Example №3. Solving of a System of Linear Equations by the Gauss elimination (No Solution)This solution was made using the calculator presented on the site. Example №1. Solving of a system of linear equations by the Gauss elimination (one solution) Example №2. Solving of a system of linear equations by the Gauss elimination (many solutions) Example №4. Solving of a system of linear equations by the Gauss Jordan elimination (one solution) Example №5. Solving of a system of linear equations by the Gauss Jordan elimination (many solutions) Please note that the coefficients will disappear which located in the "red" positions.
The equation 1 multiplied by -3 is added to the equation 2. more info ( 3 x1 + x1 * ( -3) ) + ( 3 x2 + ( -2 x2) * ( -3) ) + ( -5 x3 + 3 x3 * ( -3) ) + ( x4 + ( -4 x4) * ( -3) ) = -3 + 2 * ( -3) The "red" coefficient is zero.
The equation 1 multiplied by 2 is added to the equation 3. more info ( -2 x1 + x1 * 2 ) + ( x2 + ( -2 x2) * 2 ) + ( 2 x3 + 3 x3 * 2 ) + ( -3 x4 + ( -4 x4) * 2 ) = 5 + 2 * 2 The "red" coefficient is zero.
The equation 1 multiplied by -3 is added to the equation 4. more info ( 3 x1 + x1 * ( -3) ) - 2 x2 * ( -3) + ( 3 x3 + 3 x3 * ( -3) ) + ( -10 x4 + ( -4 x4) * ( -3) ) = 8 + 2 * ( -3) The "red" coefficient is zero.
The equation 3 and equation 2 are reversed.
The equation 2 multiplied by 3 is added to the equation 3. more info ( 9 x2 + ( -3 x2) * 3 ) + ( -14 x3 + 8 x3 * 3 ) + ( 13 x4 + ( -11 x4) * 3 ) = -9 + 9 * 3 The "red" coefficient is zero.
The equation 2 multiplied by 2 is added to the equation 4. more info ( 6 x2 + ( -3 x2) * 2 ) + ( -6 x3 + 8 x3 * 2 ) + ( 2 x4 + ( -11 x4) * 2 ) = 2 + 9 * 2 The "red" coefficient is zero.
The equation 3 multiplied by -1 is added to the equation 4. more info ( 10 x3 + 10 x3 * ( -1) ) + ( -20 x4 + ( -20 x4) * ( -1) ) = 20 + 18 * ( -1) The "red" coefficient is zero.
Equation 4 is false. The system has no solution.
© 2010-2024 If you have any comments, please write to matematika1974@yandex.ru |