Describe all solutions of $Ax=0$
The question is:
Let $A=$ begin{bmatrix}
1 & 1 & 1 & 5 \
2 & 2 & 2 & 10 \
end{bmatrix}
Describe all solutions of $Ax=0$.
Immediately I can recognize that one of the rows can be turned into a row of zeroes: begin{bmatrix}
1 & 1 & 1 & 5 \
0 & 0 & 0 & 0 \
end{bmatrix}
But I get stuck here. I get that $x=-x_2-x_3-5x_4$, but how do I go from there?
linear-algebra matrices
edited Nov 27 at 6:01
Robert Howard
1,9161822
asked Nov 27 at 4:02
Sami Jr
72
add a comment |
The question is:
Let $A=$ begin{bmatrix}
1 & 1 & 1 & 5 \
2 & 2 & 2 & 10 \
end{bmatrix}
Describe all solutions of $Ax=0$.
Immediately I can recognize that one of the rows can be turned into a row of zeroes: begin{bmatrix}
1 & 1 & 1 & 5 \
0 & 0 & 0 & 0 \
end{bmatrix}
But I get stuck here. I get that $x=-x_2-x_3-5x_4$, but how do I go from there?
linear-algebra matrices
edited Nov 27 at 6:01
Robert Howard
1,9161822
asked Nov 27 at 4:02
Sami Jr
72
add a comment |
The question is:
Let $A=$ begin{bmatrix}
1 & 1 & 1 & 5 \
2 & 2 & 2 & 10 \
end{bmatrix}
Describe all solutions of $Ax=0$.
Immediately I can recognize that one of the rows can be turned into a row of zeroes: begin{bmatrix}
1 & 1 & 1 & 5 \
0 & 0 & 0 & 0 \
end{bmatrix}
But I get stuck here. I get that $x=-x_2-x_3-5x_4$, but how do I go from there?
linear-algebra matrices
edited Nov 27 at 6:01
Robert Howard
1,9161822
asked Nov 27 at 4:02
Sami Jr
72
The question is:
Let $A=$ begin{bmatrix}
1 & 1 & 1 & 5 \
2 & 2 & 2 & 10 \
end{bmatrix}
Describe all solutions of $Ax=0$.
Immediately I can recognize that one of the rows can be turned into a row of zeroes: begin{bmatrix}
1 & 1 & 1 & 5 \
0 & 0 & 0 & 0 \
end{bmatrix}
But I get stuck here. I get that $x=-x_2-x_3-5x_4$, but how do I go from there?
linear-algebra matrices
linear-algebra matrices
edited Nov 27 at 6:01
Robert Howard
1,9161822
asked Nov 27 at 4:02
Sami Jr
72
edited Nov 27 at 6:01
Robert Howard
1,9161822
asked Nov 27 at 4:02
Sami Jr
72
edited Nov 27 at 6:01
Robert Howard
1,9161822
edited Nov 27 at 6:01
Robert Howard
1,9161822
edited Nov 27 at 6:01
Robert Howard
1,9161822
1,9161822
asked Nov 27 at 4:02
Sami Jr
72
asked Nov 27 at 4:02
Sami Jr
72
asked Nov 27 at 4:02
Sami Jr
72
72
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You can choose $x_2,x_3$ and $x_4$ freely, and they will determine $x_1$.
The solutions look like $begin{pmatrix} -s-t-5u\s\t\uend{pmatrix}$. Or ${sbegin{pmatrix}-1\1\0\0end{pmatrix}+tbegin{pmatrix}-1\0\1\0end{pmatrix}+ubegin{pmatrix}-5\0\0\1end{pmatrix}mid s,t,uin mathbb R}$.
Since there are $3$ free variables, this kernel is $3$ dimensional.
answered Nov 27 at 4:18
Chris Custer
10.8k3724
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can choose $x_2,x_3$ and $x_4$ freely, and they will determine $x_1$.
The solutions look like $begin{pmatrix} -s-t-5u\s\t\uend{pmatrix}$. Or ${sbegin{pmatrix}-1\1\0\0end{pmatrix}+tbegin{pmatrix}-1\0\1\0end{pmatrix}+ubegin{pmatrix}-5\0\0\1end{pmatrix}mid s,t,uin mathbb R}$.
Since there are $3$ free variables, this kernel is $3$ dimensional.
answered Nov 27 at 4:18
Chris Custer
10.8k3724
add a comment |
You can choose $x_2,x_3$ and $x_4$ freely, and they will determine $x_1$.
The solutions look like $begin{pmatrix} -s-t-5u\s\t\uend{pmatrix}$. Or ${sbegin{pmatrix}-1\1\0\0end{pmatrix}+tbegin{pmatrix}-1\0\1\0end{pmatrix}+ubegin{pmatrix}-5\0\0\1end{pmatrix}mid s,t,uin mathbb R}$.
Since there are $3$ free variables, this kernel is $3$ dimensional.
answered Nov 27 at 4:18
Chris Custer
10.8k3724
add a comment |
You can choose $x_2,x_3$ and $x_4$ freely, and they will determine $x_1$.
The solutions look like $begin{pmatrix} -s-t-5u\s\t\uend{pmatrix}$. Or ${sbegin{pmatrix}-1\1\0\0end{pmatrix}+tbegin{pmatrix}-1\0\1\0end{pmatrix}+ubegin{pmatrix}-5\0\0\1end{pmatrix}mid s,t,uin mathbb R}$.
Since there are $3$ free variables, this kernel is $3$ dimensional.
answered Nov 27 at 4:18
Chris Custer
10.8k3724
You can choose $x_2,x_3$ and $x_4$ freely, and they will determine $x_1$.
The solutions look like $begin{pmatrix} -s-t-5u\s\t\uend{pmatrix}$. Or ${sbegin{pmatrix}-1\1\0\0end{pmatrix}+tbegin{pmatrix}-1\0\1\0end{pmatrix}+ubegin{pmatrix}-5\0\0\1end{pmatrix}mid s,t,uin mathbb R}$.
Since there are $3$ free variables, this kernel is $3$ dimensional.
answered Nov 27 at 4:18
Chris Custer
10.8k3724
answered Nov 27 at 4:18
Chris Custer
10.8k3724
answered Nov 27 at 4:18
Chris Custer
10.8k3724
answered Nov 27 at 4:18
Chris Custer
10.8k3724
10.8k3724
add a comment |
add a comment |
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