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I just want to know how to get elementary matrices using fast and efficient way to solve it. Since I'm new in linear algebra, I hope someone able to help me

Given $$ A =\begin{pmatrix} 1 & 2 & -1 \\ 0 & 1 & 3/2 \\ 0 & -3& 1 \\ \end{pmatrix} $$

and $$ B =\begin{pmatrix} 1 & 2 & -1 \\ 0 & 1 & 3/2 \\ 0 & 0 & 11/2 \\ \end{pmatrix} $$

a) Find the elementary matrices E such that EA=B

How to solve it? Could someone guide me out by doing some steps

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Sorry I already edited the question Abhra –  maxwell Jun 10 '13 at 16:24

1 Answer 1

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See here you have to add $3 $ times the 2nd row to the third row to make the $(3,2)$ th element to be zero. When you do this you automatically have the last $(3,3)$ element of B.

So the elementary matrix will be

$\begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&3&1 \end {pmatrix}$

Here $\begin{pmatrix}1&0&0\\\end {pmatrix}$ in the first row denotes that you are adding $1$ times the first row to $0$ times the first row and $0$ times the 3rd row of A to get the 1st row of the resultant matrix B. Similarly others follow that is $\begin{pmatrix}0&1&0\\\end {pmatrix}$ in the 2nd row denotes that you have only $1$ times the 2nd row as your 2nd row of B and $\begin{pmatrix}0&3&1\\\end {pmatrix}$ in last row denotes that last row of B is comprised of $3$ times the 2nd row and 1 time the last row of A.

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But how the 3 appears in the elemenary matrix in (3,2) since you multiply 3 to 2nd row?? –  maxwell Jun 10 '13 at 16:41
    
$\begin{pmatrix}0&3&1\\\end {pmatrix}$ denotes that we are adding 0 times the first row to 3 times the 2nd row to 1 times third row of A to get the 3rd row of B. As we are adding 3 times the third 2nd row so 3 appears there. –  Abhra Abir Kundu Jun 10 '13 at 16:45
    
Now do you get it?? –  Abhra Abir Kundu Jun 10 '13 at 16:46
    
Tank you very much Abhra. I manage to get it. thnaks once again –  maxwell Jun 10 '13 at 17:21
    
you are welcome @maxwell –  Abhra Abir Kundu Jun 10 '13 at 17:22

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