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

 

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

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

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

$u_{11}=1$

$u_{12}=2$

$u_{13}=0$

$u_{21}=3$

$u_{22}=c-6$

$u_{23}=1$

$l_{31}=0$

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

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

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 $6$

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