Decomposition of the determinant into row and column elements. Theorem (decomposition of the determinant in a row or column)

ALGEBRA

1. MATRICES AND DETERMINANTS. Definitions of the determinant and its basic properties. Theorem on the decomposition of the determinant into the elements of a row (column). Matrix invertibility criterion.

Determinant or nth order determinant is a number written in the form

and calculated from given numbers (real or complex) - elements of the determinant - according to the following law:

extended to all possible different permutations from numbers. The number is equal to the number of transpositions that need to be made to go from the main permutation to the permutation n-th order . Work called member of the determinant.

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, d is expanded into elements of the i-th row

d = a i 1 A i 1 + a i 2 A i 2 +... + a i n A i n (i = )

or jth column

d = a 1 j A 1 j + a 2 j A 2 j +... + a n j A n j (j = ).

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.

Proof.

Let us verify the validity of the theorem using the example of the expansion of the 3rd order determinant, for example, in the 1st row. According to the theorem, this expansion will have the form: D= = a 11 A 11 + a 12 A 12 + a 13 A 13 = (taking into account the definition of A ij we obtain) = =a 11 (-1) 2 M 11 + a 12 (- 1) 3 M 12 + a 13 (-1) 4 M 13 = a 11 - a 12 + a 13 = a 11 (a 22 ×a 33 - a 23 ×a 32) - a 12 (a 21 ×a 33 - a 23 ×a 31) + a 13 (a 21 ×a 32 - a 22 ×a 31) = a 11 ×a 22 ×a 33 + a 12 ×a 23 ×a 31 + a 13 ×a 21 ×a 32 - a 13 ×a 22 ×a 31 - a 12 ×a 21 ×a 33 - a 11 ×a 23 ×a 32 = (according to the rule of triangles) = = D. A similar result is obtained when expanding the determinant along any row (column). Fin.

Consequence. If in the i-th row (j-th column) of the determinant D there is only one non-zero element a ij ¹ 0, then the result of decomposing the determinant along this row (column) will be the expression D = a ij ×A ij.

Determinants of nth order satisfy the following properties:

1) When transposing a determinant, its value does not change (that is, the value of the determinant does not change when replacing its rows with columns with the same numbers).

Proof:

D = = = a 11 ×a 22 - a 12 ×a 21

NB. Consequently, the rows and columns of the determinant are equal, so its properties can be formulated and proven either for rows or for columns.

2) When any two rows (columns) of the determinant are rearranged, its sign changes to the opposite.

Proof:

D = = a 11 ×a 22 - a 12 ×a 21 = - (a 12 ×a 21 - a 11 ×a 22) = -

3) A determinant with two identical rows (columns) is equal to zero.

Proof. Let the determinant D have two identical rows. If we swap them, then, on the one hand, the value of the determinant will not change, since the rows are the same, and on the other hand, the determinant must change its sign to the opposite one by property 2. Thus, we have: D = -D Þ D = 0.

4) The common factor of the elements of any row (column) can be taken beyond the sign of the determinant.

Proof:

D= = la 11 ×a 22 - la 12 ×a 21 = l(a 11 ×a 22 - a 12 ×a 21) = l.

Corollary: D = = l×m.

NB. The rule for multiplying a determinant by a number. To multiply a determinant by a number, you need all the elements of some kind one its rows (columns) multiplied by this number.

5) A determinant with a zero row (column) is equal to zero.

Proof. By property 4, we take the common factor l = 0 of the elements of the zero row (column) beyond the sign of the determinant. We get 0×D = 0.

6) A determinant with two or more proportional rows (columns) is equal to zero.

Proof. If we take the proportionality coefficient of two rows (columns) l≠0 out of the sign of the determinant, we get a determinant with two identical rows (columns), equal to zero by property 3.

7) If each element of any row (column) of the determinant is represented in

form of the sum of k terms, then such a determinant is equal to the sum of k determinants in which the elements of this row (column) are replaced by the corresponding terms, and all other elements are the same as those of the original determinant.

Proof:

D= = (a 11 + b 11)a 22 - (a 12 + b 12)a 21 = (a 11 a 22 - a 12 a 21) + (b 11 a 22 - b 12 a 21) = = + .

Def. The nth row of a determinant is called a linear combination of its remaining (n-1) rows if it can be represented as the sum of the products of these rows by the corresponding numbers l 1, l 2, …, l n - 1. For example, in the determinant

The 3rd line is a linear combination of the first two lines.

NB. A linear combination is called trivial if it has "l i = 0. Otherwise, the linear combination is called non-trivial (if $l i ¹ 0).

8 a) If one row (column) of a determinant is a linear combination of its other rows (columns), then such a determinant is equal to zero.

Proof: D =

8 b) The value of the determinant will not change if the corresponding elements of any other row (column) of the determinant, multiplied by the same number, are added to the elements of any of its rows (columns).

Proof:

Let D= Þ (to the 1st line add the 2nd line multiplied by the number l) Þ

9) The sum of the products of the elements of any row (column) of the determinant by the algebraic complements of the corresponding elements of any other row (column) of the determinant is equal to zero, that is, = 0 (if i ≠ j). For example, let

Then a 11 A 21 + a 12 A 22 + a 13 A 23 = 0, since the elements of the 1st row of the determinant are multiplied by the algebraic complements of the corresponding elements of the 2nd row.

Proof:

a 11 A 21 + a 12 A 22 + a 13 A 23 = a 11 ×(-1) 2+1 + a 12 ×(-1) 2+2 + a 13 ×(-1) 2+3 =

=(this is the expansion along the 1st row of the determinant (-1)× = 0)= 0.

If the determinant is D¹0, then by property 8 b) it is always possible to “zero” the i-th row (j-th column) to the only non-zero element and expand the determinant along this row (column). By applying this operation the required number of times, it is always possible to obtain a 2nd order determinant from the nth order determinant.

ALGEBRA

MATRICES AND DETERMINANTS. Definitions of the determinant and its basic properties. Theorem on the decomposition of the determinant into the elements of a row (column). Matrix invertibility criterion.

Determinant or determinantn-th order is a number written in the form

and calculated from these numbers (real or complex) – elements of the determinant – according to the following law:

extended to all sorts of different permutations
from numbers
. Number
equal to the number of transpositions that need to be made to move from the main permutation
to rearrangement n-th order
. Work
called member of the determinant.

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, d is expanded into elements of the i-th row

d = a i 1 A i 1 + a i 2 A i 2 +... + a i n A i n (i =
)

or jth column

d = a 1 j A 1 j + a 2 j A 2 j +... + a n j A n j (j =
).

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.

Proof.

Let us verify the validity of the theorem using the example of expansion of the 3rd order determinant, for example, in the 1st row. According to the theorem, this expansion will have the form: =
= a 11 A 11 + a 12 A 12 + a 13 A 13 = (taking into account the definition of A ij we obtain) = =a 11 (1) 2 M 11 + a 12 (1) 3 M 12 + a 13 ( 1) 4 M 13 = a 11
- a 12
+ a 13
= a 11 (a 22 a 33  a 23 a 32)  a 12 (a 21 a 33  a 23 a 31) + a 13 (a 21 a 32  a 22 a 31) = = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32  a 13 a 22 a 31  a 12 a 21 a 33  a 11 a 23 a 32 = (according to the rule of triangles) =
=. A similar result is obtained when expanding the determinant over any row (column). Fin.

Consequence. If in the i-th row (j-th column) of the determinant  there is only one non-zero element a ij  0, then the result of decomposing the determinant along this row (column) will be the expression  = a ij A ij.

Determinantsn-th order satisfy the following properties:

1) When transposing a determinant, its value does not change (that is, the value of the determinant does not change when replacing its rows with columns with the same numbers).

Proof:

 =
=
= a 11 a 22  a 12 a 21

NB. Consequently, the rows and columns of the determinant are equal, so its properties can be formulated and proven either for rows or for columns.

2) When any two rows (columns) of the determinant are rearranged, its sign changes to the opposite.

Proof:

 =
= a 11 a 22  a 12 a 21 =  (a 12 a 21  a 11 a 22) = 

3) A determinant with two identical rows (columns) is equal to zero.

Proof. Let the determinant  have two identical lines. If we swap them, then, on the one hand, the value of the determinant will not change, since the rows are the same, and on the other hand, the determinant must change its sign to the opposite one according to property 2. Thus, we have:  =    = 0.

4) The common factor of the elements of any row (column) can be taken beyond the sign of the determinant.

Proof:

=
=a 11 a 22  a 12 a 21 = (a 11 a 22  a 12 a 21) = 
.

Corollary:  =
=
.

NB. The rule for multiplying a determinant by a number. To multiply a determinant by a number, you need all the elements of some kind one its rows (columns) multiplied by this number.

5) A determinant with a zero row (column) is equal to zero.

Proof. By property 4, we take the common factor  = 0 of the elements of the zero row (column) beyond the sign of the determinant. We get 0 = 0.

6) A determinant with two or more proportional rows (columns) is equal to zero.

Proof. If we take the coefficient of proportionality of two rows (columns) ≠0 out of the sign of the determinant, we get a determinant with two identical rows (columns), equal to zero by property 3.

7) If each element of any row (column) of the determinant is represented in

Proof:

=
= (a 11 + b 11)a 22  (a 12 + b 12)a 21 = (a 11 a 22  a 12 a 21) + (b 11 a 22  b 12 a 21) = =
+
.

Def. The nth row of a determinant is called a linear combination of its remaining (n1) rows if it can be represented as the sum of the products of these rows by the corresponding numbers  1 ,  2 , …,  n  1 . For example, in the determinant

The 3rd line is a linear combination of the first two lines.

NB. A linear combination is called trivial if it has  i = 0. Otherwise, a linear combination is called non-trivial (if  i  0).

8 a) If one row (column) of a determinant is a linear combination of its other rows (columns), then such a determinant is equal to zero.

Proof:  =

Proof:

Let =
 (to the 1st line add the 2nd line multiplied by the number ) 

=
.

 =
 0

Then a 11 A 21 + a 12 A 22 + a 13 A 23 = 0, since the elements of the 1st row of the determinant are multiplied by the algebraic complements of the corresponding elements of the 2nd row.

Proof:

a 11 A 21 + a 12 A 22 + a 13 A 23 = a 11 (1) 2+1
+ a 12 (1) 2+2
+ a 13 (1) 2+3
=

=(this is the expansion along the 1st row of the determinant (1)
= 0}= 0.

If the determinant is 0, then by property 8 b) it is always possible to “zero” the i-th row (j-th column) to a single non-zero element and expand the determinant along this row (column). By applying this operation the required number of times, it is always possible to obtain a 2nd order determinant from the nth order determinant.

Inverse matrix

Def. Matrix is called adjoint (adjoint) to a square matrix A if it consists of algebraic complements of elements of the transposed matrix A m. To obtain the adjoint matrix , you should transpose the matrix A, and then replace all its elements with their algebraic complements, that is

=
(3.1)

Def. A square matrix A is called singular (singular) if its determinant |A|=0, and non-singular if its determinant |A|0.

Def. A square matrix A  1 is called inverse (inverse) to a square matrix A if the condition is satisfied

A  1 A = AA  1 = E (3.2)

NB. The inverse matrix A  1 is possible only for a non-singular matrix A.

Theorem.

For any non-singular square matrix A there is a unique inverse matrix A  1, which is found by the formula

A - 1 = (3.3)

Proof.

1) From the definition of A  1 A = AA  1 it follows that A and A - 1  are square matrices of the same order.

Let the matrix A be non-degenerate, that is, |A|0. Then, according to the rule of matrix multiplication, according to Laplace’s theorem and according to the property of 9 determinants, we obtain

A =

=
=

= |A| = |A|E

Therefore, A = |A|E. Similarly, it is proved that A = |A|E.

From A = |A|E  A  1 A = A - 1 ×|A|E  E = A  1 |A|  = A  1 |A|  A  1 = .

2) Let us prove the uniqueness of the inverse matrix. Suppose that for matrix A there is another inverse matrix B. Then, according to the definition, the product AB=E. We multiply both sides of the last equality on the left by the inverse matrix A  1 and get: A  1 AB = A  1 E  EB = A  1 E  B = A  1 . Fin.

Properties of an inverse matrix:

ALGEBRA OF POLYNOMIALS. The greatest common divisor of two polynomials (Euclidean algorithm).

Polynomial n the th degree is called a function of the form

Where are constant coefficients (real or complex), and – a complex variable that can take on any complex values
or, to put it geometrically, can be any point in the complex plane.

If
at
, then the number called root or zero polynomial
.

The following arithmetic operations are defined for polynomials:

As a result of operations 1) and 2) we again obtain a polynomial. The quotient of two polynomials may not be a polynomial.

Dividing polynomials with a remainder.

Where
- private, and
- remainder.

Bezout's theorem.

In order for a polynomial
had a (complex) root , it is necessary and sufficient that it is divisible by
, i.e. so that it can be represented as a product, where
– some polynomial of degree n-1 .

If during decomposition
, then based on Bezout’s theorem applicable to
, polynomial
not divisible by
, A
although it is divided into
, but not divisible by
. In this case they say that –simple root (zero) polynomial .

Let it now
. Then, by Bezout’s theorem, applicable to
, polynomial
divided by
, and we will get
, Where
– some polynomial of degree n-2 . If
, That
divided by
, but not divisible by
, and then the number called root (zero) of multiplicity 2.

In general, for some natural
takes place

Where
– polynomial of degree n- s, and then they say that –root (zero) of a polynomial multiplicitys.

Gauss's theorem (fundamental theorem of algebra).

Any polynomial n-th degree (non-zero, i.e.
) has at least one complex root (zero).

Corollary of Gauss's theorem.

Polynomial n th degree with the highest non-zero coefficient
has n complex roots taking into account multiplicity, in other words
is presented as a product

Where
– various roots multiplicities, respectively
.

If a polynomial with real coefficients has complex roots, then they appear in conjugate pairs, i.e. If
– root of a polynomial , then the root
will be the root of the polynomial .

Expanding the quadratic factorization of the polynomial complex roots
to conjugate ones, i.e.
we obtain the expansion of the polynomial to linear factors.

As a result, we obtain a decomposition of the form

Where
corresponds to the real root b multiplicity l, A
– complex roots And multiplicity m.

Greatest common divisor of polynomials

Let arbitrary polynomials be given
And
. The polynomial will be called common divisor For
And
, if it serves as a divisor for each of these polynomials. Property 5 shows that the number of common divisors of polynomials
And
all polynomials of degree zero belong. If these two polynomials have no other common divisors, then they are called mutually prime.

In the general case, polynomials
And
may have divisors depending on , and introduce the concept of the greatest the common divisor of these polynomials.

Greatest common divisor nonzero polynomials
And
such a polynomial is called
, which is their common divisor and, at the same time, is itself divisible by any other common divisor of these polynomials. Denotes the greatest common divisor of polynomials
And
symbol
.

This definition leaves open the question of whether there is a greatest common divisor for any polynomials
And
. The answer to this question is yes. There is a method for practically finding the greatest common divisor of given polynomials called sequential division algorithm or Euclid's algorithm.

Euclid's algorithm – a method for finding the greatest common divisor of two integers, as well as two polynomials of the same variable. It was originally stated in Euclid's Elements in geometric form as a method of finding the common measure of two segments. Euclidean's algorithm for finding the greatest common divisor, both in the ring of integers and in the ring of polynomials in one variable, is a special case of a certain general algorithm in Euclidean rings.

Euclid's algorithm for finding the greatest common divisor of two polynomials
And
consists of sequential division with a remainder
on
, then
for the first balance
, then
for the second balance
and so on. Since the degrees of residues are decreasing all the time, in this chain of successive divisions we will reach a point where the division is complete and the process stops. Last non-zero remainder
, by which the previous remainder is completely divided
, and is the greatest common divisor of the polynomials
And
.

To prove it, we write the above in the form of the following chain of equalities:

The last equality shows that
serves as a divider for
. It follows that both terms on the right side of the penultimate equality are divisible by
, and therefore
will be a divisor for
. Further, in the same way, going up, we get that
is a divisor for
, …,
,
. From here, in view of the second equality, it will follow that
serves as a divider for
, and therefore, based on the first equality, - and for
.

Let us now take an arbitrary common divisor
polynomials
And
. Since the left side and the first term of the right side of the first of the equalities are divided by
, That
will also be divisible by
. Moving on to the second and next equalities, in the same way we obtain that on
polynomials are divisible
,
, ... Finally, if it has already been proven that
And
are divided into
, then from the penultimate equality we obtain that
divided by
. Thus,
will actually be the greatest common divisor for
And
.

We have proven that any two polynomials have a greatest common factor, and we have found a way to calculate it. This method shows that if polynomials
And
have both rational or real coefficients, then the coefficients of their greatest common divisor will also be rational or, accordingly, real, although these polynomials may also have divisors, not all of whose coefficients are rational (real).

a i, j

Determinants

det(2A )= det(2E ) detA = 0 2 0 (− 2)= 23 (− 2)= − 16 . 0 0 2

(d) Likewise,

det(− 3A )= det(− 3E ) detA = (− 3)3 (− 2)= 54.

(e) First we find the matrix (A − 2E) and then its determinant:

− 1 5

A − 2 E=

−1

−3

det(A − 2E )= 0 (− 1) (− 3)= 0 .

2.4. Calculation of determinants

Here we will look at two methods for calculating determinants. The essence of one of them is to decompose the determinant into elements of a row or column, as a result of which the original determinant of the nth order is expressed through n determinants of a lower order. Another method is based on the properties of determinants and is associated with transforming the determinant to a simpler form. A combination of the two methods provides the most efficient way to calculate determinants.

2.4.1. Decomposing the determinant into elements of a row or column

Let us first introduce some concepts that are important for the subsequent presentation.

Consider a square matrix of the nth order. Let's select the i,j-th element of this matrix and cross out the i-th row and j-th column. As a result

we obtain a matrix of (n – 1)th order, the determinant of which is called the minor of the element a i, j and is denoted by the symbol M i, j.

Determinants

Algebraic complement A i, j of elementa i, j is determined by the formula

A i, j= (− 1) i + j M i, j.

It is easy to see that the algebraic complement of the i, jth element coincides with the minor of this element if the sum of the indices numbering the row and column of the element is an even number. For odd values i+j, the algebraic complement differs from the minor only in sign.

Theorem on the expansion of the determinant into elements of a string.

The determinant of matrix A is equal to the sum of the products of the row elements and their algebraic complements:

det A = a i ,1A i ,1+ a i ,2A i ,2+K+ a i ,n A i ,n =

= ∑ a i, jA i, j j= 1

Proof: By definition, the determinant of the matrix A is the sum

det A =	∑ a 1,k 1 a 2,k 2 K a i ,k i K a n ,k n (− 1) P ( k 1 , k 2 , K , k n )
	(k 1 ,k 2 ,K k i ,K k n )

according to all possible permutations of the indexes numbering the columns. Let's choose some string at random, for example, with

number i. One of the elements of this line is represented in each producta 1, k 1 a 2, k 2 K a i, k i K a n, k n. Therefore, the terms of the sum (*)

can be regrouped by combining into the first group those that contain the element a i ,1 as a common factor, into the second group - members

In other words, expression (*) can be represented as a linear combination of elements a i, j (j = 1,2,L,n),

Determinants

		∑ a 1,k 1 a 2,k 2 K a i ,j K a n ,k n (− 1) P ( k 1 , k 2 , K , k n ) =
det A = ∑
j = 1( k1 , k2 , K j, K kn )
		∑ a 1, k1 a 2, k2 K a i− 1, ki − 1 a i+ 1, ki + 1 a n, kn (− 1) P ( k 1 , k 2 , K , k n ) =
= ∑ a i , j
j = 1		(k 1 ,k 2 ,K j ,K k n )

= ∑ a i ,j A i ,j = a i ,1A i ,1+ a i ,2A i ,2+K+ a i ,n A i ,n ,
j = 1
	∑ a 1, k1 a 2, k2 L a i− 1, ki − 1 a i+ 1, ki + 1 K a n, kn (− 1) P (k 1 , L , k i − 1 , j , k i + 1 , L , k n ) .
A i, j=
(k 1 ,L ,k i − 1 ,k i = j ,k i + 1 ,L ,k n )
Let's show that		A i, j represents the algebraic	addition
element a i, j.
Consider the parity of the permutation (k 1, L, k i − 1, j, k i + 1, L, k n).
Firstly,	requires i –1 transpositions of element j with		neighboring

elements to obtain the permutation ( j , k 1 , L , k i − 1 , k i + 1 , L , k n ) .

Secondly, in the resulting permutation, element j forms j –1 inversions with other elements.

Hence,

(− 1) P (k 1 ,L ,k i − 1 ,j ,k i + 1 ,L ,k n )= (− 1) i − 1+ j − 1(− 1) P (k 1 ,L ,k i − 1 ,k i + 1 ,L ,k n )=

= (− 1) i+ j(− 1) P(k1 , L , ki − 1 , ki + 1 , L , kn )

∑ L a i− 1, ki − 1 a i+ 1, ki + 1 K (− 1) P (k 1 , L , k i − 1 , k i + 1 , L , k n ) = M i, j( k 1 , L , k i − 1 , k i + 1 , L , k n )

represents the minor of the element a i, j.

Thus, A i, j = (− 1) i + j M i, j, as required to be proved.

Since det A = det A T , the following is also true

Theorem on the expansion of the determinant into elements of a column.

The determinant of matrix A is equal to the sum of the products of the column elements and their algebraic complements:

det A = a 1,j A 1,j + a 2,j A 2,j +K+ a n ,j A n ,j

= ∑ a i, jA i, j

i = 1

Determinants

Theorems on the expansion of the determinant are important in theoretical research. They establish that the problem of calculating the determinant of the nth order reduces to the problem of calculating n determinants of the (n–1)th order.

Examples:

1) Calculate the determinant of an arbitrary matrix A = ||a ij || third

order of expansion into elements

(i) first line;

(ii) second column.

Solution:

−a

det A =

A 11(a 22a 33− a 23a 32) − a 12(a 21a 33− a 23a 31) + a 13(a 21a 32− a 22a 31)

A 11a 22a 33+ a 12a 23a 31+ a 13a 21a 32− a 11a 23a 32− a 12a 21a 33− a 13a 22a 31,

−a

det A =

= −a

= − a 12(a 12a 33− a 23a 31) + a 22(a 11a 33− a 13a 31) − a 32(a 11a 23− a 13a 21)

A 11a 22a 33+ a 12a 23a 31+ a 13a 21a 32− a 11a 23a 32− a 12a 21a 33− a 13a 22a 31.

The results obtained by different methods are identical.

Compute determinant

−5

elemental decomposition

−3

(i) first line,

(ii) second column.

Solution:

Expanding the determinant into the elements of the first row gives

−5

− (− 5)

−3

− 3 7

2 4 5 + 5 1 5+ 3(7+ 12)= 122.

(ii) The same result is obtained when expanding the determinant into elements of the second column:

Determinants

	−5


		= −(−5)			−7
−3			−3	− 3 5
−3

5(5 + 0)+ 4 (10+ 9)− 7(0− 3)= 122.

2.4.2. Calculation of determinants by the elementary method

transformations

By elementary transformations we mean the following operations.

Taking into account the equality of rows and columns of the determinant, similar operations are fully applicable to columns.

The idea of the method is to reduce the determinant to triangular form using elementary transformations of rows and columns, which solves the problem of calculating it.

You can do it a little differently: using elementary transformations, obtain a row (or column) containing only one non-zero element, and then expand the resulting determinant into the elements of this row (column). This procedure lowers the order of the determinant by one unit.

Examples.

−4

−3

Calculate det A, reducing the matrix to

1) Let A =

r 2+ 3 r 3

−3

↔r 3

→r 3

−8

−5

The determinant of a triangular matrix is equal to the product of its diagonal elements:

det A = − 1 8 9= − 72 . 2) Calculate the determinant of the matrix

			−2

				−1

Solution: First, we transform the first row using elementary operations on the columns, trying to get the maximum possible number of zeros in it. For this purpose, subtract from the second column the fifth column, previously multiplied by 5, and add double the second column to the third column:

− 2 0

c → c− 5 s

−1

→c 2

2 from 1

− 14

−1

det A =

− 35

− 15

Now let's expand the determinant into the elements of the first line:

det A =	− 14	−1
	− 14	−1
	− 35
	− 15

For determinants of the fourth and higher orders, calculation methods other than the use of ready-made formulas are usually used as for calculating determinants of the second and third orders. One of the methods for calculating determinants of higher orders is to use a corollary of Laplace’s theorem (the theorem itself can be found, for example, in the book by A.G. Kurosh “Course of Higher Algebra”). This corollary allows us to expand the determinant into elements of a certain row or column. In this case, the calculation of the determinant of the nth order is reduced to the calculation of n determinants of the (n-1) order. That is why such a transformation is called reducing the order of the determinant. For example, calculating the fourth-order determinant comes down to finding four third-order determinants.

Let's say we are given a square matrix of nth order, i.e. $A=\left(\begin(array) (cccc) a_(11) & a_(12) & \ldots & a_(1n) \\ a_(21) & a_(22) & \ldots & a_(2n) \\ \ldots & \ldots & \ldots & \ldots \\ a_(n1) & a_(n2) & \ldots & a_(nn) \\ \end(array) \right)$. The determinant of this matrix can be calculated by expanding it by row or column.

Let us fix some line whose number is $i$. Then the determinant of the matrix $A_(n\times n)$ can be expanded over the selected i-th row 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)

$A_(ij)$ denotes the algebraic complement of the element $a_(ij)$. For detailed information about this concept, I recommend looking at the topic Algebraic complements and minors. The notation $a_(ij)$ denotes the element of the matrix or determinant located at the intersection of the i-th row of the j-th column. For more complete information, you can look at the Matrix topic. Types of matrices. Basic terms.

Let's say we want to find the sum $1^2+2^2+3^2+4^2+5^2$. What phrase can describe the entry $1^2+2^2+3^2+4^2+5^2$? We can say this: this is the sum of one squared, two squared, three squared, four squared and five squared. Or we can say it more briefly: this is the sum of the squares of integers from 1 to 5. To express the sum more briefly, we can write it using the letter $\sum$ (this is the Greek letter “sigma”).

Instead of $1^2+2^2+3^2+4^2+5^2$ we can use the following notation: $\sum\limits_(i=1)^(5)i^2$. The letter $i$ is called summation index, and the numbers 1 (initial value $i$) and 5 (final value $i$) are called lower and upper summation limits respectively.

Let's decipher the entry $\sum\limits_(i=1)^(5)i^2$ in detail. If $i=1$, then $i^2=1^2$, so the first term of this sum will be the number $1^2$:

$$ \sum\limits_(i=1)^(5)i^2=1^2+\ldots $$

The next integer after one is two, so substituting $i=2$, we get: $i^2=2^2$. The amount will now be:

$$ \sum\limits_(i=1)^(5)i^2=1^2+2^2+\ldots $$

After two, the next number is three, so substituting $i=3$ we will have: $i^2=3^2$. And the sum will look like:

$$ \sum\limits_(i=1)^(5)i^2=1^2+2^2+3^2+\ldots $$

There are only two numbers left to substitute: 4 and 5. If you substitute $i=4$, then $i^2=4^2$, and if you substitute $i=5$, then $i^2=5^2$. The values $i$ have reached the upper limit of summation, so the term $5^2$ will be the last one. So, the final amount is now:

$$ \sum\limits_(i=1)^(5)i^2=1^2+2^2+3^2+4^2+5^2. $$

This amount can be calculated by simply adding the numbers: $\sum\limits_(i=1)^(5)i^2=55$.

For practice, try writing down and calculating the following sum: $\sum\limits_(k=3)^(8)(5k+2)$. The summation index here is the letter $k$, the lower summation limit is 3, and the upper summation limit is 8.

$$ \sum\limits_(k=3)^(8)(5k+2)=17+22+27+32+37+42=177. $$

An analogue of formula (1) 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 (1) and (2) can be formulated as follows: the determinant is equal to the sum of the products of the elements of a certain row or column by the algebraic complements of these elements. For clarity, consider the fourth-order determinant, written in general form:

$$\Delta=\left| \begin(array) (cccc) a_(11) & a_(12) & a_(13) & a_(14) \\ a_(21) & a_(22) & a_(23) & a_(24) \\ a_(31) & a_(32) & a_(33) & a_(34) \\ a_(41) & a_(42) & a_(43) & a_(44) \\ \end(array) \right| $$

Let's choose an arbitrary column in this determinant. Let’s take, for example, column number 4. Let’s write the formula for decomposing the determinant over the selected fourth column:

Similarly, choosing, for example, the third line, we obtain a decomposition for this line:

Example No. 1

Calculate the determinant of the matrix $A=\left(\begin(array) (ccc) 5 & -4 & 3 \\ 7 & 2 & -1 \\ 9 & 0 & 4 \end(array) \right)$ using expansion on the first row and second column.

We need to calculate the third order determinant $\Delta A=\left| \begin(array) (ccc) 5 & -4 & 3 \\ 7 & 2 & -1 \\ 9 & 0 & 4 \end(array) \right|$. To expand it along the first line you need to use the formula. Let us write this expansion in general form:

$$ \Delta A= a_(11)\cdot A_(11)+a_(12)\cdot A_(12)+a_(13)\cdot A_(13). $$

For our matrix $a_(11)=5$, $a_(12)=-4$, $a_(13)=3$. To calculate the algebraic additions $A_(11)$, $A_(12)$, $A_(13)$, we will use formula No. 1 from the topic on . So, the required algebraic complements are:

\begin(aligned) & A_(11)=(-1)^2\cdot \left| \begin(array) (cc) 2 & -1 \\ 0 & 4 \end(array) \right|=2\cdot 4-(-1)\cdot 0=8;\\ & A_(12)=( -1)^3\cdot \left| \begin(array) (cc) 7 & -1 \\ 9 & 4 \end(array) \right|=-(7\cdot 4-(-1)\cdot 9)=-37;\\ & A_( 13)=(-1)^4\cdot \left| \begin(array) (cc) 7 & 2 \\ 9 & 0 \end(array) \right|=7\cdot 0-2\cdot 9=-18. \end(aligned)

How did we find algebraic complements? show\hide

Substituting all the found values into the formula written above, we get:

$$ \Delta A= a_(11)\cdot A_(11)+a_(12)\cdot A_(12)+a_(13)\cdot A_(13)=5\cdot(8)+(-4) \cdot(-37)+3\cdot(-18)=134. $$

As you can see, we have reduced the process of finding the third-order determinant to calculating the values of three second-order determinants. In other words, we have lowered the order of the original determinant.

Usually in such simple cases they do not describe the solution in detail, separately finding algebraic additions, and only then substituting them into the formula to calculate the determinant. Most often, they simply continue writing the general formula until the answer is received. This is how we will arrange the determinant in the second column.

So, let's start expanding the determinant in the second column. We will not perform auxiliary calculations; we will simply continue the formula until we receive the answer. Please note that in the second column one element is equal to zero, i.e. $a_(32)=0$. This suggests that the term $a_(32)\cdot A_(32)=0\cdot A_(23)=0$. Using the formula for expansion in the second column, we get:

$$ \Delta A= a_(12)\cdot A_(12)+a_(22)\cdot A_(22)+a_(32)\cdot A_(32)=-4\cdot (-1)\cdot \ left| \begin(array) (cc) 7 & -1 \\ 9 & 4 \end(array) \right|+2\cdot \left| \begin(array) (cc) 5 & 3 \\ 9 & 4 \end(array) \right|=4\cdot 37+2\cdot (-7)=134. $$

The answer has been received. Naturally, the result of the expansion along the second column coincided with the result of the expansion along the first row, since we were expanding the same determinant. Notice that when we expanded in the second column, we did less calculations because one element of the second column was zero. It is on the basis of such considerations that for decomposition they try to choose the column or row that contains more zeros.

Answer: $\Delta A=134$.

Example No. 2

Calculate the determinant of the matrix $A=\left(\begin(array) (cccc) -1 & 3 & 2 & -3\\ 4 & -2 & 5 & 1\\ -5 & 0 & -4 & 0\\ 9 & 7 & 8 & -7 \end(array) \right)$ using expansion on the selected row or column.

For decomposition, it is most profitable to choose the row or column that contains the most zeros. Naturally, in this case it makes sense to expand along the third line, since it contains two elements equal to zero. Using the formula, we write the expansion of the determinant along the third line:

$$ \Delta A= a_(31)\cdot A_(31)+a_(32)\cdot A_(32)+a_(33)\cdot A_(33)+a_(34)\cdot A_(34). $$

Since $a_(31)=-5$, $a_(32)=0$, $a_(33)=-4$, $a_(34)=0$, then the formula written above will be:

$$ \Delta A= -5 \cdot A_(31)-4\cdot A_(33). $$

Let us turn to the algebraic complements $A_(31)$ and $A_(33)$. To calculate them, we will use formula No. 2 from the topic devoted to determinants of the second and third orders (in the same section there are detailed examples of the application of this formula).

\begin(aligned) & A_(31)=(-1)^4\cdot \left| \begin(array) (ccc) 3 & 2 & -3 \\ -2 & 5 & 1 \\ 7 & 8 & -7 \end(array) \right|=10;\\ & A_(33)=( -1)^6\cdot \left| \begin(array) (ccc) -1 & 3 & -3 \\ 4 & -2 & 1 \\ 9 & 7 & -7 \end(array) \right|=-34. \end(aligned)

Substituting the obtained data into the formula for the determinant, we will have:

$$ \Delta A= -5 \cdot A_(31)-4\cdot A_(33)=-5\cdot 10-4\cdot (-34)=86. $$

In principle, the entire solution can be written in one line. If you skip all the explanations and intermediate calculations, then the solution will be written as follows:

$$ \Delta A= a_(31)\cdot A_(31)+a_(32)\cdot A_(32)+a_(33)\cdot A_(33)+a_(34)\cdot A_(34)= \\= -5 \cdot (-1)^4\cdot \left| \begin(array) (ccc) 3 & 2 & -3 \\ -2 & 5 & 1 \\ 7 & 8 & -7 \end(array) \right|-4\cdot (-1)^6\cdot \left| \begin(array) (ccc) -1 & 3 & -3 \\ 4 & -2 & 1 \\ 9 & 7 & -7 \end(array) \right|=-5\cdot 10-4\cdot ( -34)=86. $$

Answer: $\Delta A=86$.

When finding determinants of the second and third order, you can use standard formulas (2 - the difference in the product of diagonal elements, 3 - the triangle rule). However, to calculate the determinant of the fourth, fifth order and higher ones, it is much faster to decompose them into the elements of the row or column containing the most zeros and reduce them to calculating several determinants per unit of lower order.

Schemes of signs for minors for determinants of the 3rd - 5th order are given below.

They are not difficult to remember if you know the following rules:
The addition to the elements of the main diagonal comes with a “+” sign, and on parallel diagonals “-”, “+”, “-”, ... alternate.
The addition to the elements of odd columns and rows begins with a “+” sign, and then “-”, “+” alternate, for paired ones they begin with a “-” sign, and then “+”, “-” alternate alternately...
The second rule is used by most students, since it is tied to the column or row by which the determinant schedule is carried out.

Let us move on to consider examples of decomposition of the determinant and study the features of this method.

Expand the third-order determinant into the elements of the first row and second column

We decompose the determinant into the elements of the first row

In a similar way, we perform calculations of the decomposition over the elements of the second column

Both values are the same, which means the calculations were carried out correctly. If it turns out that the determinants obtained by the schedule for the row and column do not match, it means that somewhere there was an error in the calculations and you need to list it or find it.

Find the fourth-order determinant using the expansion method

We decompose by elements of the third row (highlighted in red) since it contains the most zero elements.