Question

We consider the matrices below. \[ A=\left[\begin{array}{rrr} -2 & 7 & 1 \\ 3 & 4 & 1 \\ 8 & 1 & 5 \end{array}\right] \] and \[ B=\left[\begin{array}{rrr} 3 & 4 & 1 \\ -2 & 7 & 1 \\ 2 & -7 & 3 \end{array}\right] \] Find the elementary matrices $ E_{0} $ and $ E_{1} $ such that \[ B=E_{1} E_{0} A \] Then find matrices $ X $ and $ Y $ such that \[ A=(Y X) B \] Since \[ \begin{aligned} B & =\left(E_{1} E_{0}\right) A \\ \Rightarrow\left(E_{1} E_{0}\right)^{-1} B & =\left(E_{1} E_{0}\right)^{-1}\left(E_{1} E_{0}\right) A \\ \Rightarrow\left(E_{1} E_{0}\right)^{-1} B & =I_{3} A \end{aligned} \] Therefore \[ \left\{\begin{array}{r} E_{0}^{-1} E_{1}^{-1} B=A \\ Y X B=A \end{array}\right. \] By identification of terms in the above equation we obtain \[ Y=E_{0}^{-1} \text { and } X=E_{1}^{-1} \text { where } E_{1}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{array}\right] \text { and } E_{0}=\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] \] Thus \[ Y=\left[\begin{array}{rrr} 0 & -3 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] \text { and } X=I_{3}-\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1 \end{array}\right] \] Since \[ \begin{array}{l} \qquad B=\left(E_{1} E_{0}\right) A \\ \Rightarrow\left(E_{1} E_{0}\right)^{-1} B=\left(E_{1} E_{0}\right)^{-1}\left(E_{1} E_{0}\right) A \\ \Rightarrow\left(E_{1} E_{0}\right)^{-1} B=I_{3} A \\ \text { Therefore } \\ \left\{\begin{aligned} E_{0}^{-1} E_{1}^{-1} B=A \\ Y X B=A \end{aligned}\right. \end{array} \] Therefore $ \left\{\begin{aligned} E_{0}^{-1} E_{1}^{-1} B & =A \\ Y X B & =A\end{aligned}\right. $ By identification of terms in the above equation we obtain $ Y=E_{0}^{-1} $ and $ X=E_{1}^{-1} $ where $ E_{1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1\end{array}\right] $ and $ E_{0}=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] $ Thus $ Y=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] $ and $ X=\left[\begin{array}{rrr}20 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1\end{array}\right] $ S since \[ \begin{aligned} B & =\left(E_{1} E_{0}\right) A \\ \Rightarrow\left(E_{1} E_{0}\right)^{-1} B & =\left(E_{1} E_{0}\right)^{-1}\left(E_{1} E_{0}\right) A \\ \Rightarrow\left(E_{1} E_{0}\right)^{-1} B & =I_{3} A \end{aligned} \] Therefore \[ \left\{\begin{array}{r} E_{0}^{-1} E_{1}^{-1} B=A \\ Y X B=A \end{array}\right. \] By identification of terms in the above equation we obtain $ Y=E_{0}^{-1} $ and $ X=E_{1}^{-1} $ where $ E_{1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1\end{array}\right] $ and $ E_{0}=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] $ Thus $ Y=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] $ and $ X=\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1\end{array}\right] $ Since \[ \begin{array}{l} \qquad \begin{aligned} B & =\left(E_{1} E_{0}\right) A \\ \Rightarrow\left(E_{1} E_{0}\right)^{-1} B & =\left(E_{1} E_{0}\right)^{-1}\left(E_{1} E_{0}\right) A \\ \Rightarrow\left(E_{1} E_{0}\right)^{-1} B & =I_{3} A \end{aligned} \\ \text { Therefore } \\ \left\{\begin{aligned} E_{0}^{-1} E_{1}^{-1} B & =A \\ Y X B & =A \end{aligned}\right. \end{array} \] By identification of terms in the above equation we obtain $ Y=E_{0}^{-1} $ and $ X=E_{1}^{-1} $ where $ E_{1}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1\end{array}\right] $ and $ E_{0}=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] $ Thus $ Y=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] $ and $ X=\left[\begin{array}{rrr}7 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1\end{array}\right] $

          We consider the matrices below.
\[
A=\left[\begin{array}{rrr}
-2 & 7 & 1 \\
3 & 4 & 1 \\
8 & 1 & 5
\end{array}\right]
\]
and
\[
B=\left[\begin{array}{rrr}
3 & 4 & 1 \\
-2 & 7 & 1 \\
2 & -7 & 3
\end{array}\right]
\]
Find the elementary matrices \( E_{0} \) and \( E_{1} \) such that
\[
B=E_{1} E_{0} A
\]
Then find matrices \( X \) and \( Y \) such that
\[
A=(Y X) B
\]
Since
\[
\begin{aligned}
B & =\left(E_{1} E_{0}\right) A \\
\Rightarrow\left(E_{1} E_{0}\right)^{-1} B & =\left(E_{1} E_{0}\right)^{-1}\left(E_{1} E_{0}\right) A \\
\Rightarrow\left(E_{1} E_{0}\right)^{-1} B & =I_{3} A
\end{aligned}
\]
Therefore
\[
\left\{\begin{array}{r}
E_{0}^{-1} E_{1}^{-1} B=A \\
Y X B=A
\end{array}\right.
\]
By identification of terms in the above equation we obtain
\[
Y=E_{0}^{-1} \text { and } X=E_{1}^{-1} \text { where } E_{1}=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
2 & 0 & 1
\end{array}\right] \text { and } E_{0}=\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]
\]
Thus
\[
Y=\left[\begin{array}{rrr}
0 & -3 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right] \text { and } X=I_{3}-\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
-2 & 0 & 1
\end{array}\right]
\]
Since
\[
\begin{array}{l}
\qquad B=\left(E_{1} E_{0}\right) A \\
\Rightarrow\left(E_{1} E_{0}\right)^{-1} B=\left(E_{1} E_{0}\right)^{-1}\left(E_{1} E_{0}\right) A \\
\Rightarrow\left(E_{1} E_{0}\right)^{-1} B=I_{3} A \\
\text { Therefore } \\
\left\{\begin{aligned}
E_{0}^{-1} E_{1}^{-1} B=A \\
Y X B=A
\end{aligned}\right.
\end{array}
\]
Therefore
\( \left\{\begin{aligned} E_{0}^{-1} E_{1}^{-1} B & =A \\ Y X B & =A\end{aligned}\right. \)
By identification of terms in the above equation we obtain
\( Y=E_{0}^{-1} \) and \( X=E_{1}^{-1} \) where \( E_{1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1\end{array}\right] \) and \( E_{0}=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] \)
Thus
\( Y=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] \) and \( X=\left[\begin{array}{rrr}20 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1\end{array}\right] \)
S since
\[
\begin{aligned}
B & =\left(E_{1} E_{0}\right) A \\
\Rightarrow\left(E_{1} E_{0}\right)^{-1} B & =\left(E_{1} E_{0}\right)^{-1}\left(E_{1} E_{0}\right) A \\
\Rightarrow\left(E_{1} E_{0}\right)^{-1} B & =I_{3} A
\end{aligned}
\]
Therefore
\[
\left\{\begin{array}{r}
E_{0}^{-1} E_{1}^{-1} B=A \\
Y X B=A
\end{array}\right.
\]
By identification of terms in the above equation we obtain
\( Y=E_{0}^{-1} \) and \( X=E_{1}^{-1} \) where \( E_{1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1\end{array}\right] \) and \( E_{0}=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] \)
Thus
\( Y=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] \) and \( X=\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1\end{array}\right] \)
Since
\[
\begin{array}{l}
\qquad \begin{aligned}
B & =\left(E_{1} E_{0}\right) A \\
\Rightarrow\left(E_{1} E_{0}\right)^{-1} B & =\left(E_{1} E_{0}\right)^{-1}\left(E_{1} E_{0}\right) A \\
\Rightarrow\left(E_{1} E_{0}\right)^{-1} B & =I_{3} A
\end{aligned} \\
\text { Therefore } \\
\left\{\begin{aligned}
E_{0}^{-1} E_{1}^{-1} B & =A \\
Y X B & =A
\end{aligned}\right.
\end{array}
\]
By identification of terms in the above equation we obtain
\( Y=E_{0}^{-1} \) and \( X=E_{1}^{-1} \) where \( E_{1}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1\end{array}\right] \) and \( E_{0}=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] \)
Thus
\( Y=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] \) and \( X=\left[\begin{array}{rrr}7 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1\end{array}\right] \)

$We consider the matrices below. A=[ -2 7 1 3 4 1 8 1 5 ] and B=[ 3 4 1 -2 7 1 2 -7 3 ] Find the elementary matrices E0 and E1 such that B=E1 E0 A Then find matrices X and Y such that A=(Y X) B Since B =(E1 E0) A ⇒(E1 E0)^-1 B =(E1 E0)^-1(E1 E0) A ⇒(E1 E0)^-1 B =I3 A Therefore { E0^-1 E1^-1 B=A Y X B=A . By identification of terms in the above equation we obtain Y=E0^-1 and X=E1^-1 where E1=[ 1 0 0 0 1 0 2 0 1 ] and E0=[ 0 1 0 1 0 0 0 0 1 ] Thus Y=[ 0 -3 0 1 0 0 0 0 1 ] and X=I3-[ 1 0 0 0 1 0 -2 0 1 ] Since B=(E1 E0) A ⇒(E1 E0)^-1 B=(E1 E0)^-1(E1 E0) A ⇒(E1 E0)^-1 B=I3 A Therefore { E0^-1 E1^-1 B=A Y X B=A . Therefore { E0^-1 E1^-1 B =A Y X B =A. By identification of terms in the above equation we obtain Y=E0^-1 and X=E1^-1 where E1=[ 1 0 0 0 1 0 2 0 1 ] and E0=[ 0 1 0 1 0 0 0 0 1 ] Thus Y=[ 0 1 0 1 0 0 0 0 1 ] and X=[ 20 0 0 0 1 0 -2 0 1 ] S since B =(E1 E0) A ⇒(E1 E0)^-1 B =(E1 E0)^-1(E1 E0) A ⇒(E1 E0)^-1 B =I3 A Therefore { E0^-1 E1^-1 B=A Y X B=A . By identification of terms in the above equation we obtain Y=E0^-1 and X=E1^-1 where E1=[ 1 0 0 0 1 0 2 0 1 ] and E0=[ 0 1 0 1 0 0 0 0 1 ] Thus Y=[ 0 1 0 1 0 0 0 0 1 ] and X=[ 1 0 0 0 1 0 -2 0 1 ] Since B =(E1 E0) A ⇒(E1 E0)^-1 B =(E1 E0)^-1(E1 E0) A ⇒(E1 E0)^-1 B =I3 A Therefore { E0^-1 E1^-1 B =A Y X B =A . By identification of terms in the above equation we obtain Y=E0^-1 and X=E1^-1 where E1=[ 1 0 0 0 1 0 2 0 1 ] and E0=[ 0 1 0 1 0 0 0 0 1 ] Thus Y=[ 0 1 0 1 0 0 0 0 1 ] and X=[ 7 0 0 0 1 0 -2 0 1 ]$

Added by Bafana B.

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