Help understanding matrix
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I'm stuck on this one task where I should solve the matrix system, but I'm stuck on this part:
https://gyazo.com/fc02fbbe898b586d9901164a6173d766
What I don't understand is how to eliminate 1 2 0 -6 to 1 0 0 0.
Thanks in advance.
matrices matrix-equations
asked Nov 14 at 16:24
Griezy
12
add a comment |
up vote
-4
down vote
favorite
I'm stuck on this one task where I should solve the matrix system, but I'm stuck on this part:
https://gyazo.com/fc02fbbe898b586d9901164a6173d766
What I don't understand is how to eliminate 1 2 0 -6 to 1 0 0 0.
Thanks in advance.
matrices matrix-equations
asked Nov 14 at 16:24
Griezy
12
Sum to the first row the second row multiplied by $-2$.
– egreg
Nov 14 at 16:28
add a comment |
up vote
-4
down vote
favorite
up vote
-4
down vote
favorite
I'm stuck on this one task where I should solve the matrix system, but I'm stuck on this part:
https://gyazo.com/fc02fbbe898b586d9901164a6173d766
What I don't understand is how to eliminate 1 2 0 -6 to 1 0 0 0.
Thanks in advance.
matrices matrix-equations
asked Nov 14 at 16:24
Griezy
12
I'm stuck on this one task where I should solve the matrix system, but I'm stuck on this part:
https://gyazo.com/fc02fbbe898b586d9901164a6173d766
What I don't understand is how to eliminate 1 2 0 -6 to 1 0 0 0.
Thanks in advance.
matrices matrix-equations
matrices matrix-equations
asked Nov 14 at 16:24
Griezy
12
asked Nov 14 at 16:24
Griezy
12
asked Nov 14 at 16:24
Griezy
12
asked Nov 14 at 16:24
Griezy
12
asked Nov 14 at 16:24
Griezy
12
12
Sum to the first row the second row multiplied by $-2$.
– egreg
Nov 14 at 16:28
add a comment |
Sum to the first row the second row multiplied by $-2$.
– egreg
Nov 14 at 16:28
Sum to the first row the second row multiplied by $-2$.
– egreg
Nov 14 at 16:28
Sum to the first row the second row multiplied by $-2$.
– egreg
Nov 14 at 16:28
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
I rewrite the system in the normal form:
$begin{array}{lcl}
x+2y+0z & = & -6 \
0x+y+0z & = & -3 \
0x+0y+z & = & 1
end{array} rightarrow
begin{array}{lcl}
1x + 2*(-3)+0z & = & -6 \
0x+y+0z & = & -3 \
0x+0y+1z & = & 1
end{array} rightarrow
begin{array}{lcl}
1x + 0y +0z & = & 0 \
0x+1y+0z & = & -3 \
0x+0y+1z & = & 1
end{array}$
Hope it is helpful.
Moreover, plz do the research before asking, and write-right form of the question. The tutorial is here.
answered Nov 14 at 16:40
AnNg
375
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I rewrite the system in the normal form:
$begin{array}{lcl}
x+2y+0z & = & -6 \
0x+y+0z & = & -3 \
0x+0y+z & = & 1
end{array} rightarrow
begin{array}{lcl}
1x + 2*(-3)+0z & = & -6 \
0x+y+0z & = & -3 \
0x+0y+1z & = & 1
end{array} rightarrow
begin{array}{lcl}
1x + 0y +0z & = & 0 \
0x+1y+0z & = & -3 \
0x+0y+1z & = & 1
end{array}$
Hope it is helpful.
Moreover, plz do the research before asking, and write-right form of the question. The tutorial is here.
answered Nov 14 at 16:40
AnNg
375
add a comment |
up vote
0
down vote
accepted
I rewrite the system in the normal form:
$begin{array}{lcl}
x+2y+0z & = & -6 \
0x+y+0z & = & -3 \
0x+0y+z & = & 1
end{array} rightarrow
begin{array}{lcl}
1x + 2*(-3)+0z & = & -6 \
0x+y+0z & = & -3 \
0x+0y+1z & = & 1
end{array} rightarrow
begin{array}{lcl}
1x + 0y +0z & = & 0 \
0x+1y+0z & = & -3 \
0x+0y+1z & = & 1
end{array}$
Hope it is helpful.
Moreover, plz do the research before asking, and write-right form of the question. The tutorial is here.
answered Nov 14 at 16:40
AnNg
375
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I rewrite the system in the normal form:
$begin{array}{lcl}
x+2y+0z & = & -6 \
0x+y+0z & = & -3 \
0x+0y+z & = & 1
end{array} rightarrow
begin{array}{lcl}
1x + 2*(-3)+0z & = & -6 \
0x+y+0z & = & -3 \
0x+0y+1z & = & 1
end{array} rightarrow
begin{array}{lcl}
1x + 0y +0z & = & 0 \
0x+1y+0z & = & -3 \
0x+0y+1z & = & 1
end{array}$
Hope it is helpful.
Moreover, plz do the research before asking, and write-right form of the question. The tutorial is here.
answered Nov 14 at 16:40
AnNg
375
I rewrite the system in the normal form:
$begin{array}{lcl}
x+2y+0z & = & -6 \
0x+y+0z & = & -3 \
0x+0y+z & = & 1
end{array} rightarrow
begin{array}{lcl}
1x + 2*(-3)+0z & = & -6 \
0x+y+0z & = & -3 \
0x+0y+1z & = & 1
end{array} rightarrow
begin{array}{lcl}
1x + 0y +0z & = & 0 \
0x+1y+0z & = & -3 \
0x+0y+1z & = & 1
end{array}$
Hope it is helpful.
Moreover, plz do the research before asking, and write-right form of the question. The tutorial is here.
answered Nov 14 at 16:40
AnNg
375
edited Nov 14 at 16:46
answered Nov 14 at 16:40
AnNg
375
answered Nov 14 at 16:40
AnNg
375
answered Nov 14 at 16:40
AnNg
375
375
add a comment |
add a comment |
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Sum to the first row the second row multiplied by $-2$.
– egreg
Nov 14 at 16:28