Rank of a matrix. Theorem of Kronecker-Capelli

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Then the system (*) is written in a matrix representation: A(n;

Then the system (*) is written in a matrix representation: A(n;

n)⋅ X(n; 1) = B(n; 1)
X = A-1⋅ B.
A matrix A(n; n) is called regular if its determinant is not equal to zero, i.e.
A matrix A-1(n; n) is called inverse to a matrix A(n; n) if the product
A(n; n) ⋅A-1(n; n)= A-1(n; n)⋅A(n; n) = E(n; n),
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Example . Find the inverse matrix to the matrix А11 =


Example . Find the inverse matrix to the matrix
А11 = 5;

A12 = 10; A13 = 0; A21 = 4; A22 = 12; A23 = 1; A31 = -1; A32 = -3; A33 = 1.
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Rank of a matrix. Consider a matrix of the dimension :


Rank of a matrix.
Consider a matrix of the dimension :
The

rank of a matrix A is the greatest order of its non-equal to zero minors. The rank of a matrix is denoted by Rank A or r(A).
Theorem. If there is a non-equal to zero minor of the r-th order in a matrix A and all its bordering minors of the r+1-th order are equal to zero then the rank of A is equal to r, i.e. r(A)=r.
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Theorem. The rank of a matrix doesn’t change if: a) All

Theorem. The rank of a matrix doesn’t change if:
a) All

the rows are replaced by the corresponding columns and vice versa;
b) Replace two arbitrary rows (columns);
c) Multiply (divide) each element of a row (column) on the same non-zero number;
d) Add to (subtract from) elements of a row (column) the corresponding elements of any other row (column) multiplied on the same non-zero number.
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Theorem of Kronecker-Capelli. A system of linear equations is consistent if

Theorem of Kronecker-Capelli. A system of linear equations is consistent if

the rank of the basic matrix A equals the rank of the extended matrix C, i.e. Rank A = Rank C. Moreover:
1) If Rank A = Rank C = n (where n is the number of variables in the system) then the system has a unique solution.
2) If Rank A = Rank C < n then the system has infinitely many solutions.
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Solving a system of linear equations by the Gauss method Suppose

Solving a system of linear equations
by the Gauss method
Suppose that

а11 ≠ 0 (if а11 = 0 then we change the order of equations by choosing as the first equation such an equation in which the coefficient of х is not equal to zero).
I step: divide the equation (а) on а11, then multiply the obtained equation on а21 and subtract from (b); further multiply (а)/a11 on а31 and subtract from (с); at last, miltiply (а)/a11 on а41 and subtract from (d).
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where bij are obtained from aij by the following formulas: b1j


where bij are obtained from aij by the following formulas:
b1j =

a1j / a11 (j = 2, 3, 4, 5);
bij = aij – ai1 ⋅ b1j (i = 2, 3, 4; j = 2, 3, 4, 5).
II step: do the same actions with (f), (g), (i) (as with (a), (b), (c), (d)) and etc.
As a final result the initial system will be transformed to a so-called step form:
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Example 1. Interchange the first and the second equations of the

Example 1.
Interchange the first and the second equations of the system:
Subtract

from the second equation the first equation multiplied on 3; also subtract from the third equation the first equation multiplied on 4. We obtain:
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Further subtract from the third equation the second equation multiplied on

Further subtract from the third equation the second equation multiplied on

5:
Multiply the second equation on (-2), and the third – divide on
(-11):
The system of equations has a triangular form, and consequently it has a unique decision. From the last equation we have z = 2; substituting this value in the second equation, we receive у = 3 and, at last from the first equation we find х = -1.