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In the LU decomposition of the matrix $\begin{bmatrix}2 & 2 \\ 4 & 9\end{bmatrix}$, if the diagonal elements of $U$ are both $1$, then the lower diagonal entry $l_{22}$ of $L$ is_________________.
in Linear Algebra by Boss (30.8k points)
edited by | 3.2k views

3 Answers

+42 votes
Best answer
Given,$\begin{bmatrix} l_{11}&0 \\ l_{21}&l_{22}\end{bmatrix}*\begin{bmatrix} u_{11}&u_{12} \\ 0&u_{22}\end{bmatrix}$

$=\begin{bmatrix} l_{11}&0 \\ l_{21}&l_{22}\end{bmatrix}*\begin{bmatrix} 1&u_{12} \\ 0&1\end{bmatrix}$

$=\begin{bmatrix} 2&2 \\ 4&9\end{bmatrix}$

$l_{11} =2$
$l_{11}\times u_{12} =2$
Solving, $u_{12}=1$

$l_{21} = 4$
$l_{21}\times u_{12}+l_{22}=9$
Solving, $l_{22}=5$
by Active (1.2k points)
edited by
0
Can someone please  tell how LU decomposition is done
+1
Can we not take l11 and l22 as 1 rather than taking u11 and u22 as 1?
+4
isn't it a rule that L should always have principle diagonal as 1, and no such restrictions on U? Or is it that anyone one matrix should have 1 in diagonal?
0
GATERush is correct... answer is 5
+8
in dolittles method L 's diagnol taken as 1 and in crouts method U s diagnol taken as 1 both are correct
0
Can anyone please explain why we haven't taken L11& L22 as 1. ?
+2 votes
Answer is already given. I am showing the formal method to solve this question.

$A=$$\begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix}$

On applying $R_{2} \leftarrow R_{2} - 2R_1$

$\begin{bmatrix} 2 &2 \\ 0 & 5 \end{bmatrix}$ $\Leftrightarrow \begin{bmatrix} 1 &0 \\ -2&1 \end{bmatrix}\begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix}$

So, here Elimination Matrix $E_{21} $ is  $\begin{bmatrix} 1 &0 \\ -2&1 \end{bmatrix}$

It means we can write as :-

$E_{21}A = U$

$\Rightarrow \begin{bmatrix} 1 &0 \\ -2 &1 \end{bmatrix} \begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix} = \begin{bmatrix} 2 &2 \\ 0 &5 \end{bmatrix}$

Now, Since $E_{21}A = U.$ So, $A =E_{21}^{-1} U$

$\Rightarrow \begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix} =\begin{bmatrix} 1 &0 \\ -2 &1 \end{bmatrix}^{-1} \begin{bmatrix} 2 &2 \\ 0 &5 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix} =\begin{bmatrix} 1 &0 \\ 2 &1 \end{bmatrix} \begin{bmatrix} 2 &2 \\ 0 &5 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix} =\begin{bmatrix} 1 &0 \\ 2 &1 \end{bmatrix} \begin{bmatrix} 2 &0 \\ 0 &5 \end{bmatrix} \begin{bmatrix} 1 &1 \\ 0 &1 \end{bmatrix}$

It is $A= LDU$ decomposition. So, To make it $A=LU$ decomposition where diagonal entries in U are $1$, we have to multiply $1^{st}$ $2$ matrices i.e. $L$ and $D$. Both $L$ and $D$ are lower triangular matrices, So product of $2$ lower triangular matrices should also be lower triangular.

So,

$\Rightarrow \begin{bmatrix} 2 &2 \\ 4&9 \end{bmatrix} =\begin{bmatrix} 2 &0 \\ 4 &5 \end{bmatrix} \begin{bmatrix} 1 &1 \\ 0 &1 \end{bmatrix}$

Here, $\Rightarrow L =\begin{bmatrix} 2 &0 \\ 4 &5 \end{bmatrix} and \; U = \begin{bmatrix} 1 &1 \\ 0 &1 \end{bmatrix}$

So, $l_{22} = 5$
by Boss (17.1k points)
+1 vote

A better way to do LU decomposition- 

gatecse__LU_decomposition

$A=\begin{pmatrix} 2&2\\4&9\end{pmatrix}$

Since we have to make diagonal elements of U both 1 so I try to make elements above diagonal as 0 and record the multiplier..

$A=\begin{pmatrix} 2&\color{red}{(1)}\\4&5\end{pmatrix}$  $[C_2\leftarrow C_2-\color{red}1C_1 ]$

$\Rightarrow A=\begin{pmatrix} 2&0\\ 4&5\end{pmatrix}+\begin{pmatrix} 1&1\\0&1\end{pmatrix}$

$\implies L=\begin{pmatrix} 2&0\\4&5\end{pmatrix}$   

$I_{22}=5$

by Boss (12.9k points)
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