lower triangular matrix

Asked Oct 23 '13 at 12:38

Active Oct 23 '13 at 12:38

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Our current course discusses the following topics:

In one exercise there is a square matrix A given :

$$A= \begin{bmatrix} 4 & -2 & -2 \\ 3 & -1 & -1 \\ 0 & -2 & -2 \\ \end{bmatrix} $$

We should find a invertible matrix B so that BA=C and C is a lower triangular matrix.

It seems to me that this couldn't be possible because the row reduced form of the given matrix will be:

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

I am confused and I am looking for some hints

asked Oct 23 '13 at 12:38

Evgeniya

ok it's no problem to solve this. for some reason i assumed the pivot elements of the triangular matrix must be != 0 and that's wrong – Evgeniya Oct 23 '13 at 21:13
$$SA=B$$
$$\begin{bmatrix} 1 & 0 & -1 \ 0 & 2 & -1 \ 0 & -2 & -2 \end{bmatrix} * \begin{bmatrix} -4 & -2 & -2 \ 3 & -1 & -1 \ 0 & -2 & -2 \end{bmatrix} = \begin{bmatrix} 4 & 0 & 0 \ 6 & 0 & 0 \ 0 & 2 & 2 \end{bmatrix}$$

a lower triangular matrix means $1 < i < j \le n$:

$a_{12}=a_{a_13}=a_{23}=0$

I've found a invertible matrix S so that SA=B and B is a lower triangular matrix. What do you think about it ?
– Evgeniya Oct 26 '13 at 11:42

0 Answers0