 # Given the matrix A=begin{bmatrix}0 & 0&1 0 & 3&0 1&0 & -10 end{bmatrix} and suppose that we have the following row reduction to its PREF B A=begin{bma rocedwrp 2021-01-22 Answered
Given the matrix
$A=\left[\begin{array}{ccc}0& 0& 1\\ 0& 3& 0\\ 1& 0& -10\end{array}\right]$
and suppose that we have the following row reduction to its PREF B
$A=\left[\begin{array}{ccc}0& 0& 1\\ 0& 3& 0\\ 1& 0& -10\end{array}\right]⇒\left[\begin{array}{ccc}1& 0& -10\\ 0& 3& 0\\ 0& 0& 1\end{array}\right]⇒\left[\begin{array}{ccc}1& 0& -10\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]⇒\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
Write as product of elementary matrices.
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Given the matrix
$A=\left[\begin{array}{ccc}0& 0& 1\\ 0& 3& 0\\ 1& 0& -10\end{array}\right]$
Write as product of elementary matrices. Step 2 Take the matrix A and exchange the first and second rows. Let ${E}_{1}$ be the elementary row matrix corresponding to the above row operation.
${E}_{1}=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]$
Notice that ${E}_{1}A=\left[\begin{array}{ccc}1& 0& -10\\ 0& 3& 0\\ 0& 0& 1\end{array}\right]$
Step 3 Next, take the matrix $\left[\begin{array}{ccc}1& 0& -10\\ 0& 3& 0\\ 0& 0& 1\end{array}\right]$ and divide the second row by the scalar 3. Let ${E}_{2}$ be the elementary row matrix corresponding to the above row operation.
${E}_{2}=\left[\begin{array}{ccc}1& 0& 0\\ 0& \frac{1}{3}& 0\\ 0& 0& 1\end{array}\right]$
Notice that ${E}_{2}\left[\begin{array}{ccc}1& 0& -10\\ 0& 3& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& -10\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
Step 4
Next, take the matrix $\left[\begin{array}{ccc}1& 0& -10\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ and add 10 times the third row to the first row. Let ${E}_{3}$ be the elementary row matrix corresponding to the above row operation.
${E}_{3}=\left[\begin{array}{ccc}1& 0& 10\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
Notice that ${E}_{3}\left[\begin{array}{ccc}1& 0& -10\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$
Step 5
Now,
$I={E}_{3}\left[\begin{array}{ccc}1& 0& -10\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]={E}_{3}{E}_{2}\left[\begin{array}{ccc}1& 0& -10\\ 0& 3& 0\\ 0& 0& 1\end{array}\right]={E}_{3}{E}_{2}{E}_{1}A$
and so ${A}^{-1}={E}_{3}{E}_{2}{E}_{1}$ , that is
${A}^{-1}=\left[\begin{array}{ccc}1& 0& 10\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& -10\\ 0& \frac{1}{3}& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]$
Step 6
Also,
$A={E}_{1}^{-1}{E}_{2}^{-1}{E}_{3}^{-1}$

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