vyznachnikiv n GO order
1. The method of reduction to a tricot look.
a) Calculate the vyznachnik:.
Vіdnіmayuchi first row z usіh іnshih, otrimuєmo vyznachnik, which may have a tricot look and, also, dobutku diagonal elements:
. In a pouch D n = (–1)n –1 .
b) Calculate the winner: 
.
Vіdnіmaєmo the first row z reshti, and then, zі stovptsіv vyznachnik vinosim: from the first A 1 – X; from another A 2 – X; …..; h n th a n – X. We take:
D = (a 1 – x) (a 2 - x)… (a n–x) 
.
Let's write down the first element of the first column in the view: = 1 + і all the columns of the removed signifier, add to the first column. We take away the sign of a tricot look, which is a more expensive way to finish the diagonal elements. Father:
D = (a 1 – x) (a 2 – x)…(a n – x)x + + + … + .
2. The method of seeing linear multipliers.
a) Calculate the forerunner.
1. Adding to the first column, there are three more, it seems that in the first column there is a common multiplier, which is more expensive X + at + z. Otzhe, the tyrant subscribes to X + at + z.
2. Similarly, adding to the first column, the other one, and considering the third and fourth columns, it is clear that the signifier is divided into x - y – z.
3. If the first step is folded with the third and the other and the fourth, then we take into account that the sign is divided into x - y + z.
4. If you add a quarter to the first column and see another and third column, then it’s obvious that the signifier is a multiplier x - y + z. Father:
It is clear that the vyznachnik is a rich member of the 4th degree x, behind y i by z. On the right is also a rich term of the same step. Tom V= Const. The vyznachnik x 4 enter at dodanok:
a 12 a 21 a 34 a 43 = (-1) 2× X× X× X× X = X 4 .
In the right part, the senior member X: Vx 4, tobto. V= 1. Take the result:
= (x + y + z)(x – y – z)(x – y + z)(x + y – z) = x 4 + y 4 + z 4 – 2x 2 y 2 – 2x 2 z 2 – 2at 2 z 2 .
b) Calculate the winner n th order: 
.
This vyznachnik is called Vandermonde's vyznachnik. Staring at Yogo like a rich cock ( n-1) th step shodo x n it’s better that vin turns to 0 at x n = x 1, x n= x 2, … x n = x n- 1 . Todi D n = a n – 1 (x n – x 1)(x n – x 2) … (x n– x n–1), moreover a n –1 = = D n-1. Repeating this procedure, we take: D n = (x 2 – x 1)(x 3 – x 2)(x 3 – x 1)(x 4 – x 3)(x 4 – x 2)(x 4 – –x 1)… = .
3. The method of submitting a vyznachnik is like a sumi of vyznachniks.
Calculate the scapegoat: 
.
Having remembered that the elements of the first column are presented as the sum of two numbers, we will lay out the signifier for the sum of two signifiers:
.
Now the skins from the otrimanih vyznachnikiv are laid out in the bag of two vyznachnikiv, scurrying along, that the elements of other stovptsіv are presented to them in the sight of sums, and so on. Zrobivshi tse, take it away ( n> 2), so that the rows of the selected candidates will be as follows: a i, a i, … , a i or b 1, b 2, …, bn. Rows of the 1st type are proportional, of the 2nd type they are equal and, therefore, all the additions are equal to zero. Father: D n = 0 ("n > 2).
For vyznachniki of the same type, but first and another order is taken:
D 1 = | a 1 +b 1 | = a 1 +b 1 ; D 2 = =
= a 1 b 2 –a 2 b 2 +b 1 a 2 – a 1 b 1 = (a 1 – a 2)b 2 + (a 2 + a 1)b 1 = (a 1 – a 2)(b 2 – b 1).
The method of recurrent (turning) spins.
Calculate the winner n th order: 
.
Laying out the leader behind the elements of the first row, we take recurring spiving: D n= 
.
Having announced the signer at the right part of the speech on the first column, we will write down a new recurring speech: D n= 5D n –1 – 6D n –2 .
Giving tse spіvvіdnoshnja v vyglyadі: D n– 2D n –1 = 3(D n –1 – 2D n-2) and introductions of recognition:
T n= D n– 2D n-1 We take: T n= 3T n –1 – 3 2 T n-2 = ... = 3 n-2 T 2 = 3n.
Similarly, having written down the recursive spіvvіdnoshennia at the look: D n– 3D n –1 = 2(D n –1 – 3D n-2) that means: V n= D n– 3D n-1 Removable V n= 2V n = 1 = 2 2 V n –2 =…= 2n .
Vidpovіd: POWER 1. The value of the signifier does not change, so replace all rows with columns, moreover, replace the skin row with a column with the same number, so
AUTHORITY 2. Permutation of two columns or two rows of the signifier will increase the multiplier of yogo by -1. For example,
.POWER 3. If the leader can have two identical columns, or two identical rows, then wine equals zero. For example,
.POWER 5. If all the elements of a given row or row are equal to zero, then the principal itself is equal to zero. Tsya dominion є let's call it a forward slope (at k = 0). the sum of two dodankіv, then the clerk can have an idea if he sees the sum of two vouchers, of whom one n-th column otherwise nth row may first of the guessing dodankiv, and the last - others; elements, like to stand in other missions, in all three of the saints in one themselves. For example,
POWER 8. As far as the elements of the last column (or the last row), add the additional elements of the second column (or the second row), multiplied by any kind of scorching multiplier, the value of the signifier will not change. For example,
.
Away from the power of the vyznachnikіv pov'yazanі z ponyazanі z dodatty dodatku algebra and minor. The minor of the deacogo element is the name of the signer, which is taken from the given way to win the row and stand, on the retina of such rotashavanie this element. є the number of the couple and the return sign, which is the number of the unpair.
a good sum of creations of elements, whether there be any combination (of a row) with their own algebraic additions.
Vyznachnik. This is a rich term that combines the elements of a square matrix in such a rank that its value is taken when transposing and linear combinations of rows and columns. Tobto, vyznachnik characterizes the change of the matrix. Zokrema, which in the matrix is linear-deposit rows or stovpts, - the signpost is closer to zero. to a wild vapadka the matrix can be assigned over some kind of commutative ring, in which case the signifier will be an element of the same ring. The signifier of the matrix A is designated as: or Δ(A).
5.virogene matrix. the matrix is wrapped, її power, calculation, the theorem of reasoning.
Note: Virogenous, singular (singular) matrix is the square matrix A, which means that the primordial (Δ) is equal to zero. In another way, matrix A is called non-virgin.
Let's look at the problem of assigning an operation that reverses to multiplying matrices.
Come on - square matrix order. Matrix that is satisfied at once from the given matrix of equality:
It's called a badass. The matrix is called reversible, as for it it is truly reversible, otherwise it is non-reversible.
Z vznachennya slid, scho th reversal matrix іsnuє, won square ієї w order, yak i . However, for a square matrix, it is not true. If the matrix is equal to zero, then there is no turning point. In fact, zastosovuyuchi the theorem about the identity of the matrix for the single matrix, it is necessary to eliminate
Shards of the sign of a single matrix are more expensive serum matrix. It is guessed that a square matrix, which is equal to zero, is called a virogen (special), otherwise - a non-virogen (non-special).
Theorem 4.1 about the basis and unity of the pivot matrix. A square matrix, a signifier that looks like zero, can have a reverse matrix and only one before:
| (4.1) |
de - matrix, transposed for a matrix, folded from additives in the algebra of matrix elements.
The matrix is called the adjoining matrix by reference to the matrix.
Really, the matrix is mind-blowing. Slide to show that it is a return to, tobto. satisfies two minds:
Let's bring the first jealousy. Vіdpovіdno up to item 4 for respect 2.3, from the authority of the chief vyplyvaє, scho. Tom
what it was necessary to show. Similarly, another equanimity is brought. Otzhe, for the mind, the matrix may return
The unity of the pivotal matrix can be brought to the fore in the form of the opposite. Come on, cream of the matrix, there is one more turning matrix 
so what. Multiplying the hurt parts of the jealousy of the evil on the matrix 
. Zvіdsi, scho superchit pripuschenu. Otzhe, wrapped matrix єdina.
Respect 4.1
1. The choice is clear that the matrices and permutations.
2. Matrix, reversed to non-virginated diagonal, also diagonal:
3. Matrix wrapped to non-virginated lower (upper) tricot, є lower (upper) tricot.
4. Elementary matrices Mayut zvorotni, yakі є elementary (div. p. 1 for respect 1.11).
The dominance of the pivotal matrix
The operation about the matrix can have the same power:
As a rule, sensory operations are indicated at 1-4.
We bring power 2: in order to obtain non-degenerative square matrices of the same order, we have a rotation matrix, then .
To tell the truth, the primary source of matrices is not equal to zero, shards
Otzhe, vorotna matrix іsnuє єdina. It can be shown that the matrix is a reverse matrix. Diyno:
From the unity of the salutary matrix, equanimity is evident. Another power brought. Similarly, those other characteristics are reported.
Respect 4.2
1. For a complex matrix, equality is valid, similar to power 3:
De - the operation of obtaining matrices.
2. The operation of turning matrices allows you to select the number of negative steps of the matrix. For a non-virogeneous matrix, that of any natural number is significant 
.
6. systems of linear lines. Coefficients for non-domic, free members. Development of the system of linear lines. Splitting of the system of linear lines. The system of linear homogeneous equalities and singularities.
Verdict: The system of linear equalizations of algebra, which to revenge m equals and n nevіdomih, is called the system of mind
where the numbers a ij are called the coefficients of the system, the numbers b i are the free members. Calculate the value of the number xn.
Such a system can be written manually in a compact matrix form
Here A is the matrix of system coefficients, called the main matrix;
Vector stovpets z nevіdomih x j .
The vector-composition of all members b i .
Additional matrices A*X were assigned, scalars for matrix A stovptsіv style w, scalable rows for matrix X (n pieces).
The extended matrix of the system is the matrix A of the system, supplemented by the head of the free members
The solution of the system is called n value of the unknown x 1 =c 1 , x 2 =c 2 , ..., x n =c n , when substantiating such a system, the system is recreated on the correct balance. Whether or not the solution of the system can be written at the sight of the matrix-stovptsya
The system of equals is called coherent, as if it could not have one solution, and it was insane, because it could not have a solution.
A split system is called a singing one, because there is only one solution, that one is invisible, because there is more than one solution. At different times, the solution is called the private solution of the system. The succession of all other decisions is called the final decisions.
Virishiti the system - tse means z'yasuvati, spilna won't be foolish. How the system is coherent, know її blatant solution.
The two systems are called equivalent (equally strong), because the stench may be the very same solution. In other words, the systems are equivalent, as a skin solution to one of them, to other solutions, and to the contrary.
Equivalent systems appear, zokrema, with elementary transformations of the system for the mind, that transformations are no longer above the rows of the matrix.
The system of linear equalities is called homogeneous, because all the relevant terms equal to zero:
Homogeneous system is definitely coherent, shards x 1 = x 2 = x 3 = ... = x n = 0 є system solutions. The solution is called zero or trivial.
4.2. Virishennya systems of linear lines.
Kronecker-Capelli theorem
Let me give you a pretty system of n linear lines from n nevidomim
The Kronecker-Capelli theorem gives a good indication of the complexity of the system.
Theorem 4.1. The system of linear alignments of algebra is only the same, if only the rank of the extended matrix of the system is equal to the rank of the main matrix.
Acceptable without confirmation.
The rules of practical analysis of all solutions of the joint system of linear alignments are exuded from the advancing theorems.
Theorem 4.2. Since the rank of the joint system is higher than the number of non-doms, the system can only be solved.
Theorem 4.3. If the rank of the joint system is smaller for the number of non-doms, then the system may be an impersonal solution.
The rule of rozv'yazannya dovіlnoї system of linear lines
1. Find the ranks of the main and extended matrices of the system. Like r(A)≠r(A), the system is insane.
2. If r(A)=r(A)=r, the system is double. Find out what the basic minor is in the order of r (guessing: minor, the order of which determines the rank of the matrix, is called basic). Take r rivnyan, from the coefficients of which the basic minor is folded (inshі rіvnyannya vіdkinuti). It is not known, the coefficients of which are included to the basic minor, are called the head ones, and they deprive them of evil, and reshta n-r unknown call it vіlnimi and transfer it to the right part of equals.
3. To know the way of head nevіdomikh through vіlnі. Otrimano zagalne solution of the system.
4. Nadayuchi vіlnim nevіdomim dovіlnі znachennya, otrimaєmo vіdpovіdnі vіdnі vіdnі vіdnі navіdny neіdomih. In this rank, you can know the private solution of the outward system of the rivers.
Example 4.1.
4.3 Development of non-virogeneous linear systems. Cramer formulas
Let me give you a system of n linear lines from n unknowns
(4.1)
chi matrix form A * X \u003d B.
The basic matrix of such a system is square. Significant of the matrix
called the arbiter of the system. If the primate of the system is known to be zero, then the system is called non-virgin.
We know the breakdown of this system of equalities at times D¹0
Multiplying the offending parts of the equation A * X \u003d Y levoruch by the matrix A -1 is taken away
A -1 *A*X=A -1 *B Oscilki. A -1 *A=E i E*X=X, then
The solution of the system behind the formula (4.1) is called the matrix method of the solution of the system.
Matrix equality (4.1) can be written in terms of
See what follows
Ale є rozladannya vyznachnik
for the elements of the first column. Significant D 1 to exit from sign D by replacing the first column of coefficients with the leader of the free members. Otzhe,
Similarly:
de D2 subtracting D by way of replacing another stomptsіentsіv stovptsієntіv svіnіh sіmіnіv:
are called Cramer's formulas.
Also, the system of n linear alignments from n cannot be invariably found in a single solution, which can be found by the matrix method (4.1) or by Cramer's formulas (4.2).
Example 4.3.
4.4 Variation of systems of linear alignments by the Gaussian method
One of the most universal effective methods the solution of linear algebraic systems is the Gauss method, which works in the sequence of inclusions.
Let the system of equals be given
Solution process by the Gauss method and two stages. At the first stage (straight hіd), the system is directed to a stepped (zokrema, trikutny) form.
The system is shown below
The coefficients aii are called the main elements of the system.
At another stage (turnaround) and then later, the appointment of nevіdomih from the tsієї stepіnchastoy system.
We will describe the Gauss method in a report.
Let's remake the system (4.3) by including some x1 in all peers, crimson first (wicory elementary reworking of the system). For whom, we multiply the insults of the first equal parts and store it term by term with other equals of the system. Let's multiply the insults of the first part of the equal and store the third equal of the system. Continuing this process, we take an equivalent system
Here - new values of the coefficients and the right parts, as they appear after the first crochet.
An analogous rank, vvazhayuchi head element, including nevidome x 2 of all equals of the system, crim first and the other, and so on. We continue this process, as long as it is possible.
So in the process of reducing the system (4.3) to a step-wise form, zero equality is declared, so that the equality of the form 0 = 0, їх is given. 
that is to say about the inconsistency of the system.
The other stage (turning head) is near the top of the step system. Step-by-step system is equal, vzagali seeming, may be impersonal solution, In the remaining equal part of the system, it can be seen first unknown x k through other unknown (x k + 1, ..., x n). Let's introduce the value of x k before getting the system equal and turn x k-1 through (x k + 1, ..., x n). , Then we know x k-2, ..., x 1. . We hope we don't (x k + 1, ... x n). sufficient value, we take away the impersonal solution of the system.
Respect:
1. Since the step system appears to be triangular, so k = n, then the outer system can only have one solution. From the rest of the equal we know x n from the rest of the equal x n-1 , let’s go uphill with the system, we know the rest of the unknown (x n-1 ,...,x 1).
2. In practice, it is more convenient to practice not with the system (4.3), but with an expanded matrix, concentrating on all elementary transformations over її rows. It’s handy, so that the coefficient a 11 is added to 1 (replace the equalities with the points, or divide the offending parts of the equal by a 11 ¹1).
Example 4.4.
Solution: As a result of elementary transformations over the expanded matrix of the system
the output system was brought up to the stage:
That is why the solution of the system is: x 2 \u003d 5x 4 -13x 3 -3; x 1 \u003d 5x 4 -8x 3 -1 If we put, for example, x 3 \u003d 0,x 4 \u003d 0, then we know one of the private solutions of the system 1 = -1, x2 = -3, x3 = 0, x4 = 0.
butt 4.5.
Check the system using the Gauss method:
Solution: More elementary transformation over the rows of the expanded matrix system:
Otriman's matrix in vіdpovidає systems
Zdiysnyuyuchi zvorotny hіd, we know x 3 = 1, x 2 = 1, x 1 = 1.
4.5 Systems of linear uniform alignments
Let me give you a system of linear uniform rivnyan
It is obvious that a homogeneous system is always coherent, there can be a zero (trivial) solution x 1 =x 2 =x 3 =...=x n =0.
For what kind of minds is a homogeneous system possible and non-zero solutions?
Theorem 4.4. In order for the system of homogeneous equalities to be small and non-zero, it is necessary and sufficient that the rank r її of the main matrix be smaller than the number n, that is, r Necessity. Since the rank cannot change the size of the matrix, then, obviously, r<=n. Пусть r=n. Тогда один из минеров размера nхn отличен от нуля. Поэтому соответствующаясистема линейных уравнений имеет единственное решение: Otzhe, others, there are no trivial decisions. Also, as a non-trivial solution, r Availability: Come on r Theorem 4.5. In order for a homogeneous system of n linear equalities with n non-zero solutions to be invariably small, it is necessary and sufficient that її the variant D equals zero, then D=0. If the system has a non-zero solution, D=0. For at D¹0 the system is less than one, a zero solution. If D=0, then the rank r of the main matrix of the system is less than the number of non-doms, that is. r Example 4.6. Check the system If x 3 \u003d 0, we take one solution: x 1 \u003d 0, x 2 \u003d 0, x 3 \u003d 0. If x 3 \u003d 1, we can take another private solution: x 1 \u003d 2, x 2 \u003d 3, x 3 \u003d 1 etc. AUTHORITY 1. The value of the signpost does not change, so replace all rows with columns, moreover, replace the skin row with a column with the same number, so AUTHORITY 2. Permutation of two columns or two rows of the signifier will increase the multiplier of yogo by -1. For example, POWER 3. If the leader can have two identical rows, or two identical rows, then wine equals zero. POWER 4. Reproduction of all elements of one stovptsya or one row of the vyznachnik on whether the number is more than the multiplication of the vyznachnik on the whole number k. For example, AUTHORITY 5. If all the elements of a given order are equal to zero, then the leader itself is equal to zero. Tsya dominion є let's call it a forward slope (at k=0). AUTHORITY 6. If there are two equal elements of two columns, or two rows of the proportional one, then the equal one is equal to zero. POWER 7. If the skin element of the n-th column or the n-th row of the vyznachnik is the sum of two additional ones, then the vyznachnik can have representations in the view of the sum of two vyznachniks, for which one of the n-th column, or in the n-th row may be the first zgadanih dodankiv, and others - others; elements, like to stand in other missions, in all three of the saints in one themselves. For example, POWER 8. As far as the elements of the last column (or the last row), add the additional elements of the second column (or the second row), multiplied by any kind of scorching multiplier, the value of the signifier will not change. For example, Away from the power of the vyznachnikіv pov'yazanі z ponyazanі z dodatty dodatku algebra and minor. The minor of the deacogo element is the name of the signifier, which is taken from the given way to win the row and the column, on the retina of such rotashavanie this element. An addendum of the algebra of whether there is an element of the sign of the prior minor of the element, taken with its own sign, as the sum of the numbers in the row and the column, on the retina of any combinations of the element, the number of the pair, and the return sign, as the number of the unpair. Algebraic supplementation of the element will be denoted by the great letter of the same name and by the same number, which is the letter, which designates the element itself. POWER 9. Vyznachnik a good sum of creations of elements, whether there be any combination (of a row) with their own algebraic additions. Otherwise, hanging on, mowing the space of such equivalence: 6) Minority and algebraic additions. Appointment. The minor element of the signifier is th order name vyznachnik- th order, which way out of this vyznachnik vykreslyuvannyam - th row i - st stovptsya, on the pereline of which there is an element. Designation: . Appointment. Algebraic additions to the element of the primate - in the first order are called yogo minor, taking zі with a plus sign, yakscho - a guy number і zі with a minus sign in a different turn. Designation: . Theorem. (About the arrangement of the vyznachnik.) The leader of the richest sum of creations of the elements of any row (or be it of some sort) of the leader on their additions to algebra: 7) The revival matrix- taka matrix A −1
, when multiplied by yak, the output matrix A give in result single matrix E: square matrix negotiable then and only then, if it is not virogene, then it is vyznachnik does not equal zero. For nonsquare matrices, virogen matrix there are no return matrices. However, it is possible to make it easier to understand and introduce pseudoreversal matrices, similar to zvorotnі for a wealth of power. 8)Matrix rank- best of order minors matrix values, which look like zero Call the rank of the matrix is indicated by () or . Offenses have come to us from foreign language, so offenses can take root. power Theorem (about the basic minor): Let r = rang AM be the basic minor of the matrix A, then: basic rows and basic columns are linearly independent; Whether a row (stovpchik) of matrix A is a linear combination of basic rows (stovptsiv). - Let the titmouse die! I remember, the class before the 8th me was not fit for algebra. I did not like it. Bіla won me. Because I don’t understand anything there. And then everything changed, to the fact that I punched one chip: In mathematics vzagali (and algebra of science) everything will be on a literate and subsequent system of appointment. Knowing the purpose, understanding their essence - it doesn’t matter to understand otherwise. Otak i z the topic of today's lesson. We will look in detail at the kіlka of the summіzhnyh power and vznacheni, for what reason, once and for all, you will be sorted out and with matrices, and with the chiefs, and with the strength of authority. Visioners are the central concept in matrix algebra. Similar to the formulas of short multiplication, they will re-examine you in the course of higher mathematics. To that it is readable, marveling at it thoroughly. :) And let’s look at the most important thing - what is a matrix? And it is right to practice from it. A matrix is a table filled with numbers. Neo is nothing here. One of the key indicators of the matrix is tse її rozmirnіst, that. number of rows and stovptsiv, from which there are folded. Sounds like the matrix $A$ can expand $\left[ m\times n \right]$, because it has $m$ rows and $n$ columns. Write it down like this: Abo axis like this: Buvayut and іnshi znachennya - here everything should be likened to a lecturer / seminarian / author of a teacher. Ale, in any case, z usima tsimi $ \ left [m \ times n \ right] $ і $ ((a)_ (ij)) $ vinikaє one and the same problem: What is the index for what vin? Back to the row number, then stovptsya? Abo navpak? When reading lectures, those assistants seem to be obvious. But if you are sleeping in front of you - it’s less than a sheet of zavdannya, you can overdo it and rapt get lost. So let's figure it out once and for all. For the cob, let's create a perfect coordinate system for a school mathematics course: Introduction of the coordinate system on the plane Do you remember? There is an ear of coordinates (point $O=\left(0;0 \right)$) of the $x$i $y$ axis, and a skin point on the plane is uniquely assigned to the coordinates: $A=\left(1;2 \ right )$, $B=\left(3;1 \right)$ and so on. And now let's take this construction and put it in order from the matrix so that the cob of coordinates is at the left upper fold. Why are you there? The one who opens the book, we begin to read the very top left corner of the side - it’s easier to remember it for the legend. Ale where to straighten the axis? Let's direct them in such a way that our entire virtual side is covered with these axes. It's true, for which one happens to rotate our coordinate system. The only possible variant of such a distribution: Now the skin cell of the matrix can have unique coordinates $x$ and $y$. For example, the notation $((a)_(24))$ means that we go to the element with coordinates $x=2$ and $y=4$. Matrix expansions are uniquely defined by a pair of numbers: Just wonder at this picture respectfully. Play around with the coordinates (especially if you work with the right matrices and arbitrators) - and soon you will understand what to learn from the most complex theorems and designations and you will miraculously understand about what to go. Did you get it? Well, let's move on to the first part of enlightenment - the geometrical designation of the vyznachnik. We would like to point out that the vector is used only for square matrices of the form $ \ left [n \ times n \ right] $. The arbitrator is the whole number, as it respects the singing rules and one of the characteristics of this matrix (є іnshі characteristics: rank, power vectors, but about tse in other lessons). Well, what is the characteristic? What does it mean? It's simple: The signifier of the square matrix $A=\left[ n\times n \right]$ is the whole of the $n$-world parallelepiped, which makes it possible to look at the rows of the matrix as a vector, which makes the edges of that parallelepiped. For example, the signifier of the 2x2 matrix is just the area of a parallelogram, and for the 3x3 matrix it is already a 3-dimensional parallelepiped - the same one that is so played by all high school students in stereometry lessons. At first glance, the appointment may seem completely inadequate. Ale, do not hurry with the visnovki - we look at the butts. Indeed, everything is elementary, Watson: Manager. Find the key matrices: \[\left| \begin(matrix) 1 & 0 \\ 0 & 3 \\\end(matrix) \right|\quad \left| \begin(matrix) 1 & -1 \\ 2 & 2 \\end(matrix) \right|\quad \left| \begin(matrix)2 & 0 & 0 \\ 1 & 3 & 0 \\ 1 & 1 & 4 \\\end(matrix) \right|\] Solution. The first two vyznachniki may rozmir 2x2. It's just the area of parallelograms. Let's pray for them and let's hurt the square. First impulse parallelogram on vectors $((v)_(1))=\left(1;0 \right)$ i $((v)_(2))=\left(0;3 \right)$: Obviously, not just a parallelogram, but a whole rectangle. Yogo area is healthy Another parallelogram of inducements on vectors $((v)_(1))=\left(1;-1 \right)$ i $((v)_(2))=\left(2;2 \right)$. Well, so what? Tse is also a rectangle: The sides of this rectangle (in essence, two vectors) are easily understood by the Pythagorean theorem: \[\begin(align) & \left| ((v)_(1)) \right|=\sqrt(((1)^(2))+((\left(-1 \right))^(2)))=\sqrt(2); \\ & \left| ((v)_(2)) \right|=\sqrt(((2)^(2))+((2)^(2)))=\sqrt(8)=2\sqrt(2); \&S=\left| ((v)_(1)) \right|\cdot \left| ((v)_(2)) \right|=\sqrt(2)\cdot 2\sqrt(2)=4. \\end(align)\] There was no more rozіbratisya with the remaining vyznachnik there is already a 3x3 matrix. Come and check the stereometry: Looking brainy, but after the fact, solve the formula for the volume of the parallelepiped: de $S$ - the area of the base (in our view, the area of the parallelogram on the plane $OXY$), $h$ - height drawn to the center of the base (in fact, the $z$-coordinate of the vector $((v)_(3) )$). The area of the parallelogram (we christened okremo) can be easily understood: \[\begin(align) & S=2\cdot 3=6; \&V=S\cdot h=6\cdot 4=24.\\end(align)\] From i all! We write down the opinions. Suggestion: 3; 4; 24. There is little respect for the system's designation. Who, sing-song, does not deserve that I ignore the "arrows" over vectors. Nibito so you can confuse the vector with a little bit or something else. But let's be serious: the boys and the girl have already grown up with you, it’s miraculously understandable from the context, if you’re talking about a vector, and if you’re talking about a point. Arrows are less likely to smear the roses, and so it is to put the string on the back with mathematical formulas. I more. In principle, nothing cares to look at the 1x1 matrix's signifier - such a matrix is just one key, and the number written in this cell will be the key. But here is an important respect: On the vіdmіnu vіd klаsichnogo vyagu, vyznachnik give us so titles orientations”, then. obyag z urakhuvannyam sledovnostі razglyadu vektorіv-ryadkіv. And if you want to take away from the classic sensible word, have a chance to take the module of the signifier, but at the same time, don’t worry about it - all the same, in a few seconds we learn to take into account whether it’s a signifier with some signs, rozmirs, etc.: ) With all the beauty of the sharpness of the geometrical approach, we may have a serious shortcoming: we can’t tell us anything about those who are the most important. Therefore, at once we will analyze an alternative designation - algebraic. For whom we will need a short theoretical training, then on the way out we will take a tool that allows you to enter in the matrices what is always worth it. Well, it's true, a new problem will appear there ... but about everything, garazd. Let's write in a row the numbers from 1 to $ n $. Weide to the type of chogo: Now (purely for fun) we remember a couple of numbers by months. You can change the courts: And you can - not necessary judges: Do you know what? But nothing! In algebra, this crap is called a permutation. I won't get a bunch of powers. Appointment. The permutation of $n$ is a row of $n$ different numbers written in any sequence. Start looking at the first $n$ of natural numbers (these are the numbers 1, 2, ..., $n$), and then mix them up to remove the necessary permutation. Permutations are designated as such, like vectors - just a letter and a subsequent rearrangement of their elements at the arms. For example: $p=\left(1;3;2 \right)$ or $p=\left(2;5;1;4;3 \right)$. Letter can be like that, but let it be $p$.:) For the sake of simplicity, we gave a practical work with permutations of the old 5 - the stench is already taken seriously to guard against any suspected effects, and even more cushioning for a fragile brain, like a permutation of the old 6 and more. Axis apply such permutations: \[\begin(align) & ((p)_(1))=\left(1;2;3;4;5 \right) \\ & ((p)_(2))=\left(1 ;3;2;5;4 \right) \\ & ((p)_(3))=\left(5;4;3;2;1 \right) \\\end(align)\] Naturally, the permutation of $n$ can be considered as a function, as it is assigned to the impersonal $\left\( 1;2;...;n \right\)$ and it can be effectively displayed as the impersonality on itself. Turning around to write down permutations $((p)_(1))$, $((p)_(2))$ and $((p)_(3))$, we can legally write: \[((p)_(1))\left(1 \right)=1;((p)_(2))\left(3 \right)=2;((p)_(3))\ left(2\right)=4;\] The number of different permutations of the old $n$ is always demarcated and added to $n!$ - this is a fact that can be easily brought up from combinatorics. For example, if we want to write down all the permutations of dozhini 5, then we’ll be tempted, the shards of such permutations will be One of the key characteristics of any permutation is the number of inversions in them. Appointment. Inversion in permutation $p=\left(((a)_(1));((a)_(2));...;((a)_(n)) \right)$ - come on pair $ \left(((a)_(i));((a)_(j)) \right)$ such that $i \lt j$, but $((a)_(i)) \ gt ( (a)_(j))$. Seemingly simpler, inversion - as long as the number of koshtuє livishe is smaller (not obov'yazkovo sudіdny). We can signify through $N\left(p \right)$ the number of inverses in permutations of $p$, but be prepared to learn about other meanings in different authors and in different authors - there are no single standards here. The topic of inversion is already great, and there will be a lot of assignments for a lesson. At the same time, our task is simply to learn how to respect them in real tasks. For example, let's consider the number of inversions in the permutation $p=\left(1;4;5;3;2 \right)$: \[\left(4;3 \right);\left(4;2 \right);\left(5;3 \right);\left(5;2 \right);\left(3;2 \right ); ).\] So, $ N \ left (p \ right) = $5. Like bachite, there’s nothing terrible for anyone. I’ll tell you once again: they’ll give us the same number $N\left(p \right)$, but the numbers of its parity/unparity. And here we smoothly pass to the key term of the daily lesson. Give a square matrix $ A = \ left [n \ times n \ right] $. Todi: Appointment. Significant matrix $ A = \ left [n \ times n \ right] $ - the whole algebraic sum of $ n! $ dodankіv, folded in the coming rank. Skin dodanok - the cost of $n$ elements of the matrix, taken one by one from the skin row and skin stump, multiplied by (-1) step number of inversions: \[\left| A \right|=\sum\limits_(n(((\left(-1 \right))^(N\left(p \right)))\cdot ((a)_(1;p\left(1 \right)))\cdot ((a)_(2;p\left(2 \right)))\cdot ...\cdot ((a)_(n;p\left(n \right))) )\]!} The fundamental point in choosing multiples for the skin fold in the fold is the fact that every other day two multiples do not stand in the same row, but in the same column. For some reason, without interfering with cohesion, it is possible to understand that the indexes $i$ of the multipliers $((a)_(i;j))$ "pass through" the values 1, ..., $n$, and the indexes $j$ є by a certain permutation in first: And if it's a permutation of $p$, we can easily invert $N\left(p \right)$ — and the god's folder is ready. Naturally, none of them hinder the minds of the multipliers in any kind of dodanka (otherwise, in all cases - why is it flickering?), And even the first indexes will be such a permutation. But as a result, nothing changes: the total number of inversions in the $i$ and $j$ indices takes parity with similar conditions, which follows the good old rule: The type of permutation of multipliers does not change numbers. The axis only does not need to apply the rule to the multiplication of matrices - to the input of the multiplication of numbers, it is not commutative. But then I got excited. Zagal can look at the 1x1 matrix - there will be one clitin, and її vyznachnik, no matter how you guess, more than the number written in this clitin. Nothing ticky. Let's look at a square matrix with a size of 2x2: \[\left[ \begin(matrix) ((a)_(11)) & ((a)_(12)) \\ ((a)_(21)) & ((a)_(22)) \\end(matrix)\right]\] Oskіlki kіlkіst rows at nіy $n=2$, then the arbitrator of the place $n!=2!=1\cdot 2=2$ dodankіv. We write їх: \[\begin(align) & ((\left(-1 \right))^(N\left(1;2 \right)))\cdot ((a)_(11))\cdot ((a) _(22))=((\left(-1 \right))^(0))\cdot ((a)_(11))\cdot ((a)_(22))=((a)_ (11)) ((a)_(22)); \\ & ((\left(-1 \right))^(N\left(2;1 \right)))\cdot ((a)_(12))\cdot ((a)_(21)) =((\left(-1 \right))^(1))\cdot ((a)_(12))\cdot ((a)_(21))=((a)_(12))( (a)_(21)). \\end(align)\] It is clear that the permutation $\left(1;2 \right)$, that two elements have no inversions, that $N\left(1;2 \right)=0$. And the y-axis is permuted $ \ left (2; 1 \ right) $ one inversion є (vlasne, 2< 1), поэтому $N\left(2;1 \right)=1.$ The universal formula for calculating the value for a 2x2 matrix looks like this: \[\left| \begin(matrix) ((a)_(11)) & ((a)_(12)) \\ ((a)_(21)) & ((a)_(22)) \\\end( matrix) \right|=((a)_(11))((a)_(22))-((a)_(12))((a)_(21))\] Graphically, you can show how the additional elements are to stand on the head diagonal, minus the additional elements on the side: Let's take a look at the sprat of applications: \[\left| \begin(matrix) 5 & 6 \\ 8 & 9 \\end(matrix) \right|;\quad \left| \begin(matrix) 7 & 12 \\ 14 & 1 \\\end(matrix) \right|.\] Solution. Everything is put in one row. Persha matrix: І friend: Vidpovid: -3; -161. Vtіm, tse Bulo nadto simply. Let's take a look at the 3x3 matrices there already. Now let's look at a 3x3 square matrix: \[\left[ \begin(matrix) ((a)_(11)) & ((a)_(12)) & ((a)_(13)) \\ ((a)_(21)) & ((a)_(22)) & ((a)_(23)) \\ ((a)_(31)) & ((a)_(32)) & ((a)_(33) ) \\\end(matrix) \right]\] When calculating the її arbitrator, we take $ 3! \u003d 1 \ cdot 2 \ cdot 3 \u003d 6 $ dodankіv - still not enough for a panic, but still enough, so that you can start to joke about the laws. For the first time, we write down all the permutations of the three elements and, perhaps, the inversion in the skin of them: \[\begin(align) & ((p)_(1))=\left(1;2;3 \right)\Rightarrow N\left(((p)_(1)) \right)=N\ left(1; 2; 3\right) = 0; \\ & ((p)_(2))=\left(1;3;2 \right)\Rightarrow N\left(((p)_(2)) \right)=N\left(1;3 ;2 \right) = 1; \\ & ((p)_(3))=\left(2;1;3 \right)\Rightarrow N\left(((p)_(3)) \right)=N\left(2;1 ;3 \right) = 1; \\ & ((p)_(4))=\left(2;3;1 \right)\Rightarrow N\left(((p)_(4)) \right)=N\left(2;3 ;1\right) = 2; \\ & ((p)_(5))=\left(3;1;2 \right)\Rightarrow N\left(((p)_(5)) \right)=N\left(3;1 ;2\right) = 2; \\ & ((p)_(6))=\left(3;2;1 \right)\Rightarrow N\left(((p)_(6)) \right)=N\left(3;2 ;1 \right) = 3. \\end(align)\] As i was transferred, in total 6 permutations $((p)_(1))$, ... $((p)_(6))$ were written (naturally, it would be possible to write їх in a different sequence - the essence of this is not change), and the number of inversions they change from 0 to 3. Zagalom, we will have three dodanki with a “plus” (there, de $ N \ left (p \ right) $ - guy) and three more with a “minus”. And with a zagal, the vyznachnik vvazhatimetsya for the formula: \[\left| \begin(matrix) ((a)_(11)) & ((a)_(12)) & ((a)_(13)) \\ ((a)_(21)) & ((a) _(22)) & ((a)_(23)) \\ ((a)_(31)) & ((a)_(32)) & ((a)_(33)) \\\end (matrix) \right|=\begin(matrix) ((a)_(11))((a)_(22))((a)_(33))+((a)_(12))( (a)_(23))((a)_(31))+((a)_(13))((a)_(21))((a)_(32))- \\ -( (a)_(13))((a)_(22))((a)_(31))-((a)_(12))((a)_(21))((a)_ (33))-((a)_(11))((a)_(23))((a)_(32)) \\\end(matrix)\] The axis only does not need to sit at once and fiercely cramming all the indexes! Instead of unreasonable numbers, remember the following simple rule: Trickster rule. For the sign of the sign of the 3x3 matrix, it is necessary to add three creations of the elements, which stand on the head diagonal at the tops of the equal-femoral tricots on the 3rd side, parallel to the diagonal, and then see the same three creations, but on the side diagonal. Schematically, it looks like this: The very tsі trikutniks (abo pentagrams - to whom it suits them more) to love tints in some kind of assistants and manuals from algebra. Vtim, let's not talk about the sum. Let's rather take one such votive for a warm-up in front of the right badge. Manager. Calculate the signifier: \[\left| \begin(matrix) 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 1 \\\end(matrix) \right|\] Solution. Practice for the rule of tricks. On the back of the head there are three additional pieces, folded from the elements on the head diagonal and in parallel: \[\begin(align) & 1cdot 5cdot 1+2cdot 6cdot 7+3cdot 4cdot 8= \& =5+84+96=185 \end(align) \] Now we understand with a side diagonal: \[\begin(align) & 3\cdot 5\cdot 7+2\cdot 4\cdot 1+1\cdot 6\cdot 8= \\ & =105+8+48=161 \\end(align) \ ] The friend was left with less than the first day - and we take it away: From i all! Tim is no less, the masters of 3x3 matrices are still not the pinnacle of mastery. Please give us a check. :) As we know, with the increase in the size of the $n$ matrix, the number of dodankіv in the typist becomes $n!$ and quickly grows. All the same, factorial - it’s not for you to do it like a dog’s horseradish, it’s a fast-growing function. Already for 4x4 matrices, put the vyznachniki ahead (thus through permutations) as if not very good. About 5x5 and more I sang a word. To this, the deacons of the power of the chief are connected to the right, but a little theoretical training is needed for this reasoning. Are you ready? Let's go! Let's give a sufficient matrix $ A = \ left [m \ times n \ right] $. Respect: not obov'yazkovo square. On the vіdmіnu vіd vyznachnіv, minori – tse nyashki, yakі іsnuyut not only in suvori square matrices. Let's choose in the ith matrix kіlka (for example, $k$) rows i stovptsіv, moreover $1\le k\le m$ і $1\le k\le n$. Todi: Appointment. The minor of the order $k$ is the sign of a square matrix, which is used on the retina of the chosen $k$ columns and rows. Also, in a minor, we call the qiu new matrix itself. Such a minor $((M)_(k))$ is indicated. Naturally, one matrix can take the whole number of minors close to $k$. Axis butt minor order 2 for matrix $\left[5\times 6\right]$: Absolutely neobov'yazkovo, so that the chosen rows and stovpts stood as a handhold, like a looked at butt. Golovna, that the number of selected rows and columns was the same (the number of $k$). Є th іnshe vyznachennya. Possibly, to whom it is more worthy: Appointment. Give a rectangular matrix $ A = \ left [m \ times n \ right] $. Just after the Sunday in nіy one or more dekіlkoh stovptsіv i one or one dekіlkoh rows the square matrix expands $ \ left [k \ times k \ right] $, then її vyznachnik - tse i є minor $ ((M)_ (k)) $ . The matrix itself is also sometimes called a minor - it will be clear from the context. As if showing my whale, sometimes it’s better to turn once from the 11th on top of the food, lower the yap, sitting on the balcony. butt. Come on, given the matrix Selecting row 1 and stovpets 2, we take the minor of the first order: \[((M)_(1))=\left| 7\right|=7\] Selecting rows 2, 3 and columns 3, 4, we will take a minor in a different order: \[((M)_(2))=\left| \begin(matrix) 5 & 3 \\ 6 & 1 \\\end(matrix) \right|=5-18=-13\] And so choose all three rows, and also columns 1, 2, 4 will be a minor of the third order: \[((M)_(3))=\left| \begin(matrix) 1 & 7 & 0 \\ 2 & 4 & 3 \\ 3 & 0 & 1 \\\end(matrix) \right|\] It is not important for readers to know other minor orders 1, 2 or 3. Let's go to that. "Well, ok, what should we give us qi minyoni minori?" - Chant you ask you. By yourself - nothing. Ale in square matrices at the skin minor has a "companion" - an additional minor, as well as an additional algebra. І at once tsі two slaps to allow us to click the vyznachniks like peas. Appointment. Let a square matrix $A=\left[ n\times n \right]$ be given, in which the minor $((M)_(k))$ is chosen. Then the additional minor for the minor $((M)_(k))$ is the number of pieces of the $A$ matrix, which is lost when all the rows and columns are crossed out when the minor $((M)_(k))$ is folded: One point can be clarified: the additional minor is not just a “matrix piece”, but a sign of this piece. Added minors are indicated for the additional "zirochki": $M_(k)^(*)$: de operation $A\nabla ((M)_(k))$ literally means "recreate from $A$ rows and columns that go up to $((M)_(k))$". This operation is not a commonplace in mathematics - I myself have thought up a good explanation for beauty. :) Dodatkovі minori are rarely victorious by powerful forces. The stench is part of a folding construction - an algebraic addition. Appointment. Algebraic addition to the minor $((M)_(k))$ is the complementary minor of $M_(k)^(*)$, multiplications by the quantity $((\left(-1 \right))^(S))$ , where $S$ is the sum of the numbers of all rows and columns, behind the outer minor $((M)_(k))$. As a rule, the complement of the minor algebra $((M)_(k))$ is denoted by $((A)_(k))$. Tom: \[((A)_(k))=((\left(-1 \right))^(S))\cdot M_(k)^(*)\] Important? At first glance, yes. Ale, not exactly. Because it's really easy. Let's look at the example: butt. Given a 4x4 matrix: Vibermo minor in a different order \[((M)_(2))=\left| \begin(matrix) 3 & 4 \\ 15 & 16 \\\end(matrix) \right|\] Captain Obviousness stresses us that when folding the minor, rows 1 and 4, as well as columns 3 and 4, were added. Lost to know the number $S$ and take the addition of algebra. We know the numbers of the back rows (1 and 4) and the columns (3 and 4), everything is simple: \begin(align) & S=1+4+3+4=12; \\ & ((A)_(2))=((\left(-1 \right))^(S))\cdot M_(2)^(*)=((\left(-1 \right) ) )^(12))\cdot \left(-4 \right)=-4\end(align)\] Response: $((A)_(2))=-4$ From i all! In fact, the whole difference between the supplementary minor and the supplementary algebra is only a minus in front, that’s not the case. And the axis of our further, now, in the light, all the numbers in minority and algebraic additions were needed. Laplace's theorem about the layout of the arbiter. Let the matrix expand $ \ left [n \ times n \ right] $ $ k $ rows (stovptsіv) are selected, and $ 1 \ le k \ le n-1 $. The same arbiter of the matrix is the most important sum of all creations in minors in the order of $k$, which are placed in rows (stows) on their addendums of algebra: \[\left| A \right|=\sum(((M)_(k))\cdot ((A)_(k)))\] Moreover, such additions will be equal to $C_(n)^(k)$. Garazd, kindly: about $C_(n)^(k)$ - I’m already showing off, there was nothing like that in the original Laplace theorem. Ale, the combinatorics did not succumb to anything, and literally a glance at the mind will allow you to change your mind independently, so that you will add style to yourself. We will not bring, even if we do not become especially difficult - all the layouts are reduced to the good old permutations and pairing / unpairing of inversions. The proof of the proof will be given in a fine paragraph, and today we may have a daily practical lesson. Therefore, let's move on to the near future of the theorem, if the minors are the terms of the matrix matrix. Those who at once pide mov - as the main tool of the work with vyznachnikami, for the sake of which all the game was started with permutations, minors and algebraic additions. Read and enjoy: A consequence of Laplace's theorem (arrangement of the arbiter in order/stowptsyu). Let the matrix expand $ \ left [n \ times n \ right] $ one row is selected. The minors in this row will be $n$ okremy klitins: \[((M)_(1))=((a)_(ij)),\quad j=1,...,n\] Additive minors can also be easily entered: just take the matrix and replace the row and the column to avenge $((a)_(ij))$. We call it the minor $M_(ij)^(*)$. For an addition to the algebra, the number $S$ is needed, but in the case of a minor order of 1, it’s just the sum of the “coordinates” of the cell $((a)_(ij))$: And the same variable variable can be written in terms of $((a)_(ij))$ and $M_(ij)^(*)$ similarly up to Laplace's theorem: \[\left| A \right|=\sum\limits_(j=1)^(n)(((a)_(ij))\cdot ((\left(-1 \right))^(i+j))\cdot ((M)_(ij)))\] Tse i є the formula for laying out the signer behind the row. Ale te same i for stovptsіv. From this consequence, one can immediately formulate a sprat of visnovkiv: The rest of the fact is especially important. For example, the deputy of the 4x4 animal clerk will now be enough to protect the sprat of the 3x3 gnome - we already seem to fit in with them. Manager. Find the signifier: \[\left| \begin(matrix) 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\\end(matrix) \right|\] Solution. We lay out this vyznachnik in the first row: \[\begin(align)\left| A \right|=1\cdot ((\left(-1 \right))^(1+1))\cdot \left| \begin(matrix) 5 & 6 \\ 8 & 9 \\end(matrix) \right|+ & \\ 2\cdot ((\left(-1 \right))^(1+2))\cdot \ left| \begin(matrix) 4 & 6 \\ 7 & 9 \\end(matrix) \right|+ & \\ 3\cdot ((\left(-1 \right))^(1+3))\cdot \ left| \begin(matrix) 4 & 5 \\ 7 & 8 \\end(matrix) \right|= & \\end(align)\] \[\begin(align) & =1\cdot \left(45-48 \right)-2\cdot \left(36-42 \right)+3\cdot \left(32-35 \right)= \\ & =1\cdot \left(-3 \right)-2\cdot \left(-6 \right)+3\cdot \left(-3 \right)=0. \\end(align)\] Manager. Find the signifier: \[\left| \begin(matrix) 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\end(matrix) \right|\ ] Solution. For the sake of diversity, let's practice with the stovptsy once again. For example, in the rest of the column there are two zeros - obviously, it is important to speed up the calculation. Infection, why. Otzhe, we lay out the altar according to the fourth column: \[\begin(align)\left| \begin(matrix) 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\end(matrix) \right|= 0 \cdot ((\left(-1 \right))^(1+4))\cdot \left| \begin(matrix) 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\end(matrix) \right|+ & \\ +1\cdot ((\left(-1 \) right))^(2+4))\cdot \left| \begin(matrix) 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\end(matrix) \right|+ & \\ +1\cdot ((\left(-1 \) right))^(3+4))\cdot \left| \begin(matrix) 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\end(matrix) \right|+ & \\ +0\cdot ((\left(-1 \) right))^(4+4))\cdot \left| \begin(matrix) 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\\end(matrix) \right| &\\\end(align)\] And here - oh, marvelous! - two dodanki immediately tell the cat the tail, the shards in them are the multiplier "0". There are two more 3x3 vyznachniks left, which can be easily sorted out: \[\begin(align) & \left| \begin(matrix) 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\\end(matrix) \right|=0+0+1-1-1-0=-1; \\ & \left| \begin(matrix) 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\\end(matrix) \right|=0+1+1-0-0-1=1. \\end(align)\] Turn back to the weekend and we know the evidence: \[\left| \begin(matrix) 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\end(matrix) \right|= 1 \cdot \left(-1 \right)+\left(-1 \right)\cdot 1=-2\] Well, from i everything. І 4 days! = 24 dodankіv failed to enter.:) Response: -2 In the rest of the problem, we tried, as the presence of zeros in the rows (stowptsy) of the matrix, sharply simplifies the layout of the signpost and takes all the calculations into account. Blame the natural nourishment: why can't you work so that the zeros appeared to strike at your matrix, why didn't they? The proof is unambiguous: possible, possible. And here I will help us to come to the aid of the authority of the appointee: Particularly valuable to become the third power: we can subtract from one row (stovptsia). Most of the time, the rozrahunka is created before, to “zero out” the entire stovpets, crimson of one element, and then lay out the sign of that stovptsya, removing the matrix by 1 less. Let's wonder how it works in practice: Manager. Find the signifier: \[\left| \begin(matrix) 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \\ 3 & 4 & 1 & 2 \\ 2 & 3 & 4 & 1 \\end(matrix) \right|\ ] Solution. There are no zeros here, as if bi vzagali are not poster, so you can "double" in any order or stovptsyu - the calculation will be approximately the same. Let's not razminyuvatisya on drіbnitsa and "nullified" the first row: the new one has already є kіtin with one, then we just take the first row and we can see 4 times from the other, 3 times from the third and 2 times from the rest. As a result, we take away a new matrix, but the arbitrator will be ourselves: \[\begin(matrix)\left| \begin(matrix) 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \\ 3 & 4 & 1 & 2 \\ 2 & 3 & 4 & 1 \\end(matrix) \right|\ begin (matrix) \downarrow \\ -4 \\ -3 \\ -2 \\end(matrix)= \\ =\left| \begin(matrix) 1 & 2 & 3 & 4 \\ 4-4\cdot 1 & 1-4\cdot 2 & 2-4\cdot 3 & 3-4\cdot 4 \\ 3-3\cdot 1 & 4-3cdot 2 & 1-3cdot 3 & 2-3cdot 4 \ 2-2cdot 1 & 3-2cdot 2 & 4-2cdot 3 & 1-2cdot 4 \ \\end(matrix) \right|= \\ =\left | \begin(matrix) 1 & 2 & 3 & 4 \\ 0 & -7 & -10 & -13 \\ 0 & -2 & -8 & -10 \\ 0 & -1 & -2 & -7 \\ \end(matrix)\right| \\\end(matrix)\] Now, with the inviolability of the Patz, we are laying out this vyznachnik according to the first step: \[\begin(matrix) 1\cdot ((\left(-1 \right))^(1+1))\cdot \left| \begin(matrix) -7 & -10 & -13 \\ -2 & -8 & -10 \\ -1 & -2 & -7 \\end(matrix) \right|+0\cdot ((\ left (-1 \right))^(2+1))\cdot \left| ... \right|+ \\ +0\cdot ((\left(-1 \right))^(3+1))\cdot \left| ... \right|+0\cdot ((\left(-1 \right))^(4+1))\cdot \left| ... \right| \\\end(matrix)\] It dawned on me that the “live” is less than the first dodanok - at the reshti, I didn’t write out the signs, the stinks of the stink all the same multiply by zero. The coefficient before the vyznachnik is a good one, tobto. yoga can be recorded. Prote can be blamed for the "minuses" of the three rows of the leader. In fact, trichs brought the multiplier (−1): \[\left| \begin(matrix) -7 & -10 & -13 \ -2 & -8 & -10 \ -1 & -2 & -7 \\end(matrix) \right|=\cdot \left| \begin(matrix) 7 & 10 & 13 \\ 2 & 8 & 10 \\ 1 & 2 & 7 \\\end(matrix) \right|\] We took away the other 3x3 vyznachnik, which you can porahuvat according to the trikutnik rule. And let's try to lay it out on the first row - it's good to stand alone proudly in the remaining row: \[\begin(align) & \left(-1 \right)\cdot \left| \begin(matrix) 7 & 10 & 13 \\ 2 & 8 & 10 \\ 1 & 2 & 7 \\end(matrix) \right|\begin(matrix) -7 \\ -2 \\ \uparrow \ \ \end(matrix)=\left(-1 \right)\cdot \left| \begin(matrix) 0 & -4 & -36 \\ 0 & 4 & -4 \\ 1 & 2 & 7 \\end(matrix) \right|= \\ & =\cdot \left| \begin(matrix) -4 & -36 \\ 4 & -4 \\end(matrix) \right|=\left(-1 \right)\cdot \left| \begin(matrix) -4 & -36 \\ 4 & -4 \\end(matrix) \right| \\end(align)\] It is possible, obviously, to have a little more fun and lay out the 2x2 matrix in order (to the stack), but we are adequate for you, it’s just amazing to see: \[\left(-1 \right)\cdot \left| \begin(matrix) -4 & -36 \\ 4 & -4 \\end(matrix) \right|=\left(-1 \right)\cdot \left(16+144 \right)=-160\ ] Otak i break the world. All in all-160 for vіdpovіdі. :) Vidpovid: -160. A couple of respect before Tim, so let's move on to the rest of the task: Manager. Find the signifier: \[\left| \begin(matrix) 1 & 1 & 1 & 1 \\ 2 & 4 & 8 & 16 \\ 3 & 9 & 27 & 81 \\ 5 & 25 & 125 & 625 \\end(matrix) \right|\ ] Solution. Well, here the first row asks for “zeroing”. We take the first step and see exactly once from the decision: \[\begin(align) & \left| \begin(matrix) 1 & 1 & 1 & 1 \\ 2 & 4 & 8 & 16 \\ 3 & 9 & 27 & 81 \\ 5 & 25 & 125 & 625 \\end(matrix) \right|= \ \&=\left| \begin(matrix) 1 & 1-1 & 1-1 & 1-1 \\ 2 & 4-2 & 8-2 & 16-2 \\ 3 & 9-3 & 27-3 & 81-3 \\ 5 & 25-5 & 125-5 & 625-5 \\end(matrix) \right|= \\ & =\left| \begin(matrix) 1 & 0 & 0 & 0 \\ 2 & 2 & 6 & 14 \\ 3 & 6 & 24 & 78 \\ 5 & 20 & 120 & 620 \\end(matrix) \right| \\end(align)\] We lay it out in the first row, and then we blame the exuberant multipliers of the rows that are left out: \[\cdot\left| \begin(matrix) 2 & 6 & 14 \\ 6 & 24 & 78 \\ 20 & 120 & 620 \\\end(matrix) \right|=\cdot \left| \begin(matrix) 1 & 3 & 7 \\ 1 & 4 & 13 \\ 1 & 6 & 31 \\\end(matrix) \right|\] I’ll make a new poster of the “beautiful” numbers, but in the first place, we’ll lay out the sign: \[\begin(align) & 240\cdot \left| \begin(matrix) 1 & 3 & 7 \\ 1 & 4 & 13 \\ 1 & 6 & 31 \\end(matrix) \right|\begin(matrix) \downarrow \\ -1 \\ -1 \ \ \ end (matrix) = 240 \ cdot \ left | \begin(matrix) 1 & 3 & 7 \\ 0 & 1 & 6 \\ 0 & 3 & 24 \\\end(matrix) \right|= \\ & =240\cdot ((\left(-1 \ ) right))^(1+1))\cdot \left| \begin(matrix) 1 & 6 \\ 3 & 24 \\end(matrix) \right|= \\ & =240\cdot 1\cdot \left(24-18 \right)=1440 \\end( align) \] Order. The task is over. ID: 1440
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- Pash, are you in trouble?
Correct placement of indices in the matrix
Overlaying a coordinate system on a matrix 
Designation of indexes in the matrix Geometric design
Vyznachnik 2x2 - the area of the parallelogram 
One more 2x2 vyznachnik 
Vyznachnik 3x3 - the volume of the parallelepiped
Algebraic design
Permutations and inversions
What is the forerunner
Matrix 2x2
Significant Matrix 2x2
Matrix 3x3
3x3 matrix arbiter: 3x3 rule
Scheme for the calculation of the arbitrators
What is Matrix Minor
Select $k = 2$ stovptsіv and rows for molding in minor
Algebraic additions
Additional minor to minor $((M)_(2))$
Laplace's theorem
Arrangement of the vyznachnik in a row that stovptsyu
The main powers of the appointee