Given that$A=\begin{bmatrix}1&1&2\\2&2&4\\2&3&4 \end{bmatrix}$ we need to find a invertible matrix $P$ so that (also noted that $A$ is noninvertible)

$$PA = \begin{bmatrix}1&0&2\\0&1&0\\0&0&0\end{bmatrix} = M$$

Now I thought we assume $P = \begin{bmatrix}x&y&z\\a&b&c\\d&e&f\end{bmatrix}$ and form 9 equations, but is there any other way out?

I observe that If we wpply operations $(R_1\to R_1 + R_2)$,$(R_3 \to R_3 +2R_1+R_2)$, $(R_2 \to R_2 +4R_1 -R_3)$ on $M$ gives $A$ does this help me in any way?


The second route works. Reverse all your operations, i.e. operate $\boldsymbol A$ to give $\boldsymbol M$, and note that an elementary row operation is actually a left multiplication of elementary matrices, you would get one $\boldsymbol P$.

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  • $\begingroup$ Thank you xbh :D $\endgroup$ – jeea Sep 11 '18 at 14:58
  • $\begingroup$ You are welcome. Glad to help. :) $\endgroup$ – xbh Sep 11 '18 at 14:59

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