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Find the reduced row echelon form of the then: M = \begin{equation} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} \end{equation} and describe in detail the elementary row operations that you use

M = \begin{equation} \begin{pmatrix} * & * \\ * & * \end{pmatrix} \times \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} ? & ? \\ ? & ? \end{pmatrix} \end{equation}

\begin{equation} \begin{pmatrix} y & y \\ y & y \end{pmatrix} \times \begin{pmatrix} ? & ? \\ ? & ? \end{pmatrix} = \begin{pmatrix} x & x \\ x & x \end{pmatrix} \end{equation}

I've learned that you have to have a star matrix times original matrix equals ? matrix

and then ? matrix moves down to where original matrix was at. but I don't know how to determine neither * and ? matrix from the beginning. If I find out how to calculate ? and * matrices and I know I can find rest of matrices easily!! Can I get some help?

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Let $$ A=\begin{bmatrix}1&-1\\1&1\end{bmatrix} $$ Subtracting $\DeclareMathOperator{Row}{Row}\Row_1$ from $\Row_2$ gives $$ E_1A=\begin{bmatrix}1&-1\\0&2\end{bmatrix} $$ where $E_1$ is the elementary matrix $$ E_1=\begin{bmatrix}1&0\\-1&1\end{bmatrix} $$ Multiplying $\Row_2$ by $1/2$ gives $$ E_2E_1A=\begin{bmatrix}1&-1\\0&1\end{bmatrix} $$ where $E_2$ is the elementary matrix $$ E_2=\begin{bmatrix}1&0\\0&1/2\end{bmatrix} $$ Finally, adding $\Row_2$ to $\Row_1$ gives $$ E_3E_2E_1A=\begin{bmatrix}1&0\\0&1\end{bmatrix} $$ where $E_3$ is the elementary matrix $$ E_3=\begin{bmatrix}1&1\\0&1\end{bmatrix} $$ This shows that the reduced row-echelon form of $A$ is $$ \DeclareMathOperator{rref}{rref}\rref A= \begin{bmatrix}1&0\\0&1\end{bmatrix} $$

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