Value of k to give matrix infinite, 0, 1 solutions.
$begingroup$
I have a question that goes:
For which values of the constant $k$ does the system of equations below have:
- a unique solution,
- no solutions at all,
- infinitely many solutions?
$$
begin{cases}
x &- 3y & &= 6\
x & &+ 3z &= -3\
2x &+ ky &+ (3-k)z &= 1
end{cases}
$$
I tried putting the system of equations into matrix in reduced row echelon form, ended up with the last line being
$$
begin{matrix}
0 & k+6 & 3-k &| 1,
end{matrix}
$$ which I don't think would make sense.
linear-algebra
edited Dec 27 '18 at 16:26
gt6989b
35k22557
asked Dec 27 '18 at 16:09
Jamie WhitakerJamie Whitaker
162
$endgroup$
add a comment |
$begingroup$
I have a question that goes:
For which values of the constant $k$ does the system of equations below have:
- a unique solution,
- no solutions at all,
- infinitely many solutions?
$$
begin{cases}
x &- 3y & &= 6\
x & &+ 3z &= -3\
2x &+ ky &+ (3-k)z &= 1
end{cases}
$$
I tried putting the system of equations into matrix in reduced row echelon form, ended up with the last line being
$$
begin{matrix}
0 & k+6 & 3-k &| 1,
end{matrix}
$$ which I don't think would make sense.
linear-algebra
edited Dec 27 '18 at 16:26
gt6989b
35k22557
asked Dec 27 '18 at 16:09
Jamie WhitakerJamie Whitaker
162
$endgroup$
add a comment |
$begingroup$
I have a question that goes:
For which values of the constant $k$ does the system of equations below have:
- a unique solution,
- no solutions at all,
- infinitely many solutions?
$$
begin{cases}
x &- 3y & &= 6\
x & &+ 3z &= -3\
2x &+ ky &+ (3-k)z &= 1
end{cases}
$$
I tried putting the system of equations into matrix in reduced row echelon form, ended up with the last line being
$$
begin{matrix}
0 & k+6 & 3-k &| 1,
end{matrix}
$$ which I don't think would make sense.
linear-algebra
edited Dec 27 '18 at 16:26
gt6989b
35k22557
asked Dec 27 '18 at 16:09
Jamie WhitakerJamie Whitaker
162
$endgroup$
I have a question that goes:
For which values of the constant $k$ does the system of equations below have:
- a unique solution,
- no solutions at all,
- infinitely many solutions?
$$
begin{cases}
x &- 3y & &= 6\
x & &+ 3z &= -3\
2x &+ ky &+ (3-k)z &= 1
end{cases}
$$
I tried putting the system of equations into matrix in reduced row echelon form, ended up with the last line being
$$
begin{matrix}
0 & k+6 & 3-k &| 1,
end{matrix}
$$ which I don't think would make sense.
linear-algebra
linear-algebra
edited Dec 27 '18 at 16:26
gt6989b
35k22557
asked Dec 27 '18 at 16:09
Jamie WhitakerJamie Whitaker
162
edited Dec 27 '18 at 16:26
gt6989b
35k22557
asked Dec 27 '18 at 16:09
Jamie WhitakerJamie Whitaker
162
edited Dec 27 '18 at 16:26
gt6989b
35k22557
edited Dec 27 '18 at 16:26
gt6989b
35k22557
edited Dec 27 '18 at 16:26
gt6989b
35k22557
35k22557
asked Dec 27 '18 at 16:09
Jamie WhitakerJamie Whitaker
162
asked Dec 27 '18 at 16:09
Jamie WhitakerJamie Whitaker
162
asked Dec 27 '18 at 16:09
Jamie WhitakerJamie Whitaker
162
162
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have
$$
begin{pmatrix}
1 & -3 & 0 & 6 \
1 & 0 & 3 & -3 \
2 & k & 3-k & 1
end{pmatrix}
to
begin{pmatrix}
1 & -3 & 0 & 6 \
0 & 3 & 3 & -9 \
0 & k+6 & 3-k & -11
end{pmatrix}
to\
begin{pmatrix}
1 & 0 & 3 & -3 \
0 & 1 & 1 & -3 \
0 & 0 & 3-k-(k+6) & -11 + 3(k+6)
end{pmatrix}
to \
begin{pmatrix}
1 & 0 & 3 & -3 \
0 & 1 & 1 & -3 \
0 & 0 & -3-2k& 7 + 3k
end{pmatrix}
$$
What happens in $-3-2k=0$? Can you finish this?
answered Dec 27 '18 at 16:23
gt6989bgt6989b
35k22557
$endgroup$
add a comment |
$begingroup$
Hint: From the first equation we get
$$x=6+3y$$ plug this in the second equation we obtain
$$y+z=-3$$
and with the third equation we obtain
$$(6+k)y+(3-k)z=1$$
with $$y=-3-z$$ we obtain $$(6+k)(-3-z)+(3-k)z=-11$$
so we get
$$z(-k-3)=9+3k$$
Can you finish?
answered Dec 27 '18 at 16:25
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
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active
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active
oldest
votes
$begingroup$
You have
$$
begin{pmatrix}
1 & -3 & 0 & 6 \
1 & 0 & 3 & -3 \
2 & k & 3-k & 1
end{pmatrix}
to
begin{pmatrix}
1 & -3 & 0 & 6 \
0 & 3 & 3 & -9 \
0 & k+6 & 3-k & -11
end{pmatrix}
to\
begin{pmatrix}
1 & 0 & 3 & -3 \
0 & 1 & 1 & -3 \
0 & 0 & 3-k-(k+6) & -11 + 3(k+6)
end{pmatrix}
to \
begin{pmatrix}
1 & 0 & 3 & -3 \
0 & 1 & 1 & -3 \
0 & 0 & -3-2k& 7 + 3k
end{pmatrix}
$$
What happens in $-3-2k=0$? Can you finish this?
answered Dec 27 '18 at 16:23
gt6989bgt6989b
35k22557
$endgroup$
add a comment |
$begingroup$
You have
$$
begin{pmatrix}
1 & -3 & 0 & 6 \
1 & 0 & 3 & -3 \
2 & k & 3-k & 1
end{pmatrix}
to
begin{pmatrix}
1 & -3 & 0 & 6 \
0 & 3 & 3 & -9 \
0 & k+6 & 3-k & -11
end{pmatrix}
to\
begin{pmatrix}
1 & 0 & 3 & -3 \
0 & 1 & 1 & -3 \
0 & 0 & 3-k-(k+6) & -11 + 3(k+6)
end{pmatrix}
to \
begin{pmatrix}
1 & 0 & 3 & -3 \
0 & 1 & 1 & -3 \
0 & 0 & -3-2k& 7 + 3k
end{pmatrix}
$$
What happens in $-3-2k=0$? Can you finish this?
answered Dec 27 '18 at 16:23
gt6989bgt6989b
35k22557
$endgroup$
add a comment |
$begingroup$
You have
$$
begin{pmatrix}
1 & -3 & 0 & 6 \
1 & 0 & 3 & -3 \
2 & k & 3-k & 1
end{pmatrix}
to
begin{pmatrix}
1 & -3 & 0 & 6 \
0 & 3 & 3 & -9 \
0 & k+6 & 3-k & -11
end{pmatrix}
to\
begin{pmatrix}
1 & 0 & 3 & -3 \
0 & 1 & 1 & -3 \
0 & 0 & 3-k-(k+6) & -11 + 3(k+6)
end{pmatrix}
to \
begin{pmatrix}
1 & 0 & 3 & -3 \
0 & 1 & 1 & -3 \
0 & 0 & -3-2k& 7 + 3k
end{pmatrix}
$$
What happens in $-3-2k=0$? Can you finish this?
answered Dec 27 '18 at 16:23
gt6989bgt6989b
35k22557
$endgroup$
You have
$$
begin{pmatrix}
1 & -3 & 0 & 6 \
1 & 0 & 3 & -3 \
2 & k & 3-k & 1
end{pmatrix}
to
begin{pmatrix}
1 & -3 & 0 & 6 \
0 & 3 & 3 & -9 \
0 & k+6 & 3-k & -11
end{pmatrix}
to\
begin{pmatrix}
1 & 0 & 3 & -3 \
0 & 1 & 1 & -3 \
0 & 0 & 3-k-(k+6) & -11 + 3(k+6)
end{pmatrix}
to \
begin{pmatrix}
1 & 0 & 3 & -3 \
0 & 1 & 1 & -3 \
0 & 0 & -3-2k& 7 + 3k
end{pmatrix}
$$
What happens in $-3-2k=0$? Can you finish this?
answered Dec 27 '18 at 16:23
gt6989bgt6989b
35k22557
answered Dec 27 '18 at 16:23
gt6989bgt6989b
35k22557
answered Dec 27 '18 at 16:23
gt6989bgt6989b
35k22557
answered Dec 27 '18 at 16:23
gt6989bgt6989b
35k22557
35k22557
add a comment |
add a comment |
$begingroup$
Hint: From the first equation we get
$$x=6+3y$$ plug this in the second equation we obtain
$$y+z=-3$$
and with the third equation we obtain
$$(6+k)y+(3-k)z=1$$
with $$y=-3-z$$ we obtain $$(6+k)(-3-z)+(3-k)z=-11$$
so we get
$$z(-k-3)=9+3k$$
Can you finish?
answered Dec 27 '18 at 16:25
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
add a comment |
$begingroup$
Hint: From the first equation we get
$$x=6+3y$$ plug this in the second equation we obtain
$$y+z=-3$$
and with the third equation we obtain
$$(6+k)y+(3-k)z=1$$
with $$y=-3-z$$ we obtain $$(6+k)(-3-z)+(3-k)z=-11$$
so we get
$$z(-k-3)=9+3k$$
Can you finish?
answered Dec 27 '18 at 16:25
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
add a comment |
$begingroup$
Hint: From the first equation we get
$$x=6+3y$$ plug this in the second equation we obtain
$$y+z=-3$$
and with the third equation we obtain
$$(6+k)y+(3-k)z=1$$
with $$y=-3-z$$ we obtain $$(6+k)(-3-z)+(3-k)z=-11$$
so we get
$$z(-k-3)=9+3k$$
Can you finish?
answered Dec 27 '18 at 16:25
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
Hint: From the first equation we get
$$x=6+3y$$ plug this in the second equation we obtain
$$y+z=-3$$
and with the third equation we obtain
$$(6+k)y+(3-k)z=1$$
with $$y=-3-z$$ we obtain $$(6+k)(-3-z)+(3-k)z=-11$$
so we get
$$z(-k-3)=9+3k$$
Can you finish?
answered Dec 27 '18 at 16:25
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Dec 27 '18 at 16:25
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Dec 27 '18 at 16:25
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Dec 27 '18 at 16:25
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
77.9k42866
add a comment |
add a comment |
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