Prove that the reduced row echelon forms of the matrices(1 1 4 8 0 −1 −11 2 3 9 0 −5 −20 −2 2 −2 1 14 31 4 1 11 0 −13 −4) (0 −3 3 1 50 1 −1 0 00 2 −2 0 −3) are the two matrices. The point of the reduced row echelon form is that the corresponding system of linear equations is in a particularly simple form, from which the solutions to the system AX=C in (4) can be determined immediately.

Question

Prove that the reduced row echelon forms of the matrices(1 1 4 8 0 −1 −11 2 3 9 0 −5 −20 −2 2 −2 1 14 31 4 1 11 0 −13 −4) (0 −3 3 1 50 1 −1 0 00 2 −2 0 −3)  are the two matrices.   The point of the reduced row echelon form is that the corresponding system of linear equations is in a particularly simple form, from which the solutions to the system AX=C in (4) can be determined immediately.

coffentw · Accepted Answer

Step 1  First augmented matrix is: A=[11480−1−112390−5−20−22−21143141110−13−4]  Now we use the fact that matrix A is row equivqlent to a matrix in reduced roe echelon form.  A=[11480−1−112390−5−20−22−21143141110−13−4] R2 → R2 − R1→ [11480−1−101−110−4−10−22−21143141110−13−4]     R4 → R4 − R1→ [11480−1−101−110−4−10−22−2114303−330−12−3]     R1 → R1 − R2→ [105703001−110−4−10−22−2114303−330−12−3]  Step 2   Now we will choose lower right submatrix and apply row operations. It is matrix:  A=[1−110−4−1−22−211433−330−12−3]  Then:  A=[1−110−4−1−22−211433−330−12−3] R2 → R1 + R2→ [1−110−4−100−011613−330−12−3]   R3 → R3 − 3R1→ [1−110−4−100

Prove that the reduced row echelon forms of the matricesleft

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