Reduction of a matrix: can't figure out all the steps

thelazyman

Junior Member
Joined
Jan 14, 2006
Messages
58
Can someone please help me figure out how this matrix reduces? The matrix is:

Code:
[ 1  -1/2    1    1   3 ]
[ 0     1   -2    0   0 ]
[ 0  13/2  -13  -6  -18 ]
[ 0   7/2   -7  -6  -18 ]
I don't know how it becomes this matrix:

Code:
[ 1  0   0   1    3 ]
[ 0  1  -2   0    0 ]
[ 0  0   0  -6  -18 ]
[ 0  0   0  15   45 ]
I understand how they get the 2 zeros in rows 3 and 4 and columns 1 and 2, but I do not understand how they got the 15 and 45 or how they managed to get the 2 zeros on row 1.

Thank you!
 
1 -1/2 1 1 3
0 1 -2 0 0
0 13/2 -13 -6 -18
0 7/2 -7 -6 -18
multiply 2nd row by 1/2 and add to 1st
" " " " -13/2 " " " 3rd
" " " " -7/2 " " " " 4th

1 0 0 1 3
0 1 -2 0 0
0 0 0 -6 -18
0 0 0 -6 -18
multiply 3rd row by -21/6 and add to 4th row
[ tricky instructor not just reducing 4th row to all 0's]

1 0 0 1 3
0 1 -2 0 0
0 0 0 -6 -18
0 0 0 15 45

Arthur
 
Hello, thelazyman!

I have no idea where the 15 and 45 came from!
. . But row 1 has a simple explanation.


\(\displaystyle \L\begin{bmatrix}1 & -\frac{1}{2} & 1 & 1 & 3 \\ \\ 0 & 1 & -2 & 0 & 0 \\ \\ 0 & \frac{13}{2} & -13 & -6 & -18 \\ \\ 0 & \frac{7}{2} & -7 & -6 & -18 \end{bmatrix}\)

This is what I got . . .

\(\displaystyle \begin{array}{cccccc}R1+\frac{1}{2}R2 \\ \\ .\\ \\ R3-\frac{13}{2}R2 \\ R4-\frac{7}{2}R2\end{array}\;
\L\begin{bmatrix}1 & 0 & 0 & 1 & 3 \\ \\ 0 & 1 & -2 & 0 & 0 \\ \\ 0 & 0 & 0 & -6 & -18 \\ \\ 0 & 0 & 0 & -6 & -18\end{bmatrix}\)

Edit: Too fast for me, Arthur!
 
take row 3; 0 0 0 -6 -18
multiply it by -21/6 and add it to 4th row
row 3 times -21/6 is: 0 0 0 21 63
add this to row 4 : 0 0 0 -6 -18

row 4 becomes 0 0 0 [21-6] [63-18]
row 4 becomes 0 0 0 15 45

Arthur
 
soroban-
If you multiply row 3 by -1 and add to row 4 you get
0 0 0 0 0 which is what you would normally do

if you multiply row 3 by -2 and add to row 4 you get
0 0 0 6 18

if you multiply row 3 by -3.5 and add to row 4 you get
0 0 0 15 45

Arthur
 
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