Given that the matrix [2,-2,16|-4], [-1,0,-3|5],[1,-1,8|-2] can be reduced (using elementary row operations) to the equivalent linear system [1,0,3|-5],[0,1,-5|-3],[0,0,0|0],solve the following system of linear equations {2x−2y+16z = −4, −x−3z = 5, x−y+8z = −2. If necessary, parametrise your answer using the free variables of the system. (You must set the right-most non-leading variable to t if using a single parameter, and if a second parameter is required, then set it to s.)

se Gauss-Jordan elimination to solve the following linear system.x2 + x3 − 2x4 = −8x1 + 2x2 − x3 − x4 = 22x1 + 4x2 + x3 − 3x4 = −7x1 + 3x2 − 7x3 − x4 = 16

Consider the following system of linear equations:x+2y−3z=2𝑥+2𝑦−3𝑧=22x−3y+4z=−72𝑥−3𝑦+4𝑧=−73x+y−2z=−23𝑥+𝑦−2𝑧=−2then, by using the guassian elimination method, the solution of the system is:

Given the following, find the reduced row echelon form of the augmented matrix [A | b]. (Enter your answer as one augmented matrix.)A = 2 2 6 14 4 4 15 34 and b = 16 38RREF = Find the solution of Ax = b, where s and t are arbitrary real numbers. (Assume x2 = s and x4 = t. If a solution does not exist, enter DNE into any cell of each matrix.)

Put the following system in reduced row echelon form−3w+6x+2y+7z−2w+4x−3y−4z−w+2x−3y−5z=9=6=3

Solve the system of linear equations by matrix method2x – 3y + 5z = 11 , 3x +2y –4z = –5 , x + y –2z = – 3