In the general case, the rule for calculating $n$th order determinants is quite cumbersome. For second- and third-order determinants, there are rational ways to calculate them.
Calculations of second order determinants
To calculate the determinant of a second-order matrix, you need to subtract the product of the elements of the secondary diagonal from the product of the elements of the main diagonal:
$$\left| \begin(array)(ll)(a_(11)) & (a_(12)) \\ (a_(21)) & (a_(22))\end(array)\right|=a_(11) \ cdot a_(22)-a_(12) \cdot a_(21)$$
Example
Exercise. Compute the second order determinant $\left| \begin(array)(rr)(11) & (-2) \\ (7) & (5)\end(array)\right|$
Solution.$\left| \begin(array)(rr)(11) & (-2) \\ (7) & (5)\end(array)\right|=11 \cdot 5-(-2) \cdot 7=55+14 =69$
Answer.$\left| \begin(array)(rr)(11) & (-2) \\ (7) & (5)\end(array)\right|=69$
Methods for calculating third-order determinants
The following rules exist for calculating third-order determinants.
Triangle rule
Schematically, this rule can be depicted as follows:
The product of elements in the first determinant that are connected by straight lines is taken with a plus sign; similarly, for the second determinant, the corresponding products are taken with a minus sign, i.e.
$$\left| \begin(array)(ccc)(a_(11)) & (a_(12)) & (a_(13)) \\ (a_(21)) & (a_(22)) & (a_(23)) \\ (a_(31)) & (a_(32)) & (a_(33))\end(array)\right|=a_(11) a_(22) a_(33)+a_(12) a_( 23) a_(31)+a_(13) a_(21) a_(32)-$$
$$-a_(11) a_(23) a_(32)-a_(12) a_(21) a_(33)-a_(13) a_(22) a_(31)$$
Example
Exercise. Compute the determinant of $\left| \begin(array)(rrr)(3) & (3) & (-1) \\ (4) & (1) & (3) \\ (1) & (-2) & (-2)\end (array)\right|$ using the triangle method.
Solution.$\left| \begin(array)(rrr)(3) & (3) & (-1) \\ (4) & (1) & (3) \\ (1) & (-2) & (-2)\end (array)\right|=3 \cdot 1 \cdot(-2)+4 \cdot(-2) \cdot(-1)+$
$$+3 \cdot 3 \cdot 1-(-1) \cdot 1 \cdot 1-3 \cdot(-2) \cdot 3-4 \cdot 3 \cdot(-2)=54$$
Answer.
Sarrus rule
To the right of the determinant, add the first two columns and take the products of elements on the main diagonal and on the diagonals parallel to it with a plus sign; and the products of the elements of the secondary diagonal and the diagonals parallel to it, with a minus sign:
$$-a_(13) a_(22) a_(31)-a_(11) a_(23) a_(32)-a_(12) a_(21) a_(33)$$
Example
Exercise. Compute the determinant of $\left| \begin(array)(rrr)(3) & (3) & (-1) \\ (4) & (1) & (3) \\ (1) & (-2) & (-2)\end (array)\right|$ using Sarrus' rule.
Solution. 
$$+(-1) \cdot 4 \cdot(-2)-(-1) \cdot 1 \cdot 1-3 \cdot 3 \cdot(-2)-3 \cdot 4 \cdot(-2)= 54$$
Answer.$\left| \begin(array)(rrr)(3) & (3) & (-1) \\ (4) & (1) & (3) \\ (1) & (-2) & (-2)\end (array)\right|=54$
Expanding the determinant by row or column
The determinant is equal to the sum of the products of the elements of the row of the determinant and their algebraic complements. Usually the row/column that contains zeros is selected. The row or column along which the decomposition is carried out will be indicated by an arrow.
Example
Exercise. Expanding along the first row, calculate the determinant $\left| \begin(array)(lll)(1) & (2) & (3) \\ (4) & (5) & (6) \\ (7) & (8) & (9)\end(array) \right|$
Solution.$\left| \begin(array)(lll)(1) & (2) & (3) \\ (4) & (5) & (6) \\ (7) & (8) & (9)\end(array) \right| \leftarrow=a_(11) \cdot A_(11)+a_(12) \cdot A_(12)+a_(13) \cdot A_(13)=$
$1 \cdot(-1)^(1+1) \cdot \left| \begin(array)(cc)(5) & (6) \\ (8) & (9)\end(array)\right|+2 \cdot(-1)^(1+2) \cdot \left | \begin(array)(cc)(4) & (6) \\ (7) & (9)\end(array)\right|+3 \cdot(-1)^(1+3) \cdot \left | \begin(array)(cc)(4) & (5) \\ (7) & (8)\end(array)\right|=-3+12-9=0$
Answer.
This method allows the calculation of the determinant to be reduced to the calculation of a determinant of a lower order.
Example
Exercise. Compute the determinant of $\left| \begin(array)(lll)(1) & (2) & (3) \\ (4) & (5) & (6) \\ (7) & (8) & (9)\end(array) \right|$
Solution. Let us perform the following transformations on the rows of the determinant: from the second row we subtract the first four, and from the third the first row multiplied by seven, as a result, according to the properties of the determinant, we obtain a determinant equal to the given one.
$$\left| \begin(array)(ccc)(1) & (2) & (3) \\ (4) & (5) & (6) \\ (7) & (8) & (9)\end(array) \right|=\left| \begin(array)(ccc)(1) & (2) & (3) \\ (4-4 \cdot 1) & (5-4 \cdot 2) & (6-4 \cdot 3) \\ ( 7-7 \cdot 1) & (8-7 \cdot 2) & (9-7 \cdot 3)\end(array)\right|=$$
$$=\left| \begin(array)(rrr)(1) & (2) & (3) \\ (0) & (-3) & (-6) \\ (0) & (-6) & (-12)\ end(array)\right|=\left| \begin(array)(ccc)(1) & (2) & (3) \\ (0) & (-3) & (-6) \\ (0) & (2 \cdot(-3)) & (2 \cdot(-6))\end(array)\right|=0$$
The determinant is zero because the second and third rows are proportional.
Answer.$\left| \begin(array)(lll)(1) & (2) & (3) \\ (4) & (5) & (6) \\ (7) & (8) & (9)\end(array) \right|=0$
To calculate determinants of fourth order and higher, either row/column expansion, or reduction to triangular form, or using Laplace's theorem are used.
Decomposing the determinant into elements of a row or column
Example
Exercise. Compute the determinant of $\left| \begin(array)(llll)(9) & (8) & (7) & (6) \\ (5) & (4) & (3) & (2) \\ (1) & (0) & (1) & (2) \\ (3) & (4) & (5) & (6)\end(array)\right|$ , decomposing it into the elements of some row or some column.
Solution. Let us first perform elementary transformations on the rows of the determinant, making as many zeros as possible either in the row or in the column. To do this, first subtract nine thirds from the first line, five thirds from the second, and three thirds from the fourth, we get:
$$\left| \begin(array)(cccc)(9) & (8) & (7) & (6) \\ (5) & (4) & (3) & (2) \\ (1) & (0) & (1) & (2) \\ (3) & (4) & (5) & (6)\end(array)\right|=\left| \begin(array)(cccc)(9-1) & (8-0) & (7-9) & (6-18) \\ (5-5) & (4-0) & (3-5) & (2-10) \\ (1) & (0) & (1) & (2) \\ (0) & (4) & (2) & (0)\end(array)\right|=\ left| \begin(array)(rrrr)(0) & (8) & (-2) & (-12) \\ (0) & (4) & (-2) & (-8) \\ (1) & (0) & (1) & (2) \\ (0) & (4) & (2) & (0)\end(array)\right|$$
Let us decompose the resulting determinant into the elements of the first column:
$$\left| \begin(array)(rrrr)(0) & (8) & (-2) & (-12) \\ (0) & (4) & (-2) & (-8) \\ (1) & (0) & (1) & (2) \\ (0) & (4) & (2) & (0)\end(array)\right|=0+0+1 \cdot(-1)^( 3+1) \cdot \left| \begin(array)(rrr)(8) & (-2) & (-12) \\ (4) & (-2) & (-8) \\ (4) & (2) & (0)\ end(array)\right|+0$$
We will also expand the resulting third-order determinant into row and column elements, having previously obtained zeros, for example, in the first column. To do this, subtract the second two lines from the first line, and the second from the third:
$$\left| \begin(array)(rrr)(8) & (-2) & (-12) \\ (4) & (-2) & (-8) \\ (4) & (2) & (0)\ end(array)\right|=\left| \begin(array)(rrr)(0) & (2) & (4) \\ (4) & (-2) & (-8) \\ (0) & (4) & (8)\end( array)\right|=4 \cdot(-1)^(2+2) \cdot \left| \begin(array)(ll)(2) & (4) \\ (4) & (8)\end(array)\right|=$$
$$=4 \cdot(2 \cdot 8-4 \cdot 4)=0$$
Answer.$\left| \begin(array)(cccc)(9) & (8) & (7) & (6) \\ (5) & (4) & (3) & (2) \\ (1) & (0) & (1) & (2) \\ (3) & (4) & (5) & (6)\end(array)\right|=0$
Comment
The last and penultimate determinants could not be calculated, but immediately conclude that they are equal to zero, since they contain proportional rows.
Reducing the determinant to triangular form
Using elementary transformations over rows or columns, the determinant is reduced to a triangular form and then its value, according to the properties of the determinant, is equal to the product of the elements on the main diagonal.
Example
Exercise. Calculate the determinant $\Delta=\left| \begin(array)(rrrr)(-2) & (1) & (3) & (2) \\ (3) & (0) & (-1) & (2) \\ (-5) & ( 2) & (3) & (0) \\ (4) & (-1) & (2) & (-3)\end(array)\right|$ reducing it to triangular form.
Solution. First we make zeros in the first column under the main diagonal. All transformations will be easier to perform if the element $a_(11)$ is equal to 1. To do this, we will swap the first and second columns of the determinant, which, according to the properties of the determinant, will cause it to change its sign to the opposite:
$$\Delta=\left| \begin(array)(rrrr)(-2) & (1) & (3) & (2) \\ (3) & (0) & (-1) & (2) \\ (-5) & ( 2) & (3) & (0) \\ (4) & (-1) & (2) & (-3)\end(array)\right|=-\left| \begin(array)(rrrr)(1) & (-2) & (3) & (2) \\ (0) & (3) & (-1) & (2) \\ (2) & (- 5) & (3) & (0) \\ (-1) & (4) & (2) & (-3)\end(array)\right|$$
$$\Delta=-\left| \begin(array)(rrrr)(1) & (-2) & (3) & (2) \\ (0) & (3) & (-1) & (2) \\ (0) & (- 1) & (-3) & (-4) \\ (0) & (2) & (5) & (-1)\end(array)\right|$$
Next, we get zeros in the second column in place of the elements under the main diagonal. Again, if the diagonal element is equal to $\pm 1$ , then the calculations will be simpler. To do this, swap the second and third lines (and at the same time change to the opposite sign of the determinant):
$$\Delta=\left| \begin(array)(rrrr)(1) & (-2) & (3) & (2) \\ (0) & (-1) & (-3) & (-4) \\ (0) & (3) & (-1) & (2) \\ (0) & (2) & (5) & (-1)\end(array)\right|$$
Let there be a square matrix A of size n x n.
Definition. The determinant is the algebraic sum of all possible products of elements, taken one from each column and each row of the matrix A. If in each such product (term of the determinant) the factors are arranged in the order of the columns (i.e., the second indices of the elements a ij in the product are arranged in ascending order), then with the sign (+) those products are taken whose permutation of the first indices is even, and with a sign (-) – those for which it is odd.
.
Here is the number of inversions in the permutation of indices i 1, i 2, ..., i n.
Methods for finding determinants
- Determinant of a matrix by row and column decomposition through minors.
- Determinant by reduction method to triangular form (Gauss method)
Property of determinants
- When a matrix is transposed, its determinant does not change.
- If you swap two rows or two columns of a determinant, the determinant will change sign, but will not change in absolute value.
- Let C = AB where A and B are square matrices. Then detC = detA ∙ detB.
- A determinant with two identical rows or two identical columns is equal to 0. If all elements of a certain row or column are equal to zero, then the determinant itself is equal to zero.
- A determinant with two proportional rows or columns is 0.
- The determinant of a triangular matrix is equal to the product of the diagonal elements. The determinant of a diagonal matrix is equal to the product of the elements on the main diagonal.
- If all elements of a row (column) are multiplied by the same number, then the determinant will be multiplied by this number.
- If each element of a certain row (column) of a determinant is presented as the sum of two terms, then the determinant is equal to the sum of two determinants, in which all rows (columns) except this one are the same, and in this row (column) the first determinant is the first, and in the second - the second terms.
- Jacobi's theorem: If to the elements of a certain column of the determinant we add the corresponding elements of another column, multiplied by an arbitrary factor λ, then the value of the determinant will not change.
- transpose matrix;
- add to any string another string multiplied by any number.
Task 1. Calculate the determinant by expanding it by row or column.
Solution:xls
Example 1:xls
Task 2. Calculate the determinant in two ways: a) using the “triangles” rule; b) expansion along a line.
Solution.
a) The terms included in the minus sign are constructed in the same way with respect to the side diagonal.
| = |
b) We write the matrix in the form:
| A= |
|
Main determinant:
∆ = 2 (0 0-2 4)-(-1 (2 0-2 1))+(-2 (2 4-0 1)) = -34
Task 3. Indicate what the determinant of a fourth-order square matrix A is equal to if its rank r(A)=1.
Answer: det(A) = 0.
Determinant: det, ||, determinant.
The determinant is not a matrix, but a number.
How to find the determinant of a matrix?
To find the determinant of a matrix, introduce the concept "minor". Designation: M ij - minor, M ij 2 - second order minor (determinant of the 2*2 matrix), etc.
To find the minor for the element a ij , we delete A from the matrix i-th line And jth column. We obtain a matrix of dimension n-1*m-1, we find determinant of this matrix.
Example: find the second order minor for element a 12 of matrix A:
We cross out the 1st row and 2nd column from matrix A. We obtain a matrix of dimension 2*2, find determinant of this matrix:
Thus, minor is not a matrix, but a number.
Example: find the determinant (in general view) 2*2 matrices by decomposition along 1) rows; 2) column:
By line: det A = a 11 *(-1) 1+1 *M 11 +a 12 *(-1) 1+2 *M 12 = a 11 *1*a 22 +a 12 *(-1)* a 21 =
= a 11 *a 22 -a 12 *a 21
By column: det A = a 11 *(-1) 1+1 *M 11 +a 21 *(-1) 2+1 *M 21 = a 11 *1*a 22 +a 21 *(-1)* a 12 =
= a 11 *a 22 -a 21 *a 12
It is easy to see that the same result is obtained.
Thus, to find the determinant of a 2*2 matrix, it is enough to subtract the product of the elements of the secondary diagonal from the product of the elements of the main diagonal:
How to quickly calculate the third order determinant?
To calculate the third order determinant, use triangle rule(or "stars").
1. Multiply the elements of the main diagonal: det(A)=11*22*33...
2. To the resulting product we add the product of “triangles with bases parallel to the main diagonal”: det(A)=11*22*33+31*12*23+13*21*32...
3. We take everything connected with the secondary diagonal with a “-” sign. We multiply the elements of the secondary diagonal and subtract: det(A)=11*22*33+31*12*23+13*21*32-13*22*31...
4. Similarly to the “main triangles”, we multiply the secondary ones and subtract: det(A)=11*22*33+31*12*23+13*21*32-13*22*31-11*23*32-33*12 *21.
det(A)=11*22*33+31*12*23+13*21*32-13*22*31-11*23*32-33*12*21=
=7986+8556+8736-8866-8096-8316=0
Properties of the determinant of a matrix.
- When two parallel rows or columns of a determinant are swapped, its sign is reversed;
- The determinant containing two identical rows or columns is equal to zero;
- If one of the lines of the determinant is multiplied by any number, the result is a determinant equal to the original determinant multiplied by this number;
- When a matrix is transposed, its determinant does not change its value;
- If in the determinant, instead of any line, we write the sum of this line and any other line, multiplied by a certain number, then the resulting new determinant will be equal to the original one;
- If each element of any row or column of a determinant is represented as a sum of two terms, then this determinant can be decomposed into the sum of two corresponding determinants;
- The common factor of the elements of any row or column of the determinant can be taken out of the sign of the determinant.
Determinants and their properties. Rearrangement numbers 1, 2,..., n is any arrangement of these numbers in a certain order. IN elementary algebra it is proved that the number of all permutations that can be formed from n numbers is equal to 12...n = n!. For example, from three numbers 1, 2, 3 you can form 3!=6 permutations: 123, 132, 312, 321, 231, 213. They say that in this permutation the numbers i and j are inversion(disorder) if i>j, but i comes before j in this permutation, that is, if the larger number is to the left of the smaller one.
The permutation is called even(or odd), if it has an even (odd) total number of inversions. The operation by which one goes from one permutation to another, composed of the same n numbers, is called substitution nth degree.
A substitution that transforms one permutation into another is written in two lines in common brackets, and the numbers occupying the same places in the permutations under consideration are called relevant and are written one under the other. For example, the symbol represents a substitution in which 3 becomes 4, 1 → 2, 2 → 1, 4 → 3. The substitution is called even(or odd), if the total number of inversions in both substitution strings is even (odd). Any substitution of the nth degree can be written in the form, i.e. with natural numbers in the top line.
Let us be given a square matrix of order n
Let us consider all possible products of n elements of this matrix, taken one and only one from each row and each column, i.e. works of the form:
, (4.4)
where the indices q 1, q 2,...,q n constitute some permutation of numbers
1, 2,..., n. The number of such products is equal to the number of different permutations of n symbols, i.e. equals n!. The sign of the product (4.4) is equal to (- 1) q, where q is the number of inversions in the permutation of the second indices of the elements.
Determinant The nth order corresponding to matrix (4.3) is called the algebraic sum n! members of the form (4.4). To write a determinant, use the symbol 
or detA = 
(determinant, or determinant, of matrix A).
Properties of determinants
1. The determinant does not change during transposition.
2. If one of the lines of the determinant consists of zeros, then the determinant is equal to zero.
3. If two lines in the determinant are rearranged, the determinant will change sign.
4. A determinant containing two identical strings is equal to zero.
5. If all elements of a certain row of the determinant are multiplied by some number k, then the determinant itself will be multiplied by k.
6. A determinant containing two proportional lines is equal to zero.
7. If all elements of the i-th row of the determinant are presented as the sum of two terms a i j = b j + c j (j = 1,...,n), then the determinant is equal to the sum of determinants for which all rows except the i-th are - are the same as in the given determinant, and the i-th row in one of the terms consists of elements b j , in the other - of elements c j .
8. The determinant does not change if the corresponding elements of another row are added to the elements of one of its rows, multiplied by the same number.
Comment. All properties remain valid if we take columns instead of rows.
Minor M i j of the element a i j of the determinant d of the nth order is called the determinant of order n-1, which is obtained from d by deleting the row and column containing this element.
Algebraic complement element a i j of the determinant d is called its minor M i j , taken with the sign (-1) i + j . The algebraic complement of an element a i j will be denoted by A i j . Thus, A i j = (-1) i + j M i j .
Methods for practical calculation of determinants, based on the fact that a determinant of order n can be expressed in terms of determinants of lower orders, is given by the following theorem.
Theorem (decomposition of the determinant in a row or column).
The determinant is equal to the sum of the products of all elements of its arbitrary row (or column) by their algebraic complements. In other words, there is an expansion of d in elements of the i-th lines
d = a i 1 A i 1 + a i 2 A i 2 +... + a i n A i n (i = 1,...,n)
or jth column
d = a 1 j A 1 j + a 2 j A 2 j +... + a n j A n j (j =1,...,n).
In particular, if all but one element of a row (or column) is zero, then the determinant is equal to that element multiplied by its algebraic complement.
Formula for calculating the third order determinant.
To make this formula easier to remember:
Example 2.4. Without calculating the determinant, show that it is equal to zero.
Solution. Subtracting the first from the second line, we obtain a determinant equal to the original one. If we also subtract the first from the third line, we get a determinant in which the two lines are proportional. This determinant is equal to zero.
Example 2.5. Calculate the determinant D = by expanding it into the elements of the second column.
Solution. Let's expand the determinant into the elements of the second column:
D = a 12 A 12 + a 22 A 22 +a 32 A 32 =
.
Example 2.6. Compute determinant
,
in which all elements on one side of the main diagonal are equal to zero.
Solution. Let us expand the determinant of A along the first line:
.
The determinant on the right can be expanded again along the first line, then we get:
.
Example 2.7. Compute determinant 
.
Solution. If you add the first line to each line of the determinant, starting from the second, you will get a determinant in which all elements below the main diagonal will be equal to zero. Namely, we get the determinant: 
, equal to the original one.
Reasoning as in the previous example, we find that it is equal to the product of the elements of the main diagonal, i.e. n!. The method by which this determinant is calculated is called the method of reduction to triangular form.
Here we will outline those properties that are usually used to calculate determinants in a standard course higher mathematics. This is an auxiliary topic that we will refer to from other sections as necessary.
So, let a certain square matrix $A_(n\times n)=\left(\begin(array) (cccc) a_(11) & a_(12) & \ldots & a_(1n) \\ a_(21) be given & a_(22) & \ldots & a_(2n) \\ \ldots & \ldots & \ldots & \ldots \\ a_(n1) & a_(n2) & \ldots & a_(nn) \\ \end( array) \right)$. Every square matrix has a characteristic called a determinant (or determinant). I will not go into the essence of this concept here. If it requires clarification, then please write about it on the forum, and I will touch on this issue in more detail.
The determinant of the matrix $A$ is denoted as $\Delta A$, $|A|$, or $\det A$. Determinant order equal to the number of rows (columns) in it.
- The value of the determinant will not change if its rows are replaced by the corresponding columns, i.e. $\Delta A=\Delta A^T$.
show\hide
Let’s replace the rows with columns in it according to the principle: “there was a first row - there was a first column”, “there was a second row - there was a second column”:
Let's calculate the resulting determinant: $\left| \begin(array) (cc) 2 & 9 \\ 5 & 4 \end(array) \right|=2\cdot 4-9\cdot 5=-37$. As you can see, the value of the determinant has not changed due to the replacement.
- If you swap two rows (columns) of the determinant, the sign of the determinant will change to the opposite.
Example of using this property: show\hide
Consider the determinant $\left| \begin(array) (cc) 2 & 5 \\ 9 & 4 \end(array) \right|$. Let's find its value using formula No. 1 from the topic of calculating determinants of the second and third orders:
$$\left| \begin(array) (cc) 2 & 5 \\ 9 & 4 \end(array) \right|=2\cdot 4-5\cdot 9=-37.$$
Now let's swap the first and second lines. We obtain the determinant $\left| \begin(array) (cc) 9 & 4 \\ 2 & 5 \end(array) \right|$. Let's calculate the resulting determinant: $\left| \begin(array) (cc) 9 & 4 \\ 2 & 5 \end(array) \right|=9\cdot 5-4\cdot 2=37$. So, the value of the original determinant was (-37), and the value of the determinant with the changed row order is $-(-37)=37$. The sign of the determinant has changed to the opposite.
- A determinant for which all elements of a row (column) are equal to zero is equal to zero.
Example of using this property: show\hide
Since in the determinant $\left| \begin(array) (ccc) -7 & 10 & 0\\ -9 & 21 & 0\\ 2 & -3 & 0 \end(array) \right|$ all elements of the third column are zero, then the determinant is zero , i.e. $\left| \begin(array) (ccc) -7 & 10 & 0\\ -9 & 21 & 0\\ 2 & -3 & 0 \end(array) \right|=0$.
- The determinant in which all elements of a certain row (column) are equal to the corresponding elements of another row (column) is equal to zero.
Example of using this property: show\hide
Since in the determinant $\left| \begin(array) (ccc) -7 & 10 & 0\\ -7 & 10 & 0\\ 2 & -3 & 18 \end(array) \right|$ all elements of the first row are equal to the corresponding elements of the second row, then the determinant is equal to zero, i.e. $\left| \begin(array) (ccc) -7 & 10 & 0\\ -7 & 10 & 0\\ 2 & -3 & 18 \end(array) \right|=0$.
- If in a determinant all elements of one row (column) are proportional to the corresponding elements of another row (column), then such a determinant is equal to zero.
Example of using this property: show\hide
Since in the determinant $\left| \begin(array) (ccc) -7 & 10 & 28\\ 5 & -3 & 0\\ -15 & 9 & 0 \end(array) \right|$ The second and third rows are proportional, i.e. $r_3=-3\cdot(r_2)$, then the determinant is equal to zero, i.e. $\left| \begin(array) (ccc) -7 & 10 & 28\\ 5 & -3 & 0\\ -15 & 9 & 0 \end(array) \right|=0$.
- If all elements of a row (column) have a common factor, then this factor can be taken out of the determinant sign.
Example of using this property: show\hide
Consider the determinant $\left| \begin(array) (cc) -7 & 10 \\ -9 & 21 \end(array) \right|$. Notice that all elements in the second row are divisible by 3:
$$\left| \begin(array) (cc) -7 & 10 \\ -9 & 21 \end(array) \right|=\left| \begin(array) (cc) -7 & 10 \\ 3\cdot(-3) & 3\cdot 7 \end(array) \right|$$
The number 3 is the common factor of all elements of the second row. Let's take the three out of the determinant sign:
$$\left| \begin(array) (cc) -7 & 10 \\ -9 & 21 \end(array) \right|=\left| \begin(array) (cc) -7 & 10 \\ 3\cdot(-3) & 3\cdot 7 \end(array) \right|= 3\cdot \left| \begin(array) (cc) -7 & 10 \\ -3 & 7 \end(array) \right| $$
- The determinant will not change if to all the elements of a certain row (column) we add the corresponding elements of another row (column), multiplied by an arbitrary number.
Example of using this property: show\hide
Consider the determinant $\left| \begin(array) (ccc) -7 & 10 & 0\\ -9 & 21 & 4 \\ 2 & -3 & 1 \end(array) \right|$. Let's add to the elements of the second line the corresponding elements of the third line, multiplied by 5. This action is written as follows: $r_2+5\cdot(r_3)$. The second line will be changed, the remaining lines will remain unchanged.
$$\left| \begin(array) (ccc) -7 & 10 & 0\\ -9 & 21 & 4 \\ 2 & -3 & 1 \end(array) \right| \begin(array) (l) \phantom(0)\\ r_2+5\cdot(r_3)\\ \phantom(0) \end(array)= \left| \begin(array) (ccc) -7 & 10 & 0\\ -9+5\cdot 2 & 21+5\cdot (-3) & 4+5\cdot 1 \\ 2 & -3 & 1 \end (array) \right|= \left| \begin(array) (ccc) -7 & 10 & 0\\ 1 & 6 & 9 \\ 2 & -3 & 1 \end(array) \right|. $$
- If a certain row (column) in a determinant is a linear combination of other rows (columns), then the determinant is equal to zero.
Example of using this property: show\hide
Let me immediately explain what the phrase “linear combination” means. Let us have s rows (or columns): $A_1$, $A_2$,..., $A_s$. Expression
$$ k_1\cdot A_1+k_2\cdot A_2+\ldots+k_s\cdot A_s, $$
where $k_i\in R$ is called a linear combination of rows (columns) $A_1$, $A_2$,..., $A_s$.
For example, consider the following determinant:
$$\left| \begin(array) (cccc) -1 & 2 & 3 & 0\\ -2 & -4 & -5 & 1\\ 5 & 0 & 7 & 10 \\ -13 & -8 & -16 & -7 \end(array) \right| $$
In this determinant, the fourth row can be expressed as a linear combination first three lines:
$$ r_4=2\cdot(r_1)+3\cdot(r_2)-r_3 $$
Therefore, the determinant in question is equal to zero.
- If each element of a certain k-th row (k-th column) of a determinant is equal to the sum of two terms, then such a determinant is equal to the sum of determinants, the first of which has kth line(kth column) contains the first terms, and the second determinant in the kth row (kth column) contains the second terms. Other elements of these determinants are the same.
Example of using this property: show\hide
Consider the determinant $\left| \begin(array) (ccc) -7 & 10 & 0\\ -9 & 21 & 4 \\ 2 & -3 & 1 \end(array) \right|$. Let's write the elements of the second column like this: $\left| \begin(array) (ccc) -7 & 3+7 & 0\\ -9 & 21+0 & 4 \\ 2 & 5+(-8) & 1 \end(array) \right|$. Then such a determinant is equal to the sum of two determinants:
$$\left| \begin(array) (ccc) -7 & 10 & 0\\ -9 & 21 & 4 \\ 2 & -3 & 1 \end(array) \right|= \left| \begin(array) (ccc) -7 & 3+7 & 0\\ -9 & 21+0 & 4 \\ 2 & 5+(-8) & 1 \end(array) \right|= \left| \begin(array) (ccc) -7 & 3 & 0\\ -9 & 21 & 4 \\ 2 & 5 & 1 \end(array) \right|+ \left| \begin(array) (ccc) -7 & 7 & 0\\ -9 & 0 & 4 \\ 2 & -8 & 1 \end(array) \right| $$
- The determinant of the product of two square matrices of the same order is equal to the product of the determinants of these matrices, i.e. $\det(A\cdot B)=\det A\cdot \det B$. From this rule we can obtain the following formula: $\det \left(A^n \right)=\left(\det A \right)^n$.
- If the matrix $A$ is non-singular (i.e. its determinant is not equal to zero), then $\det \left(A^(-1)\right)=\frac(1)(\det A)$.
Formulas for calculating determinants
For determinants of the second and third orders, the following formulas are correct:
\begin(equation) \Delta A=\left| \begin(array) (cc) a_(11) & a_(12) \\ a_(21) & a_(22) \end(array) \right|=a_(11)\cdot a_(22)-a_( 12)\cdot a_(21) \end(equation) \begin(equation) \begin(aligned) & \Delta A=\left| \begin(array) (ccc) a_(11) & a_(12) & a_(13) \\ a_(21) & a_(22) & a_(23) \\ a_(31) & a_(32) & a_(33) \end(array) \right|= a_(11)\cdot a_(22)\cdot a_(33)+a_(12)\cdot a_(23)\cdot a_(31)+a_(21 )\cdot a_(32)\cdot a_(13)-\\ & -a_(13)\cdot a_(22)\cdot a_(31)-a_(12)\cdot a_(21)\cdot a_(33 )-a_(23)\cdot a_(32)\cdot a_(11)\end(aligned)\end(equation)
Examples of using formulas (1) and (2) are in the topic "Formulas for calculating determinants of the second and third orders. Examples of calculating determinants".
The determinant of the matrix $A_(n\times n)$ can be expanded in i-th line using the following formula:
\begin(equation)\Delta A=\sum\limits_(j=1)^(n)a_(ij)A_(ij)=a_(i1)A_(i1)+a_(i2)A_(i2)+\ ldots+a_(in)A_(in) \end(equation)
An analogue of this formula also exists for columns. The formula for expanding the determinant in the jth column is as follows:
\begin(equation)\Delta A=\sum\limits_(i=1)^(n)a_(ij)A_(ij)=a_(1j)A_(1j)+a_(2j)A_(2j)+\ ldots+a_(nj)A_(nj) \end(equation)
The rules expressed by formulas (3) and (4) are illustrated in detail with examples and explained in the topic Reducing the order of the determinant. Decomposition of the determinant in a row (column).
Let us indicate another formula for calculating the determinants of upper triangular and lower triangular matrices (for an explanation of these terms, see the topic “Matrixes. Types of matrices. Basic terms”). The determinant of such a matrix is equal to the product of the elements on the main diagonal. Examples:
\begin(aligned) &\left| \begin(array) (cccc) 2 & -2 & 9 & 1 \\ 0 & 9 & 8 & 0 \\ 0 & 0 & 4 & -7 \\ 0 & 0 & 0 & -6 \end(array) \right|= 2\cdot 9\cdot 4\cdot (-6)=-432.\\ &\left| \begin(array) (cccc) -3 & 0 & 0 & 0 \\ -5 & 0 & 0 & 0 \\ 8 & 2 & 1 & 0 \\ 5 & 4 & 0 & 10 \end(array) \ right|= -3\cdot 0\cdot 1 \cdot 10=0. \end(aligned)
