Gauss-Jordan row reduction?
Standard form: x+y+6z=4; 1/3x-1/3y+2/3z=2; 1/2x+z=0; substitute/eliminate x = -1y-6z+4 substitute/eliminate y = -2z-1 substitute/eliminate z = +5/2 Solution: x = -5; y = -6; z = 5/2; (x, y, z) = (-5, -6, 2.5)
Use Gauss-Jordan row reduction to solve the given system of equations.?
Standard form: -7w+x-6y+z=1; 5w+x+7y+4z=2; -2w+2x+y+5z=3; =0; -7, 1, -6, 1, 1 5, 1, 7, 4, 2 -2, 2, 1, 5, 3 0, 0, 0, 0, 0 ----------------------- substitute/eliminate w = +1/7x -6/7y +1/7z-1/7 -7, 1, -6, 1, 1 0, 12/7, 19/7, 33/7, 19/7 0, 12/7, 19/7, 33/7, 19/7 0, 0, 0, 0, 0 ----------------------- substitute/eliminate x = -19/12y -11/4z+19/12 -7, 0, -91/12, -7/4, -7/12 0, 12/7, 19/7, 33/7, 19/7 0, 0, 0, 0, 0 0, 0, 0, 0, 0 ----------------------- -7, 0, -91/12, -7/4, -7/12 0, 12/7, 19/7, 33/7, 19/7 0, 0, 0, 0, 0 0, 0, 0, 0, 0 ----------------------- dependent system - infinitely many solutions. y = -0.92307692w -0.23076923z +0.07692308 x = 1.46153846w -2.38461539z +1.46153846
Use Gauss-Jordan row reduction to solve the given system of equations. (If there is no solution, enter NO SOLUTION. If the system is?
Standard form: -6w+x-7y+z=1; 2w+x+8y+6z=2; -4w+2x+y+7z=3; -6, 1, -7, 1, 1 2, 1, 8, 6, 2 -4, 2, 1, 7, 3 0, 0, 0, 0, 0 ----------------------- substitute/eliminate w = +1/6x -7/6y +1/6z-1/6 -6, 1, -7, 1, 1 0, 4/3, 17/3, 19/3, 7/3 0, 4/3, 17/3, 19/3, 7/3 0, 0, 0, 0, 0 ----------------------- substitute/eliminate x = -17/4y -19/4z+7/4 -6, 0, -45/4, -15/4, -3/4 0, 4/3, 17/3, 19/3, 7/3 0, 0, 0, 0, 0 0, 0, 0, 0, 0 ----------------------- -6, 0, -45/4, -15/4, -3/4 0, 4/3, 17/3, 19/3, 7/3 0, 0, 0, 0, 0 0, 0, 0, 0, 0 ----------------------- dependent system - infinitely many solutions. solutions: w=-1.875y-0.625z+0.125; x=-4.25y-4.75z+1.75; solving for y: w=-1.875y-0.625z+0.125; w=-15/8y-5/8z+1/8 15/8y = -w -5/8z +1/8 y = -8/15w -1/3z +1/15 solving for x: x=-4.25y-4.75z+1.75; x=-4.25(-8/15w -1/3z +1/15)-4.75z+1.75; x=34/15w+17/12z-17/60-19/4z+7/4 x=34/15w-10/3z+22/15
Use Gauss-Jordan Row reduction to solve system of equations?
Use Gauss-Jordan row reduction to solve the given system of equations. (If there is no solution, enter NO SOLUTION. If the system is dependent, express your answer in terms of x, where y = y(x) and z = z(x).) x-y+8z=9 x-y+9z=3 I'm not sure if im doing it correctly but I think that z=6? But i'm not sure if that's correct and if so i don't know how to put that in terms of x. If anyone can explain this it would be awesome!
Help with solving matrices and Gauss-Jordan row reduction?
Help please!! 1) Use Gauss-Jordan row reduction to solve the given system of equations. HINT [See Examples 1-6.] (If there is no solution, enter NO SOLUTION. If the system is dependent, express your answer in terms of x, where y = y(x) and z = z(x).) x + y − 3z = 2 x − y − 5z = 0 2/5x − 8/5z = 6 (x, y, z) = ? 2) Use technology to solve the system of equations. Express all solutions as fractions. 4x − 2y + z + w = 20 3y + 3z − 4w = −2 2x + 4y − w = −3 x + 3y + 3z = 1 (x, y, z, w) = ?
How to spot a "no solution" when doing Gauss Jordan row reduction?
The best and the most systematic way is to try to solve it by doing row reduction until you either find the solution or you can conclude that there isn't a solution. There are same cases, though, where you can show that no solution exists. The easiest way to spot it is if one equation is a multiple of another; other cases are not so easy to spot. For your first system, you can solve the third equation for z to get: z = -x - 1. Substituting this into the second equation yields: y - (-x - 1) = 6 ==> x + y = 5. Now, -x + y = -1 and x + y = 5 do intersect as the slopes of the lines are different. So, there is a solution to this system. The second system, however, you can subtract the second equation from the third to get: (x + z) - (y + z) = 1 - 4 ==> x - y = -3. The LHS is actually half of the LHS of the first equation, but the RHS do not follow this pattern. Thus, no solution exists for this one. I hope this helps!
Help with Gauss-Jordan method of elimination?
Solve the system of linear equations, using the Gauss-Jordan method of elimination. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions involving one parameter, enter the solution using t for the last variable. If the solution involves two parameters, add s for the second to last variable.) x1 + 2x2 + 16x3 = 6 x1 + x2 + 8x3 = 3 (x1, x2, x3) = I don't understand Gauss-Jordan..but I really don't get it with these subscripts and all x's. Please help me. Can this be done all the calculator as well?
How do you tell if a matrix equation has no solution?
Say we have an equation of the form [math]Ax = a[/math].If the matrix is small (say 5 or 6 in maximal dimension) I would approach it by reducing it to row-echelon form using elementary row operations, applying those operation to [math]a[/math] as well.A matrix has row-echelon form if it satisfies both:All non-zero rows (rows with at least one non-zero element) are above all zero rows.The leading coefficient (first non-zero element in a row) of a row is strictly to the right of the leading coefficient of the row above.For examples, refer to Row echelon form at Wikipedia.An elementary operation is one of three:Row switching – swapping two rows,Row multiplication – multiplying a row by a non-zero scalar,Row addition – adding a non-zero multiple of one row to another.Note that all those operation are linear and can be represented by multiplying the transformed matrix by elementary matrices (Elementary matrix at Wikipedia).The process of reducing a matrix to it's row-echelon form using elementary operations is well known as Gaussian elimination.So we end up with a row-echelon matrix [math]B = EA[/math], where [math]E[/math] is the cummulative transformation, i.e. the product of all elementary matrices used to transform [math]A[/math] into [math]B[/math]. We also get [math]b = Ea[/math] since we also applied those transformations to [math]a[/math].Note that it is of course not necessary to keep track of the operations and their respective matrices. I use these matrices solely for explanation purposes.Consider this equation:[math]Bx = b[/math].It has exactly as many solutions as the first one, since [math]E[/math] has full rank.If the number of the last non-zero element in [math]b[/math] is strictly greater than the rank of [math]B[/math], i.e. the number of non-zero rows in [math]B[/math], which is btw. also the rank of [math]A[/math], then there are obviously no solutions.If it's not the case, there are certainly some solutions. Because you could simply "cut off" all dimensions of the equation beyound the rank of [math]B[/math] and note that the columns of the resulting equation matrix span [math]\mathbb{R}^k[/math], where [math]k[/math] is the rank.For greater matrices, this check can be performed by computers as efficient as Gaussian elimination.