# Thread: Another matrix as elementary matrices

1. ## Another matrix as elementary matrices

Hey everyone

I am trying to write this matrix as a product elementary matrices:

$
\begin{bmatrix}
-3 & 1 \\
2 & 2
\end{bmatrix}
$

I keep on getting this answer:
$
\begin{bmatrix}
-3 & 0 \\
0 & 1
\end{bmatrix}

\begin{bmatrix}
1 & 0 \\
2 & 1
\end{bmatrix}

\begin{bmatrix}
1 & 0 \\
0 & \frac4{3}
\end{bmatrix}

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

I've did it 3 times and keep getting the same answer. Can someone point me to where I am going wrong.

Thanks

2. Can this system be solved?

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

\begin{bmatrix}
l_{11} & 0 \\
l_{21} & l_{22}
\end{bmatrix}
\times
\begin{bmatrix}
u_{11} & u_{12} \\
0 & u_{22}
\end{bmatrix}$

$\displaystyle -3=l_{11}u_{11}$

$\displaystyle 1= l_{11}u_{12}$

$\displaystyle 2=l_{21}u_{11}$

$\displaystyle 2=l_{21}u_{12}+l_{12}u_{22}$

3. Originally Posted by pickslides
Can this system be solved?

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

\begin{bmatrix}
l_{11} & 0 \\
l_{21} & l_{22}
\end{bmatrix}
\times
\begin{bmatrix}
u_{11} & u_{12} \\
0 & u_{22}
\end{bmatrix}$

$\displaystyle -3=l_{11}u_{11}$

$\displaystyle 1= l_{11}u_{12}$

$\displaystyle 2=l_{21}u_{11}$

$\displaystyle 2=l_{21}u_{12}+l_{12}u_{22}$
This the LU Factorization which I informed you of yesterday evant8950.

pickslides, did you mean to write this

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

\begin{bmatrix}
l_{11} & 0 \\
l_{21} & l_{22}
\end{bmatrix}
\times
\begin{bmatrix}
u_{11} & u_{12} \\
0 & u_{22}
\end{bmatrix}\text{?}$