Gilbert_Strang_LU_Decomposition

Question

For which numbers c is $A=LU$ impossible?

 

$\begin{bmatrix} 1 & 2 &0 \\ 3 & c &1 \\ 0 &1 &1 \end{bmatrix}$

Comments

@akash.dinkar12 

How $c=1?$

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Lakshman Patel RJIT

U were saying that c cannot be 1 but what question is asking, he is asking about for what value of c, LU decomposition is not possible but if we put c=1 then we can decompose this matrix into L and U..

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check

$(1)|A|_{1\times1}\neq0$

$(2)|A|_{2\times2}\neq0$

$(3)|A|_{3\times3}\neq0$

if all are true then $A=LU$ is possible.

https://gateoverflow.in/168014/matrices-lu-decompostion

 

Answer 1

$\begin{bmatrix} 1 & 2 &0 \\ 3 & c &1 \\ 0 &1 &1 \end{bmatrix}$

$=\begin{bmatrix} 1 & 0 & 0\\ l_{21}& 1 &0 \\ l_{31}& l_{32} & 1 \end{bmatrix}$.$\begin{bmatrix} u_{11} & u_{12} & u_{13}\\ 0& u_{22} &u_{23} \\ 0& 0 & u_{33} \end{bmatrix}$

$u_{11}=1$

$u_{12}=2$

$u_{13}=0$

$u_{21}=3$

$u_{22}=c-6$

$u_{23}=1$

$l_{31}=0$

$l_{32}=\frac{1}{c-6}$

$u_{33}=1-\frac{1}{c-6}$

So, Now lower triangular matrix is

$=\begin{bmatrix} 1 & 0 & 0\\ 3& 1 &0 \\ 0& \frac{1}{c-6}& 1 \end{bmatrix}$

And upper triangular matrix

$\begin{bmatrix} 1 & 2 & 0\\ 0& c-6 &1 \\ 0& 0 & 1-\frac{1}{c-6} \end{bmatrix}$

So, denominator cannot be $0$, So, c cannot be $6$

Comments

Yes c cannot be 6, otherwise, the pivot in the second row would disappear and then this matrix won't have an LU decomposition possible.

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see my comment also $C\neq1,6,7$

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@Lakshman elaborate more

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@srestha ma'am

see my  comment (above in the answer)

https://gateoverflow.in/168014/matrices-lu-decompostion