Find elementary matrices E1 and E2 such that...

[Prodigy]

C=E1*E2*A

Where A=
1 2
-1 1

C=
-1 1
2 1

These are both 2 x 2 matrices.

Anyways, I know that R=UA where in this case U=E1E2 but what I don't understand is that R does not equal C...Because when I row reduced echlon form for A...I got an identity matrix which does not equal C...And I used more than 2 elementary steps in order to row reduce it. Then I tried to do this..
C*A^-1=E1*E2
But that does not work either as it takes more than 2 elementary steps...

I'd really appreciate it if someone can give me some hints how to solve this so I can do part b lol

Thanks

 

[soroban]

Hello, Prodigy!

I don't understand your difficulty,
. . nor do I understand the question itself.

\(\displaystyle \text{Find elementary matrices }E_1\text{ and }E_2\text{ such that: }\; C\:=\:E_1\!\cdot\! E_2\!\cdot\!A\)

\(\displaystyle \text{where: }\:A \:=\:\begin {pmatrix}1 & 2 \\ \text{-}1 & 1\end{pmatrix} \;\text{ and }\;C \:=\:\begin{pmatrix}\text{-}1 & 1 \\ 2 & 1 \end{pmatrix}\)

\(\displaystyle \text{We have: }\;E_1\!\cdot\!E_2\!\cdot\!A \;=\;C \quad\Rightarrow\quad\left(E_1\!\cdot\!E_2\right)\begin{pmatrix}1&2\\\text{-}1&1\end{pmatrix} \;=\;\begin{pmatrix}\text{-}1&1\\2&1\end{pmatrix}\)

. . \(\displaystyle \text{We find that: }\:A^{-1} \:=\:\begin{pmatrix}\frac{1}{3} & \text{-}\frac{2}{3} \\ \\[-3mm] \frac{1}{3} & \frac{1}{3} \end{pmatrix}\)


\(\displaystyle \text{Right-multiply by }A^{-1}\!:\;\;(E_1\!\cdot\!E_2)\!\cdot\!A\!\cdot\!A^{-1} \;=\;C\!\cdot\!A^{-1}\)

. . \(\displaystyle (E_1\!\cdot\!E_2)\begin{pmatrix}1&2\\\text{-}1&1\end{pmatrix}\begin{pmatrix}\frac{1}{3} & \text{-}\frac{2}{3}\\ \\[-3mm] \frac{1}{3} & \frac{1}{3}\end{pmatrix} \;=\;\begin{pmatrix}\text{-}1&1\\2&1\end{pmatrix}\begin{pmatrix}\frac{1}{3} & \text{-}\frac{2}{3}\\ \\[-3mm] \frac{1}{3} & \frac{1}{3} \end{pmatrix}\)


\(\displaystyle \text{And we have: }\;E_1\!\cdot\!E_2 \;=\;\begin{pmatrix}0&1\\1&\text{-}1 \end{pmatrix}\)


. . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


\(\displaystyle \text{We want two matrices, }E_1\text{ and }E_2\text{, whose product is: }\:\begin{pmatrix}0&1\\1&\text{-}1\end{pmatrix}\)

\(\displaystyle \text{There is an infinite number of solutions.}\)

. . \(\displaystyle \text{If }E_1\:=\:\begin{pmatrix}a&b\\c&d\end{pmatrix},\text{ then: }\:E_2 \;=\;\frac{1}{ad-bc}\begin{pmatrix}\text{-}b & b+d \\ a & \text{-}a-c \end{pmatrix}\)


Could this be what they expected?

 

[tkhunny]

I believe an "Elementary" matrix normally would be a matrix that represents a single "Elementary" row operation.

This makes a possible solution:

\(\displaystyle E_{2} = \begin {pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\) -- Swapping Rows.

\(\displaystyle E_{1} = \begin {pmatrix}1 & 0 \\ \text{-}1 & 1\end{pmatrix}\) -- Subtract Row 1 from Row 2 and replace Row 2 with the result.