Suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given row echelon form. Solve the system by back substitution. Assume that the variables are named x1,x2,from left to right. [1053201−24−700113]

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oppturf · Accepted Answer

The linear system corresponding to the augmented matrix is x1+5x3+3x4=2  x2−2x3+4x4=−7 x3+x4=3 Solve for the leading variables. x3+x4=3 x2=−7+2x3−4x4 x3=3−x4 Substitute x3=3−x4 into the first two equations. x1=−13+2x4 x2=−1−6x4 x3=3−x4 Assign arbitrary value to free variable x4, say x4=t. Then the solution is described by the parametric equations. x1=−13+2t, x2=−1−6t, x3=3−t, x4=t

RizerMix · Accepted Answer

Given the row echelon form:[1053201−24−700113]We can see that the third row corresponds to the equation:0x1+0x2+1x3+1x4=3This simplifies to:x3+x4=3Now, moving to the second row, we have:0x1+1x2−2x3+4x4=−7Substituting the value of x3 from the third row into this equation, we get:x2−2(1x4)+4x4=−7x2+2x4=−7Finally, considering the first row, we have:1x1+0x2+5x3+3x4=2Substituting the values of x2 and x3 from the second and third rows, respectively, we get:x1+5(1x4)+3x4=2x1+8x4=2Therefore, the solution to the system of linear equations is:x1=2−8x4x2=−7−2x4x3=3−x4 where x4 is a free variable.

nick1337 · Accepted Answer

To solve the system of linear equations using back substitution, we start from the last equation and work our way up. Let&#039;s go step by step.The given row echelon form is:[1053201−24−700113]From the third equation, we have:x3+x4=3Let&#039;s solve for x3 in terms of x4:x3=3−x4Now, substituting this value of x3 in the second equation, we get:x2−2x3+4x4=−7Replacing x3 with 3−x4, we have:x2−2(3−x4)+4x4=−7Simplifying this equation gives us:x2−6+2x4+4x4=−7Combining like terms:x2+6x4=−1From the first equation, we have:x1+5x3+3x4=2Substituting the value of x3 as 3−x4:x1+5(3−x4)+3x4=2Expanding and simplifying:x1+15−5x4+3x4=2Combining like terms:x1−2x4=−13Finally, we have the equations:x1−2x4=−13x2+6x4=−1x3=3−x4These are the solutions to the system of linear equations in terms of x1, x2, and x4.

Don Sumner · Accepted Answer

Result:x1=−7−2x3, x2=−19+6x3, x3, x4=3−x3Solution:From the given row echelon form:x3+x4=3x2−2x3+4x4=−7x1+5x3+3x4=2Solving the first equation for x4:x4=3−x3Substituting this value of x4 into the second equation:x2−2x3+4(3−x3)=−7Simplifying:x2−2x3+12−4x3=−7x2−6x3=−19Solving the equation for x2:x2=−19+6x3Substituting the values of x2 and x4 into the third equation:x1+5x3+3(3−x3)=2Simplifying:x1+5x3+9−3x3=2x1+2x3=−7Solving the equation for x1:x1=−7−2x3Thus, the solution to the system of linear equations is:x1=−7−2x3x2=−19+6x3x3 is a free variable (it can take any real value)x4=3−x3

Suppose that the augmented matrix for a system of linear equations has been reduced by row operations to the given row echelon form. Solve the system

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