Is there any way to obtain a diagonal matrix $D$ from a square matrix $A$ applying only row operations $R_{ij}(k)$ with $i\neq j$, $k\in\mathbb R$, which means add to the $i$-th row the $j$-th row multiplied by $k$.
For example, if
$$A = \begin{pmatrix}a&b\\c&d\end{pmatrix},\quad P = \begin{pmatrix}-\frac{\lambda d}{b} & \lambda \\ -\frac{\mu c}{a} & \mu\end{pmatrix} \quad (\lambda,\mu\in\mathbb R) $$
$P A = \begin{pmatrix} * & 0 \\ 0 & * \end{pmatrix}$, but how can $P$ be expressed as product of elementary matrices of that type?
Thanks in advance.
[Math] Applying row operations to a matrix in order to get a diagonal matrix
linear algebramatrices
Related Solutions
This another approach, however; Don's way explains it to you in a brief solid form.
$A =\begin{pmatrix} 1& 2& 1& 0\\ -1 &0 &3 &5\\ 1& -2& 1& 1\end{pmatrix}\xrightarrow{R_1+R_2\mapsto R_2}\begin{pmatrix} 1& 2& 1& 0\\ 0 &2 &4 &5\\ 1& -2& 1& 1\end{pmatrix} \xrightarrow{-R_1+R_3\mapsto R_3}\begin{pmatrix} 1& 2& 1& 0\\ 0 &2 &4 &5\\ 0& -4& 0& 1\end{pmatrix}\\\xrightarrow{-R_2+R_1\mapsto R_1}\begin{pmatrix} 1& 0& -3& -5\\ 0 &2 &4 &5\\ 0& -4& 0& 1\end{pmatrix}\xrightarrow{2R_2+R_3\mapsto R_3}\begin{pmatrix} 1& 0& -3& -5\\ 0 &2 &4 &5\\ 0& 0& 8& 11\end{pmatrix}\xrightarrow{\frac{-1}{2}R_3+R_2\mapsto R_2}\begin{pmatrix} 1& 0& -3& -5\\ 0 &2 &0 &-0.5\\ 0& 0& 8& 11\end{pmatrix}\xrightarrow{\frac{3}{8}R_3+R_1\mapsto R_1}\begin{pmatrix} 1& 0& 0& \frac{-13}{8}\\ 0 &2 &0 &-0.5\\ 0& 0& 8& 11\end{pmatrix}$
and the last matrix is the row-reduced echelon form of $A$.
We are changing the system of equations, what we are not changing is the set of solutions to the system of equations.
The following figure shows the equations \begin{align*} -x+2y &= 2 \\ x-y &= 0 \end{align*} and what happens after we add the first equation to the second, i.e., the system of equations \begin{align*} -x+2y &= 2 \\ y &= 2. \end{align*}
We can see that the system of equations has indeed changed, but the set of solutions (in this case $\{(2,2)\}$) is preserved.
To prove this: If $\mathbf{x}$ satisfies equations $E_1$ and $E_2$, then it satisfies equations $E_1$ and $E_1+E_2$. Conversely, if $\mathbf{x}$ satisfies equation $E_1$ and $E_1+E_2$, then it satisfies equations $E_1$ and $(E_1+E_2)-E_1$ (which is $E_2$).
We conclude that $\mathbf{x}$ satisfies equations $E_1$ and $E_2$ if and only if $\mathbf{x}$ satisfies equations $E_1$ and $E_1+E_2$.
Thus the set of solutions is preserved.
Best Answer
You can't always, e.g. there's no way to get a diagonal matrix from $$\pmatrix{1 & 1\cr 0 & 0\cr}$$ by these operations. But if you don't run into any inconvenient zeros, it is possible. First add appropriate multiples of row $1$ to each other row to make all $a_{i1} = 0$ for $i \ne 1$. Then add appropriate multiples of row $2$ to each other row to make $a_{i2} = 0$ for $i \ne 2$. Continue...