Algebraic complement examples. Algebraic addition. Compute decide matrices

Matrix minors

Let given a square matrix A, nth order. Minor some element a ij , determinant of the matrix nth order is called determinant(n - 1)th order, obtained from the original one by crossing out the row and column at the intersection of which the selected element a ij is located. Denoted by M ij.

Let's look at an example determinant of the matrix 3 - its order:

Then according to the definition minor, minor M 12, corresponding to element a 12, will be determinant:

At the same time, with the help minors can make the calculation task easier determinant of the matrix. We need to spread it out matrix determinant along some line and then determinant will be equal to the sum of all elements of this line by their minors. Decomposition determinant of the matrix 3 - its order will look like this:

The sign in front of the product is (-1) n, where n = i + j.

Algebraic additions:

Algebraic complement element a ij is called its minor, taken with a "+" sign if the sum (i + j) is an even number, and with a "-" sign if this sum is an odd number. Denoted by A ij. A ij = (-1) i+j × M ij.

Then we can reformulate the property stated above. Matrix determinant equal to the sum of the product of the elements of a certain row (row or column) matrices to their corresponding algebraic additions. Example:

4. Inverse matrix and its calculation.

Let A be square matrix nth order.

Square matrix A is called non-degenerate if matrix determinant(Δ = det A) is not zero (Δ = det A ≠ 0). Otherwise (Δ = 0) matrix A is called degenerate.

Matrix, allied to matrix Ah, it's called matrix

Where A ij - algebraic complement element a ij given matrices(it is defined in the same way as algebraic complement element determinant of the matrix).

Matrix A -1 is called inverse matrix A, if the condition is met: A × A -1 = A -1 × A = E, where E is unit matrix same order as matrix A. Matrix A -1 has the same dimensions as matrix A.

Inverse matrix

If there are square matrices X and A, satisfying the condition: X × A = A × X = E, where E is the unit matrix of the same order, then matrix X is called inverse matrix to the matrix A and is denoted by A -1. Any non-degenerate matrix has inverse matrix and, moreover, only one, i.e., in order to have a square matrix A had inverse matrix, it is necessary and sufficient for it determinant was different from zero.

To receive inverse matrix use the formula:

Where M ji is additional minor element a ji matrices A.

5. Matrix rank. Calculating rank using elementary transformations.

Consider a rectangular matrix mxn. Let us select some k rows and k columns in this matrix, 1 £ k £ min (m, n) . From the elements located at the intersection of the selected rows and columns, we compose a k-th order determinant. All such determinants are called matrix minors. For example, for a matrix you can compose second-order minors and first order minors 1, 0, -1, 2, 4, 3.

Definition. The rank of a matrix is ​​the highest order of the non-zero minor of this matrix. Denote the rank of the matrix r(A).

In the example given, the rank of the matrix is ​​two, since, for example, minor

It is convenient to calculate the rank of a matrix using the method of elementary transformations. Elementary transformations include the following:

1) rearrangement of rows (columns);

2) multiplying a row (column) by a number other than zero;

3) adding to the elements of a row (column) the corresponding elements of another row (column), previously multiplied by a certain number.

These transformations do not change the rank of the matrix, since it is known that 1) when the rows are rearranged, the determinant changes sign and, if it was not equal to zero, then it will no longer be; 2) when multiplying a string of a determinant by a number that is not equal to zero, the determinant is multiplied by this number; 3) the third elementary transformation does not change the determinant at all. Thus, by performing elementary transformations on a matrix, one can obtain a matrix for which it is easy to calculate the rank of it and, consequently, of the original matrix.

Definition. A matrix obtained from a matrix using elementary transformations is called equivalent and is denoted A IN.

Theorem. The rank of the matrix does not change during elementary matrix transformations.

Using elementary transformations, you can reduce the matrix to the so-called step form, when calculating its rank is not difficult.

Matrix is called stepwise if it has the form:

Obviously, the rank of the echelon matrix is ​​equal to the number of non-zero rows , because there is a minor of order not equal to zero:

.

Example. Determine the rank of a matrix using elementary transformations.

The rank of the matrix is ​​equal to the number of non-zero rows, i.e. .

Algebraic complement- concept of matrix algebra; in relation to the element aij of the square matrix A is formed by multiplying the minor of the element aij by (1)i+j; is denoted by Аij: Aij=(1)i+jMij, where Mij is the minor of the element aij of the matrix A=, i.e. determinant... ... Economic-mathematical dictionary

algebraic complement- The concept of matrix algebra; in relation to the element aij of the square matrix A is formed by multiplying the minor of the element aij by (1)i+j; is denoted by Аij: Aij=(1)i+jMij, where Mij is the minor of the element aij of the matrix A=, i.e. matrix determinant,... ... Technical Translator's Guide

See Art. Determinant... Great Soviet Encyclopedia

For a minor M, a number equal to where M is a minor of order k, located in rows with numbers and columns with numbers of some square matrix A of order n; determinant of a matrix of order n k obtained from the matrix A by deleting the rows and columns of the minor M;... ... Mathematical Encyclopedia

Wiktionary has an entry for "addition" Addition can mean... Wikipedia

The operation puts a subset of the given set X in correspondence with another subset so that if Mi N are known, then the set X can be restored in one way or another. Depending on what structure the set X is endowed with,... ... Mathematical Encyclopedia

Or a determinant, in mathematics, a recording of numbers in the form of a square table, in correspondence with which another number is placed (the value of the determinant). Very often, the concept of determinant means both the meaning of the determinant and the form of its recording.… … Collier's Encyclopedia

For a theorem from probability theory, see the article Local theorem of Moivre-Laplace. Laplace's theorem is one of the theorems of linear algebra. Named after the French mathematician Pierre Simon Laplace (1749 1827), who is credited with formulating ... ... Wikipedia

- (Laplacian matrix) one of the representations of a graph using a matrix. The Kirchhoff matrix is ​​used to count the spanning trees of a given graph (matrix tree theorem) and is also used in spectral graph theory. Contents 1... ...Wikipedia

An equation is a mathematical relationship that expresses the equality of two algebraic expressions. If an equality is true for any admissible values ​​of the unknowns included in it, then it is called an identity; for example, a ratio of the form... ... Collier's Encyclopedia

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• Discrete mathematics, A. V. Chashkin. 352 pp. The textbook consists of 17 chapters on the main sections of discrete mathematics: combinatorial analysis, graph theory, Boolean functions, computational complexity and coding theory. Contains...

Matrix minors

Let given a square matrix A, n - th order. Minor some element aij, the determinant of a matrix of nth order is called determinant(n - 1) - th order, obtained from the original one by crossing out the row and column at the intersection of which the selected element aij is located. Denoted by Mij.

Let's look at an example determinant of the matrix 3 - its order:
Minors and algebraic complements, the determinant of the matrix 3 is its order, then according to the definition minor, minor M12, corresponding to element a12, will be determinant:At the same time, with the help minors can make the calculation task easier determinant of the matrix. We need to spread it out matrix determinant along some line and then determinant will be equal to the sum of all elements of this line by their minors. Decomposition determinant of the matrix 3 - its order will look like this:

, the sign in front of the product is (-1) n, where n = i + j.

Algebraic additions:

Algebraic complement element aij is called its minor, taken with a “+” sign if the sum (i + j) is an even number, and with a “-” sign if this sum is an odd number. Denoted by Aij.
Аij = (-1)i+j × Мij.

Then we can reformulate the property stated above. Matrix determinant equal to the sum of the product of the elements of a certain row (row or column) matrices to their corresponding algebraic additions. Example.

Task 1.

For a given determinant

find minors and algebraic complements of the elements α 12, α 32. Compute determinant : a) decomposing it into the elements of the first row and second column; b) having previously received zeros in the first line.

We find:

M 12 =
= –8–16+6+12+4–16 = –18,

M 32 =
= –12+12–12–8 = –20.

The algebraic complements of the elements a 12 and a 32 are respectively equal:

A 12 = (–1) 1+2 M 12 = –(–18) = 18,

A 32 = (–1) 3+2 M 32 = –(–20) = 20.

a) Let’s calculate the determinant by expanding it into the elements of the first row:

A 11 A 11 + a 12 A 12 + a 13 A 13 + a 14 A 14 = –3
–2 +

1
= – 3(8 + 2 + 4 – 4) – 2(– 8 – 16 + 6 + 12 + 4 – 16) + (16 – 12 – – 4 + 32) = 38;

Let's expand the determinant into the elements of the second column:

= – 2 – 2
+ 1
= – 2(– 8 + 6 – 16 + + 12 + 4 – 16) – 2(12 + 6 – 6 – 16) + (– 6 + 16 – 12 – 4) = 38;

b) Let's calculate , having first obtained zeros in the first line. We use the corresponding property of determinants. Let's multiply the third column of the determinant by 3 and add it to the first, then multiply by –2 and add it to the second. Then in the first line all elements except one will be zeros. Let us decompose the determinant obtained in this way into the elements of the first row and calculate it:

= =
=
=
=

= – (– 56 + 18) = 38.

(In the third-order determinant, we got zeros in the first column due to the same property of determinants as above.) ◄

Task 2.

A system of linear inhomogeneous algebraic equations is given

Check whether this system is compatible, and if so, solve it: a) using Cramer’s formulas; b) using an inverse matrix (matrix method); c) Gaussian method.

We will check the compatibility of this system using the Kronecker–Capelli theorem. Using elementary transformations, we find the rank of the matrix

A =

given system and the rank of the extended matrix

IN =

.

To do this, multiply the first row of matrix B by –2 and add it with the second, then multiply the first row by –3 and add it with the third, swap the second and third columns. We get

IN =

~

~
.

Therefore, rank A= rank IN= 3 (i.e. the number of unknowns). This means that the original system is consistent and has a unique solution.

a) According to Cramer’s formulas

x = x/ , y = y/ , z = z/ ,

=
= – 16;

x =
= 64;

y =
= – 16;

z=
= 32,

we find: x = 64/(– 16) = – 4, y = – 16/(– 16) = 1, z = 32/(– 16)= – 2;

b) To find a solution to the system using the inverse matrix, we write the system of equations in matrix form AH = . The solution of the system in matrix form has the form x = A –1 . Using the formula, we find the inverse matrix A –1 (it exists because = det A = – 16 ≠ 0):

A 11 =
= – 15, A 21 = –
= 16, A 31 =
= – 11,

A 12 = –
= – 3, A 22 =
= 0, A 32 = –
= 1,

A 13 =
= – 14, A 23 = –
= 16, A 33 =
= – 6,

A –1 =

.

System solution:

X = =
=
=

.

So, x = –4, y = 1, z = –2;

c) Let's solve the system using the Gaussian method. Let's exclude x from the second and third equations. To do this, multiply the first equation by 2 and subtract it from the second, then multiply the first equation by 3 and subtract it from the third:

From the resulting system we find x = – 4, y = 1, z = –2. ◄

Task 5.

The vertices of the pyramid are at the points A(2; 3; 4), B(4; 7; 3), C(1; 2; 2) And D(– 2; 0; – 1). Calculate: a) area of ​​the face ABC; b) cross-sectional area passing through the middle of the ribs AB, A.C., AD; c) volume of the pyramid ABCD.

A) It is known that S ABC =
. We find:
= (2; 4; – 1) ,

= (– 1; – 1; – 2) ,

=
= – 9 i + 5 j + 2 k.

Finally we have:

S ABC =
=
;

b) Midpoints of ribs AB, Sun And AD are at points K (3; 5; 3.5),

M (1.5; 2.5; 3),N (0; 1,5; 1,5) . Next we have:

S slaughter =
,
= (– 1,5; – 2,5; – 0,5),
= (– 3; – 3,5; – 2),

=
= 3.25i – 1.5j – 2.25k,

S slaughter =
=
;

c) Since V feast =
,
= (– 4; – 3; – 5),

=
= 11, That V = 11/6 . ◄

Problem 6

Strength F = (2; 3;– 5) applied to a point A(1; – 2; 2). Calculate: a) work of force F in the case when the point of its application, moving rectilinearly, moves from the position A to position B(1; 4; 0); b) moment modulus F relative to the point IN.

A) Since A =F · s , s =
= (0; 6; – 2) ,

That F · = 2·0 + 3·6 + (– 5)(– 2) = 28; A = 28;

b) Moment of force M =
,
= (0; – 6; 2) ,

=
= 24 i + 4 j + 12 k .

Hence, =
= 4
. ◄

Task 8.

Known peaks O(0; 0),A(– 2; 0) parallelogram OASD and the point of intersection of its diagonals B(2;–2). Write down the equations of the parallelogram sides.

Side equation OA you can immediately write: y = 0 . Further, since the point IN is the midpoint of the diagonal AD(Fig. 1), then using the formulas for dividing a segment in half, you can calculate the coordinates of the vertex D(x; y) :

2 =
, –2 =
,

where x = 6 , y = –4 .

Now you can find the equations for all other sides. Considering the parallelism of the sides O.A. And CD, we compose the equation of the side CD: y = –4 . Side equation O.D. is compiled from two known points:

=
,

where y = – x, 2 x + 3 y = 0 .

Finally, we find the equation of the side A.C., given the fact that it passes through a known point A (– 2; 0) parallel to a known line O.D.:

y – 0 = – (x + 2) or 2 x + 3 y + 4 = 0 . ◄

Task 9.

Given the vertices of a triangle ABC: A(4; 3), B(– 3; – 3), C(2; 7) . Find:

a) side equation AB;

b) height equation CH;

c) median equation A.M.;

d) point N median intersection A.M. and heights CH;

e) equation of a line passing through a vertex C parallel to the side AB;

e) distance from the point C to a straight line AB.

A) Using the equation straight line passing through two points, we obtain the equation of the side AB:

=
,

where 6(x – 4) = 7(y – 3) or 6 x – 7 y – 3 = 0 ;

b) According to the equation

y = kx + b (k = tg α ) ,

straight line slope AB k 1 =6/7 . Taking into account conditions for perpendicularity of lines AB And CH height slope CH k 2 = –7/6 (k 1∙ k 2 = –1). By point C(2; 7) and slope k 2 = –7/6 make up the height equation CH: (y – y 0 = k(x – x 0 ) )

y – 7 = – (x – 2) or 7 x + 6 y – 56 = 0 ;

c) Using known formulas we find the coordinates x, y middle M segment B.C.:

x = (– 3 + 2)/2 = –1/2, y = (– 3 + 7)/2 = 2.

Now for two known points A And M compose the median equation A.M.:

=
or 2 x – 9 y + 19 = 0 ;

d) To find the coordinates of a point N median intersection A.M. and heights CH compose a system of equations

Solving it, we get N (26/5; 49/15) ;

e) Since the line passing through the vertex C, parallel to the side AB, then their angular coefficients are equal k 1 =6/7 . Then, according to the equation:

y – y 0 = k(x – x 0 ) , by point C and slope k 1 compose equations of a straight line CD:

y – 7 = (x – 2) or 6 x – 7 y + 37 = 0 ;

f) Distance from point C to a straight line AB calculated using the well-known formula:

d = | CH| =

The solution to this problem is illustrated in Fig. 2 ◄

Problem 10.

Given four points A 1 (4; 7; 8), A 2 (– 1;13; 0), A 3 (2; 4; 9), A 4 (1; 8; 9) . Make up equations:

a) planes A 1 A 2 A 3 ; b) straight A 1 A 2 ;

c) straight A 4 M, perpendicular to the plane A 1 A 2 A 3 ;

d) straight A 4 N, parallel to the line A 1 A 2 .

Calculate:

e) sine of the angle between the straight line A 1 A 4 and plane A 1 A 2 A 3 ;

e) cosine of the angle between the coordinate plane ABOUTxy and plane A 1 A 2 A 3 .

A) Using the formula plane equations from three points, we compose the equation of the plane A 1 A 2 A 3 :

where 6x – 7y – 9z + 97 = 0;

b) Considering equations of a line passing through two points, straight line equations A 1 A 2 can be written in the form

=
=
;

c) From conditions for perpendicularity of a line A 4 M and planes A 1 A 2 A 3 it follows that as the direction vector of the straight line s you can take a normal vector n = (6; – 7; – 9) plane A 1 A 2 A 3 . Then the equation of the line A 4 M taking into account canonical equations of the straight line will be written in the form

=
=
;

d) Since it is straight A 4 N parallel to the line A 1 A 2 , then their direction vectors s 1 And s 2 can be considered identical: s 1 =s 2 = (5; – 6; 8) . Therefore, the equation of the line A 4 N looks like

=
=
;

d) According to the formula for finding the magnitude of the angle between a straight line and a plane

sin φ =

e) In accordance with the formula for finding angle between planes

cos φ =
=
◄

Problem 11.

Write an equation for a plane passing through the points M(4; 3; 1) And

N(– 2; 0; – 1) parallel to the line drawn through the points A(1; 1; – 1) And

B(– 3; 1; 0).

According to the formula equations of a line in space passing through two points, the equation of a line AB looks like

=
=
.

If the plane passes through a point M(4; 3; 1) , then its equation can be written in the form A(x – 4) + B(y – 3) + C(z – 1) = 0 . Since this plane also passes through the point N(– 2; 0; – 1) , then the condition is satisfied

A(– 2 – 4) + B(0 – 3) + C(– 1 – 1) = 0 or 6A + 3B + 2C = 0.

Since the desired plane is parallel to the found line AB, then taking into account the formulas conditions for parallelism of a line and a plane we have:

–4A + 0B + 1C = 0 or 4A – C = 0.

Solving the system

we find that C = 4 A, B = – A. Let's substitute the obtained values WITH And B into the equation of the desired plane, we have

A(x – 4) – A(y – 3) + 4A(z – 1) = 0.

Because A ≠ 0 , then the resulting equation is equivalent to the equation

3(x – 4) – 14(y – 3) + 12(z – 1) = 0. ◄

Problem 12.

Find coordinates x 2 , y 2 , z 2 points M 2 , symmetrical point M 1 (6; – 4; – 2) relative to the plane x + y + z – 3 = 0 .

Let us write down the parametric equations of the straight line M 1 M 2 , perpendicular to this plane: x = 6 + t, y = – 4 + t, z = – 2 + t. Having solved them together with the equation of the given plane, we find t = 1 and therefore the point M intersection of a straight line M 1 M 2 with this plane: M (7; – 3; – 1) . Since the point M is the midpoint of the segment M 1 M 2 , then the equalities are true.; c) a parabola with directrix b