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Let $A$ =

$$ \begin{align} \begin{bmatrix} -4 & 3\\ 1 & 0 \end{bmatrix} \end{align} $$

Find $2 \times 2$ elementary matrices $E_1$,$E_2$,$E_3$ such that $A$ = $E_1 E_2 E_3$

I figured out the operations which need to be performed which are;

$E_1$ = $R_2 \leftrightarrow R_1$

$E_2$ = $R_2$ = $R_2$ + $4R_1$

$E_3$ = $R_2$ * $\frac{1}{3}$

My question is how would I go about writing the elementary matrices? The solution says that they are;

$E_1$ = $ \begin{align} \begin{bmatrix} 1 & -4\\ 0 & 1 \end{bmatrix} \end{align} $ $E_2$ = $ \begin{align} \begin{bmatrix} 3 & 0\\ 0 & 1 \end{bmatrix} \end{align} $ $E_3$ = $ \begin{align} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \end{align} $

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Hint: what do elementary matrices correspond to? Can you some how form a correspondence between the row operations you used to reduce the matrix and elementary matrices? In other words, the elementary matrices are related to how $R_{1}$ and $R_{2}$ are manipulated in each row reduction step.

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Note that the solutions are not unique. With your elementary row operations, we have $$ \pmatrix{1&0\\ 0&\tfrac13}\pmatrix{1&0\\ 4&1}\pmatrix{0&1\\ 1&0}A = I_2. $$ Therefore, by performing the reverse row operations (and also in reverse order) on $I_2$, we get $$ A = \pmatrix{0&1\\ 1&0}\pmatrix{1&0\\ -4&1}\pmatrix{1&0\\ 0&3}I_2. $$

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