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I am studying Linear Algebra part-time and would like to know if anyone has advice on solving the following type of questions:

Considering the matrix:

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

Find elementary Matrices $E_1$ and $E_2$ such that $E_2E_1A = I$

Firstly can this be re-written as?

$$E_2E_1 = IA^{-1}$$

and that is the same as?

$$E_2E_1 = A^{-1}$$

So I tried to find $E_1$ and $E_2$ such that $E_2E_1 = A^{-1}$:

My solution: $$A^{-1} = \begin{bmatrix}1 & 0 & \\{\frac {5}{2}} & {\frac {1}{2}}\end{bmatrix}$$ $$E_2 = \begin{bmatrix}1 & 0 & \\0 & {\frac {5}{2}}\end{bmatrix}$$ $$E_1 = \begin{bmatrix}1 & 0 & \\1 & {\frac {1}{5}}\end{bmatrix}$$

This is the incorrect answer. Any help as to what I did wrong as well as suggestions on how to approach these questions would be aprpeciated.

Thanks

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1  
Is your $E_1$ really an elementary matrix? –  Dylan Moreland Jul 12 '12 at 6:24
4  
To get the matrices. Apply the elementary row operations on $A$. Each elementary row operation is equivalent to multiplying the corresponding elementary matrix. Two row operations get you $I$, so you get two elementary matrices corresponding to them. –  Host-website-on-iPage Jul 12 '12 at 6:29

3 Answers 3

Just look at what needs to be done.

First, eliminate the $-5$ using $E_1 = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$. This gives $$ E_1 A = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}.$$ Can you figure out what $E_2$ must be?

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You can reduce A to I in two elementary row transformation,

  1. $R_2 + 5R_1 \rightarrow R_2$
  2. $\frac{1}{2} R_2 \rightarrow R_2$

An elementary matrix is a matrix that differs from I in exactly one elementary transformation. It turns out that you just need matrix corresponding to each of the row transformation above to come up with your elementary matrices. For example, the elementary matrix corresponding to the first row transformation is, $$\begin{bmatrix}1 & 0\\5&1\end{bmatrix}$$ Notice that when you multiply this matrix with A, it does exactly the first elementary row transformation. So, this is $E_1$. Similarly, figure out $E_2$ by getting the elementary matrix corresponding to the second transformation.

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up vote 1 down vote accepted

Thanks for the advice and hints TenaliRaman and copper.hat. I'll post the solution here for anyone else who needs to solve something similar.

So we have to find the elementary matrices $E_1$ and $E_2$ such that $E_2E_1A = I$:

Start by eliminating the -5 in matrix A.

So find $E_1$ so that

$$E_1A = \begin{bmatrix}1 & 0 & \\0 & 2\end{bmatrix}$$

the matrix $E_1$ is thus:

$$E_1 = \begin{bmatrix}1 & 0 & \\5 & 1\end{bmatrix}$$

If we rewrite the original equation like so: $$E_2(E_1A) = I$$

Find $E_2$ (since we know what $E_1A$ is)

Solving this we get:

$$E_2 = \begin{bmatrix}1 & 0 & \\0 & {\frac {1}{2}}\end{bmatrix}$$

Thanks everyone!

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