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I've been working on matrices lately. Currently, I am stuck on solving systems of linear equations using matrices. I've read the following article which has proved very helpful in understanding the basics of matrices intuitively: An Intuitive Guide to Linear algebra, however, one thing still confuses me. In the following illustrations, pic A picture

One thing still confuses me. If the "operations" matrix and the "input data" matrix are switched, then the output changes. Therefore, I conclude they matrices are not commutative, but in the following example:

$$\begin{bmatrix} 3 &2 &1 \\ 2& 3 &4 \\ 2 & 6 & 1 \end{bmatrix} \begin{bmatrix} x\\ y \\z \end{bmatrix}=\begin{bmatrix}5 \\ 31 \\ 43 \end{bmatrix}$$ But say I want to multiply both sides of the matrix by a certain matrix. $$\begin{bmatrix} 2 \\5\\ 1 \end{bmatrix}$$

are both these operations valid?

this: $$\begin{bmatrix} 2 \\5\\ 1 \end{bmatrix}\begin{bmatrix} 3 &2 &1 \\ 2& 3 &4 \\ 2 & 6 & 1 \end{bmatrix} \begin{bmatrix} x\\ y \\z \end{bmatrix}=\begin{bmatrix} 2 \\5\\ 1 \end{bmatrix}\begin{bmatrix}5 \\ 31 \\ 43 \end{bmatrix}$$

...and this:

$$\begin{bmatrix} 3 &2 &1 \\ 2& 3 &4 \\ 2 & 6 & 1 \end{bmatrix} \begin{bmatrix} x\\ y \\z \end{bmatrix}\begin{bmatrix} 2 \\5\\ 1 \end{bmatrix}=\begin{bmatrix}5 \\ 31 \\ 43 \end{bmatrix}\begin{bmatrix} 2 \\5\\ 1 \end{bmatrix}$$

Anyways, can anyone explain to me how I can simplify,$$\begin{bmatrix} 3 &2 &1 \\ 2& 3 &4 \\ 2 & 6 & 1 \end{bmatrix}$$ into an identity matrix.

In other words, how can I simplify a system of linear equations?

Thanks.

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1 Answer 1

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are both these operations valid

No. I think what you want to do is right-multiply both sides by a $3\times 3$ matrix whose diagonal elements are $2,5,1$.

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