Find $lambda$ and solve the matrix
$begingroup$
Find $lambda$ and solve the matrix. I have 4 equations:
begin{cases}
x+y+z-t=2\
x+y-z+t=2\
3x+y+z+t=lambda\
x-y+z+t=2\
end{cases}
I've got $$t=frac{lambda -6}{4}; z=frac{6-lambda}{4}; y=frac{lambda -6}{4}; x=frac{2+lambda}{4}$$
What I did wrong?
matrices matrix-equations
edited Jan 7 at 10:06
Lorenzo B.
1,8402520
asked Jan 7 at 9:39
J.DoeJ.Doe
899
$endgroup$
add a comment |
$begingroup$
Find $lambda$ and solve the matrix. I have 4 equations:
begin{cases}
x+y+z-t=2\
x+y-z+t=2\
3x+y+z+t=lambda\
x-y+z+t=2\
end{cases}
I've got $$t=frac{lambda -6}{4}; z=frac{6-lambda}{4}; y=frac{lambda -6}{4}; x=frac{2+lambda}{4}$$
What I did wrong?
matrices matrix-equations
edited Jan 7 at 10:06
Lorenzo B.
1,8402520
asked Jan 7 at 9:39
J.DoeJ.Doe
899
$endgroup$
$begingroup$
Yes, it's wrong. Substitute it in the second equation.
$endgroup$
– Michael Rozenberg
Jan 7 at 9:43
add a comment |
$begingroup$
Find $lambda$ and solve the matrix. I have 4 equations:
begin{cases}
x+y+z-t=2\
x+y-z+t=2\
3x+y+z+t=lambda\
x-y+z+t=2\
end{cases}
I've got $$t=frac{lambda -6}{4}; z=frac{6-lambda}{4}; y=frac{lambda -6}{4}; x=frac{2+lambda}{4}$$
What I did wrong?
matrices matrix-equations
edited Jan 7 at 10:06
Lorenzo B.
1,8402520
asked Jan 7 at 9:39
J.DoeJ.Doe
899
$endgroup$
Find $lambda$ and solve the matrix. I have 4 equations:
begin{cases}
x+y+z-t=2\
x+y-z+t=2\
3x+y+z+t=lambda\
x-y+z+t=2\
end{cases}
I've got $$t=frac{lambda -6}{4}; z=frac{6-lambda}{4}; y=frac{lambda -6}{4}; x=frac{2+lambda}{4}$$
What I did wrong?
matrices matrix-equations
matrices matrix-equations
edited Jan 7 at 10:06
Lorenzo B.
1,8402520
asked Jan 7 at 9:39
J.DoeJ.Doe
899
edited Jan 7 at 10:06
Lorenzo B.
1,8402520
asked Jan 7 at 9:39
J.DoeJ.Doe
899
edited Jan 7 at 10:06
Lorenzo B.
1,8402520
edited Jan 7 at 10:06
Lorenzo B.
1,8402520
edited Jan 7 at 10:06
Lorenzo B.
1,8402520
1,8402520
asked Jan 7 at 9:39
J.DoeJ.Doe
899
asked Jan 7 at 9:39
J.DoeJ.Doe
899
asked Jan 7 at 9:39
J.DoeJ.Doe
899
899
$begingroup$
Yes, it's wrong. Substitute it in the second equation.
$endgroup$
– Michael Rozenberg
Jan 7 at 9:43
add a comment |
$begingroup$
Yes, it's wrong. Substitute it in the second equation.
$endgroup$
– Michael Rozenberg
Jan 7 at 9:43
$begingroup$
Yes, it's wrong. Substitute it in the second equation.
$endgroup$
– Michael Rozenberg
Jan 7 at 9:43
$begingroup$
Yes, it's wrong. Substitute it in the second equation.
$endgroup$
– Michael Rozenberg
Jan 7 at 9:43
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
After adding of the first, second and forth equations we'll get $$3x+y+z+t=6,$$ which gives $lambda=6.$
answered Jan 7 at 9:45
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
$begingroup$
But using Gauss I can't get this solution?
$endgroup$
– J.Doe
Jan 7 at 10:00
$begingroup$
@J.Doe Which says that there is a mistake in your computations.
$endgroup$
– Michael Rozenberg
Jan 7 at 10:05
add a comment |
$begingroup$
Hint: It is $$x=2-t,y=t,z=t$$ and $$lambda=6$$
answered Jan 7 at 9:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
$endgroup$
add a comment |
$begingroup$
Subtracting eq. 2 from both eq. 1 and eq. 4 gives
begin{cases}
2z-2t=0\
x+y-z+t=2\
3x+y+z+t=lambda\
-2y+2z=0\
end{cases}
Which implies
$$begin{cases}
z=t\
x=2-y\
6-3y+3y=lambda\
y=z=t\
end{cases}implies lambda = 6$$
answered Jan 7 at 10:14
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
After adding of the first, second and forth equations we'll get $$3x+y+z+t=6,$$ which gives $lambda=6.$
answered Jan 7 at 9:45
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
$begingroup$
But using Gauss I can't get this solution?
$endgroup$
– J.Doe
Jan 7 at 10:00
$begingroup$
@J.Doe Which says that there is a mistake in your computations.
$endgroup$
– Michael Rozenberg
Jan 7 at 10:05
add a comment |
$begingroup$
After adding of the first, second and forth equations we'll get $$3x+y+z+t=6,$$ which gives $lambda=6.$
answered Jan 7 at 9:45
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
$begingroup$
But using Gauss I can't get this solution?
$endgroup$
– J.Doe
Jan 7 at 10:00
$begingroup$
@J.Doe Which says that there is a mistake in your computations.
$endgroup$
– Michael Rozenberg
Jan 7 at 10:05
add a comment |
$begingroup$
After adding of the first, second and forth equations we'll get $$3x+y+z+t=6,$$ which gives $lambda=6.$
answered Jan 7 at 9:45
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
After adding of the first, second and forth equations we'll get $$3x+y+z+t=6,$$ which gives $lambda=6.$
answered Jan 7 at 9:45
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 7 at 9:45
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 7 at 9:45
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 7 at 9:45
Michael RozenbergMichael Rozenberg
103k1891195
103k1891195
$begingroup$
But using Gauss I can't get this solution?
$endgroup$
– J.Doe
Jan 7 at 10:00
$begingroup$
@J.Doe Which says that there is a mistake in your computations.
$endgroup$
– Michael Rozenberg
Jan 7 at 10:05
add a comment |
$begingroup$
But using Gauss I can't get this solution?
$endgroup$
– J.Doe
Jan 7 at 10:00
$begingroup$
@J.Doe Which says that there is a mistake in your computations.
$endgroup$
– Michael Rozenberg
Jan 7 at 10:05
$begingroup$
But using Gauss I can't get this solution?
$endgroup$
– J.Doe
Jan 7 at 10:00
$begingroup$
But using Gauss I can't get this solution?
$endgroup$
– J.Doe
Jan 7 at 10:00
$begingroup$
@J.Doe Which says that there is a mistake in your computations.
$endgroup$
– Michael Rozenberg
Jan 7 at 10:05
$begingroup$
@J.Doe Which says that there is a mistake in your computations.
$endgroup$
– Michael Rozenberg
Jan 7 at 10:05
add a comment |
$begingroup$
Hint: It is $$x=2-t,y=t,z=t$$ and $$lambda=6$$
answered Jan 7 at 9:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
$endgroup$
add a comment |
$begingroup$
Hint: It is $$x=2-t,y=t,z=t$$ and $$lambda=6$$
answered Jan 7 at 9:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
$endgroup$
add a comment |
$begingroup$
Hint: It is $$x=2-t,y=t,z=t$$ and $$lambda=6$$
answered Jan 7 at 9:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
$endgroup$
Hint: It is $$x=2-t,y=t,z=t$$ and $$lambda=6$$
answered Jan 7 at 9:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
answered Jan 7 at 9:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
answered Jan 7 at 9:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
answered Jan 7 at 9:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
75.4k42865
add a comment |
add a comment |
$begingroup$
Subtracting eq. 2 from both eq. 1 and eq. 4 gives
begin{cases}
2z-2t=0\
x+y-z+t=2\
3x+y+z+t=lambda\
-2y+2z=0\
end{cases}
Which implies
$$begin{cases}
z=t\
x=2-y\
6-3y+3y=lambda\
y=z=t\
end{cases}implies lambda = 6$$
answered Jan 7 at 10:14
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
add a comment |
$begingroup$
Subtracting eq. 2 from both eq. 1 and eq. 4 gives
begin{cases}
2z-2t=0\
x+y-z+t=2\
3x+y+z+t=lambda\
-2y+2z=0\
end{cases}
Which implies
$$begin{cases}
z=t\
x=2-y\
6-3y+3y=lambda\
y=z=t\
end{cases}implies lambda = 6$$
answered Jan 7 at 10:14
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
add a comment |
$begingroup$
Subtracting eq. 2 from both eq. 1 and eq. 4 gives
begin{cases}
2z-2t=0\
x+y-z+t=2\
3x+y+z+t=lambda\
-2y+2z=0\
end{cases}
Which implies
$$begin{cases}
z=t\
x=2-y\
6-3y+3y=lambda\
y=z=t\
end{cases}implies lambda = 6$$
answered Jan 7 at 10:14
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
Subtracting eq. 2 from both eq. 1 and eq. 4 gives
begin{cases}
2z-2t=0\
x+y-z+t=2\
3x+y+z+t=lambda\
-2y+2z=0\
end{cases}
Which implies
$$begin{cases}
z=t\
x=2-y\
6-3y+3y=lambda\
y=z=t\
end{cases}implies lambda = 6$$
answered Jan 7 at 10:14
Lorenzo B.Lorenzo B.
1,8402520
answered Jan 7 at 10:14
Lorenzo B.Lorenzo B.
1,8402520
answered Jan 7 at 10:14
Lorenzo B.Lorenzo B.
1,8402520
answered Jan 7 at 10:14
Lorenzo B.Lorenzo B.
1,8402520
1,8402520
add a comment |
add a comment |
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$begingroup$
Yes, it's wrong. Substitute it in the second equation.
$endgroup$
– Michael Rozenberg
Jan 7 at 9:43