Matrix minors
Let a square matrix A, n -th order. Minor some element aij , the determinant of a matrix of the nth order is called determinant(n - 1) - th order, obtained from the original by deleting the row and column at the intersection of which the selected element aij is located. Designated Mij.
Let's look at an example matrix determinant 3 - its order:
Minors and algebraic additions, 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 it easier to calculate matrix determinant. Need to decompose matrix determinant along some line and then determinant will be equal to the sum of all elements of this row and their minors. Decomposition matrix determinant 3 - its order will look like this:
, the sign before the product is (-1) n , where n = i + j.
Algebraic additions:
Algebraic addition 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 Aij.
Аij = (-1)i+j × Мij.
Then we can reformulate the above property. Matrix determinant is equal to the sum of the product of the elements of some series (row or column) matrices to their respective algebraic additions. Example.
Definition. If we choose arbitrarily k rows and k columns in the nth order determinant, then the elements at the intersection of the specified rows and columns form a square matrix of order k. The determinant of such a square matrix is called k-th order minor .
Denoted by M k . If k=1, then the first-order minor is an element of the determinant.
The elements at the intersection of the remaining (n-k) rows and (n-k) columns make up a square matrix of order (n-k). The determinant of such a matrix is called the minor, additional to the minor M k . Denoted M n-k .
Algebraic complement of the minor M k we will call it an additional minor, taken with the “+” or “-” sign, depending on whether the sum of the numbers of all rows and columns in which the minor M k is located is even or odd.
If k=1, then the algebraic complement to the element aik calculated by the formula
A ik =(-1) i+k M ik, where M ik- minor (n-1) order.
Theorem. The product of a k-th order minor and its algebraic complement is equal to the sum of a certain number of terms of the determinant D n .
Proof
1. Consider a special case. Let the minor M k occupy the upper left corner of the determinant, that is, it is located in rows with numbers 1, 2, ..., k, then the minor M n-k will occupy rows k+1, k+2, ..., n.
Let us calculate the algebraic complement to the minor M k . By definition,
A n-k=(-1)s M n-k, where s=(1+2+...+k) +(1+2+...+k)= 2(1+2+...+k), then
(-1) s=1 and A n-k = M n-k. Get
M k A n-k = M k M n-k. (*)
We take an arbitrary term of the minor M k
where s is the number of inversions in the substitution
and an arbitrary term of the minor M n-k
where s * is the number of inversions in the substitution
Multiplying (1) and (3), we get
The product consists of n elements located in different rows and columns of the determinant D. Therefore, this product is a member of the determinant D. The sign of the product (5) is determined by the sum of the inversions in substitutions (2) and (4), and the sign of the analogous product in the determinant D is determined by number of inversions s k in the substitution
Obviously, s k =s+s * .
Thus, returning to equality (*), we obtain that the product M k A n-k consists only of the terms of the determinant.
2. Let the minor M k located in rows with numbers i 1 , i 2 , ..., i k and in columns with numbers j 1 , j 2 , ..., j k , and i 1< i 2 < ...< i k and j1< j 2 < ...< j k .
Using the properties of determinants, with the help of transpositions, we shift the minor to the upper left corner. We obtain a determinant D ¢ in which the minor M k occupies the upper left corner, and the complementary minor M¢ n-k is the lower right corner, then, according to what was proved in paragraph 1, we get that the product M k M¢ n-k is the sum of some number of elements of the determinant D ¢ taken with its own sign. But D ¢ is obtained from D with ( i 1 -1)+(i 2 -2)+ ...+(i k -k)=(i 1 + i 2 + ...+ i k)-(1+2+...+k) string transpositions and ( j 1 -1)+(j 2 -2)+ ...+(j k -k)=(j 1 + j 2 + ...+ j k)- (1+2+...+k) column transpositions. That is, everything was done
(i 1 + i 2 + ...+ i k)-(1+2+...+k)+ (j 1 + j 2 + ...+ j k)- (1+2+...+k )= (i 1 + i 2 + ...+ i k)+ (j 1 + j 2 + ...+ j k)- 2(1+2+...+k)=s-2(1+2 +...+k). Therefore, the terms of the determinants D and D ¢ differ in sign (-1) s-2(1+2+...+k) =(-1) s , therefore, the product (-1) s M k M¢ n-k will consist of a certain number of terms of the determinant D, taken with the same signs as they have in this determinant.
Laplace's theorem. If we choose arbitrarily k rows (or k columns) 1£k£n-1 in the nth order determinant, then the sum of the products of all the kth order minors contained in the chosen rows and their algebraic complements is equal to the determinant D.
Proof
Pick random rows i 1 , i 2 , ..., i k and prove that
It was proved earlier that all elements on the left side of the equality are contained as terms in the determinant D. Let us show that each term of the determinant D falls into only one of the terms . Indeed, every t s has the form t s =. if in this product we mark the factors whose first indices i 1 , i 2 , ..., i k, and compose their product , then you can see that the resulting product belongs to the k-th order minor. Therefore, the remaining terms, taken from the remaining n-k rows and n-k columns, form an element that belongs to the complementary minor, and, taking into account the sign, to the algebraic complement, therefore, any t s falls into only one of the products , which proves the theorem.
Consequence(theorem about the expansion of the determinant in a row) . The sum of the products of the elements of some row of the determinant and the corresponding algebraic additions is equal to the determinant.
(Proof as an exercise.)
Theorem. The sum of the products of the elements of the i-th row of the determinant and the corresponding algebraic complements to the elements of the j-th row (i¹j) is equal to 0.
Comment. It is convenient to apply the corollary of Laplace's theorem to a determinant transformed using properties in such a way that in one of the rows (or in one of the columns) all elements except one are equal to 0.
Example. Compute determinant
12 -14 +35 -147 -20 -2= -160.
Matrix minors
Let a square matrix A, nth order. Minor some element a ij , matrix determinant nth order is called determinant(n - 1) -th order, obtained from the original one by deleting the row and column at the intersection of which the selected element a ij is located. Denoted M ij .
Let's look at an example matrix determinant 3 - its order:
Then according to the definition minor, minor M 12 corresponding to the element a 12 will be determinant:
At the same time, with the help minors can make it easier to calculate matrix determinant. Need to decompose matrix determinant along some line and then determinant will be equal to the sum of all elements of this row and their minors. Decomposition matrix determinant 3 - its order will look like this:
The sign before the product is (-1) n , where n = i + j.
Algebraic additions:
Algebraic addition 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 A ij . A ij \u003d (-1) i + j × M ij.
Then we can reformulate the above property. Matrix determinant is equal to the sum of the product of the elements of a certain row (row or column) matrices to their respective algebraic additions. Example:
4. Inverse matrix and its calculation.
Let A be a square matrix nth order.
Square matrix A is called non-degenerate if matrix determinant(Δ = det A) is not equal to zero (Δ = det A ≠ 0). Otherwise (Δ = 0) matrix It's called degenerate.
Matrix, allied to matrix Ah, it's called matrix
Where A ij - algebraic addition element a ij given matrices(it is defined in the same way as algebraic addition element matrix determinant).
Matrix A -1 is called inverse matrix A, if the condition is met: A × A -1 \u003d A -1 × A \u003d E, where E is a single matrix the same order as matrix BUT. Matrix A -1 has the same dimensions as matrix BUT.
inverse matrix
If there are square matrices X and A satisfying the condition: X × A \u003d A × X \u003d E, where E is the unit matrix the same order, then matrix X is called inverse matrix to the matrix A and is denoted by A -1 . Any non-degenerate matrix It has inverse matrix and moreover, only one, i.e., in order to square matrix A had inverse matrix, it is necessary and sufficient that it determinant was different from zero.
For getting inverse matrix use the formula:
Where M ji is optional minor element a ji matrices BUT.
5. Matrix rank. Calculation of the rank using elementary transformations.
Consider a rectangular matrix mxn. Let's single out some k rows and k columns in this matrix, 1 £ k £ min (m, n) . From the elements at the intersection of the selected rows and columns, we will compose the determinant of the kth order. 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 above example, the rank of the matrix is two, since, for example, the minor
The rank of a matrix is conveniently calculated by the method of elementary transformations. The elementary transformations include the following:
1) permutations of rows (columns);
2) multiplying a row (column) by a non-zero number;
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 permuted, the determinant changes sign and, if it was not equal to zero, then it will not; 2) when multiplying the row of the 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 the 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 BUT AT.
Theorem. The rank of a matrix does not change under elementary matrix transformations.
With the help of elementary transformations, one can bring the matrix to the so-called step form, when the calculation of its rank is not difficult.
Matrix is called stepped if it has the form:
Obviously, the rank of the step matrix is equal to the number of non-zero rows , because there is a minor of the th order, not equal to zero:
.
Example. Determine the rank of a matrix using elementary transformations.
The rank of a matrix is equal to the number of non-zero rows, i.e. .
Minor M ij element aij determinant n order determinant is called ( n-1 ) obtained from the given determinant by deleting the row and column containing this element ( i -th line and j -th column).
Algebraic addition element aij is given by:
Order determinants n>3 are calculated using the theorem on the expansion of the determinant by the elements of a row or column:
Theorem. The determinant is equal to the sum of the products of the elements of any row or any column and the algebraic complements corresponding to these elements, i.e.
Example.
Calculate the determinant by expanding it over the elements of a row or column:
Solution
1. If there is only one non-zero element in any one row or one column, then there is no need to convert the determinant. Otherwise, before applying the theorem on the expansion of the determinant, we transform it using the following property: if we add the corresponding elements of another row (column) multiplied by an arbitrary factor to the elements of a row (column), then the value of the determinant will not change.
From the elements of line 3 subtract the corresponding elements of line 2 .
From the elements of column 4 we subtract the corresponding elements of column 3, multiplied by 2.
Expanding the determinant by the elements of the third row
2. The resulting 3rd order determinant can be calculated using the rule of triangles or the Sarrus rule (see above). However, the elements of the determinant are rather large numbers, so let's expand the determinant by first transforming it:
From the elements of the second row, subtract the corresponding elements of the first row, multiplied by 3.
From the elements of the first row subtract the corresponding elements of the third row.
To the elements of line 1 we add the corresponding elements of line 2
The null row determinant is 0.
So the order determinants n>3 calculated:
transformation of the determinant to a triangular form using the properties of determinants;
Decomposition of the determinant by elements of a term or column, thereby lowering its order.
Matrix rank.
The rank of a matrix is an important numerical characteristic. The most typical problem that requires finding the rank of a matrix is checking the compatibility of a system of linear algebraic equations.
Let's take a matrix BUT
order p x n
. Let k
- some natural number not exceeding the smallest of the numbers p
and n
, that is, 
Minor k-th order matrices BUT is called the determinant of the square matrix of order k x k , composed of elements of the matrix BUT , which are in preselected k lines and k columns, and the arrangement of matrix elements BUT is saved.
Consider the matrix:
Let us write down several first-order minors of this matrix. For example, if we select the third row and second column of the matrix BUT , then our choice corresponds to the first-order minor det(-4)=-4. In other words, to obtain this minor, we crossed out the first and second rows, as well as the first, third and fourth columns from the matrix BUT , and the determinant was made from the remaining element.
Thus, the first-order minors of a matrix are the matrix elements themselves.
Let us show several minors of the second order. Select two rows and two columns. For example, take the first and second rows, and the third and fourth columns. With this choice, we have a second-order minor 
.
Another second-order minor of the matrix BUT is a minor 
The third-order minors of the matrix can be found similarly BUT . Since in the matrix BUT only three lines, then select them all. If we select the first three columns for these rows, we get a third-order minor:
Another third-order minor is:
For a given matrix BUT There are no minors of order higher than three, since
How many minors are there k -wow matrix order BUT order p x n ? Not a lot!
Number of order minors k can be calculated using the formula:
Matrix rank is the highest order of a matrix minor other than zero.
Matrix rank BUT denoted as rang(A). From the definitions of the rank of a matrix and the minor of a matrix, we can conclude that the rank of a zero matrix is equal to zero, and the rank of a nonzero matrix is at least one.
So, the first method for finding the rank of a matrix is minor enumeration method . This method is based on determining the rank of the matrix.
Suppose we need to find the rank of a matrix BUT order p x n .
If there is at least one matrix element that is non-zero, then the rank of the matrix is at least equal to one (since there is a first-order minor that is not equal to zero).
Next, we iterate over the minors of the second order. If all second-order minors are equal to zero, then the rank of the matrix is equal to one. If there exists at least one non-zero second-order minor, then we pass to the enumeration of third-order minors, and the rank of the matrix is at least equal to two.
Similarly, if all third-order minors are zero, then the rank of the matrix is two. If there is at least one non-zero third-order minor, then the rank of the matrix is at least three, and we proceed to the enumeration of fourth-order minors.
Note that the rank of a matrix cannot exceed the smallest of the numbers p and n .
Example.
Find the rank of a matrix 
.
Solution.
1. Since the matrix is non-zero, its rank is not less than one.
2. One of the minors of the second order 
is different from zero, therefore, the rank of the matrix BUT
at least two.
3. Minors of the third order 
All third-order minors are equal to zero. Therefore, the rank of the matrix is two.
rank(A) = 2.
There are other methods for finding the rank of a matrix that allow you to get the result with less computational work.
One of these methods is fringing minor method . When using this method, the calculations are somewhat reduced, and yet they are rather cumbersome.
There is another way to find the rank of a matrix - using elementary transformations (Gauss method).
The following matrix transformations are called elementary :
swapping the rows (or columns) of the matrix;
multiplication of all elements of any row (column) of a matrix by an arbitrary number k, different from zero;
addition to the elements of any row (column) of the corresponding elements of another row (column) of the matrix, multiplied by an arbitrary number k.
Matrix B is called equivalent to matrix A, if AT derived from BUT with the help of a finite number of elementary transformations. The equivalence of matrices is denoted by the symbol « ~ » , that is, written A~B.
Finding the rank of a matrix using elementary matrix transformations is based on the statement: if the matrix AT obtained from the matrix BUT using a finite number of elementary transformations, then r ang(A) = rank(B) , i.e. the ranks of the equivalent matrices are .
The essence of the method of elementary transformations is to bring the matrix, the rank of which we need to find, to a trapezoid (in a particular case, to an upper triangular one) using elementary transformations.
The rank of matrices of this kind is very easy to find. It is equal to the number of rows containing at least one non-null element. And since the rank of the matrix does not change during elementary transformations, the resulting value will be the rank of the original matrix.
Example.
Using the method of elementary transformations, find the rank of the matrix
.
Solution.
1. Swap the first and second rows of the matrix BUT , since the element a 11 =0, and the element a 21 different from zero:
~ 
In the resulting matrix, the element is equal to one. Otherwise, it was necessary to multiply the elements of the first row by . Let's make all elements of the first column, except for the first one, zero. The second line already has zero, to the third line we add the first one, multiplied by 2:
The element in the resulting matrix is nonzero. Multiply the elements of the second row by 
The second column of the resulting matrix has the desired form, since the element is already equal to zero.
Because 
, a 
, then swap the third and fourth columns and multiply the third row of the resulting matrix by :
The original matrix is reduced to a trapezoid, its rank is equal to the number of rows containing at least one non-zero element. There are three such rows, so the rank of the original matrix is three. r ang(A)=3.
Inverse matrix.
Let's have a matrix BUT .
Matrix inverse to matrix A , is called the matrix A-1 such that A -1 A = A A -1 = E .
An inverse matrix can only exist for a square matrix. Moreover, it itself is of the same dimension as the original matrix.
For a square matrix to have an inverse, it must be nonsingular (i.e. Δ ≠0 ). This condition is also sufficient for the existence A-1 to the matrix BUT . So, every non-singular matrix has an inverse, and, moreover, a unique one.
Algorithm for finding the inverse matrix using the example of a matrix BUT :
1. Find the matrix determinant. If a Δ ≠0 , then the matrix A-1 exists.
2. Compose the matrix B of algebraic complements of the elements of the original matrix BUT . Those. in the matrix AT element i - oh line and j -th column will be the algebraic complement A ij element aij original matrix.
3. Transpose the matrix AT and get B t .
4. Find the inverse matrix by multiplying the resulting matrix B t per number .
Example.
For a given matrix, find the inverse and check:
Solution
We use the previously described algorithm for finding the inverse matrix.
1. To find out the existence of an inverse matrix, it is necessary to calculate the determinant of this matrix. Let's use the rule of triangles:
The matrix is non-degenerate, hence it is invertible.
Find the algebraic complements of all elements of the matrix:
From the found algebraic complements, a matrix is compiled:
and transposed 
Dividing each element of the resulting matrix by the determinant, we obtain a matrix inverse to the original one:
The check is carried out by multiplying the resulting matrix by the original one. If the inverse matrix is found correctly, the result of multiplication will be the identity matrix.
To find the inverse matrix for a given one, you can use the Gaussian method (of course, you first need to make sure that the matrix is invertible), which I leave for independent work.
Task 1.
For this determinant
find minors and algebraic complements of elements α 12 , α 32 . Compute determinant 
: a) decomposing it into 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, respectively, are:
A 12 \u003d (-1) 1 + 2 M 12 \u003d - (-18) \u003d 18,
A 32 \u003d (-1) 3 + 2 M 32 \u003d - (-20) \u003d 20.
a) Calculate the determinant by expanding it over 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 by 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) Calculate , having previously obtained zeros in the first row. We use the corresponding property of determinants. Multiply the third column of the determinant by 3 and add to the first, then multiply by -2 and add to the second. Then in the first line all elements, except for one, will be zeros. We expand the determinant obtained in this way by the elements of the first row and calculate it:
=
=
=
=
=
= – (– 56 + 18) = 38.
(In the third-order determinant, zeros were obtained in the first column by the same property of determinants as above.) ◄
Task 2.
Given a system of linear nonhomogeneous algebraic equations
Check whether this system is compatible and, if it is compatible, solve it: a) using Cramer's formulas; b) using the 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
BUT =
of this system and the rank of the extended matrix
AT =
.
To do this, we multiply the first row of matrix B by -2 and add it to the second one, then we multiply the first row by -3 and add it to the third one, and swap the second and third columns. Get
AT =
~
~
.
Therefore, rank BUT= rank AT= 3 (i.e., the number of unknowns). Hence, 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 BUT –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) We solve the system using the Gauss method. Exclude x from the second and third equations. To do this, multiply the first equation by 2 and subtract from the second, then multiply the first equation by 3 and subtract 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) face area ABC; b) the area of the section passing through the middle of the ribs AB, AC, AD; c) the 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) The middle of the ribs AB, Sun and BUTD are at the points K (3; 5; 3.5),
M (1.5; 2.5; 3),N (0; 1,5; 1,5) . Next we have:
S sech
=
,
= (– 1,5; – 2,5; – 0,5),
= (– 3; – 3,5; – 2),
=
= 3.25i - 1.5j - 2.25k,
S sech
=
=
;
c) Because V feast
=
,
= (– 4; – 3; – 5),
=
= 11,
then
V
= 11/6
. ◄
Task 6
Strength F = (2; 3;– 5) attached to a point A(1; - 2; 2). Calculate: a) the work of the force F in the case when the point of its application, moving rectilinearly, moves from the position BUT into position B(1; 4; 0); b) the modulus of the moment of force F relative to the point AT.
A) because A =F
·
s
,
s
=
= (0; 6; – 2)
,
then F = 2 0 + 3 6 + (– 5)(– 2) = 28; A = 28;
b) Moment of force M
=
,
= (0; – 6; 2)
,
=
= 24
i
+ 4
j
+ 12
k
.
Consequently,
=
= 4
.
◄
Task 8.
Peaks known O(0; 0),A(– 2; 0) parallelogram SLAD and the point of intersection of its diagonals B(2;–2). Write the equations for the sides of the parallelogram.
side equation OA can be written immediately: y = 0 . Further, since the point AT is the midpoint of the diagonal AD(Fig. 1), then using the formulas for dividing the 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 of all other sides. Given the parallelism of the sides OA and CD, compose the equation of the side CD: y = –4 . side equation OD is compiled from two known points:
=
,
where y = – x, 2 x + 3 y = 0 .
Finally, we find the side equation AC, given the fact that it passes through a known point A (-2; 0) parallel to a known line OD:
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 AM;
d) point N median intersections AM and height CH;
e) equation of a straight line passing through the vertex C parallel to side AB;
f) distance from point C to straight AB.
A) using the equation straight line passing through two points, we get the side equation 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 ) ,
slope of straight line AB k 1 =6/7 . Taking into account perpendicularity conditions AB and CH height slope CH k 2 = –7/6 (k 1∙ k 2 = –1). By point C(2; 7) and angular coefficient k 2 = –7/6 write the height equation CH: (y – y 0 = k(x – x 0 ) )
y – 7 = – (x – 2) or 7 x + 6 y – 56 = 0 ;
c) According to known formulas, we find the coordinates x, y middle M segment BC:
x = (– 3 + 2)/2 = –1/2, y = (– 3 + 7)/2 = 2.
Now for two known points A and M write the median equation AM:
=
or 2
x
– 9
y
+ 19 = 0
;
d) To find the coordinates of a point N median intersections AM and height 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 slopes are k 1 =6/7 . Then, according to the equation:
y – y 0 = k(x – x 0 ) , by point C and angular coefficient k 1 compose the equations of a straight line CD:
y – 7 = (x – 2) or 6 x – 7 y + 37 = 0 ;
f) Distance from point C to straight AB calculated according to the well-known formula:
d
= |
CH|
=
The solution of this problem is illustrated in fig. 2◄
Task 10.
Given four points A 1 (4; 7; 8), A 2 (– 1; 13; 0), A 3 (2; 4; 9), A 4 (1; 8; 9) . Write 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) the sine of the angle between the line A 1 A 4 and plane A 1 A 2 A 3 ;
f) cosine of the angle between the coordinate plane Oxy and plane BUT 1 BUT 2 BUT 3 .
A) Using the formula three-point plane equations, compose the equation of the plane BUT 1 BUT 2 BUT 3 :
where 6x - 7y - 9z + 97 = 0;
b) Considering equation of a straight line passing through two points, equations of the straight line BUT 1 BUT 2 can be written in the form
=
=
;
c) From perpendicularity conditions BUT 4 M and plane BUT 1 BUT 2 BUT 3 it follows that as a direction vector, the line s you can take a normal vector n = (6; – 7; – 9) plane BUT 1 BUT 2 BUT 3 . Then the equation of the line BUT 4 M taking into account canonical equations of the straight line will be written in the form
=
=
;
d) Since the line A 4 N parallel to a straight line BUT 1 BUT 2 , then their direction vectors s 1 and s 2 can be considered as matching: s 1 =s 2 = (5; – 6; 8) . Therefore, the equation of the line A 4 N has the form
=
=
;
e) By the formula of finding the angle between the line and the plane
sin φ =
f) In accordance with the formula for finding angle values between planes
cos phi
=
=
◄
Task 11.
Write an equation for a plane passing through the points M(4; 3; 1) and
N(– 2; 0; – 1) parallel to the line through the points A(1; 1; – 1) and
B(– 3; 1; 0).
According to the formula equations of a straight line in space passing through two points, the equation of a straight line AB has the form
=
=
.
If the plane passes through a point M(4; 3; 1) , then its equation can be written as 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
A(-2-4) + B(0-3) + C(-1-1) = 0 or 6A + 3B + 2C = 0.
Since the desired plane is parallel to the line found AB, then taking into account the formulas parallelism conditions for 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. Substitute the obtained values FROM 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. ◄
Task 12.
Find coordinates x 2 , y 2 , z 2 points M 2 , symmetric point M 1 (6; – 4; – 2) relative to the plane x + y + z – 3 = 0 .
Let us write the parametric equations of the straight line M 1 M 2 , perpendicular to the given plane: x = 6 + t, y = – 4 + t, z = – 2 + t. Solving them together with the equation of the given plane, we find t = 1 and hence the point M line intersection 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 equalities are true.; c) a parabola with directrix b
Elements of Linear Algebra This section includes the main types of problems that are considered in the topic "Linear Algebra": calculation of determinants, actions on
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Guidelines for the implementation of extracurricular independent work of the student Discipline "Mathematics" for the specialty
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