It will help not only teapots, but even those who first heard the word "determinant". Two years have passed since the site had only ten pages, and now, after my long, long journey into the world of matan, everything is back to normal.
Imagine that you need to calculate a third-order determinant by expanding it over the elements of a row (column). Although what is there to represent - you need to =) You can sit on it for 5 minutes, or you can sit for 2-3 minutes. Or even in the region of one minute. The time you spend depends not only on your experience, but also on your knowledge of the properties of determinants. It is not uncommon when the solution process is quite realistic to reduce to a matter of seconds, and sometimes you can immediately see the result! “Nonsense, why save on matches, and so we will decide everything,” some will say. Let's say. And we will not allow mistakes ;-) But what about the determinant of the 4th order, which is quite common in practice? It will take 10-20 minutes to fight with this pepper. And it will not even be a battle, but a massacre, since the probability of a computational error is very high, which will “wrap” you into the second round of the decision. And if the determinant of the fifth order? Save only lowering the order of the determinant. Yes, such examples are also found in control papers.
The materials on this page will significantly improve your technique for solving determinants and simplify the further development of higher mathematics.
Efficient Methods for Calculating the Determinant
First of all, we will not touch on the properties of the determinant, but just on the methods of its rational calculation. These methods of solution lie on the surface and are clear to many, but nevertheless we will dwell on them in more detail. It is assumed that the reader already knows how to confidently reveal the third-order determinant. As is known, this determinant can be expanded 6 in standard ways: on any row or any column. It would seem that it does not matter, because the answer will be the same. But are all methods equally easy? No. In most cases there is less profitable ways and more profitable ways solutions.
Consider the determinant, which I abundantly covered with tattoos in the first lesson. In that article, we laid it out in detail, with pictures, on the first line. The first line is good and academic, but is it possible to achieve a result faster? There is a zero in the determinant, and by expanding it by the second row or by the second column, the calculations will noticeably decrease!
Let's expand the determinant in the second column: 
In practice, zero elements are ignored, and the solution takes a more compact form:
Exercise 1
Expand the given determinant along the second line using the shortened notation.
Solution at the end of the lesson.
If there are two zeros in a row (or column), then this is generally a real gift. Let's consider the determinant. There are two zeros in the third line, and we open it on it:
That's the whole solution!
The special case when the determinant has the so-called stepped or triangular view, for example: - in such a determinant, all numbers located below main diagonal, are equal to zero.
Let's expand it by the first column: 
In practical tasks, it is convenient to be guided by the following rule - step determinant is equal to the product of the numbers of its main diagonal:
A similar principle is also valid for step determinants of other orders, for example: 
Triangular determinants appear in some linear algebra problems, and their solution is most often framed in this way.
And if the row (column) of the determinant contains only zeros? The answer, I think, is clear. We will return to this issue in the properties of the determinant.
Now let's imagine that the long-awaited bagels are not included in the New Year's gift. So let's gut the bad Santa Claus!
There are no zeros here, but there is still a way to make your life easier. This determinant is best expanded in the third column, since there are the smallest numbers. In this case, the solution entry takes a very concise form:
Summarizing the paragraph, we formulate the golden rule of calculations:
It is more profitable to open the determinant by THAT line (column), where:
1) more zeros;
2) smaller numbers.
Naturally, this is also true for determinants of higher orders.
A small example to consolidate the material:
Task 2
Calculate the determinant by expanding it by row or column, using the most rational way
This is an example for a do-it-yourself solution, optimal solution and the answer is at the end of the lesson.
And one more important advice: do not complex! There is no need to "go in cycles" in the traditional expansion by the first row or the first column. In short, so be it!
Determinant properties
Consider the old acquaintances of the first lesson: the matrix 
and its determinant 
.
Just in case, I repeat the elementary difference between the concepts: matrix is a table of elements, a determinant is a number.
When transposing a matrix, the value of its determinant does not change
We transpose the matrix: 
According to the property, the determinant of the transposed matrix is equal to the same value: 
. Those who wish can verify this for themselves.
A more simple formulation of this property is also in use: if the determinant is transposed, then its value will not change.
We write both determinants side by side and analyze one important point:
As a result of the transposition, the first row became the first column, the second row became the second column, and the third row became the third column. The rows became columns, but the result did not change. From which follows an important fact: rows and columns of the determinant are equal. In other words, if a property is true for a row, then a similar property is true for a column! In fact, we have already encountered this for a long time - after all, the determinant can be expanded both in a row and equally in a column.
Don't like numbers in strings? Transpose the determinant! There is only one question, why? The practical meaning of the considered property is small, but it is useful to throw it into the baggage of knowledge in order to better understand other problems of higher mathematics. For example, it immediately becomes clear why the study of vectors for coplanarity their coordinates can be written both in the rows of the determinant and in the columns.
If two rows (or two columns) of the determinant are interchanged,
then the determinant changes sign
! Remember , we are talking about the determinant! You cannot rearrange anything in the matrix itself!
Let's play a Rubik's cube with a determinant 
.
Let's swap the first and third lines: 
The determinant has changed sign.
Now, in the resulting determinant, rearrange the second and third rows: 
The determinant changed sign again.
Rearrange the second and third columns: 
That is, any pair permutation of rows (columns) entails a change in the sign of the determinant to the opposite.
Games are games, but in practice such actions are better do not use. There is not much sense from them, but it is not difficult to get confused and make a mistake. However, I will give one of the few situations where this really makes sense. Suppose that in the course of solving some example, you have drawn a determinant with a minus sign:
Let's expand it, say, by the first line:
The obvious inconvenience is that I had to perform unnecessary curtsies - to put big brackets, and then open them (by the way, I strongly do not recommend performing such actions “in one sitting” orally).
To get rid of the "minus", it is more rational to swap any two rows or any two columns. Let's rearrange, for example, the first and second lines: 
It looks stylish, but in most cases it is more expedient to deal with a negative sign in a different way (read on).
The considered action again helps to better understand, for example, some properties cross product of vectors or a mixed product of vectors.
Now this is more interesting:
From the row (column) of the determinant, you can take out the common factor
!!! Attention! The rule is about ONE line or about ONE determinant column. Please do not confuse with matrices, in the matrix the multiplier is taken out / brought in by ALL numbers at once.
Let's start with a special case of the rule - the removal of "minus one" or simply "minus".
We meet another patient: .
There are too many minuses in this determinant and it would be nice to reduce their number.
Take out -1 from the first line: 
Or shorter: 
The minus in front of the determinant, as already demonstrated, is not convenient. We look at the second line of the determinant and notice that there are too many minuses there.
We take out the "minus" from the second line: 
What else can be done? All numbers in the second column are divisible by 4 without a remainder. Take out 4 from the second column: 
The opposite rule is also true - multiplier can not only endure, but also contribute, moreover, in ANY row or in ANY column of the determinant.
For fun, let's multiply the third row of the determinant by 4: 
Meticulous minds can verify the equality of the original and received determinants (correct answer: -216).
In practice, the introduction of a minus is often performed. Let's consider the determinant. The negative sign before the determinant can be entered in ANY row or in ANY column. The best candidate is the third column, and we will add a minus to it: 
We also notice that all the numbers in the first column are divisible by 2 without a remainder, but is it worth taking out the “two”? If you are going to lower the order of the determinant (which will be discussed in the final section), then it is definitely worth it. But if you open the determinant in a row (column), then the “two” in front will only lengthen the record of the solution.
However, if the multiplier is large, for example, 13, 17, etc., then, of course, it is more profitable to take it out anyway. Let's get acquainted with the little monster:. From the first line we take out -11, from the second line we take out -7:
You say, calculations are already clicking so fast on a regular calculator? It's true. But, firstly, it may not be at hand, and secondly, if a determinant of the 3rd or 4th order with large numbers is given, then you don’t really want to knock on the buttons.
Task 3
Calculate the determinant by factoring out the rows and columns
This is a do-it-yourself example.
A couple more useful rules:
If two rows (columns) of the determinant are proportional
(as a special case, they are the same), then this determinant is equal to zero
Here, the corresponding elements of the first and second rows are proportional:
It is sometimes said that the strings of the determinant linearly dependent. Since the value of the determinant does not change during transposition, then the linear dependence of the columns follows from the linear dependence of the rows.
You can put a geometric meaning into the example - if we assume that the coordinates are written in the lines vectors space, then the first two vectors with proportional coordinates will be collinear, which means that all three vectors - linearly dependent, that is, coplanar.
In the following example, three columns are proportional (and, by the way, three rows too): 
Here the second and third columns are the same, this is a special case - when the proportionality coefficient is equal to one
These properties can be used in practice. But remember, an increased level of knowledge is sometimes punishable ;-) Therefore, it may be better to reveal such determinants in the usual way (knowing in advance that it will turn out to be zero).
It should be noted that the converse is not true in general- if the determinant is equal to zero, then from this not yet to be that its rows (columns) are proportional. That is, the linear dependence of rows / columns may not be explicit.
There is also a more obvious sign, when you can immediately say that the determinant is zero:
A determinant with a zero row (column) is zero
The "amateur" check is elementary, let's expand the determinant by the first column: 
However, the result will not change if the determinant is expanded on any row or any column.
Squeeze out the second glass of orange juice:
What properties of determinants are useful to know?
1) The value of the determinant does not change when transposing. We remember the property.
2) Any pair permutation of rows (columns) changes the sign of the determinant to the opposite. We also remember the property and try not to use it in order to avoid confusion.
3) From the row (column) of the determinant, you can take out the multiplier (and bring it back). We use it where it's beneficial.
4) If the rows (columns) of the determinant are proportional, then it is equal to zero. A determinant with a zero row (column) is equal to zero.
During the lesson, an elementary pattern was repeatedly observed - the more zeros in a row (column), the easier it is to calculate the determinant. The question arises, is it possible to organize the zeros specially with the help of some kind of transformation? Can! Let's get acquainted with another very powerful property:
Reducing the order of the determinant
It is very good if you have already dealt with Gauss method and have experience in solving systems of linear equations in this manner. In fact, the property formulated below duplicates one of elementary transformations.
To work up an appetite, let's crush a little frog: 
You can add another string multiplied by a non-zero number to the determinant string. In this case, the value of the determinant will not change
Example: in the determinant we get zero at the top left.
To do this, the second line mentally or in draft multiply by 3: (–3, 6) and add the second row to the first row multiplied by 3:
We write the result to the first line:
Examination: 
Now in the same determinant we get zero at the bottom right. For this to the second line, add the first line, multiplied (mentally) by -2):
We write the result to the second line:
note: when elementary transformation changes TA the line to which we add UT.
Let's formulate a mirror rule for columns:
To the determinant column, you can add another column multiplied by a non-zero number. In this case, the value of the determinant will not change
Let's take the animal by the paws and, using this transformation, we get zero at the top left. To do this, mentally or on a draft, multiply the second column by -3: and add the second column to the first column, multiplied by -3:
We write the result to the first column:
And, finally, in the determinant we get zero at the bottom right. For this to the second column we add the first column, multiplied (mentally) by 2(look and count from right to left):
The result is placed to the second column:
Under an elementary transformation, it changes THAT the column to which we add UT.
Try to digest the following example qualitatively.
Let's send the grown amphibian to the soup:
The task is to reduce the order of the determinant using elementary transformations up to the second order.
Where to begin? First, in the determinant, you need to choose the number - "target". The "target" is almost always one or -1. We look at the determinant and notice that there is even a choice here. Let the element be the “target” number: 
Note : the meaning of double subscripts can be found in the article Cramer's rule. Matrix method. V this case the indexes of an element tell us that it is located in the second row, third column.
The idea is to get two zeros in the third column: 
Or get two zeros in the second line: 
In the second line, the numbers are smaller (do not forget the golden rule), so it is more profitable to take it. And the third column with the “target” number will remain unchanged: 
Add a third column to the second column:
There was no need to multiply anything.
The result is written in the second column: 
To the first column we add the third column, multiplied (mentally) by -2:
We write the result in the first column, expand the determinant along the second line: 
How do we lower the order of the determinant? Got two zeros in the second line.
Let's solve the example in the second way, organize the zeros in the third column: 
The second line with the "target" number will remain unchanged: 
To the first line, add the second line, multiplied (mentally) by -4: 
To the third line, add the second line, multiplied (mentally) by 3 (look and count from the bottom up):
We write the result in the third line, expand the determinant in the third column: 
Notice that there is no need to rearrange rows or columns. Elementary transformations work fine both from left to right and from right to left. Both top down and bottom up.
Task 4
Calculate the same determinant by choosing the element as the “target” number. Lower its order in two ways: by getting zeros in the second row and by getting zeros in the second column.
This is a do-it-yourself example. Full solution and brief comments at the end of the lesson.
Sometimes there is no unit or -1 in the determinant, for example: . In this case, the "target" should be organized using an additional elementary transformation. This can be done most often in several ways. For example: to the first line, add the second line, multiplied by -1: 
The result is written in the first line: 
! Attention : NO NEED from the first line subtract second line, this greatly increases the chance of error. We just fold! Therefore, to the first line we add the second line, multiplied by -1. Exactly!
The unit was received, which was required to be achieved. Then you can get two zeros in the first row or in the first column. Those who wish can complete the solution (correct answer: -176).
It is worth noting that the finished “target” is most often present in the original determinant, and for a determinant of the 4th order and higher, an additional transformation is extremely unlikely.
Let's chop a few large toads into goulash:
Task
Solve the system linear equations according to Cramer's formulas 
It's okay if you haven't read it yet. Cramer's method, in this case, you can simply see how the order of the determinant "four by four" decreases. And the rule itself will become clear if you delve a little into the course of the solution.
Solution: first calculate main determinant systems: 
It is possible to go the standard way by expanding the given determinant by row or column. Recalling the algorithm of the first lesson, and using the matrix of signs that I invented , let's expand the determinant, for example, by the "classic" first line:
I don't see your enthusiasm =) Of course, you can sit for ten minutes and carefully and carefully give birth to the correct answer. But the trouble is that in the future we have to calculate 4 more determinants of the fourth order. Therefore, the only reasonable way out is to lower the order of the determinant.
There are many units in the determinant, and our task is to choose the best way. We recall the golden rule: there should be more zeros in a row (column) and fewer numbers. For this reason, the second row or the fourth column is quite suitable. The fourth column looks more attractive, moreover, there are two ones. As a "target" select the element: 
The first line will not change. And the second one too - there is already a necessary zero: 
To the third row, add the first row, multiplied by -1 (look and count from the bottom up):
! Attention again : No need from the third line subtract first line. We just fold!
The result is written in the third line: 
To the fourth line, add the first line, multiplied by 3 (look and count from the bottom up):
The result is written in the fourth line: 
(1) Expand the determinant in the fourth column. Do not forget that you need to add a “minus” to the element (see the matrix of signs).
(2) The order of the determinant is lowered to 3rd. In principle, it can be decomposed into a row (column), but it is better to work out the properties of the determinant. We enter a minus in the second line.
(3) To the second row we add the first row multiplied by 3. To the third row we add the first row multiplied by 7.
(4) We expand the determinant by the second column, thereby further lowering its order to two.
Notice how the solution has shrunk! The main thing is to get a little "hands on" with elementary transformations, and such an opportunity will present itself right now. In addition, you have at your disposal a calculator that counts determinants (in particular, it can be found on the page Mathematical formulas and tables). With the help of the calculator, it is easy to control the actions performed. Got a determinant 
at the first step - and immediately checked whether it is equal to the original determinant.
(1) Expand the determinant by the third line. The order of the determinant is reduced to three.
(2) Enter "minus" in the first column.
(3) To the second row we add the first row multiplied by 3. To the third row we add the first row multiplied by 5.
(4) Expand the determinant by the second column, lowering the order of the determinant to two.
We get amazing complex lunch, and it's time for dessert: 
It's not even a toad anymore, it's Godzilla himself. Let's take a prepared glass of orange juice and see how the order of the determinant decreases. The algorithm, I think, is clear: from the fifth order we lower it to the fourth, from the fourth to the third, and from the third to the second:
(1) To the first, third, fourth and fifth lines, add the second line.
(2) Expand the determinant in the 3rd column. The order of the determinant has dropped to four.
(3) We take out 2 from the 4th column. We multiply the first row by -1, and so that the determinant does not change, we put a “minus” in front of it. This transformation performed in order to simplify further calculations.
(4) Add the first line to the second and third lines. To the fourth line, add the first line, multiplied by 3.
(5) Expand the determinant in the 4th column. The order is downgraded to three.
(6) Expand the determinant in the 2nd column. The order is downgraded to two.
(7) We take out the "minus" from the 1st column.
Everything turned out easier than it seemed, all monsters have weak points!
Tireless readers can try to solve the fifth-order determinant in some other way, fortunately, there are a lot of units in it.
The second column multiplied by 2 was added to the first column. The second column was added to the third column. The determinant was opened on the second line.
Lower the order of the determinant, getting zeros in the second column:
The second line was added to the first line, multiplied by -2. The second row, multiplied by 2, was added to the third row. The determinant was opened according to the second column.
Task 5: Solution:
(1) To the first row we add the third row multiplied by 3. To the second row we add the third row multiplied by 5. To the 4th row we add the third row multiplied by 2.
(2) Expand the determinant by the first column.
(3) To the second column, add the third column multiplied by 9. To the first column, add the third column.
(4) Expand the determinant by the third line.
(1) Add the second column to the first column. Add the second column to the third column
(2) Expand the determinant by the third line.
(3) Enter "minus" in the first line.
(4) To the second line add the first line multiplied by 6. To the third line add the first line
(5) Expand the determinant by the first column.
In the general case, the rule for calculating $n$-th order determinants is rather cumbersome. For determinants of the second and third order, there are rational ways to calculate them.
Calculations of second-order determinants
To calculate the determinant of a matrix of the second order, it is necessary 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. Calculate 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
There are rules for computing third-order determinants.
triangle rule
Schematically, this rule can be represented as follows:
The product of elements in the first determinant that are connected by 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. Calculate the $\left| \begin(array)(rrr)(3) & (3) & (-1) \\ (4) & (1) & (3) \\ (1) & (-2) & (-2)\end (array)\right|$ by 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, the first two columns are added and the products of the elements on the main diagonal and on the diagonals parallel to it are taken 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. Calculate the $\left| \begin(array)(rrr)(3) & (3) & (-1) \\ (4) & (1) & (3) \\ (1) & (-2) & (-2)\end (array)\right|$ using the 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.$\left| \begin(array)(rrr)(3) & (3) & (-1) \\ (4) & (1) & (3) \\ (1) & (-2) & (-2)\end (array)\right|=54$
Row or column expansion of determinant
The determinant is equal to the sum of the products of the elements of the row of the determinant and their algebraic complements. Usually choose the row/column in which/th there are zeros. The row or column on which the decomposition is carried out will be indicated by an arrow.
Example
Exercise. Expanding over the first line, 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. Calculate the $\left| \begin(array)(lll)(1) & (2) & (3) \\ (4) & (5) & (6) \\ (7) & (8) & (9)\end(array) \right|$
Solution. Let's 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 get 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 the determinants of the fourth order and higher, either row / column expansion, or reduction to a triangular form, or using Laplace's theorem is used.
Decomposition of the determinant in terms of the elements of a row or column
Example
Exercise. Calculate the $\left| \begin(array)(llll)(9) & (8) & (7) & (6) \\ (5) & (4) & (3) & (2) \\ (1) & (0) & (1) & (2) \\ (3) & (4) & (5) & (6)\end(array)\right|$ , expanding it into 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 a row or in a column. To do this, first we subtract nine thirds from the first line, five thirds from the second, and three thirds from the fourth, we get:
$$\left| \begin(array)(cccc)(9) & (8) & (7) & (6) \\ (5) & (4) & (3) & (2) \\ (1) & (0) & (1) & (2) \\ (3) & (4) & (5) & (6)\end(array)\right|=\left| \begin(array)(cccc)(9-1) & (8-0) & (7-9) & (6-18) \\ (5-5) & (4-0) & (3-5) & (2-10) \\ (1) & (0) & (1) & (2) \\ (0) & (4) & (2) & (0)\end(array)\right|=\ left| \begin(array)(rrrr)(0) & (8) & (-2) & (-12) \\ (0) & (4) & (-2) & (-8) \\ (1) & (0) & (1) & (2) \\ (0) & (4) & (2) & (0)\end(array)\right|$$
We expand the resulting determinant by the elements of the first column:
$$\left| \begin(array)(rrrr)(0) & (8) & (-2) & (-12) \\ (0) & (4) & (-2) & (-8) \\ (1) & (0) & (1) & (2) \\ (0) & (4) & (2) & (0)\end(array)\right|=0+0+1 \cdot(-1)^( 3+1) \cdot \left| \begin(array)(rrr)(8) & (-2) & (-12) \\ (4) & (-2) & (-8) \\ (4) & (2) & (0)\ end(array)\right|+0$$
The resulting third-order determinant is also expanded by the elements of the row and column, having previously obtained zeros, for example, in the first column. To do this, we subtract two second 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.
Bringing the determinant to a triangular form
With the help of 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 $\Delta=\left| \begin(array)(rrrr)(-2) & (1) & (3) & (2) \\ (3) & (0) & (-1) & (2) \\ (-5) & ( 2) & (3) & (0) \\ (4) & (-1) & (2) & (-3)\end(array)\right|$ bringing 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 swap the first and second columns of the determinant, which, according to the properties of the determinant, will cause it to change 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. And again, if the diagonal element is equal to $\pm 1$ , then the calculations will be simpler. To do this, we swap the second and third lines (and at the same time change to the opposite sign of the determinant):
$$\Delta=\left| \begin(array)(rrrr)(1) & (-2) & (3) & (2) \\ (0) & (-1) & (-3) & (-4) \\ (0) & (3) & (-1) & (2) \\ (0) & (2) & (5) & (-1)\end(array)\right|$$
PROPERTY 1. The value of the determinant will not change if all its rows are replaced by columns, and each row is replaced by a column with the same number, that is
PROPERTY 2. Permuting two columns or two rows of a determinant is equivalent to multiplying it by -1. For instance,
.
PROPERTY 3. If the determinant has two identical columns or two identical rows, then it is equal to zero.
PROPERTY 4. Multiplying all elements of one column or one row of a determinant by any number k is equivalent to multiplying the determinant by this number k. For instance,
.
PROPERTY 5. If all elements of some column or some row are equal to zero, then the determinant itself is equal to zero. This property is special case the previous one (for k=0).
PROPERTY 6. If the corresponding elements of two columns or two rows of a determinant are proportional, then the determinant is equal to zero.
PROPERTY 7. If each element of the nth column or nth row of the determinant is the sum of two terms, then the determinant can be represented as the sum of two determinants, of which one in the nth column or, respectively, in the nth row has the first from the mentioned terms, and the other - the second; the elements in the remaining places are the same for the milestones of the three determinants. For instance,
PROPERTY 8. If we add to the elements of some column (or some row) the corresponding elements of another column (or another row), multiplied by any common factor, then the value of the determinant will not change. For instance,
.
Further properties of determinants are connected with the concept of algebraic complement and minor. The minor of some element is the determinant obtained from the given one by deleting the row and column at the intersection of which this element is located.
The algebraic complement of any element of the determinant is equal to the minor of this element, taken with its sign, if the sum of the row and column numbers at the intersection of which the element is located is an even number, and with the opposite sign if this number is odd.
We will denote the algebraic complement of an element by a capital letter of the same name and the same number as the letter that denotes the element itself.
PROPERTY 9. Determinant
is equal to the sum of the products of the elements of any column (or row) and their algebraic complements.
In other words, the following equalities hold:
, ,
, .
6) Minors and algebraic additions.
Definition. The minor element of the determinant is th order called determinant-th order, which is obtained from the given determinant by deleting the -th row and -th column, at the intersection of which is the element .
Designation: .
Definition. The algebraic complement of an element of the determinant of the -th order is its minor, taken with a plus sign if - an even number and with a minus sign otherwise.
Designation: .
Theorem. (On the expansion of the determinant.)
The determinant is equal to the sum of the products of the elements of any row (or any column) of the determinant and their algebraic complements:
7) Inverse matrix- such matrix A −1 , when multiplied by which, the original matrix A gives as a result identity matrix E:
square matrix is invertible if and only if it is non-degenerate, that is, its determinant is not equal to zero. For non-square matrices and degenerate matrices inverse matrices do not exist. However, it is possible to generalize this concept and introduce pseudoinverse matrices, similar to inverses in many properties.
8)Matrix rank- highest order minors this matrix, non-zero
Usually the rank of a matrix is denoted by () or . Both designations came to us from foreign languages, and therefore both can be used.
Properties
Theorem (on the basis minor): Let r = rang A M be the basis minor of the matrix A, then:
basic rows and basic columns are linearly independent;
any row (column) of matrix A is a linear combination of basic rows (columns).
Here, those properties that are usually used to calculate determinants in the standard course of higher mathematics will be stated. This is a secondary topic, which we will refer to from the remaining sections as needed.
So, given some square matrix $A_(n\times n)=\left(\begin(array) (cccc) a_(11) & a_(12) & \ldots & a_(1n) \\ a_(21) & a_(22) & \ldots & a_(2n) \\ \ldots & \ldots & \ldots & \ldots \\ a_(n1) & a_(n2) & \ldots & a_(nn) \\ \end( array)\right)$. Each 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 this issue more details.
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 in it with columns according to the principle: "there was the first row - the first column became", "there was the second row - the second column became":
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 from the replacement.
- If you swap two rows (columns) of the determinant, then the sign of the determinant will change to the opposite.
An example of using this property: show\hide
Consider the $\left| \begin(array) (cc) 2 & 5 \\ 9 & 4 \end(array) \right|$. Let's find its value using formula No. 1 from the topic of calculating second and third order determinants:
$$\left| \begin(array) (cc) 2 & 5 \\ 9 & 4 \end(array) \right|=2\cdot 4-5\cdot 9=-37.$$
Now let's swap the first and second lines. Get 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 in which all elements of a row (column) are equal to zero is equal to zero.
An example of using this property: show\hide
Since in the $\left| \begin(array) (ccc) -7 & 10 & 0\\ -9 & 21 & 0\\ 2 & -3 & 0 \end(array) \right|$ all elements of the third column are zero, then the determinant is zero , i.e. $\left| \begin(array) (ccc) -7 & 10 & 0\\ -9 & 21 & 0\\ 2 & -3 & 0 \end(array) \right|=0$.
- A determinant in which all elements of a certain row (column) are equal to the corresponding elements of another row (column) is equal to zero.
An example of using this property: show\hide
Since in the $\left| \begin(array) (ccc) -7 & 10 & 0\\ -7 & 10 & 0\\ 2 & -3 & 18 \end(array) \right|$ all elements of the first row are equal to the corresponding elements of the second row, then the determinant is zero, i.e. $\left| \begin(array) (ccc) -7 & 10 & 0\\ -7 & 10 & 0\\ 2 & -3 & 18 \end(array) \right|=0$.
- If in the 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.
An example of using this property: show\hide
Since in the $\left| \begin(array) (ccc) -7 & 10 & 28\\ 5 & -3 & 0\\ -15 & 9 & 0 \end(array) \right|$ the second and third lines 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 sign of the determinant.
An example of using this property: show\hide
Consider the $\left| \begin(array) (cc) -7 & 10 \\ -9 & 21 \end(array) \right|$. Note that all elements of 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 triple 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 does not change if all the elements of a certain row (column) are added to the corresponding elements of another row (column), multiplied by an arbitrary number.
An example of using this property: show\hide
Consider the $\left| \begin(array) (ccc) -7 & 10 & 0\\ -9 & 21 & 4 \\ 2 & -3 & 1 \end(array) \right|$. Let's add to the elements of the second line the corresponding elements of the third line, multiplied by 5. Write this action as follows: $r_2+5\cdot(r_3)$. The second line will be changed, the rest of the 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 the determinant is a linear combination of other rows (columns), then the determinant is equal to zero.
An example of using this property: show\hide
I will immediately explain what the phrase "linear combination" means. Let we 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 of the first three rows:
$$ r_4=2\cdot(r_1)+3\cdot(r_2)-r_3 $$
Therefore, the determinant under consideration 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 the determinants, the first of which has in the k-th row ( k-th column) have the first terms, and the second determinant in the kth row (kth column) has the second terms. Other elements of these determinants are the same.
An example of using this property: show\hide
Consider the $\left| \begin(array) (ccc) -7 & 10 & 0\\ -9 & 21 & 4 \\ 2 & -3 & 1 \end(array) \right|$. 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, you can get the following formula: $\det \left(A^n \right)=\left(\det A \right)^n$.
- If the matrix $A$ is nonsingular (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 true:
\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 applying formulas (1) and (2) are in the topic "Formulas for calculating second and third order determinants. Examples of calculating determinants" .
The determinant of the matrix $A_(n\times n)$ can be expanded in terms of 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 j-th 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 Decreasing the order of a determinant. Decomposition of the determinant in a row (column).
We indicate one more formula for calculating the determinants of upper triangular and lower triangular matrices (for an explanation of these terms, see the topic "Matrices. 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)
