Prove that the reduced row echelon forms of the matricesleft

Question

Prove that the reduced row echelon forms of the matrices $(\begin{matrix} 1 & 1 & 4 & 8 & 0 & - 1 & - 1 \\ 1 & 2 & 3 & 9 & 0 & - 5 & - 2 \\ 0 & - 2 & 2 & - 2 & 1 & 14 & 3 \\ 1 & 4 & 1 & 11 & 0 & - 13 & - 4 \end{matrix}) (\begin{matrix} 0 & - 3 & 3 & 1 & 5 \\ 0 & 1 & - 1 & 0 & 0 \\ 0 & 2 & - 2 & 0 & - 3 \end{matrix})$

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 $A X = C i n (4)$ can be determined immediately.

Answer 1

Step 1

First augmented matrix is: $A = [\begin{matrix} 1 & 1 & 4 & 8 & 0 & - 1 & - 1 \\ 1 & 2 & 3 & 9 & 0 & - 5 & - 2 \\ 0 & - 2 & 2 & - 2 & 1 & 14 & 3 \\ 1 & 4 & 1 & 11 & 0 & - 13 & - 4 \end{matrix}]$

Now we use the fact that matrix A is row equivqlent to a matrix in reduced roe echelon form.

$A = [\begin{matrix} 1 & 1 & 4 & 8 & 0 & - 1 & - 1 \\ 1 & 2 & 3 & 9 & 0 & - 5 & - 2 \\ 0 & - 2 & 2 & - 2 & 1 & 14 & 3 \\ 1 & 4 & 1 & 11 & 0 & - 13 & - 4 \end{matrix}] \vec{R_{2} \to R_{2} - R_{1}} [\begin{matrix} 1 & 1 & 4 & 8 & 0 & - 1 & - 1 \\ 0 & 1 & - 1 & 1 & 0 & - 4 & - 1 \\ 0 & - 2 & 2 & - 2 & 1 & 14 & 3 \\ 1 & 4 & 1 & 11 & 0 & - 13 & - 4 \end{matrix}]$

$\vec{R_{4} \to R_{4} - R_{1}} [\begin{matrix} 1 & 1 & 4 & 8 & 0 & - 1 & - 1 \\ 0 & 1 & - 1 & 1 & 0 & - 4 & - 1 \\ 0 & - 2 & 2 & - 2 & 1 & 14 & 3 \\ 0 & 3 & - 3 & 3 & 0 & - 12 & - 3 \end{matrix}]$

$\vec{R_{1} \to R_{1} - R_{2}} [\begin{matrix} 1 & 0 & 5 & 7 & 0 & 3 & 0 \\ 0 & 1 & - 1 & 1 & 0 & - 4 & - 1 \\ 0 & - 2 & 2 & - 2 & 1 & 14 & 3 \\ 0 & 3 & - 3 & 3 & 0 & - 12 & - 3 \end{matrix}]$

Step 2

Now we will choose lower right submatrix and apply row operations. It is matrix:

$A = [\begin{matrix} 1 & - 1 & 1 & 0 & - 4 & - 1 \\ - 2 & 2 & - 2 & 1 & 14 & 3 \\ 3 & - 3 & 3 & 0 & - 12 & - 3 \end{matrix}]$

Then:

$A = [\begin{matrix} 1 & - 1 & 1 & 0 & - 4 & - 1 \\ - 2 & 2 & - 2 & 1 & 14 & 3 \\ 3 & - 3 & 3 & 0 & - 12 & - 3 \end{matrix}] \vec{R_{2} \to R_{1} + R_{2}} [\begin{matrix} 1 & - 1 & 1 & 0 & - 4 & - 1 \\ 0 & 0 & - 0 & 1 & 16 & 1 \\ 3 & - 3 & 3 & 0 & - 12 & - 3 \end{matrix}]$

$\vec{R_{3} \to R_{3} - 3 R_{1}} [\begin{matrix} 1 & - 1 & 1 & 0 & - 4 & - 1 \\ 0 & 0 \end{matrix}$

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Prove that the reduced row echelon forms of the matricesleft

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