In this topic we will consider the concepts of algebraic complement and minor. The presentation of the material is based on the terms explained in the topic "Matrixes. Types of matrices. Basic terms". We will also need some formulas for calculating determinants. Since this topic contains a lot of terms related to minors and algebraic complements, I will add a brief summary to make it easier to navigate the material.
Minor $M_(ij)$ of element $a_(ij)$
$M_(ij)$ element$a_(ij)$ matrices $A_(n\times n)$ name the determinant of the matrix obtained from matrix $A$ by deleting i-th line and the j-th column (i.e. the row and column at the intersection of which the element $a_(ij)$ is located).
For example, consider a fourth-order square matrix: $A=\left(\begin(array) (cccc) 1 & 0 & -3 & 9\\ 2 & -7 & 11 & 5 \\ -9 & 4 & 25 & 84 \\ 3 & 12 & -5 & 58 \end(array) \right)$. Let's find the minor of the element $a_(32)$, i.e. let's find $M_(32)$. First, let's write down the minor $M_(32)$ and then calculate its value. In order to compose $M_(32)$, we delete the third row and second column from the matrix $A$ (it is at the intersection of the third row and the second column that the element $a_(32)$ is located). We will obtain a new matrix, the determinant of which is the required minor $M_(32)$:
This minor is easy to calculate using formula No. 2 from the calculation topic:
$$ M_(32)=\left| \begin(array) (ccc) 1 & -3 & 9\\ 2 & 11 & 5 \\ 3 & -5 & 58 \end(array) \right|= 1\cdot 11\cdot 58+(-3) \cdot 5\cdot 3+2\cdot (-5)\cdot 9-9\cdot 11\cdot 3-(-3)\cdot 2\cdot 58-5\cdot (-5)\cdot 1=579. $$
So, the minor of the element $a_(32)$ is 579, i.e. $M_(32)=579$.
Often, instead of the phrase “matrix element minor” in the literature, “determinant element minor” is found. The essence remains the same: to get the minor of the element $a_(ij)$ you need to cross out from the original determinant i-th line And jth column. The remaining elements are written into a new determinant, which is the minor of the element $a_(ij)$. For example, let's find the minor of the element $a_(12)$ of the determinant $\left| \begin(array) (ccc) -1 & 3 & 2\\ 9 & 0 & -5 \\ 4 & -3 & 7 \end(array) \right|$. To write down the required minor $M_(12)$ we need to delete the first row and second column from the given determinant:
To find the value of this minor, we use formula No. 1 from the topic of calculating determinants of the second and third orders:
$$ M_(12)=\left| \begin(array) (cc) 9 & -5\\ 4 & 7 \end(array) \right|=9\cdot 7-(-5)\cdot 4=83. $$
So, the minor of the element $a_(12)$ is 83, i.e. $M_(12)=83$.
Algebraic complement $A_(ij)$ of element $a_(ij)$
Let a square matrix $A_(n\times n)$ be given (i.e., a square matrix of nth order).
Algebraic complement$A_(ij)$ element$a_(ij)$ of matrix $A_(n\times n)$ is found by the following formula: $$ A_(ij)=(-1)^(i+j)\cdot M_(ij), $$
where $M_(ij)$ is the minor of element $a_(ij)$.
Let us find the algebraic complement of element $a_(32)$ of the matrix $A=\left(\begin(array) (cccc) 1 & 0 & -3 & 9\\ 2 & -7 & 11 & 5 \\ -9 & 4 & 25 & 84\\ 3 & 12 & -5 & 58 \end(array) \right)$, i.e. let's find $A_(32)$. We previously found the minor $M_(32)=579$, so we use the result obtained:
Usually, when finding algebraic complements, the minor is not calculated separately, and only then the complement itself. The minor note is omitted. For example, let's find $A_(12)$ if $A=\left(\begin(array) (ccc) -5 & 10 & 2\\ 6 & 9 & -4 \\ 4 & -3 & 1 \end( array)\right)$. According to the formula $A_(12)=(-1)^(1+2)\cdot M_(12)=-M_(12)$. However, to get $M_(12)$ it is enough to cross out the first row and second column of the matrix $A$, so why introduce an extra notation for the minor? Let’s immediately write down the expression for the algebraic complement $A_(12)$:
Minor of the kth order of the matrix $A_(m\times n)$
If in the previous two paragraphs we talked only about square matrices, then here we will also talk about rectangular matrices, in which the number of rows does not necessarily equal the number of columns. So, let the matrix $A_(m\times n)$ be given, i.e. a matrix containing m rows and n columns.
Minor kth order matrix $A_(m\times n)$ is a determinant whose elements are located at the intersection of k rows and k columns of matrix $A$ (it is assumed that $k≤ m$ and $k≤ n$).
For example, consider this matrix:
$$A=\left(\begin(array) (cccc) -1 & 0 & -3 & 9\\ 2 & 7 & 14 & 6 \\ 15 & -27 & 18 & 31\\ 0 & 1 & 19 & 8\\ 0 & -12 & 20 & 14\\ 5 & 3 & -21 & 9\\ 23 & -10 & -5 & 58 \end(array) \right) $$
Let us write some third order minor for it. To write a third-order minor, we need to select any three rows and three columns of this matrix. For example, take rows No. 2, No. 4, No. 6 and columns No. 1, No. 2, No. 4. At the intersection of these rows and columns the elements of the required minor will be located. In the figure, the minor elements are shown in blue:
$$ \left(\begin(array) (cccc) -1 & 0 & -3 & 9 \\ \boldblue(2) & \boldblue(7) & 14 & \boldblue(6) \\ 15 & -27 & 18 & 31\\ \boldblue(0) & \boldblue(1) & 19 & \boldblue(8)\\ 0 & -12 & 20 & 14\\ \boldblue(5) & \boldblue(3) & -21 & \boldblue(9)\\ 23 & -10 & -5 & 58 \end(array) \right);\; M=\left|\begin(array) (ccc) 2 & 7 & 6 \\ 0 & 1 & 8 \\ 5 & 3 & 9 \end(array) \right|. $$
First order minors are found at the intersection of one row and one column, i.e. first order minors are equal to the elements of a given matrix.
The kth order minor of the matrix $A_(m\times n)=(a_(ij))$ is called main, if on the main diagonal of a given minor there are only the main diagonal elements of the matrix $A$.
Let me remind you that the main diagonal elements are those elements of the matrix whose indices are equal: $a_(11)$, $a_(22)$, $a_(33)$ and so on. For example, for the matrix $A$ considered above, such elements will be $a_(11)=-1$, $a_(22)=7$, $a_(33)=18$, $a_(44)=8$. They are highlighted in green in the figure:
$$\left(\begin(array) (cccc) \boldgreen(-1) & 0 & -3 & 9\\ 2 & \boldgreen(7) & 14 & 6 \\ 15 & -27 & \boldgreen(18 ) & 31\\ 0 & 1 & 19 & \boldgreen(8)\\ 0 & -12 & 20 & 14\\ 5 & 3 & -21 & 9\\ 23 & -10 & -5 & 58 \end( array)\right)$$
For example, if in the matrix $A$ we cross out the rows and columns numbered 1 and 3, then at their intersection there will be elements of a minor of the second order, on the main diagonal of which there will be only diagonal elements of the matrix $A$ (elements $a_(11) =-1$ and $a_(33)=18$ of matrix $A$). Therefore, we get a second-order principal minor:
$$ M=\left|\begin(array) (cc) \boldgreen(-1) & -3 \\ 15 & \boldgreen(18) \end(array) \right| $$
Naturally, we could take other rows and columns, for example, with numbers 2 and 4, thereby obtaining a different principal minor of the second order.
Let some minor $M$ of the kth order of the matrix $A_(m\times n)$ be not equal to zero, i.e. $M\neq 0$. In this case, all minors whose order is higher than k are equal to zero. Then the minor $M$ is called basic, and the rows and columns on which the elements of the basic minor are located are called base strings And base columns.
For example, consider the following matrix:
$$A=\left(\begin(array) (ccc) -1 & 0 & 3 & 0 & 0 \\ 2 & 0 & 4 & 1 & 0\\ 1 & 0 & -2 & -1 & 0\ \ 0 & 0 & 0 & 0 & 0 \end(array) \right) $$
Let's write down the minor of this matrix, the elements of which are located at the intersection of rows No. 1, No. 2, No. 3 and columns No. 1, No. 3, No. 4. We get a third-order minor (its elements are highlighted in purple in the matrix $A$):
$$ \left(\begin(array) (ccc) \boldpurple(-1) & 0 & \boldpurple(3) & \boldpurple(0) & 0 \\ \boldpurple(2) & 0 & \boldpurple(4) & \boldpurple(1) & 0\\ \boldpurple(1) & 0 & \boldpurple(-2) & \boldpurple(-1) & 0\\ 0 & 0 & 0 & 0 & 0 \end(array) \ right);\; M=\left|\begin(array) (ccc) -1 & 3 & 0 \\ 2 & 4 & 1 \\ 1 & -2 & -1 \end(array) \right|. $$
Let's find the value of this minor using formula No. 2 from the topic of calculating determinants of the second and third orders:
$$ M=\left| \begin(array) (ccc) -1 & 3 & 0\\ 2 & 4 & 1 \\ 1 & -2 & -1 \end(array) \right|=4+3+6-2=11. $$
So, $M=11\neq 0$. Now let's try to compose any minor whose order is higher than three. To make a fourth-order minor, we have to use the fourth row, but all the elements of this row are zero. Therefore, any fourth-order minor will have a zero row, which means that all fourth-order minors are equal to zero. We cannot create minors of the fifth and higher orders, since the matrix $A$ has only 4 rows.
We have found a third order minor that is not equal to zero. In this case, all minors of higher orders are equal to zero, therefore, the minor we considered is basic. The rows of the matrix $A$ on which the elements of this minor are located (the first, second and third) are the basic rows, and the first, third and fourth columns of the matrix $A$ are the basic columns.
This example, of course, is trivial, since its purpose is to clearly show the essence of the basic minor. In general, there can be several basic minors, and usually the process of searching for such a minor is much more complex and extensive.
Let's introduce another concept - bordering minor.
Let a certain kth order minor $M$ of the matrix $A_(m\times n)$ be located at the intersection of k rows and k columns. Let's add another row and column to the set of these rows and columns. The resulting minor of (k+1)th order is called edge minor for minor $M$.
For example, let's look at the following matrix:
$$A=\left(\begin(array) (ccccc) -1 & 2 & 0 & -2 & -14\\ 3 & -17 & -3 & 19 & 29\\ 5 & -6 & 8 & - 9 & 41\\ -5 & 11 & 19 & -20 & -98\\ 6 & 12 & 20 & 21 & 54\\ -7 & 10 & 14 & -36 & 79 \end(array) \right) $ $
Let's write a second-order minor, the elements of which are located at the intersection of rows No. 2 and No. 5, as well as columns No. 2 and No. 4. These elements are highlighted in red in the matrix:
$$ \left(\begin(array) (ccccc) -1 & 2 & 0 & -2 & -14\\ 3 & \boldred(-17) & -3 & \boldred(19) & 29\\ 5 & -6 & 8 & -9 & 41\\ -5 & 11 & 19 & -20 & -98\\ 6 & \boldred(12) & 20 & \boldred(21) & 54\\ -7 & 10 & 14 & -36 & 79 \end(array) \right);\; M=\left|\begin(array) (ccc) -17 & 19 \\ 12 & 21 \end(array) \right|. $$
Let's add another row No. 1 to the set of rows on which the elements of the minor $M$ lie, and column No. 5 to the set of columns. We obtain a new minor $M"$ (already of the third order), the elements of which are located at the intersection of rows No. 1, No. 2, No. 5 and columns No. 2, No. 4, No. 5. The elements of the minor $M$ in the figure are highlighted in red, and the elements we add to the minor $M$ are blue:
$$ \left(\begin(array) (ccccc) -1 & \boldblue(2) & 0 & \boldblue(-2) & \boldblue(-14)\\ 3 & \boldred(-17) & -3 & \boldred(19) & \boldblue(29)\\ 5 & -6 & 8 & -9 & 41\\ -5 & 11 & 19 & -20 & -98\\ 6 & \boldred(12) & 20 & \boldred(21) & \boldblue(54)\\ -7 & 10 & 14 & -36 & 79 \end(array) \right);\; M"=\left|\begin(array) (ccc) 2 & -2 & -14 \\ -17 & 19 & 29 \\ 12 & 21 & 54 \end(array) \right|. $$
The minor $M"$ is the bordering minor for the minor $M$. Similarly, adding row No. 4 to the set of rows on which the elements of the minor $M$ lie, and column No. 3 to the set of columns, we obtain the minor $M""$ (third order minor):
$$ \left(\begin(array) (ccccc) -1 & 2 & 0 & -2 & -14\\ 3 & \boldred(-17) & \boldblue(-3) & \boldred(19) & 29 \\ 5 & -6 & 8 & -9 & 41\\ -5 & \boldblue(11) & \boldblue(19) & \boldblue(-20) & -98\\ 6 & \boldred(12) & \ boldblue(20) & \boldred(21) & 54\\ -7 & 10 & 14 & -36 & 79 \end(array) \right);\; M""=\left|\begin(array) (ccc) -17 & -3 & 19 \\ 11 & 19 & -20 \\ 12 & 20 & 21 \end(array) \right|. $$
The minor $M""$ is also a bordering minor for the minor $M$.
Minor of the kth order of the matrix $A_(n\times n)$. Additional minor. Algebraic complement to the minor of a square matrix.
Let's return to square matrices again. Let us introduce the concept of an additional minor.
Let a certain minor $M$ of the kth order of the matrix $A_(n\times n)$ be given. A determinant of (n-k)th order, the elements of which are obtained from the matrix $A$ after deleting the rows and columns containing the minor $M$, is called a minor, complementary to minor$M$.
For example, consider a fifth-order square matrix:
$$ A=\left(\begin(array)(ccccc) -1 & 2 & 0 & -2 & -14\\ 3 & -17 & -3 & 19 & 29\\ 5 & -6 & 8 & - 9 & 41\\ -5 & 11 & 16 & -20 & -98\\ -7 & 10 & 14 & -36 & 79 \end(array) \right) $$
Let's select rows No. 1 and No. 3, as well as columns No. 2 and No. 5. At the intersection of these rows and columns there will be elements of the minor $M$ of the second order. These elements are highlighted in green in the $A$ matrix:
$$ \left(\begin(array)(ccccc) -1 & \boldgreen(2) & 0 & -2 & \boldgreen(-14)\\ 3 & -17 & -3 & 19 & 29\\ 5 & \boldgreen(-6) & 8 & -9 & \boldgreen(41)\\ -5 & 11 & 16 & -20 & -98\\ -7 & 10 & 14 & -36 & 79 \end(array)\ right);\; M=\left|\begin(array)(cc) 2 & -14 \\ -6 & 41 \end(array) \right|. $$
Now let’s remove from the matrix $A$ rows No. 1 and No. 3 and columns No. 2 and No. 5, at the intersection of which there are elements of the minor $M$ (the elements of the removed rows and columns are shown in red in the figure below). The remaining elements form the minor $M"$:
$$ \left(\begin(array)(ccccc) \boldred(-1) & \boldred(2) & \boldred(0) & \boldred(-2) & \boldred(-14)\\ 3 & \ boldred(-17) & -3 & 19 & \boldred(29)\\ \boldred(5) & \boldred(-6) & \boldred(8) & \boldred(-9) & \boldred(41)\ \ -5 & \boldred(11) & 16 & -20 & \boldred(-98)\\ -7 & \boldred(10) & 14 & -36 & \boldred(79) \end(array) \right) ;\; M"=\left|\begin(array) (ccc) 3 & -3 & 19 \\ -5 & 16 & -20 \\ -7 & 14 & -36 \end(array)\right|. $$
The minor $M"$, whose order is $5-2=3$, is the complementary minor to the minor $M$.
Algebraic complement to a minor$M$ of a square matrix $A_(n\times n)$ is called the expression $(-1)^(\alpha)\cdot M"$, where $\alpha$ is the sum of the row and column numbers of the matrix $A$, on which the elements of the minor $M$ are located, and $M"$ is the minor complementary to the minor $M$.
The phrase "algebraic complement to the minor $M$" is often replaced by the phrase "algebraic complement to the minor $M$".
For example, consider the matrix $A$, for which we found the second-order minor $ M=\left| \begin(array) (ccc) 2 & -14 \\ -6 & 41 \end(array) \right| $ and its additional third-order minor: $M"=\left| \begin(array) (ccc) 3 & -3 & 19\\ -5 & 16 & -20 \\ -7 & 14 & -36 \end (array) \right|$. Let us denote the algebraic complement of the minor $M$ as $M^*$. Then, according to the definition:
$$ M^*=(-1)^\alpha\cdot M". $$
The $\alpha$ parameter is equal to the sum of the row and column numbers on which the minor $M$ is located. This minor is located at the intersection of rows No. 1, No. 3 and columns No. 2, No. 5. Therefore, $\alpha=1+3+2+5=11$. So:
$$ M^*=(-1)^(11)\cdot M"=-\left| \begin(array) (ccc) 3 & -3 & 19\\ -5 & 16 & -20 \\ -7 & 14 & -36 \end(array) \right|.
In principle, using formula No. 2 from the topic of calculating determinants of the second and third orders, you can complete the calculations, obtaining the value $M^*$:
$$ M^*=-\left| \begin(array) (ccc) 3 & -3 & 19\\ -5 & 16 & -20 \\ -7 & 14 & -36 \end(array) \right|=-30. $$
determinant by elements of a row or column
Further properties are related to the concepts of minor and algebraic complement
Definition. Minor element is called a determinant made up of elements remaining after crossing outi-th drains andjth column at the intersection of which this element is located. Minor of the element of the determinant n-th order has order ( n- 1). We will denote it by .
Example 1. Let 
, Then 
.
This minor is obtained from A by crossing out the second row and third column.
Definition.
Algebraic complement
element is called the corresponding minor, multiplied by nat.e 
, Wherei–line number andj-columns at the intersection of which this element is located.
VІІІ. (Decomposition of the determinant into elements of a certain string). The determinant is equal to the sum of the products of the elements of a certain row and their corresponding algebraic complements.
.
Example 2. Let it be then
.
Example 3. Let's find the determinant of the matrix by expanding it into the elements of the first row.
Formally, this theorem and other properties of determinants are applicable only for determinants of matrices of no higher than third order, since we have not considered other determinants. The following definition will allow us to extend these properties to determinants of any order.
Definition. Determinant matrices A nth order is a number calculated by sequential application of the expansion theorem and other properties of determinants.
You can check that the result of the calculations does not depend on the order in which the above properties are applied and for which rows and columns. Using this definition, the determinant is uniquely found.
Although this definition does not contain an explicit formula for finding the determinant, it allows one to find it by reducing it to the determinants of matrices of lower order. Such definitions are called recurrent.
Example 4. Calculate the determinant: .
Although the factorization theorem can be applied to any row or column of a given matrix, fewer computations are obtained by factoring along the column that contains as many zeros as possible.
Since the matrix does not have zero elements, we obtain them using property 7). Multiply the first line sequentially by the numbers (–5), (–3) and (–2) and add it to the 2nd, 3rd and 4th lines and get:
Let's expand the resulting determinant along the first column and get:
(we take (–4) from the 1st line, (–2) from the 2nd line, (–1) from the 3rd line according to property 4) 
(since the determinant contains two proportional columns).
§ 1.3. Some types of matrices and their determinants
Definition. Square m an matrix with zero elements below or above the main diagonal(=0 at i j, or =0 at i j) calledtriangular .
Their schematic structure accordingly looks like: 
or 
.
Here 0 means zero elements, and means arbitrary elements.
Theorem. The determinant of a square triangular matrix is equal to the product of its elements located on the main diagonal, i.e.
.
For example:
.
Definition. A square matrix with zero elements outside the main diagonal is calleddiagonal .
Its schematic view: 
A diagonal matrix with only unit elements on the main diagonal is called single matrix. It is denoted by:
The determinant of the identity matrix is 1, i.e. E=1.
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 A -1. Any non-degenerate matrix It has inverse matrix and, moreover, only one, i.e., in order for it to be square matrix A had inverse matrix, it is necessary and sufficient for it determinant was different from zero.
For getting 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 mхn. 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. .
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 previously received 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.
Let us 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
Force 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 altitude 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
=
=
;
e) 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
Elements of linear algebra This section includes the main types of problems that are discussed in the topic “Linear algebra”: calculation of determinants, actions
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Document... element matrix". Definition. Algebraic addition elementаік matrix A is called minor Mik of this matrix, multiplied by (-1) and + k: Algebraic addition element...method. Example 1. Given a matrix Find det A. Solution. Let's transform...
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Methodological recommendations for performing extracurricular independent work of a student in the discipline “Mathematics” for the specialty
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Or a determinant, in mathematics, a recording of numbers in the form of a square table, in correspondence with which another number (the value of the determinant) is placed. 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
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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...
