begin(array)(*(35)(r)) 1 \\ 2 \\ -1 \\\end(array) \right]=\left[ \begin(array)(*(35 )(r)) 2\cdot 1+\left(-1 \right)\cdot 2+3\cdot \left(-1 \right) \\ 4\cdot 1+2\cdot 2+0\cdot 2 \ \ -1\cdot 1+1\cdot 2+1\cdot \left(-1 \right) \\\end(array) \right]=\left[ \begin(array)(*(35)(r) ) -3 \\ 8 \\ 0 \\\end(array) \right]\]

Answer: $\left[ \begin(array)(*(35)(r))-3 \\ 8 \\ 0 \\\end(array) \right]$.

Task 6. Perform the multiplication:

\[\left[ \begin(array)(*(35)(r)) 1 & 2 & -3 \\\end(array) \right]\cdot \left[ \begin(array)(*(35) (r)) 3 & 1 & -1 \\ 4 & -1 & 3 \\ 2 & 6 & 0 \\\end(array) \right]\]

Solution. Again everything is consistent: $\left[ 1\times 3 \right]\cdot \left[ 3\times 3 \right]=\left[ 1\times 3 \right]$. We consider the work:

\[\left[ \begin(array)(*(35)(r)) 1 & 2 & -3 \\\end(array) \right]\cdot \left[ \begin(array)(*(35) (r)) 3 & 1 & -1 \\ 4 & -1 & 3 \\ 2 & 6 & 0 \\\end(array) \right]=\left[ \begin(array)(*(35)( r))5 & -19 & 5 \\\end(array) \right]\]

Answer: $\left[ \begin(matrix) 5 & -19 & 5 \\\end(matrix) \right]$.

As you can see, when multiplying a row vector and a column vector by a square matrix, the output is always a row or column of the same size. This fact has many applications - from solving linear equations to all sorts of coordinate transformations (which in the end also come down to systems of equations, but let's not talk about sad things).

I think everything was obvious here. Let's move on to the final part of today's lesson.

Matrix exponentiation

Among all multiplication operations, exponentiation deserves special attention - this is when we multiply the same object by itself several times. Matrices are no exception, they can also be raised to various powers.

Such works are always coordinated:

\\cdot \left[ n\times n \right]=\left[ n\times n \right]\]

And they are designated in the same way as ordinary degrees:

\[\begin(align) & A\cdot A=((A)^(2)); \\ & A\cdot A\cdot A=((A)^(3)); \\ & \underbrace(A\cdot A\cdot \ldots \cdot A)_(n)=((A)^(n)). \\ \end(align)\]

At first glance, everything is simple. Let's see how it looks in practice:

Task 7. Raise the matrix to the specified power:

$((\left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right])^(3))$

Solution. OK, let's build. Let's square it first:

\[\begin(align) & ((\left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right])^(2))=\left[ \begin(matrix ) 1 & 1 \\ 0 & 1 \\\end(matrix) \right]\cdot \left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right]= \\ & =\left[ \begin(array)(*(35)(r)) 1\cdot 1+1\cdot 0 & 1\cdot 1+1\cdot 1 \\ 0\cdot 1+1\cdot 0 & 0\cdot 1+1\cdot 1 \\\end(array) \right]= \\ & =\left[ \begin(array)(*(35)(r)) 1 & 2 \\ 0 & 1 \ \\end(array) \right] \end(align)\]

\[\begin(align) & ((\left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right])^(3))=((\left[ \begin (matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right])^(3))\cdot \left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\end( matrix) \right]= \\ & =\left[ \begin(array)(*(35)(r)) 1 & 2 \\ 0 & 1 \\\end(array) \right]\cdot \left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right]= \\ & =\left[ \begin(array)(*(35)(r)) 1 & 3 \\ 0 & 1 \\\end(array) \right] \end(align)\]

That's all.:)

Answer: $\left[ \begin(matrix)1 & 3 \\ 0 & 1 \\\end(matrix) \right]$.

Problem 8. Raise the matrix to the specified power:

\[((\left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right])^(10))\]

Solution. Just don’t cry now about the fact that “the degree is too high”, “the world is not fair” and “the teachers have completely lost their banks”. In fact, everything is easy:

\[\begin(align) & ((\left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right])^(10))=((\left[ \begin (matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right])^(3))\cdot ((\left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\ end(matrix) \right])^(3))\cdot ((\left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right])^(3))\ cdot \left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right]= \\ & =\left(\left[ \begin(matrix) 1 & 3 \\ 0 & 1 \\\end(matrix) \right]\cdot \left[ \begin(matrix) 1 & 3 \\ 0 & 1 \\\end(matrix) \right] \right)\cdot \left(\left[ \begin(matrix) 1 & 3 \\ 0 & 1 \\\end(matrix) \right]\cdot \left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right ] \right)= \\ & =\left[ \begin(matrix) 1 & 6 \\ 0 & 1 \\\end(matrix) \right]\cdot \left[ \begin(matrix) 1 & 4 \\ 0 & 1 \\\end(matrix) \right]= \\ & =\left[ \begin(matrix) 1 & 10 \\ 0 & 1 \\\end(matrix) \right] \end(align)\ ]

Note that in the second line we used multiplication associativity. Actually, we used it in the previous task, but there it was implicit.

Answer: $\left[ \begin(matrix) 1 & 10 \\ 0 & 1 \\\end(matrix) \right]$.

As you can see, there is nothing complicated in raising a matrix to a power. The last example can be summarized:

\[((\left[ \begin(matrix) 1 & 1 \\ 0 & 1 \\\end(matrix) \right])^(n))=\left[ \begin(array)(*(35) (r)) 1 & n \\ 0 & 1 \\\end(array) \right]\]

This fact is easy to prove through mathematical induction or direct multiplication. However, it is far from always possible to catch such patterns when raising to a power. Therefore, be careful: it is often easier and faster to multiply several matrices "blank" than to look for some patterns there.

In general, do not look for a higher meaning where there is none. Finally, let's consider the exponentiation of a larger matrix - as much as $\left[ 3\times 3 \right]$.

Problem 9. Raise the matrix to the specified power:

\[((\left[ \begin(matrix) 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\\end(matrix) \right])^(3))\]

Solution. Let's not look for patterns. We work "through":

\[((\left[ \begin(matrix) 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\\end(matrix) \right])^(3))=(( \left[ \begin(matrix) 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\\end(matrix) \right])^(2))\cdot \left[ \begin (matrix)0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\\end(matrix) \right]\]

Let's start by squaring this matrix:

\[\begin(align) & ((\left[ \begin(matrix) 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\\end(matrix) \right])^( 2))=\left[ \begin(matrix) 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\\end(matrix) \right]\cdot \left[ \begin(matrix ) 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\\end(matrix) \right]= \\ & =\left[ \begin(array)(*(35)(r )) 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \\\end(array) \right] \end(align)\]

Now let's cube it:

\[\begin(align) & ((\left[ \begin(matrix) 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\\end(matrix) \right])^( 3))=\left[ \begin(array)(*(35)(r)) 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \\\end(array) \right] \cdot \left[ \begin(matrix) 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\\end(matrix) \right]= \\ & =\left[ \begin( array)(*(35)(r)) 2 & 3 & 3 \\ 3 & 2 & 3 \\ 3 & 3 & 2 \\\end(array) \right] \end(align)\]

That's all. Problem solved.

Answer: $\left[ \begin(matrix) 2 & 3 & 3 \\ 3 & 2 & 3 \\ 3 & 3 & 2 \\\end(matrix) \right]$.

As you can see, the amount of calculations has become larger, but the meaning has not changed at all. :)

This lesson can end. Next time we will consider the inverse operation: we will look for the original multipliers using the existing product.

As you probably already guessed, we will talk about the inverse matrix and methods for finding it.

Lecture 6. Parallel numerical algorithms for solving typical problems of computational mathematics: matrix multiplication.

Multiplication of a matrix by a vector. Achieve the highest possible speed. Use of middle level parallelism. Organization of parallel computing for p = n. Use of a limited set of processors. Matrix multiplication. Macrooperational analysis of problem solving algorithms. Organization of parallelism based on data sharing.

Multiplication of a matrix by a vector

The problem of multiplying a matrix by a vector is defined by the relations

Thus, obtaining the resulting vector involves repeating the same type of operations for multiplying the rows of the matrix and vector . Obtaining each such operation includes element-by-element multiplication of the elements of the row of the matrix and vector and the subsequent summation of the resulting products. The total number of required scalar operations is estimated by the value

As follows from the actions performed when multiplying a matrix and a vector, parallel methods for solving the problem can be obtained based on parallel summation algorithms (see paragraph 4.1). In this section, the analysis of parallelization methods will be supplemented by consideration of the organization of parallel computing depending on the number of processors available for use. In addition, using the example of the problem of multiplying a matrix by a vector, attention will be drawn to the need to choose the most appropriate topology of a computing system (existing communication channels between processors) to reduce costs for organizing interprocessor interaction.

Achieving the fastest possible performance ()

Let's perform an analysis of information dependencies in the algorithm of matrix-vector multiplication to select possible ways of parallelization. As you can see, the operations of multiplying individual rows of a matrix by a vector performed during calculations are independent and can be performed in parallel;

Multiplying each row by a vector involves independent element-wise multiplications and can also be performed in parallel;

The summation of the products obtained in each operation of multiplying a row of a matrix by a vector can be performed using one of the previously considered variants of the summation algorithm (serial algorithm, conventional and modified cascade schemes).

Thus, the maximum required number of processors is determined by the value

The use of such a number of processors can be represented as follows. The set of processors is divided into groups

,

each of which represents a set of processors for performing the operation of multiplying a single row of a matrix by a vector. At the beginning of calculations, each processor of the group receives an element of the row of the matrix and the corresponding element of the vector . Next, each processor performs the multiplication operation. Subsequent then calculations are performed according to the cascade summation scheme. For illustration in fig. 6.1 shows the computational scheme for the processors of the group with the dimension of the matrix .

Rice. 6.1. Computational scheme for multiplying a matrix row by a vector

The execution time of a parallel algorithm when using processors is determined by the execution time of the parallel multiplication operation and the execution time of the cascade scheme

As a result, the performance indicators of the algorithm are determined by the following relationships:

For the considered problem of multiplication of a matrix by a vector, the most suitable topologies are structures that provide fast data transfer (paths of unit length) in a cascade summation scheme (see Fig. 4.5). Such topologies are a structure with a complete system of connections ( complete graph) and hypercube. Other topologies result in increased communication time due to longer data paths. So, with a linear ordering of processors with a system of connections only with the nearest neighbors on the left and on the right ( ruler or ring) for the cascade scheme, the length of the transmission path of each received partial sum at the iteration , , is equal to . If we accept that data transmission along a path of length in topologies with a linear structure requires the execution of data transmission operations, the total number of parallel operations (total length of paths) of data transmission is determined by the value

(excluding data transfers for bootstrapping processors).

Application of a computing system with a rectangular topology two-dimensional lattice size leads to a simple and visual interpretation of the calculations performed (the network structure corresponds to the structure of the processed data). For such a topology, it is most expedient to place the rows of the matrix along the horizontal lines of the lattice; in this case, the elements of the vector must be sent along the verticals of the computing system. The execution of calculations with this arrangement of data can be carried out in parallel along the lines of the lattice; as a result, the total number of data transfers is the same as the results for ruler().

The communication actions performed in solving the problem are to transfer data between pairs of MCS processors. A detailed analysis of the duration of the implementation of such operations is carried out in paragraph 3.3.

4. Recommendations for the implementation of the parallel algorithm. When implementing a parallel algorithm, it is advisable to single out the initial stage of loading the used processors with initial data. Such initialization is most simply provided for the topology of a computing system with a topology in the form complete graph(loading is provided with a single parallel data transfer operation). When organizing a set of processors in the form hypercube It may be useful to have a two-level control of the bootstrap process, in which the central control processor distributes the matrix and vector rows to the control processors of processor groups , , which, in turn, distribute the elements of the matrix and vector rows to the executive processors. For topologies in the form rulers or rings sequential data transfer operations are required with a sequentially decreasing amount of data transferred from to elements.

Using Middle Level Parallelism()

1. Choice of parallel computing method. With a decrease in the available number of processors used (), the usual cascade summation scheme when performing operations of multiplying matrix rows by a vector becomes inapplicable. For simplicity of presentation of the material, we assume and use a modified cascade scheme. The initial load of each processor in this case increases and the processor is loaded () by parts of the rows of the matrix and vector . The execution time of the operation of multiplying a matrix by a vector can be estimated as the value

When using the number of processors required to implement the modified cascade scheme, i.e. at , this expression gives an estimate of the execution time (at ).

With the number of processors , when the execution time of the algorithm is estimated as , a new scheme for parallel execution of calculations can be proposed, in which for each iteration of cascaded summations are used non-overlapping processor sets. With this approach, the available number of processors is sufficient to implement only one operation of multiplying a row of a matrix and a vector. In addition, when performing the next iteration of the cascade summation, the processors responsible for the execution of all previous iterations are free. However, this disadvantage of the proposed approach can be turned into an advantage by using idle processors to process the next rows of the matrix. As a result, the following scheme can be formed conveyor perform matrix and vector multiplication:

The set of processors is divided into non-overlapping processor groups

,

the group , , consists of processors and is used to iterate the cascade algorithm (the group is used to implement element-wise multiplication); total number of processors;

Calculation initialization consists in element-by-element loading of the processors of the group with the values ​​1 of the row of the matrix and the vector ; after the bootstrap, a parallel operation of element-wise multiplication and subsequent implementation of the conventional cascade summation circuit is performed;

When performing calculations, each time after the completion of the operation of element-wise multiplication, the processors of the group are loaded with elements of the next row of the matrix and the calculation process is initiated for the newly loaded data.

As a result of applying the described algorithm, a plurality of processors implements a pipeline for performing the operation of multiplying a matrix row by a vector. On such a pipeline, several individual rows of the matrix can simultaneously be at different stages of processing. So, for example, after the element-wise multiplication of the elements of the first row and the vector, the group processors will perform the first iteration of the cascade algorithm for the first row of the matrix, and the group processors will perform the element-wise multiplication of the values ​​of the second row of the matrix, and so on. For illustration in fig. 6.2 shows the situation of the computation process after 2 pipeline iterations at .

Rice. 6.2. The state of the pipeline for the operation of multiplying a row of a matrix by a vector after performing 2 iterations

2. Evaluation of algorithm performance indicators. The multiplication of the first row by the vector according to the cascade scheme will be completed, as usual, after the execution of () parallel operations. For other rows, in accordance with the pipeline scheme of organization of calculations, the results of multiplication of each successive row will appear after the completion of each subsequent iteration of the pipeline. As a result, the total execution time of the matrix-vector multiplication operation can be expressed as

This estimate is slightly longer than the execution time of the parallel algorithm described in the previous paragraph (), however, the newly proposed method requires less data to be transmitted (the vector is sent only once). In addition, the use of a pipeline scheme leads to an earlier appearance of some of the calculation results (which can be useful in a number of data processing situations).

As a result, the performance indicators of the algorithm are determined by the following relations:

3. Choice of computer system topology. The expedient topology of a computing system is completely determined by the computing scheme - this is a complete binary tree height . The number of data transfers with such a network topology is determined by the total number of iterations performed by the pipeline, i.e.

The initialization of calculations starts from the leaves of the tree, the summation results are accumulated in the root processor.

The analysis of the complexity of the communication actions performed for computer systems with other topologies of interprocessor communications is supposed to be carried out as an independent task (see also Section 3.4).

Organization of parallel computing with

1. Choice of parallel computing method. When using processors for multiplying a matrix by a vector, the parallel row-by-row multiplication algorithm already discussed in the manual can be used, in which the rows of the matrix are distributed row by row among the processors and each processor implements the operation of multiplying any individual row of the matrix by the vector . Another possible way to organize parallel computing can be to build pipeline scheme for the operation of multiplying a row of a matrix by a vector(dot product of vectors) by arranging all available processors in a linear sequence ( rulers).

Such a calculation scheme can be defined as follows. Let's represent the set of processors as a linear sequence (see Fig. 4.7):

each processor , , is used to multiply the matrix column elements and the vector element . The execution of calculations on each processor , , consists of the following:

The next element of the matrix column is requested;

The elements and are multiplied;

The result of the calculations of the previous processor is requested;

Values ​​are added ;

The result is sent to the next processor.

Rice. 6.3. The state of the linear pipeline for the operation of multiplying a row of a matrix by a vector after performing two iterations

When initializing the described scheme, it is necessary to perform a number of additional actions:

During the first iteration, each processor additionally requests an element of the vector ;

To synchronize calculations (during the execution of the next iteration of the circuit, the result of the calculation of the previous processor is requested) at the initialization stage, the processor , , executes () a waiting loop.

In addition, for the uniformity of the described scheme for the first processor , which has no previous processor, it is advisable to introduce an empty addition operation ( ).

For illustration in fig. 6.3 shows the state of the computation process after the second iteration of the pipeline at .

2. Evaluation of algorithm performance indicators. The multiplication of the first row by the vector according to the described pipeline scheme will be completed after the execution of () parallel operations. The result of the multiplication of the following rows will occur after the completion of each next iteration of the pipeline (recall, the iteration of each processor includes the execution of multiplication and addition operations). As a result, the total execution time of the matrix-vector multiplication operation can be expressed as:

This estimate is also greater than the minimum possible execution time of the parallel algorithm for . The usefulness of using a pipeline computing scheme is, as noted in the previous paragraph, in reducing the amount of transmitted data and in the earlier appearance of part of the calculation results.

The performance indicators of this computational scheme are determined by the relations:

, ,

3. Choice of computer system topology. The required topology of the computing system for the implementation of the described algorithm is uniquely determined by the proposed computational scheme - this is a linearly ordered set of processors ( ruler).

Using a limited set of processors ()

1. Choice of parallel computing method. When the number of processors is reduced to a value, a parallel computational scheme for matrix-vector multiplication can be obtained as a result of adapting the row-by-row multiplication algorithm. In this case, the cascade scheme for summing the results of elementwise multiplication degenerates and the operation of multiplying a matrix row by a vector is completely performed on a single processor. The computational scheme obtained with this approach can be specified as follows:

A vector and matrix rows are sent to each of the available processors;

The operation of multiplying rows of a matrix by a vector is performed using the usual sequential algorithm.

It should be noted that the size of the matrix may not be a multiple of the number of processors, and then the rows of the matrix cannot be divided equally between the processors. In these situations, it is possible to deviate from the requirement of processor load uniformity and, in order to obtain a simpler computational scheme, accept the rule that data is placed on processors only row by row (i.e., elements of one row of a matrix cannot be shared between several processors). A different number of rows results in a different computational load on the processors; thus, the completion of calculations (the total duration of the task solution) will be determined by the operating time of the most loaded processor (at the same time, some processors may idle part of this total time due to the exhaustion of their share of calculations). The uneven loading of processors reduces the efficiency of using the MCS and, as a result of considering this example, we can conclude that balancing problem

3. Choice of computer system topology. In accordance with the nature of the interprocessor interactions performed in the proposed computational scheme, the organization of processors in the form stars(see fig. 1.1). A control processor of such a topology can be used to load computing processors with initial data and to receive the results of performed calculations.

Matrix multiplication

The problem of multiplying a matrix by a matrix is ​​defined by the relations

.

(for simplicity, we will assume that the multiplied matrices and are square and have order ).

The analysis of possible ways of parallel execution of this task can be carried out by analogy with the consideration of the problem of multiplying a matrix by a vector. Leaving such an analysis for independent study, we will show, using the example of the problem of matrix multiplication, the use of several general approaches that allow us to form parallel methods for solving complex problems.

Definition 1

The product of matrices (C=AB) is an operation only for consistent matrices A and B, in which the number of columns of matrix A is equal to the number of rows of matrix B:

C ⏟ m × n = A ⏟ m × p × B ⏟ p × n

Example 1

Matrix data:

• A = a (i j) of dimensions m × n;
• B = b (i j) p × n

Matrix C , whose elements c i j are calculated by the following formula:

c i j = a i 1 × b 1 j + a i 2 × b 2 j + . . . + a i p × b p j , i = 1 , . . . m , j = 1 , . . . m

Example 2

Let's calculate the products AB=BA:

A = 1 2 1 0 1 2 , B = 1 0 0 1 1 1

Solution using the matrix multiplication rule:

A ⏟ 2 × 3 × B ⏟ 3 × 2 = 1 2 1 0 1 2 × 1 0 0 1 1 1 = 1 × 1 + 2 × 0 + 1 × 1 1 × 0 + 2 × 1 + 1 × 1 0 × 1 + 1 × 0 + 2 × 1 0 × 0 + 1 × 1 + 2 × 1 = = 2 3 2 3 ⏟ 2 × 2

B ⏟ 3 × 2 × A ⏟ 2 × 3 = 1 0 0 1 1 1 × 1 2 1 0 1 2 = 1 × 1 + 0 × 0 1 × 2 + 0 × 1 1 × 1 + 0 × 2 0 × 1 + 1 × 0 0 × 2 + 1 × 1 0 × 1 + 1 × 2 1 × 1 + 1 × 0 1 × 2 + 1 × 1 1 × 1 + 1 × 2 = 1 2 1 0 1 2 1 3 3 ⏟ 3×3

The product A B and B A are found, but they are matrices of different sizes: A B is not equal to B A.

Properties of matrix multiplication

Matrix multiplication properties:

• (A B) C = A (B C) - associativity of matrix multiplication;
• A (B + C) \u003d A B + A C - distributive multiplication;
• (A + B) C \u003d A C + B C - distributivity of multiplication;
• λ (A B) = (λ A) B

Example 1

Check property #1: (A B) C = A (B C) :

(A × B) × A = 1 2 3 4 × 5 6 7 8 × 1 0 0 2 = 19 22 43 50 × 1 0 0 2 = 19 44 43 100,

A (B × C) = 1 2 3 4 × 5 6 7 8 1 0 0 2 = 1 2 3 4 × 5 12 7 16 = 19 44 43 100 .

Example 2

We check property No. 2: A (B + C) \u003d A B + A C:

A × (B + C) = 1 2 3 4 × 5 6 7 8 + 1 0 0 2 = 1 2 3 4 × 6 6 7 10 = 20 26 46 58,

A B + A C \u003d 1 2 3 4 × 5 6 7 8 + 1 2 3 4 × 1 0 0 2 \u003d 19 22 43 50 + 1 4 3 8 \u003d 20 26 46 58 .

Product of three matrices

The product of three matrices A B C is calculated in 2 ways:

• find A B and multiply by C: (A B) C;
• or find first B C, and then multiply A (B C) .

Example 3

Multiply matrices in 2 ways:

4 3 7 5 × - 28 93 38 - 126 × 7 3 2 1

Action algorithm:

• find the product of 2 matrices;
• then again find the product of 2 matrices.

one). A B \u003d 4 3 7 5 × - 28 93 38 - 126 \u003d 4 (- 28) + 3 × 38 4 × 93 + 3 (- 126) 7 (- 28) + 5 × 38 7 × 93 + 5 (- 126 ) = 2 - 6 - 6 21

2). A B C = (A B) C = 2 - 6 - 6 21 7 3 2 1 = 2 × 7 - 6 × 2 2 × 3 - 6 × 1 - 6 × 7 + 21 × 2 - 6 × 3 + 21 × 1 = 2 0 0 3 .

We use the formula A B C \u003d (A B) C:

one). B C = - 28 93 38 - 126 7 3 2 1 = - 28 × 7 + 93 × 2 - 28 × 3 + 93 × 1 38 × 7 - 126 × 2 38 × 3 - 126 × 1 = - 10 9 14 - 12

2). A B C \u003d (A B) C \u003d 7 3 2 1 - 10 9 14 - 12 \u003d 4 (- 10) + 3 × 14 4 × 9 + 3 (- 12) 7 (- 10) + 5 × 14 7 × 9 + 5 (- 12) = 2 0 0 3

Answer: 4 3 7 5 - 28 93 38 - 126 7 3 2 1 = 2 0 0 3

Multiplying a Matrix by a Number

Definition 2

The product of the matrix A by the number k is the matrix B \u003d A k of the same size, which is obtained from the original by multiplying by a given number of all its elements:

b i , j = k × a i , j

Properties of multiplying a matrix by a number:

• 1 × A = A
• 0 × A = zero matrix
• k(A + B) = kA + kB
• (k + n) A = k A + n A
• (k×n)×A = k(n×A)

Example 4

Find the product of the matrix A \u003d 4 2 9 0 by 5.

5 A = 5 4 2 9 0 5 × 4 5 × 2 5 × 9 5 × 0 = 20 10 45 0

Multiplication of a matrix by a vector

Definition 3

To find the product of a matrix and a vector, you need to multiply according to the row-by-column rule:

• if you multiply a matrix by a column vector, the number of columns in the matrix must match the number of rows in the column vector;
• the result of multiplication of a column vector is only a column vector:

A B = a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋯ ⋯ ⋯ ⋯ a m 1 a m 2 ⋯ a m n b 1 b 2 ⋯ b 1 n = a 11 × b 1 + a 12 × b 2 + ⋯ + a 1 n × b n a 21 × b 1 + a 22 × b 2 + ⋯ + a 2 n × b n ⋯ ⋯ ⋯ ⋯ a m 1 × b 1 + a m 2 × b 2 + ⋯ + a m n × b n = c 1 c 2 ⋯ c 1 m

• if you multiply a matrix by a row vector, then the matrix to be multiplied must be exclusively a column vector, and the number of columns must match the number of columns in the row vector:

A B = a a ⋯ a b b ⋯ b = a 1 × b 1 a 1 × b 2 ⋯ a 1 × b n a 2 × b 1 a 2 × b 2 ⋯ a 2 × b n ⋯ ⋯ ⋯ ⋯ a n × b 1 a n × b 2 ⋯ a n × b n = c 11 c 12 ⋯ c 1 n c 21 c 22 ⋯ c 2 n ⋯ ⋯ ⋯ ⋯ c n 1 c n 2 ⋯ c n n

Example 5

Find the product of matrix A and column vector B:

A B \u003d 2 4 0 - 2 1 3 - 1 0 1 1 2 - 1 \u003d 2 × 1 + 4 × 2 + 0 × (- 1) - 2 × 1 + 1 × 2 + 3 × (- 1) - 1 × 1 + 0 × 2 + 1 × (- 1) = 2 + 8 + 0 - 2 + 2 - 3 - 1 + 0 - 1 = 10 - 3 - 2

Example 6

Find the product of matrix A and row vector B:

A \u003d 3 2 0 - 1, B \u003d - 1 1 0 2

A B = 3 2 0 1 × - 1 1 0 2 = 3 × (- 1) 3 × 1 3 × 0 3 × 2 2 × (- 1) 2 × 1 2 × 0 2 × 2 0 × (- 1) 0 × 1 0 × 0 0 × 2 1 × (- 1) 1 × 1 1 × 0 1 × 2 = - 3 3 0 6 - 2 2 0 4 0 0 0 0 - 1 1 0 2

Answer: A B \u003d - 3 3 0 6 - 2 2 0 4 0 0 0 0 - 1 1 0 2

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

Each vector can be viewed as a one-column or one-row matrix. A one-column matrix will be called a column vector, and a one-row matrix will be called a row vector.

If A is a matrix of size m*n, then the column vector b has size n, and the row vector b has size m.

Thus, in order to multiply a matrix by a vector, one must treat the vector as a column vector. When multiplying a vector by a matrix, it must be treated as a row vector.

multiply matrix

to the complex vector

We get the result

As you can see, with the dimension of the vector unchanged, we can have two solutions.

I would like to draw your attention to the fact that the matrix in the first and second versions, despite the same values, is completely different (they have different dimensions)

In the first case, the vector is considered as a column and then it is necessary multiply matrix by vector, and in the second case we have a row vector and then we have the product of a vector and a matrix.

This bot also multiplies vectors and matrices that have complex values. Based on a more complete calculator Multiplication of Matrices with Complex Values ​​Online

Properties of matrix-vector multiplication

Matrix

Vector column

Row vector

Arbitrary number

1. The product of a matrix by the sum of the column vectors is equal to the sum of the products of the matrix by each of the vectors

2. The product of the sum of row vectors by the matrix is ​​equal to the sum of the products of vectors by the matrix

3. The common factor of a vector can be taken out of the product of a matrix by a vector / a vector by a matrix

4. The product of a row vector by the product of a matrix and a column vector is equivalent to the product of the product of a row vector by a matrix and a column vector.