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I have a question that says...

Let $A$ be a generic $4$ by $n$ matrix. Find a matrix $E$ such that the matrix $EA$ is the matrix obtained from $A$ by scaling the third row of $A$ by $-9$ and leaving all other rows unchanged.

I think I understand what this question is asking me to do, but I don't really understand how to go about it. Am I suppose to choose some values for the matrix A and then solve from there to get a specific matrix E? Or am I suppose to find an abstract matrix E? Could someone please explain how one would go about solving this.

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  • $\begingroup$ You have to assume nothing about $A$, other than that it is a $4 \times n$ matrix. The matrix $E$ can be explicitly written down, in terms of numbers, and has to be found (including the dimensions of $E$). $\endgroup$ – астон вілла олоф мэллбэрг Nov 5 '16 at 3:06
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I'm going to assume we're working over $\mathbb{R}$. Let $A$ be a $4\times n$ matrix with entries in $\mathbb{R}$. Then we can think of $A$ as \begin{equation} A = \begin{pmatrix} a_{11} & a_{12} &\dots & a_{1n}\\ a_{21} & a_{22} &\dots & a_{2n}\\ a_{31} & a_{32} &\dots & a_{3n}\\ a_{41} & a_{42} &\dots & a_{4n}\\ \end{pmatrix} \end{equation} where each entry $a_{ij}$ is some real number. Now, 'scaling' a row by $\alpha\in\mathbb{R}$ just means multiplying each element in that row by $\alpha$. Let's write $A'$ for the matrix you get by scaling the third row of $A$ by $-9$. Then $A'$ looks like: \begin{equation} A' = \begin{pmatrix} a_{11} & a_{12} &\dots & a_{1n}\\ a_{21} & a_{22} &\dots & a_{2n}\\ -9a_{31} & -9a_{32} &\dots & -9a_{3n}\\ a_{41} & a_{42} &\dots & a_{4n}\\ \end{pmatrix} \end{equation} So now the question is asking us to find a matrix $E$ such that $EA = A'$. To find what $E$ is, we would do well to first think about its size. We know that to multiply a $4\times n$ matrix on the left, we need a $y\times 4$ matrix, where $y$ can be any integer. Hence $E$ must be of size $y\times 4$. We also know that the size of the resulting matrix will be $y\times n$. Our matrix $A'$ is a $4\times n$ matrix (its the same size as $A$ since all we did was multiply the third row by $-9$) and so we need $y=4$ to get the right size, so we can conclude that $E$ is a $4\times 4$ matrix.

From here, it doesn't take too much to see what the entries of $E$ must be. Multiplying on the left by $E$ must keep the 1st, 2nd and 4th rows the same. There is only one way to do this, whilst scaling the third row by $-9$. It is: \begin{equation} E=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -9 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \end{equation} In fact, $E$ is what is sometimes called an elementary matrix, corresponding to the elementary row operation of scaling a row. There are several other kinds of elementary matrices, which you can read about here: https://en.wikipedia.org/wiki/Elementary_matrix

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