Example №5. Solving of a System of Linear Equations by the Gauss Jordan Elimination (Many Solutions)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 №3. Solving of a system of linear equations by the Gauss elimination (no solution) Example №4. Solving of a system of linear equations by the Gauss Jordan elimination (one solution) Please note that the coefficients will disappear which located in the "red" positions.
The equation 1 multiplied by 4 is added to the equation 2. more info ( 8 x_{1} + 2 x_{1} * ( 4) ) + ( 12 x_{2} + 3 x_{2} * ( 4) ) + ( 9 x_{3} + (  x_{3}) * ( 4) ) + ( 8 x_{4} + x_{4} * ( 4) ) = 3 + 1 * ( 4) The "red" coefficient is zero.
The equation 1 multiplied by 2 is added to the equation 3. more info ( 4 x_{1} + 2 x_{1} * ( 2) ) + ( 6 x_{2} + 3 x_{2} * ( 2) ) + ( 3 x_{3} + (  x_{3}) * ( 2) ) + ( 2 x_{4} + x_{4} * ( 2) ) = 3 + 1 * ( 2) The "red" coefficient is zero.
The equation 1 multiplied by 1 is added to the equation 4. more info ( 2 x_{1} + 2 x_{1} * ( 1) ) + ( 3 x_{2} + 3 x_{2} * ( 1) ) + ( 9 x_{3} + (  x_{3}) * ( 1) ) + ( 7 x_{4} + x_{4} * ( 1) ) = 3 + 1 * ( 1) The "red" coefficient is zero.
The equation 2 is added to the equation 3. more info ( 5 x_{3} + ( 5 x_{3}) ) + ( 4 x_{4} + 4 x_{4} ) = 1 + ( 1) The "red" coefficient is zero.
The equation 2 multiplied by 2 is added to the equation 4. more info ( 10 x_{3} + ( 5 x_{3}) * 2 ) + ( 8 x_{4} + 4 x_{4} * 2 ) = 2 + ( 1) * 2 The "red" coefficient is zero.
The equation 2 is divided by 5.
The equation 2 is added to the equation 1. more info 2 x_{1} + 3 x_{2} + (  x_{3} + x_{3} ) + ( x_{4} + ( 4/5 x_{4}) ) = 1 + 1/5 The "red" coefficient is zero.
The equation 1 is divided by 2.
Result: x_{1} = 3/5  3/2 x_{2}  1/10 x_{4} x_{3} = 1/5 + 4/5 x_{4}
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