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Let A = \begin{bmatrix}1&2\\-1&1\end{bmatrix}

Give an elementary matrix M such that MA= \begin{bmatrix}1&2\\1&5\end{bmatrix}

I'm trying to figure out what the values for M would be, but I just don't understand where I'm supposed to start. I know the rules for turning a matrix into RREF form and how I can interchange rows, multiply a row by a nonzero constant, and add a constant multiple of a row to another row, but I don't understand how I would reverse the steps to find the values for the M matrix stated in the solution?

The solution that is stated in the textbook is M = \begin{bmatrix}1&0\\2&1\end{bmatrix}

I really don't know where to start. Any help or a push in the right direction would be appreciated.

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    $\begingroup$ Whatever row operation you do to achieve that matrix, do the same operation to the identity matrix: this will give you the elementary matrix. In your example, to get the desired $MA$, you would add $2$ times the first row to the second row. And the $M$ given in your textbook is exactly what you get if you take the identity matrix and add $2$ times the first row to the second row. $\endgroup$ – André 3000 Mar 4 at 4:15
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Let $$B=\begin{bmatrix}1&2\\1&5\end{bmatrix}$$ You're looking for a matrix $M$ such that $$MA=B\tag1$$ Since $A$ is invertible (exercise: why?), we know that the solution is $M=BA^{-1}$.

To find $M$, one way is to explicitly compute $A^{-1}$ and then $BA^{-1}$

Another way is to write $M$ as a matrix whose 4 components are 4 unknown numbers: $$M=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ And $(1)$ is equivalent to $$\left(\begin{align} a-b &=1\\ 2a+b &=2\\ c-d&=1\\ 2c+d&=5 \end{align}\right.$$ And now you can apply Gaussian elimination to solve for $a,b,c$ and $d$.

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  • $\begingroup$ Hi! This might be a dumb question but I don't quite understand how MA is equivalent to the four system of equations you stated. Could you explain? $\endgroup$ – Eagerissac Mar 4 at 4:28
  • $\begingroup$ No dumb question. I simply computed the product of the matrix $M=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ with $A$. The result is a 2x2 matrix whose coefficients are $a-b$, $2a+b$, $c-d$ and $2c+d$. I then equated this to $B$. $\endgroup$ – Stefan Lafon Mar 4 at 4:34
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When you left-multiply a matrix by a row vector, the result will be a linear combination of the rows of the matrix with the elements of the vector as the coefficients. This extends to left-multiplication by a matrix: each row of the product is the product of that row of the left-hand matrix with the right-hand one. This is a straightforward consequence of the definition of matrix multiplication.

The first row of $MA$ is identical to that of $A$, so the first row of $M$ must be $\small{\begin{bmatrix}1&0\end{bmatrix}}$. It’s not hard to work out that the second row of $MA$ is equal to $2$ times the first row of $A$ plus the second row of $A$, therefore, the second row of $M$ is $\small{\begin{bmatrix}2&1\end{bmatrix}}$.

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Expanding on the operate-on-the-identity-matrix trick in the comments:

We know multiplying by the identity matrix does not change a matrix $A$: $IA=A$.

So asking what matrix $M$ encodes a particular row operation on $A$ is the same as asking for the matrix encoding $M$ of a row operation on $IA$.

Since matrix multiplication is associative, we have $MA=M(IA)=(MI)A$.

You can see that constructing $M$ by row operating on the identity matrix works because $M$ is by definition the encoding of the row operation, so $MI$ is the row operation executed on the identity matrix.

See this helpful reference: https://stattrek.com/matrix-algebra/elementary-operations.aspx

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