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 x1 + 2 x1 * ( -4) ) + ( 12 x2 + 3 x2 * ( -4) ) + ( -9 x3 + ( - x3) * ( -4) ) + ( 8 x4 + x4 * ( -4) ) = 3 + 1 * ( -4) The "red" coefficient is zero.
The equation 1 multiplied by -2 is added to the equation 3. more info ( 4 x1 + 2 x1 * ( -2) ) + ( 6 x2 + 3 x2 * ( -2) ) + ( 3 x3 + ( - x3) * ( -2) ) + ( -2 x4 + x4 * ( -2) ) = 3 + 1 * ( -2) The "red" coefficient is zero.
The equation 1 multiplied by -1 is added to the equation 4. more info ( 2 x1 + 2 x1 * ( -1) ) + ( 3 x2 + 3 x2 * ( -1) ) + ( 9 x3 + ( - x3) * ( -1) ) + ( -7 x4 + x4 * ( -1) ) = 3 + 1 * ( -1) The "red" coefficient is zero.
The equation 2 is added to the equation 3. more info ( 5 x3 + ( -5 x3) ) + ( -4 x4 + 4 x4 ) = 1 + ( -1) The "red" coefficient is zero.
The equation 2 multiplied by 2 is added to the equation 4. more info ( 10 x3 + ( -5 x3) * 2 ) + ( -8 x4 + 4 x4 * 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 x1 + 3 x2 + ( - x3 + x3 ) + ( x4 + ( -4/5 x4) ) = 1 + 1/5 The "red" coefficient is zero.
The equation 1 is divided by 2.
Result: x1 = 3/5 - 3/2 x2 - 1/10 x4 x3 = 1/5 + 4/5 x4
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