Question

The matrix $\left[A\right]=\left[\begin{array}{cc}2& 1\\ 4& -1\end{array}\right]$ is decomposed into a product of a lower triangular matrix [L] and an upper triangular matrix[U]. The properly decomposed [L] and [U] matrices respectively are

• A. $\left[\begin{array}{cc}1& 0\\ 4& -1\end{array}\right]$ and $\left[\begin{array}{cc}1& 1\\ 0& -2\end{array}\right]$
• B. $\left[\begin{array}{cc}2& 0\\ 4& -1\end{array}\right]$ and $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right]$ and $\left[\begin{array}{cc}2& 1\\ 0& -1\end{array}\right]$
• D. $\left[\begin{array}{cc}2& 0\\ 4& -3\end{array}\right]$ and $\left[\begin{array}{cc}1& 0.5\\ 0& 1\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{cc}2& 0\\ 4& -3\end{array}\right]$ and $\left[\begin{array}{cc}1& 0.5\\ 0& 1\end{array}\right]$

Matrix, $\left[A\right]=\left[\begin{array}{cc}2& 1\\ 4& -1\end{array}\right]=\left[L\right]\left[U\right]$
Let $\left[L\right]=\left[\begin{array}{cc}a& 0\\ b& c\end{array}\right]$
and $\left[\begin{array}{cc}1& d\\ 0& 1\end{array}\right]$ (using Crout's Method)

$\Rightarrow \left[\begin{array}{cc}2& 1\\ 4& -1\end{array}\right]=\left[\begin{array}{cc}a& 0\\ b& c\end{array}\right]\left[\begin{array}{cc}1& d\\ 0& 1\end{array}\right]$
$\left[\begin{array}{cc}2& 1\\ 4& -1\end{array}\right]=\left[\begin{array}{cc}a& ad\\ b& bd+c\end{array}\right]$
$a=2,b=4$
$ad=1$
$d=\frac{1}{2}$
$bd+c=1$
$2+c=-1$
$c=-3$
$\therefore L=\left[\begin{array}{cc}2& 0\\ 4& -3\end{array}\right]$
$U=\left[\begin{array}{cc}1& 1/2\\ 0& 1\end{array}\right]$