Kramer's method represents the algorithm allowing to solve the system of the linear equations by means of a matrix. The author of a method is Gabriel Kramer living in the first half of the 18th century.
Instruction
1. Let some system of the linear equations be set. It needs to be written down in a matrix look. Coefficients before variables will go to the main matrix. For record of additional matrixes also the free members who are settling down usually to the right of an equal-sign will be necessary.
2. Each of variables has to have "serial number". For example, in all equations of a system x1, on the second – x2, on the third – x3, etc. is on the first place. Then to each of these variables there will correspond the column in a matrix.
3. It is necessary for application of a method of Kramer that the turned-out matrix was square. To this condition there corresponds equality numbersof unknown and quantity of the equations in a system.
4. Find a determinant of the main matrix Δ. It has to be nonzero: only in this case the solution of a system will be only and unambiguously certain.
5. To write down an additional determinant Δ(i), replace i-y a column with a column of free members. The number of additional determinants will equal to number of variables in a system. Calculate all determinants.
6. From the received determinants it was necessary only to find value of unknown. In a general view, the formula for finding of variables looks so: x (i) = Δ(i)/Δ.
7. Example. The system consisting of three linear equations, containing three unknown x1, x2 and x3 has an appearance: a11 • x1 + a12 • x2 + a13 • x3 = b1, a21 • x1 + a22 • x2 + a23 • x3 = b2, a31 • x1 + a32 • x2 + a33 • x3 = b3.
8. From coefficients before unknown write down the main determinant: a11 a12 a13a21 a22 a23a31 a32 a33
9. Calculate it: Δ = a11 • a22 • a33 + a31 • a12 • a23 + a13 • a21 • a32 – a13 • a22 • a31 – a11 • a32 • a23 – a33 • a12 • a21.
10. Having replaced the first column with free members, make the first additional determinant: b1 a12 a13b2 a22 a23b3 a32 a33
11. Carry out the similar procedure with the second and third columns: a11 b1 a13a21 b2 a23a31 b3 a33a11 a12 b1a21 a22 b2a31 a32 b3
12. Calculate additional determinants: Δ (1) = b1 • a22 • a33 + b3 • a12 • a23 + a13 • b2 • a32 – a13 • a22 • b3 – b1 • a32 • a23 – a33 • a12 • b2.Δ(2) = a11 • b2 • a33 + a31 • b1 • a23 + a13 • a21 • b3 – a13 • b2 • a31 – a11 • b3 • a23 – a33 • b1 • a21.Δ(3) = a11 • a22 • b3 + a31 • a12 • b2 + b1 • a21 • a32 – b1 • a22 • a31 – a11 • a32 • b2 – b3 • a12 • a21.
13. Find unknown, write down the answer: x1 = Δ (1)/Δ, x2 = Δ (2)/Δ, x3 = Δ (3)/Δ.