# Row reduction as matrix multiplication help?

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.

• 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. – André 3000 Mar 4 at 4:15

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 $$d$$.

• 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? – Eagerissac Mar 4 at 4:28
• 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 $B$. – Stefan Lafon Mar 4 at 4:34

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 $$\small{\begin{bmatrix}2&1\end{bmatrix}}$$.

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.