Newton's binomial properties of the binomial coefficients of Pascal's triangle. Newton's binomial using factorial notation

Obviously, for a system of n linear equations With n unknowns we obtain a matrix of coefficients of size:

Let us introduce the concept of determinant n-th order.

Definition 4.1:

Determinant n-th order is a number equal to

Amount n! terms;

Each term is a product n matrix elements taken one from each row and each column;

Each term is taken with a “+” sign if the permutation of the second indices is even, and with a “-” sign if the permutation of the second indices is odd, provided that the first indices form a natural series of numbers.

That.

Here å is taken over all possible permutations composed of the numbers 1,2,…, n.

5. Basic properties of determinants.

Let us establish the basic properties of determinants, which for simplicity we will show using a 2nd order determinant.

1. When replacing rows with corresponding columns (called transposition) the determinant remains unchanged. Really:

Hence, , which was what needed to be proven.

Note: The result obtained above gives us the right to assert that the rows and columns of the determinant, hereinafter referred to as rows, are equal.

2. When two rows are rearranged, the determinant changes sign to the opposite one.

Really, Let's swap the lines and calculate the determinant

Q.E.D.

3. If two parallel series in the determinant are identical, then it equal to zero. Indeed, let's swap two identical lines. Then the value of the determinant will not change, but the sign, due to property 2, will change. Singular, which does not change when the sign changes – zero.

4. Total multiplier members of any series can be taken out of the determinant sign.

Q.E.D.

5. If all elements of any series are sums the same number terms, then the determinant equal to the sum determinants in which the elements of the series under consideration are individual terms.

Q.E.D.

6. The determinant will not change if the corresponding elements of a parallel series are added to the elements of any series, multiplied by a certain number.

Multiply the second line by and add it to the first line:

Indeed, due to properties 3,4,5

=

Q.E.D.

6. Minors and algebraic additions elements of the determinant.

Consider the determinant n-th order:

.

Let us highlight in the determinant i th line and j th column. At the intersection of these rows there is an element

If in the determinant we cross out i- adjustment and j-th column, then we get the order determinant n-1 (i.e., having an order one less than the original determinant), called minor element determinant We will denote minor element symbol.

Definition 6.1. Aalgebraic complement element The determinant is called a minor, taken with a sign, and is denoted by the symbol. According to the definition, we get

.

Example 6.1. Find the minor and algebraic complement of the determinant

orthogonal unitary matrix multilinear

Calculation of 2nd and 3rd order determinants.

We obtain formulas for calculating determinants of the second and third orders. By definition, when

When we cross out the first row and one column, we get a matrix containing one element, so

Substituting these values ​​into the right side, we obtain the formula for calculating the second-order determinant

Second order determinant equal to the difference the product of the elements on the main diagonal and the product of the elements on the secondary diagonal (Fig. 2.1).

For the third order determinant we have

By deleting the first row and one column, we obtain the determinants of second-order square matrices:

We write these second-order determinants using formula (2.2) and obtain the formula for calculating the third-order determinant

The determinant (2.3) is the sum of six terms, each of which is the product of three elements of the determinant, located in different rows and different columns. Moreover, three terms are taken with a plus sign, and the other three - with a minus sign.

To remember formula (2.3), the rule of triangles is used: you need to add three products of three elements standing on the main diagonal and at the vertices of two triangles having a side parallel to the main diagonal (Fig. 2.2a), and subtract three products of elements standing on the side diagonals and at the vertices of two triangles having a side parallel to the side diagonal (Fig. 2.2,6).

You can also use the calculation scheme shown in Fig. 2.3 (Sarrus’s rule): add the first and second columns to the right of the matrix, calculate the products of elements on each of the six lines indicated, and then find the algebraic sum of these products, while the product of elements on lines parallel to the main diagonal is taken with a plus sign , and the product of elements on straight lines parallel to the side diagonal is with a minus sign (according to the notation in Fig. 2.3).

Calculation of determinants of order N>3.

So, we have obtained formulas for calculating the determinants of the second and third orders. You can continue the calculations using formula (2.1) for and obtain formulas for calculating the fourth, fifth, etc. determinants. orders of magnitude. Consequently, the inductive determination allows one to calculate the determinant of any order. Another thing is that the formulas will be cumbersome and inconvenient for practical calculations. Therefore the determinants high order(fourth or more), as a rule, are calculated based on the properties of the determinants.

Example 2.1. Compute determinants

Solution. Using formulas (2.2) and (2.3) we find;

Formula for decomposing the determinant into row (column) elements

Let a square matrix of order be given.

An additional minor of an element is the determinant of an order matrix obtained from a matrix by deleting i-th line and jth column.

The algebraic complement of a matrix element is the additional minor of this element multiplied by

Theorem 2.1 formula for decomposing the determinant into the elements of a row (column). The determinant of the matrix is ​​equal to the sum of the products of the elements of an arbitrary row (column) and their algebraic complements:

(decomposition along the i-th row);

(expansion in the jth column).

Notes 2.1.

1. The proof of the formula is carried out using the method of mathematical induction.

2. In the inductive definition (2.1), the formula for decomposing the determinant into the elements of the first row was actually used.

Example 2.2. Find the determinant of the matrix

Solution. Let's expand the determinant along the 3rd line:

Now let's expand the third-order determinant in the last column:

The second-order determinant is calculated using formula (2.2):

Matrix determinant triangular in appearance

Let us apply the decomposition formula to find the determinant of the upper triangular matrix

Let's expand the determinant along the last line (nth line):

where is an additional minor element. Let's denote Then. Note that when we cross out the last row and the last column of the determinant, we obtain the determinant of the upper triangular matrix of the same type as, but of (n-1)th order. Expanding the determinant along the last row ((n-1)th row), we get. Continuing in a similar way and taking into account that, we arrive at the formula.e. determinant of an upper triangular matrix equal to the product elements on the main diagonal.

Notes 2.2

1. The determinant of a lower triangular matrix is ​​equal to the product of the elements on the main diagonal.

2. The determinant of the identity matrix is ​​1.

3. The determinant of a matrix of triangular form will be called the determinant of triangular form. As shown above, the determinant of a triangular matrix (the determinant of an upper or lower triangular matrix, in particular a diagonal one) is equal to the product of the elements on the main diagonal.

Basic properties of determinants (determinants)

1. For anyone square matrix, i.e. When transposed, the determinant does not change. From this property it follows that the columns and rows of the determinant are “equal”: any property that is true for columns will be true for rows.

2. If in the determinant one of the columns is zero (all elements of the column are equal to zero), then the determinant is equal to zero:.

3. When rearranging two columns, the determinant changes sign to the opposite (antisymmetry property):

4. If the determinant has two identical columns, then it is equal to zero:

5. If the determinant has two proportional columns, then it is equal to zero:

6. When multiplying all elements of one column of the determinant by a number, the determinant is multiplied by this number:

7. If jth column determinant is represented as the sum of two columns, then the determinant is equal to the sum of two determinants whose j-th columns are and , respectively, and the remaining columns are the same:

8. The determinant is linear in any column:

9. The determinant will not change if the corresponding elements of another column are added to the elements of one column, multiplied by the same number:

10. The sum of the products of elements of any column of the determinant by the algebraic complements of the corresponding elements of another column is equal to zero:

Notes 2.3

1. The first property of the determinant is proven by induction. Proofs of other properties are carried out using the formula for decomposing the determinant into column elements. For example, to prove the second property, it is enough to expand the determinant into the elements of the zero column (assume that the jth column is zero, i.e.):

To prove property 10, you need to read the formula for decomposing the determinant from right to left, namely, the sum of the products of the elements of the i-th column by the algebraic complements of the elements of the j-th column is represented as an expansion in the j-th column of the determinant

which in place of the elements of the j-ro column are the corresponding elements of the i-th column. According to the fourth property, such a determinant is equal to zero.

2. From the first property it follows that all properties 2-10 formulated for the columns of the determinant will also be valid for its rows.

3. Using the formulas for decomposing the determinant into the elements of a row (column) and property 10, we conclude that

4. Let be a square matrix. A square matrix of the same order as is said to be adjoint of if each of its elements is equal to the algebraic complement of an element of the matrix. In other words, to find the adjoint matrix one should:

a) replace each element of the matrix with its algebraic complement, and we obtain a matrix;

b) find the adjoint matrix by transposing the matrix.

From formulas (2.4) it follows that, where is the identity matrix of the same order as.

Example 2.5. Find the determinant of a block-diagonal matrix, where is an arbitrary square matrix, is the identity matrix, and is a zero matrix of the corresponding order, is transposed.

Solution. Let's expand the determinant over the last column. Since all elements in this column are zero, with the exception of the last one, which is equal to 1, we obtain a determinant of the same form as the original one, but of a lower order. By expanding the resulting determinant along the last column, we reduce its order. Continuing in the same way, we obtain the determinant of the matrix. Hence,

Methods for calculating nth order determinants.

Let an ordered set be given n elements. Any arrangement n elements in in a certain order called rearrangement from these elements.

Since each element is determined by its number, we will say that given n natural numbers.

Number of different permutations from n numbers are equal to n!

If in some permutation of n numbers number i costs earlier j, But i > j, i.e. larger number stands before the smaller one, then they say that the pair i, j amounts to inversion.

Example 1. Determine the number of inversions in the permutation (1, 5, 4, 3, 2)

Solution.

The numbers 5 and 4, 5 and 3, 5 and 2, 4 and 3, 4 and 2, 3 and 2 form inversions. The total number of inversions in this permutation is 6.

The permutation is called even, If total number its inversions are even, otherwise it is called odd. In the example discussed above, an even permutation is given.

Let some permutation be given..., i, …, j, … (*) . Transformation in which numbers i And j change places, and the rest remain in their places, is called transposition. After number transposition i And j in permutation (*) there will be a rearrangement..., j, …, i, ..., where all elements except i And j, remained in their places.

From any permutation from n numbers, you can go to any other permutation of these numbers using several transpositions.

Every transposition changes the parity of the permutation.

At n ≥ 2 number of even and odd permutations from n numbers are the same and equal.

Let M– ordered set of n elements. Every bijective transformation of a set M called substitutionnth degree.

Substitutions are written like this: https://pandia.ru/text/78/456/images/image005_119.gif" width="27" height="19"> and that's all ik are different.

Substitution called even, if both of its rows (permutations) have the same parities, i.e., either both even or both odd. Otherwise substitution called odd.

At n ≥ 2 number of even and odd substitutions nth degrees the same and equal to .

The determinant of a square matrix A of second order A= is the number equal to = a11a22–a12a21.

The determinant of a matrix is ​​also called determinant. For the determinant of matrix A, the following notation is used: det A, ΔA.

Determinant square matrices A= third order call the number equal to │A│= a11a22a33+a12a23a31+a21a13a32-a13a22a31-a21a12a33-a32a23a11

Each term algebraic sum on the right side of the last formula is the product of the matrix elements, taken one and only one from each column and each row. To determine the sign of the product, it is useful to know the rule (it is called the triangle rule), schematically depicted in Fig. 1:

«+» «-»

https://pandia.ru/text/78/456/images/image012_64.gif" width="73" height="75 src=">.

Solution.

Let A be an nth order matrix with complex elements:

A=https://pandia.ru/text/78/456/images/image015_54.gif" width="112" height="27 src="> (1) ..gif" width="111" height="51"> (2) .

The determinant of the nth order, or the determinant of the square matrix A=(aij) for n>1, is the algebraic sum of all possible products of the form (1) , and the work (1) is taken with a “+” sign if the corresponding substitution (2) even, and with a “‑” sign if the substitution is odd.

Minor Mij element aij determinant is a determinant obtained from the original by deleting i th line and j- th column.

Algebraic complement Aij element aij the determinant is called a number Aij=(–1) i+ jMij, Where Mij – element minor aij.

Properties of determinants

1. The determinant does not change when replacing all rows with the corresponding columns (the determinant does not change when transposing).

2. When two rows (columns) are rearranged, the determinant changes sign.

3. A determinant with two identical (proportional) rows (columns) is equal to zero.

4. The factor common to all elements of a row (column) can be taken beyond the sign of the determinant.

5. The determinant will not change if the corresponding elements of another row (column) are added to the elements of a certain row (column), multiplied by the same number other than zero.

6. If all elements of a certain row (column) of a determinant are equal to zero, then it is equal to zero.

7. The determinant is equal to the sum of the products of the elements of any row (column) by their algebraic complements (the property of decomposition of the determinant in a row (column)).

Let's look at some methods for calculating order determinants n .

1. If in an nth-order determinant at least one row (or column) consists of zeros, then the determinant is equal to zero.

2. Let some row in the nth order determinant contain non-zero elements. The calculation of the nth order determinant can be reduced in this case to the calculation of the n-1 order determinant. Indeed, using the properties of the determinant, you can make all elements of a row, except one, zero, and then expand the determinant along the specified row. For example, let us rearrange the rows and columns of the determinant so that in place a11 there was an element different from zero.

https://pandia.ru/text/78/456/images/image018_51.gif" width="32 height=37" height="37">.gif" width="307" height="101 src=">

Note that it is not necessary to rearrange rows (or columns). You can get zeros in any row (or column) of the determinant.

There is no general method for calculating determinants of order n, other than calculating the determinant given order directly by definition. To the determinant of this or that special type apply various methods calculations leading to simpler determinants.

3. Let's take it to triangular form. Using the properties of the determinant, we reduce it to the so-called triangular form, when all elements standing on one side of the main diagonal are equal to zero. The resulting triangular determinant is equal to the product of the elements on the main diagonal. If it is more convenient to get zeros on one side of the secondary diagonal, then it will be equal to the product of the elements of the secondary diagonal, taken with the sign https://pandia.ru/text/78/456/images/image022_48.gif" width="49" height= "37">.

Example 3. Calculate determinant by row expansion

https://pandia.ru/text/78/456/images/image024_44.gif" width="612" height="72">

Example 4. Calculate the fourth order determinant

https://pandia.ru/text/78/456/images/image026_45.gif" width="373" height="96 src=">.

2nd method(calculating the determinant by expanding it along the line):

Let us calculate this determinant by row expansion, having previously transformed it so that in some of its rows all elements except one become zero. To do this, add the first line of the determinant to the third. Then multiply the third column by (-5) and add it to the fourth column. We expand the transformed determinant along the third line. We reduce the third-order minor to a triangular form relative to the main diagonal.

https://pandia.ru/text/78/456/images/image028_44.gif" width="202" height="121 src=">

Solution.

Let's subtract the second from the first line, the third from the second, etc., and finally, the last from the penultimate (the last line remains unchanged).

https://pandia.ru/text/78/456/images/image030_39.gif" width="445" height="126 src=">

The first determinant in the sum is triangular with respect to the main diagonal, so it is equal to the product of the diagonal elements, i.e. (n–1)n. We transform the second determinant in the sum by adding last line to all previous lines of the determinant. The determinant obtained from this transformation will be triangular with respect to the main diagonal, so it will be equal to the product of the diagonal elements, i.e. nn-1:

=(n–1)n+ (n–1)n + nn-1.

4. Calculation of the determinant using Laplace's theorem. If in the determinant we select k rows (or columns) (1 £ k £ n-1), then the determinant is equal to the sum of the products of all minors of the kth order located in the selected k rows (or columns) and their algebraic complements.

Example 6. Compute determinant

https://pandia.ru/text/78/456/images/image033_36.gif" width="538" height="209 src=">

INDIVIDUAL TASK No. 2

“CALCULATION OF NTH ORDER DETERMINANTS”

Option 1

Compute determinants

https://pandia.ru/text/78/456/images/image035_39.gif" width="114" height="94 src=">

algebraic formula, discovered by Newton, expressing any degree of binomial, namely:

(x + a) n = x n + n/1(ax n-1) + (a 2 x n-2) + …(a n x n-m) + …

or, in compact form, using the symbol n! = 1.2.3…n:

(x + a) n = ∑ m (!x n-m a m

This formula was first given by Newton in 1676 without proof. It is carved on Newton's tomb, in Westminster Abbey, in London, although it cannot be considered one of most important discoveries Newton.

The proof of B.'s formula for an integer exponent is easy, as special case from more general formula, expressing the work any number binomials. It is easy to verify by direct multiplication that for the case n = 2 or n = 3 the formula holds:

(x + a 1)(x + a 2)…(x + a n) = x n + S n 1 x n-l + S n 2 x n-2 + … + S n n

where S n 1 is the sum of these quantities a 1 , a 2 . . . and n, S n 2 is the sum of their products by two, - S n n is the product of all these quantities. And then you can prove that if it is true for n, then it is also true for n + 1 factors. For, adding one factor x + a n+1, we obtain by direct multiplication

(x + a 1)(x + a 2)…(x + a n-1) = x n-1 + (S n 1 + a n+1)x n + (S n 2 + S n 1 a n- 1)x n-1 + … + S n n a n

and at the same time it is obvious that

S n 1 + a n+1 + 1 = S 1 n+1

S n 2 + S n 1 a n+1 = S 2 n+1

etc., so right side the last equality is

x n+1 + S 1 n+1 x n + S 2 n+1 x n-1 + … + (S n+1) n+1

etc. Now let everything A equal to each other and equal, for example, A, Then:

S 2 = a 2 ...

and we get (x + a) n = x n + nax n-1 + (a 2 x n-2) + ...

Thus, the validity of Newton's formula for n is a positive integer is proven. But Newton himself already showed that it is true for both fractional and negative. Let us present Euler's proof for any n. Consider the expression:

1+nx + + x 3 + …

For n integer it is equal to (1 + x) n. Let for every n it be generally f(n). In the same way, let a similar expression with n replaced by m be f(m). Multiplying, we find, on the one hand, f(n)f(m), on the other hand, an expression whose coefficient composition law is known to us from the case of n, m integers, namely:

f(n)f(m) = 1 + [(n + m)/1]x + [(n + m)(n + m - 1)/1.2]x 2 + [(n + m)(n + m - 1)(n + m - 2)/1.2.3]x 3 + …

and this is obviously f(n+m). So we got f(n)f(m) = f(n + m); in the same way for an arbitrary number of factors f(n 1)f(n 2)... f(n μ) = f(n 1 +n 2 +…+n μ); putting n 1 = n 2 =…= n μ = λ/μ, we have

f(n)f(–n) = f(0) = 1, i.e. f(–n) = 1/f(n) or

f(–n) = (1 + x) –l = nx + x 2 - x 3 + … etc.

• - binomial, the sum or difference of two algebras. expressions called members of B., for example. , etc. About the powers of B., that is, expressions yes, see Newton's binomial...

  Mathematical Encyclopedia

• - an algebraic expression consisting of the sum or difference of two quantities, for example axm +...
• - an algebraic formula discovered by Newton, expressing any degree of binomial, namely: n = xn + n/1 + + … + … or, in compact form, using the symbol n! = 1.2...

  Encyclopedic Dictionary Brockhaus and Euphron

• - and lat. nomen - name) binomial, the sum or difference of two algebraic expressions called terms of the equation; for example a + b, etc. On powers of B., that is, expressions of the form n, see Newton's binomial...
• - the name of a formula expressing any integer positive degree the sum of two terms through the powers of these terms, namely: where n is an integer positive number, a and b - whatever...

  Great Soviet Encyclopedia

• - the name of a formula that allows you to write out the decomposition of an algebraic sum of two terms of an arbitrary degree...

  Collier's Encyclopedia

• - the same as binomial. For a binomial of the form n, see Art. Newton binomial...
• - a formula expressing the positive integer power of the sum of two terms through the powers of these terms (their coefficients are called binomial coefficients...

  Large encyclopedic dictionary

• - Borrowing. in the first half of the 19th century. from French lang., where binôme is the addition of lat. bi and Greek nomē “part, share”. Wed. the derivational calque of this word is binomial...

  Etymological dictionary Russian language

• - From the novel “The Master and Margarita” by Mikhail Afanasyevich Bulgakov. The words of Koroviev-Fagot, commenting on the dialogue between Woland and the bartender Andrei Fokich Sokov...

  Dictionary winged words and expressions

• - ; pl. bino/we, R....

  Spelling dictionary Russian language

• - husband. binomy female in literal notation: numerical expression, consisting of two members; binomial, binomial quantity...

  Dictionary Dahl

• - BINOM, husband. In mathematics: binomial...

  Ozhegov's Explanatory Dictionary

• - binomial m. Algebraic expression, representing the sum or difference of two monomials; binomial...

  Explanatory Dictionary by Efremova

• - Talk. Joking. About smb. complex, confusing. Elistratov, 41...

  Big dictionary Russian sayings

• - BINOM, -a, m. Iron. About smth. seemingly complex and confusing. Poss. spread under the influence of M. Bulgakov’s novel “The Master and Margarita”...

  Dictionary of Russian argot

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