Raising a matrix to a power. Calculating the results of matrix expressions. Finding the inverse of a matrix Degrees of a square matrix

In July 2020, NASA launches an expedition to Mars. The spacecraft will deliver to Mars an electronic medium with the names of all registered expedition participants.

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One of these code options needs to be copied and pasted into the code of your web page, preferably between tags and or immediately after the tag. According to the first option, MathJax loads faster and slows down the page less. But the second option automatically monitors and loads the latest versions of MathJax. If you insert the first code, it will need to be updated periodically. If you insert the second code, the pages will load more slowly, but you will not need to constantly monitor MathJax updates.

The easiest way to connect MathJax is in Blogger or WordPress: in the site control panel, add a widget designed to insert third-party JavaScript code, copy the first or second version of the download code presented above into it, and place the widget closer to the beginning of the template (by the way, this is not at all necessary , since the MathJax script is loaded asynchronously). That's all. Now learn the markup syntax of MathML, LaTeX, and ASCIIMathML, and you are ready to insert mathematical formulas into your site's web pages.

Another New Year's Eve... frosty weather and snowflakes on the window glass... All this prompted me to write again about... fractals, and what Wolfram Alpha knows about it. There is an interesting article on this subject, which contains examples of two-dimensional fractal structures. Here we will look at more complex examples of three-dimensional fractals.

A fractal can be visually represented (described) as a geometric figure or body (meaning that both are a set, in this case, a set of points), the details of which have the same shape as the original figure itself. That is, this is a self-similar structure, examining the details of which when magnified, we will see the same shape as without magnification. Whereas in the case of an ordinary geometric figure (not a fractal), upon magnification we will see details that have a simpler shape than the original figure itself. For example, at a high enough magnification, part of an ellipse looks like a straight line segment. This does not happen with fractals: with any increase in them, we will again see the same complex shape, which will be repeated again and again with each increase.

Benoit Mandelbrot, the founder of the science of fractals, wrote in his article Fractals and Art in the Name of Science: “Fractals are geometric shapes that are as complex in their details as in their overall form. That is, if part of the fractal will be enlarged to the size of the whole, it will appear as a whole, either exactly, or perhaps with a slight deformation."

Matrix A -1 is called the inverse matrix with respect to matrix A if A*A -1 = E, where E is the identity matrix of the nth order. An inverse matrix can only exist for square matrices.

Purpose of the service. Using this service online you can find algebraic complements, transposed matrix A T, allied matrix and inverse matrix. The decision is carried out directly on the website (online) and is free. The calculation results are presented in a report in Word and Excel format (i.e., it is possible to check the solution). see design example.

Instructions. To obtain a solution, it is necessary to specify the dimension of the matrix. Next, fill out matrix A in the new dialog box.

See also Inverse matrix using the Jordano-Gauss method

Algorithm for finding the inverse matrix

Finding the transposed matrix A T .

Definition of algebraic complements. Replace each element of the matrix with its algebraic complement.

Compiling an inverse matrix from algebraic additions: each element of the resulting matrix is ​​divided by the determinant of the original matrix. The resulting matrix is ​​the inverse of the original matrix.

Next algorithm for finding the inverse matrix similar to the previous one except for some steps: first the algebraic complements are calculated, and then the allied matrix C is determined.

Determine whether the matrix is ​​square. If not, then there is no inverse matrix for it.

Calculation of the determinant of the matrix A. If it is not equal to zero, we continue the solution, otherwise the inverse matrix does not exist.

Definition of algebraic complements.

Filling out the union (mutual, adjoint) matrix C .

Compiling an inverse matrix from algebraic additions: each element of the adjoint matrix C is divided by the determinant of the original matrix. The resulting matrix is ​​the inverse of the original matrix.

They do a check: they multiply the original and the resulting matrices. The result should be an identity matrix.

Example No. 1. Let's write the matrix in the form:

Algebraic additions. ∆ 1.2 = -(2·4-(-2·(-2))) = -4 ∆ 2.1 = -(2 4-5 3) = 7 ∆ 2.3 = -(-1 5-(-2 2)) = 1 ∆ 3.2 = -(-1·(-2)-2·3) = 4

A -1 =

0,6 -0,4 0,8 0,7 0,2 0,1 -0,1 0,4 -0,3

Another algorithm for finding the inverse matrix We present another scheme for finding the inverse matrix.

Find the determinant of a given square matrix A.

We find algebraic complements to all elements of the matrix A.

We write algebraic additions of row elements to columns (transposition).

We divide each element of the resulting matrix by the determinant of the matrix A.

As we see, the transposition operation can be applied both at the beginning, on the original matrix, and at the end, on the resulting algebraic additions.

A special case: The inverse of the identity matrix E is the identity matrix E.

Here we will continue the topic of operations on matrices started in the first part and look at a couple of examples in which several operations will need to be applied at once.

Raising a matrix to a power.

Let k be a non-negative integer. For any square matrix $A_(n\times n)$ we have: $$ A^k=\underbrace(A\cdot A\cdot \ldots \cdot A)_(k \; times) $$

In this case, we assume that $A^0=E$, where $E$ is the identity matrix of the corresponding order.

Example No. 4

Given a matrix $ A=\left(\begin(array) (cc) 1 & 2 \\ -1 & -3 \end(array) \right)$. Find matrices $A^2$ and $A^6$.

According to the definition, $A^2=A\cdot A$, i.e. to find $A^2$ we just need to multiply the matrix $A$ by itself. The operation of matrix multiplication was discussed in the first part of the topic, so here we will simply write down the solution process without detailed explanations:

$$ A^2=A\cdot A=\left(\begin(array) (cc) 1 & 2 \\ -1 & -3 \end(array) \right)\cdot \left(\begin(array) (cc) 1 & 2 \\ -1 & -3 \end(array) \right)= \left(\begin(array) (cc) 1\cdot 1+2\cdot (-1) & 1\cdot 2 +2\cdot (-3) \\ -1\cdot 1+(-3)\cdot (-1) & -1\cdot 2+(-3)\cdot (-3) \end(array) \right )= \left(\begin(array) (cc) -1 & -4 \\ 2 & 7 \end(array) \right). $$

To find the matrix $A^6$ we have two options. Option one: it’s trivial to continue multiplying $A^2$ by the matrix $A$:

$$ A^6=A^2\cdot A\cdot A\cdot A\cdot A. $$

However, you can take a slightly simpler route, using the associativity property of matrix multiplication. Let's place parentheses in the expression for $A^6$:

$$ A^6=A^2\cdot A\cdot A\cdot A\cdot A=A^2\cdot (A\cdot A)\cdot (A\cdot A)=A^2\cdot A^2 \cdot A^2. $$

If solving the first method would require four multiplication operations, then the second method would require only two. Therefore, let's go the second way:

$$ A^6=A^2\cdot A^2\cdot A^2=\left(\begin(array) (cc) -1 & -4 \\ 2 & 7 \end(array) \right)\ cdot \left(\begin(array) (cc) -1 & -4 \\ 2 & 7 \end(array) \right)\cdot \left(\begin(array) (cc) -1 & -4 \\ 2 & 7 \end(array) \right)=\\= \left(\begin(array) (cc) -1\cdot (-1)+(-4)\cdot 2 & -1\cdot (-4 )+(-4)\cdot 7 \\ 2\cdot (-1)+7\cdot 2 & 2\cdot (-4)+7\cdot 7 \end(array) \right)\cdot \left(\ begin(array) (cc) -1 & -4 \\ 2 & 7 \end(array) \right)= \left(\begin(array) (cc) -7 & -24 \\ 12 & 41 \end( array) \right)\cdot \left(\begin(array) (cc) -1 & -4 \\ 2 & 7 \end(array) \right)=\\= \left(\begin(array) (cc ) -7\cdot(-1)+(-24)\cdot 2 & -7\cdot (-4)+(-24)\cdot 7 \\ 12\cdot (-1)+41\cdot 2 & 12 \cdot (-4)+41\cdot 7 \end(array) \right)= \left(\begin(array) (cc) -41 & -140 \\ 70 & 239 \end(array) \right). $$

Answer: $A^2=\left(\begin(array) (cc) -1 & -4 \\ 2 & 7 \end(array) \right)$, $A^6=\left(\begin(array) (cc) -41 & -140 \\ 70 & 239 \end(array) \right)$.

Example No. 5

Given matrices $ A=\left(\begin(array) (cccc) 1 & 0 & -1 & 2 \\ 3 & -2 & 5 & 0 \\ -1 & 4 & -3 & 6 \end(array) \right)$, $ B=\left(\begin(array) (ccc) -9 & 1 & 0 \\ 2 & -1 & 4 \\ 0 & -2 & 3 \\ 1 & 5 & 0 \end (array) \right)$, $ C=\left(\begin(array) (ccc) -5 & -20 & 13 \\ 10 & 12 & 9 \\ 3 & -15 & 8 \end(array) \ right)$. Find the matrix $D=2AB-3C^T+7E$.

We begin calculating the matrix $D$ by finding the result of the product $AB$. Matrices $A$ and $B$ can be multiplied, since the number of columns of matrix $A$ is equal to the number of rows of matrix $B$. Let's denote $F=AB$. In this case, the matrix $F$ will have three columns and three rows, i.e. will be square (if this conclusion does not seem obvious, see the description of matrix multiplication in the first part of this topic). Let's find the matrix $F$ by calculating all its elements:

$$ F=A\cdot B=\left(\begin(array) (cccc) 1 & 0 & -1 & 2 \\ 3 & -2 & 5 & 0 \\ -1 & 4 & -3 & 6 \ end(array) \right)\cdot \left(\begin(array) (ccc) -9 & 1 & 0 \\ 2 & -1 & 4 \\ 0 & -2 & 3 \\ 1 & 5 & 0 \ end(array) \right)\\ \begin(aligned) & f_(11)=1\cdot (-9)+0\cdot 2+(-1)\cdot 0+2\cdot 1=-7; \\ & f_(12)=1\cdot 1+0\cdot (-1)+(-1)\cdot (-2)+2\cdot 5=13; \\ & f_(13)=1\cdot 0+0\cdot 4+(-1)\cdot 3+2\cdot 0=-3;\\ \\ & f_(21)=3\cdot (-9 )+(-2)\cdot 2+5\cdot 0+0\cdot 1=-31;\\ & f_(22)=3\cdot 1+(-2)\cdot (-1)+5\cdot (-2)+0\cdot 5=-5;\\ & f_(23)=3\cdot 0+(-2)\cdot 4+5\cdot 3+0\cdot 0=7;\\ \\ & f_(31)=-1\cdot (-9)+4\cdot 2+(-3)\cdot 0+6\cdot 1=23; \\ & f_(32)=-1\cdot 1+4\cdot (-1)+(-3)\cdot (-2)+6\cdot 5=31;\\ & f_(33)=-1 \cdot 0+4\cdot 4+(-3)\cdot 3+6\cdot 0=7. \end(aligned) $$

So $F=\left(\begin(array) (ccc) -7 & 13 & -3 \\ -31 & -5 & 7 \\ 23 & 31 & 7 \end(array) \right)$. Let's go further. Matrix $C^T$ is the transposed matrix for matrix $C$, i.e. $ C^T=\left(\begin(array) (ccc) -5 & 10 & 3 \\ -20 & 12 & -15 \\ 13 & 9 & 8 \end(array) \right) $. As for the matrix $E$, it is the identity matrix. In this case, the order of this matrix is ​​three, i.e. $E=\left(\begin(array) (ccc) 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end(array) \right)$.

In principle, we can continue to go step by step, but it is better to consider the remaining expression as a whole, without being distracted by auxiliary actions. In fact, we are left with only the operations of multiplying matrices by a number, as well as the operations of addition and subtraction.

$$ D=2AB-3C^T+7E=2\cdot \left(\begin(array) (ccc) -7 & 13 & -3 \\ -31 & -5 & 7 \\ 23 & 31 & 7 \ end(array) \right)-3\cdot \left(\begin(array) (ccc) -5 & 10 & 3 \\ -20 & 12 & -15 \\ 13 & 9 & 8 \end(array) \ right)+7\cdot \left(\begin(array) (ccc) 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end(array) \right) $$

Let's multiply the matrices on the right side of the equality by the corresponding numbers (i.e. by 2, 3 and 7):

$$ 2\cdot \left(\begin(array) (ccc) -7 & 13 & -3 \\ -31 & -5 & 7 \\ 23 & 31 & 7 \end(array) \right)-3\ cdot \left(\begin(array) (ccc) -5 & 10 & 3 \\ -20 & 12 & -15 \\ 13 & 9 & 8 \end(array) \right)+7\cdot \left(\ begin(array) (ccc) 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end(array) \right)=\\= \left(\begin(array) (ccc) - 14 & 26 & -6 \\ -62 & -10 & 14 \\ 46 & 62 & 14 \end(array) \right)-\left(\begin(array) (ccc) -15 & 13 & 9 \\ -60 & 36 & -45 \\ 39 & 27 & 24 \end(array) \right)+\left(\begin(array) (ccc) 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end(array) \right) $$

Let's perform the last steps: subtraction and addition:

$$ \left(\begin(array) (ccc) -14 & 26 & -6 \\ -62 & -10 & 14 \\ 46 & 62 & 14 \end(array) \right)-\left(\begin (array) (ccc) -15 & 30 & 9 \\ -60 & 36 & -45 \\ 39 & 27 & 24 \end(array) \right)+\left(\begin(array) (ccc) 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end(array) \right)=\\ =\left(\begin(array) (ccc) -14-(-15)+7 & 26-30+0 & -6-9+0 \\ -62-(-60)+0 & -10-36+7 & 14-(-45)+0 \\ 46-39+0 & 62-27 +0 & 14-24+7 \end(array) \right)= \left(\begin(array) (ccc) 8 & -4 & -15 \\ -2 & -39 & 59 \\ 7 & 35 & -3 \end(array) \right). $$

Problem solved, $D=\left(\begin(array) (ccc) 8 & -4 & -15 \\ -2 & -39 & 59 \\ 7 & 35 & -3 \end(array) \right)$ .

Answer: $D=\left(\begin(array) (ccc) 8 & -4 & -15 \\ -2 & -39 & 59 \\ 7 & 35 & -3 \end(array) \right)$.

Example No. 6

Let $f(x)=2x^2+3x-9$ and matrix $ A=\left(\begin(array) (cc) -3 & 1 \\ 5 & 0 \end(array) \right) $. Find the value of $f(A)$.

If $f(x)=2x^2+3x-9$, then $f(A)$ is understood as the matrix:

$$ f(A)=2A^2+3A-9E. $$

This is how a polynomial from a matrix is ​​defined. So, we need to substitute the matrix $A$ into the expression for $f(A)$ and get the result. Since all the actions were discussed in detail earlier, here I will simply give the solution. If the process of performing the operation $A^2=A\cdot A$ is unclear to you, then I advise you to look at the description of matrix multiplication in the first part of this topic.

$$ f(A)=2A^2+3A-9E=2A\cdot A+3A-9E=2 \left(\begin(array) (cc) -3 & 1 \\ 5 & 0 \end(array) \right)\cdot \left(\begin(array) (cc) -3 & 1 \\ 5 & 0 \end(array) \right)+3 \left(\begin(array) (cc) -3 & 1 \\ 5 & 0 \end(array) \right)-9\left(\begin(array) (cc) 1 & 0 \\ 0 & 1 \end(array) \right)=\\ =2 \left( \begin(array) (cc) (-3)\cdot(-3)+1\cdot 5 & (-3)\cdot 1+1\cdot 0 \\ 5\cdot(-3)+0\cdot 5 & 5\cdot 1+0\cdot 0 \end(array) \right)+3 \left(\begin(array) (cc) -3 & 1 \\ 5 & 0 \end(array) \right)-9 \left(\begin(array) (cc) 1 & 0 \\ 0 & 1 \end(array) \right)=\\ =2 \left(\begin(array) (cc) 14 & -3 \\ - 15 & 5 \end(array) \right)+3 \left(\begin(array) (cc) -3 & 1 \\ 5 & 0 \end(array) \right)-9\left(\begin(array ) (cc) 1 & 0 \\ 0 & 1 \end(array) \right) =\left(\begin(array) (cc) 28 & -6 \\ -30 & 10 \end(array) \right) +\left(\begin(array) (cc) -9 & 3 \\ 15 & 0 \end(array) \right)-\left(\begin(array) (cc) 9 & 0 \\ 0 & 9 \ end(array) \right)=\left(\begin(array) (cc) 10 & -3 \\ -15 & 1 \end(array) \right). $$

Answer: $f(A)=\left(\begin(array) (cc) 10 & -3 \\ -15 & 1 \end(array) \right)$.