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So

$$A = \begin{bmatrix} 1 & 2 \\ 3 & 1 \end{bmatrix}$$

and the matrix can be reduced in these steps:

$$\begin{bmatrix} 1 & 2 \\ 0 & -5 \end{bmatrix}$$

via an elementary matrix that looks like this:

$$ E_1 = \begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix}$$

next:

$$\left [ \begin{matrix} 1 & 0 \\ 0 & -5 \end{matrix} \right ] $$

via an elementary matrix that looks like this:

$$ E_2 = \left [ \begin{matrix} 1 & \frac{2}{5} \\ 0 & 1 \end{matrix} \right ] $$

next:

$$\left [ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right ] $$

via an elementary matrix that looks like this:

$$ E_1 = \left [ \begin{matrix} 1 & 0 \\ 0 & \frac{-1}{5} \end{matrix} \right ] $$

So...

$$E_1^{-1} = \left [ \begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix} \right ] $$ $$E_2^{-1} = \left [ \begin{matrix} 1 & \frac{-2}{5} \\ 0 & 1 \end{matrix} \right ] $$ $$E_3^{-1} = \left [ \begin{matrix} 1 & 0 \\ 0 & -5 \end{matrix} \right ] $$

And if I multiply them together I get:

$$E_1^{-1} * E_2^{-2} = \left [ \begin{matrix} 1 & \frac{-2}{5} \\ 3 & \frac{-1}{5} \end{matrix} \right ] = C$$

and

$$C * E_3^{-1} = \left [ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right ] $$

So this works out. Is this right?

Also question, the underlying premise of all of this is that $A$ is invertible right? And if $A$ is invertible, that means that a series of row operations can change it to the identity matrix. Why is this? This doesn't make intuitive sense to me.

Also, why does the product of elementary matrices equal $A$? What is the underlying theorem?

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  • $\begingroup$ Multiplying a matrix $A$ by an elementary matrix $E$ is equivalent to applying elementary row operations to $A$. The underlying theorem that you seek says that $A$ is invertible if and only if it is row-equivalent and column equivalent to the identity matrix; it is commonly called "the invertible matrix theorem". You can search up a proof in any elementary linear algebra textbook. $\endgroup$ – amarney Aug 6 '18 at 19:18
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Given a matrix, the steps involved in determining a sequence of elementary matrices which, when multiplied together, give the original matrix is the same work involved in performing row reduction on the matrix.

For example, in your case you have \begin{gather} E_1 = \begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix} \end{gather}

This is equivalent to performing the operation $R_2 -3R_1$ on $I_2$, the 2x2 identity matrix. This row operation would be the first operation performed to reduce this matrix. Personally, when I have to compute the sequence of elementary matrices I always do the reduction first and then compute the corresponding matrices.

Now, knowing that these are equivalent, we can clearly see why there is a relation to invertibility. We know that if a matrix is invertible, it can be reduced to the identity matrix. If it can be reduced to the identity matrix, then this means that there is a sequence of row operations involved in this reduction. This in turn means that there is a corresponding sequence of elementary matrices of which the product forms the original matrix.

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  • $\begingroup$ What do you mean by equivalent? Are the matrices equal? $\endgroup$ – Jwan622 Aug 7 '18 at 15:59
  • $\begingroup$ Yes -- by equivalent I meant that you end up with the same matrix, it's just two different ways of expressing the same work. $\endgroup$ – emarteca Aug 7 '18 at 17:56

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