Mixing
Solving Ax=b Using the Basis of the Nullspace
$begingroup$
Let B = begin{bmatrix}2&3&1&-1\1&2&1&2\3&5&2&1\4&7&3&3end{bmatrix}
Find the complete solution to the nonhomogenous system Bx=begin{bmatrix}6\-4\2\-2end{bmatrix} by first computing a basis for the nullspace of B.
x = $c_1$ + $sc_2$ + $tc_3$
Where $c_1$, $c_2$ and $c_3$ are vectors.
I know how to normally solve an Ax=b equation, by taking the inverse of A and multiplying by b, but I'm really not understanding the method that's being used here.
linear-algebra
asked Mar 19 '17 at 0:38
JanoyCresvaJanoyCresva
18811
$endgroup$
add a comment |
$begingroup$
Let B = begin{bmatrix}2&3&1&-1\1&2&1&2\3&5&2&1\4&7&3&3end{bmatrix}
Find the complete solution to the nonhomogenous system Bx=begin{bmatrix}6\-4\2\-2end{bmatrix} by first computing a basis for the nullspace of B.
x = $c_1$ + $sc_2$ + $tc_3$
Where $c_1$, $c_2$ and $c_3$ are vectors.
I know how to normally solve an Ax=b equation, by taking the inverse of A and multiplying by b, but I'm really not understanding the method that's being used here.
linear-algebra
asked Mar 19 '17 at 0:38
JanoyCresvaJanoyCresva
18811
$endgroup$
$begingroup$
That only works when the matrix is nonsingular. The idea is that if $c$ any solution to the equation, then so is $c+d$, where $d$ is any vector such that $Bd=0$, i.e., an element of the nullspace.
$endgroup$
– amd
Mar 19 '17 at 0:56
$begingroup$
Your method is for invertible matrices (i. e. Cramer systems of linear equations). In the general case, you have to perform a full row reduction on the augmented matrix.
$endgroup$
– Bernard
Mar 19 '17 at 0:57
$begingroup$
Since A is singular, $A^{-1}$ does not exist. So, how will you find a solution? Once you find 1 solution, call it $P$ and and find a basis for the kernel, then $P + $ any combination of vectors in the kernel is your solution set.
$endgroup$
– Doug M
Mar 19 '17 at 1:00
$begingroup$
So how would you find the first solution? I reduced the augmented matrix and got two free variables.
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:04
add a comment |
$begingroup$
Let B = begin{bmatrix}2&3&1&-1\1&2&1&2\3&5&2&1\4&7&3&3end{bmatrix}
Find the complete solution to the nonhomogenous system Bx=begin{bmatrix}6\-4\2\-2end{bmatrix} by first computing a basis for the nullspace of B.
x = $c_1$ + $sc_2$ + $tc_3$
Where $c_1$, $c_2$ and $c_3$ are vectors.
I know how to normally solve an Ax=b equation, by taking the inverse of A and multiplying by b, but I'm really not understanding the method that's being used here.
linear-algebra
asked Mar 19 '17 at 0:38
JanoyCresvaJanoyCresva
18811
$endgroup$
Let B = begin{bmatrix}2&3&1&-1\1&2&1&2\3&5&2&1\4&7&3&3end{bmatrix}
Find the complete solution to the nonhomogenous system Bx=begin{bmatrix}6\-4\2\-2end{bmatrix} by first computing a basis for the nullspace of B.
x = $c_1$ + $sc_2$ + $tc_3$
Where $c_1$, $c_2$ and $c_3$ are vectors.
I know how to normally solve an Ax=b equation, by taking the inverse of A and multiplying by b, but I'm really not understanding the method that's being used here.
linear-algebra
linear-algebra
asked Mar 19 '17 at 0:38
JanoyCresvaJanoyCresva
18811
asked Mar 19 '17 at 0:38
JanoyCresvaJanoyCresva
18811
asked Mar 19 '17 at 0:38
JanoyCresvaJanoyCresva
18811
asked Mar 19 '17 at 0:38
JanoyCresvaJanoyCresva
18811
asked Mar 19 '17 at 0:38
JanoyCresvaJanoyCresva
18811
18811
$begingroup$
That only works when the matrix is nonsingular. The idea is that if $c$ any solution to the equation, then so is $c+d$, where $d$ is any vector such that $Bd=0$, i.e., an element of the nullspace.
$endgroup$
– amd
Mar 19 '17 at 0:56
$begingroup$
Your method is for invertible matrices (i. e. Cramer systems of linear equations). In the general case, you have to perform a full row reduction on the augmented matrix.
$endgroup$
– Bernard
Mar 19 '17 at 0:57
$begingroup$
Since A is singular, $A^{-1}$ does not exist. So, how will you find a solution? Once you find 1 solution, call it $P$ and and find a basis for the kernel, then $P + $ any combination of vectors in the kernel is your solution set.
$endgroup$
– Doug M
Mar 19 '17 at 1:00
$begingroup$
So how would you find the first solution? I reduced the augmented matrix and got two free variables.
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:04
add a comment |
$begingroup$
That only works when the matrix is nonsingular. The idea is that if $c$ any solution to the equation, then so is $c+d$, where $d$ is any vector such that $Bd=0$, i.e., an element of the nullspace.
$endgroup$
– amd
Mar 19 '17 at 0:56
$begingroup$
Your method is for invertible matrices (i. e. Cramer systems of linear equations). In the general case, you have to perform a full row reduction on the augmented matrix.
$endgroup$
– Bernard
Mar 19 '17 at 0:57
$begingroup$
Since A is singular, $A^{-1}$ does not exist. So, how will you find a solution? Once you find 1 solution, call it $P$ and and find a basis for the kernel, then $P + $ any combination of vectors in the kernel is your solution set.
$endgroup$
– Doug M
Mar 19 '17 at 1:00
$begingroup$
So how would you find the first solution? I reduced the augmented matrix and got two free variables.
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:04
$begingroup$
That only works when the matrix is nonsingular. The idea is that if $c$ any solution to the equation, then so is $c+d$, where $d$ is any vector such that $Bd=0$, i.e., an element of the nullspace.
$endgroup$
– amd
Mar 19 '17 at 0:56
$begingroup$
That only works when the matrix is nonsingular. The idea is that if $c$ any solution to the equation, then so is $c+d$, where $d$ is any vector such that $Bd=0$, i.e., an element of the nullspace.
$endgroup$
– amd
Mar 19 '17 at 0:56
$begingroup$
Your method is for invertible matrices (i. e. Cramer systems of linear equations). In the general case, you have to perform a full row reduction on the augmented matrix.
$endgroup$
– Bernard
Mar 19 '17 at 0:57
$begingroup$
Your method is for invertible matrices (i. e. Cramer systems of linear equations). In the general case, you have to perform a full row reduction on the augmented matrix.
$endgroup$
– Bernard
Mar 19 '17 at 0:57
$begingroup$
Since A is singular, $A^{-1}$ does not exist. So, how will you find a solution? Once you find 1 solution, call it $P$ and and find a basis for the kernel, then $P + $ any combination of vectors in the kernel is your solution set.
$endgroup$
– Doug M
Mar 19 '17 at 1:00
$begingroup$
Since A is singular, $A^{-1}$ does not exist. So, how will you find a solution? Once you find 1 solution, call it $P$ and and find a basis for the kernel, then $P + $ any combination of vectors in the kernel is your solution set.
$endgroup$
– Doug M
Mar 19 '17 at 1:00
$begingroup$
So how would you find the first solution? I reduced the augmented matrix and got two free variables.
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:04
$begingroup$
So how would you find the first solution? I reduced the augmented matrix and got two free variables.
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Start with the reduced row echelon form:
$$
begin{align}
mathbf{B} &mapsto mathbf{E}_{mathbf{B}} \
%
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
%
&mapsto
%
left[
begin{array}{rrrr}
color{blue}{1} & 0 & -1 & -8 \
0 & color{blue}{1} & 1 & 5 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
end{array}
right]
end{align}
$$
The message is that the fundamental columns are the first two, highlighted by blue coloring.
Your question is look for the values of $alpha$ and $beta$ which solve
$$
left[
begin{array}{r}
6 \ -4 \ 2 \ -2
end{array}
right]
%
=
%
alpha
left[
begin{array}{r}
2 \ 1 \ 3 \ 4
end{array}
right]
%
+
%
beta
left[
begin{array}{r}
3 \ 2 \ 5 \ 7
end{array}
right]
$$
We are look for two unknowns, so solve using two rows. Let's pick rows 1 and 3, liberating us from the chance of sign errors. The problem is
now
$$
%
left[
begin{array}{rr}
2 & 3 \
3 & 5
end{array}
right]
%
left[
begin{array}{c}
alpha \
beta
end{array}
right]
%
=
left[
begin{array}{c}
6 \
2
end{array}
right]
%%
quad Rightarrow quad
%
left[
begin{array}{c}
alpha \
beta
end{array}
right]
=
%
left[
begin{array}{rr}
5 & -3 \
-3 & 2
end{array}
right]
%
left[
begin{array}{c}
6 \
2
end{array}
right]
%
=
%
left[
begin{array}{r}
24 \
-14
end{array}
right]
$$
It doesn't seem that we need to resolve the nullspace. But it is a byproduct of the augmented reduction producing $mathbf{E}_{mathbf{B}}$:
$$
mathcal{N} left( mathbf{B}^{mathrm{T}} right) =
text{span } left{ ,
%
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right], ,
%
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
%
, right}
$$
Solution
The solution to
$$
begin{align}
mathbf{B} x &= y \
%%
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
%
left[
begin{array}{c}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{array}
right]
%
&=
%
left[
begin{array}{r}
6 \
-4 \
2 \
-2
end{array}
right]
%
end{align}
$$
is
$$
%
left[
begin{array}{c}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{array}
right]
%
=
left[
begin{array}{r}
24 \
-14 \
0 \
0
end{array}
right]
%
+
%
s
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right]
%
+
%
t
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
$$
where the constants $s$ and $t$ are arbitrary.
To emphasize:
$$
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
left(
left[
begin{array}{r}
24 \
-14 \
0 \
0
end{array}
right]
%
+
%
s
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right]
%
+
%
t
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
right)
%
=
%
left[
begin{array}{r}
6 \
-4 \
2 \
-2
end{array}
right]
$$
answered Mar 19 '17 at 1:15
dantopadantopa
6,46942243
$endgroup$
$begingroup$
I'm confused what you are solving for because the question was asking for 3 vectors in $R^4$. I don't think I made it clear but it was asking for $c_1, c_2$ and $c_3$
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:23
$begingroup$
@JanoyCresva: Edited solution to be more explicit.
$endgroup$
– dantopa
Mar 19 '17 at 1:42
$begingroup$
It's saying these answers are incorrect, I don't know why
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:44
$begingroup$
So to get the solution you'd multiply the three vectors in the parentheses by the original matrix?
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:49
$begingroup$
Why didn’t you simply augment the matrix before row-reduction instead of going through all those extra steps? This would’ve given you the correct $alpha$ and $beta$ immediately.
$endgroup$
– amd
Mar 19 '17 at 2:38
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Start with the reduced row echelon form:
$$
begin{align}
mathbf{B} &mapsto mathbf{E}_{mathbf{B}} \
%
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
%
&mapsto
%
left[
begin{array}{rrrr}
color{blue}{1} & 0 & -1 & -8 \
0 & color{blue}{1} & 1 & 5 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
end{array}
right]
end{align}
$$
The message is that the fundamental columns are the first two, highlighted by blue coloring.
Your question is look for the values of $alpha$ and $beta$ which solve
$$
left[
begin{array}{r}
6 \ -4 \ 2 \ -2
end{array}
right]
%
=
%
alpha
left[
begin{array}{r}
2 \ 1 \ 3 \ 4
end{array}
right]
%
+
%
beta
left[
begin{array}{r}
3 \ 2 \ 5 \ 7
end{array}
right]
$$
We are look for two unknowns, so solve using two rows. Let's pick rows 1 and 3, liberating us from the chance of sign errors. The problem is
now
$$
%
left[
begin{array}{rr}
2 & 3 \
3 & 5
end{array}
right]
%
left[
begin{array}{c}
alpha \
beta
end{array}
right]
%
=
left[
begin{array}{c}
6 \
2
end{array}
right]
%%
quad Rightarrow quad
%
left[
begin{array}{c}
alpha \
beta
end{array}
right]
=
%
left[
begin{array}{rr}
5 & -3 \
-3 & 2
end{array}
right]
%
left[
begin{array}{c}
6 \
2
end{array}
right]
%
=
%
left[
begin{array}{r}
24 \
-14
end{array}
right]
$$
It doesn't seem that we need to resolve the nullspace. But it is a byproduct of the augmented reduction producing $mathbf{E}_{mathbf{B}}$:
$$
mathcal{N} left( mathbf{B}^{mathrm{T}} right) =
text{span } left{ ,
%
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right], ,
%
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
%
, right}
$$
Solution
The solution to
$$
begin{align}
mathbf{B} x &= y \
%%
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
%
left[
begin{array}{c}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{array}
right]
%
&=
%
left[
begin{array}{r}
6 \
-4 \
2 \
-2
end{array}
right]
%
end{align}
$$
is
$$
%
left[
begin{array}{c}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{array}
right]
%
=
left[
begin{array}{r}
24 \
-14 \
0 \
0
end{array}
right]
%
+
%
s
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right]
%
+
%
t
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
$$
where the constants $s$ and $t$ are arbitrary.
To emphasize:
$$
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
left(
left[
begin{array}{r}
24 \
-14 \
0 \
0
end{array}
right]
%
+
%
s
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right]
%
+
%
t
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
right)
%
=
%
left[
begin{array}{r}
6 \
-4 \
2 \
-2
end{array}
right]
$$
answered Mar 19 '17 at 1:15
dantopadantopa
6,46942243
$endgroup$
$begingroup$
I'm confused what you are solving for because the question was asking for 3 vectors in $R^4$. I don't think I made it clear but it was asking for $c_1, c_2$ and $c_3$
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:23
$begingroup$
@JanoyCresva: Edited solution to be more explicit.
$endgroup$
– dantopa
Mar 19 '17 at 1:42
$begingroup$
It's saying these answers are incorrect, I don't know why
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:44
$begingroup$
So to get the solution you'd multiply the three vectors in the parentheses by the original matrix?
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:49
$begingroup$
Why didn’t you simply augment the matrix before row-reduction instead of going through all those extra steps? This would’ve given you the correct $alpha$ and $beta$ immediately.
$endgroup$
– amd
Mar 19 '17 at 2:38
|
show 2 more comments
$begingroup$
Start with the reduced row echelon form:
$$
begin{align}
mathbf{B} &mapsto mathbf{E}_{mathbf{B}} \
%
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
%
&mapsto
%
left[
begin{array}{rrrr}
color{blue}{1} & 0 & -1 & -8 \
0 & color{blue}{1} & 1 & 5 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
end{array}
right]
end{align}
$$
The message is that the fundamental columns are the first two, highlighted by blue coloring.
Your question is look for the values of $alpha$ and $beta$ which solve
$$
left[
begin{array}{r}
6 \ -4 \ 2 \ -2
end{array}
right]
%
=
%
alpha
left[
begin{array}{r}
2 \ 1 \ 3 \ 4
end{array}
right]
%
+
%
beta
left[
begin{array}{r}
3 \ 2 \ 5 \ 7
end{array}
right]
$$
We are look for two unknowns, so solve using two rows. Let's pick rows 1 and 3, liberating us from the chance of sign errors. The problem is
now
$$
%
left[
begin{array}{rr}
2 & 3 \
3 & 5
end{array}
right]
%
left[
begin{array}{c}
alpha \
beta
end{array}
right]
%
=
left[
begin{array}{c}
6 \
2
end{array}
right]
%%
quad Rightarrow quad
%
left[
begin{array}{c}
alpha \
beta
end{array}
right]
=
%
left[
begin{array}{rr}
5 & -3 \
-3 & 2
end{array}
right]
%
left[
begin{array}{c}
6 \
2
end{array}
right]
%
=
%
left[
begin{array}{r}
24 \
-14
end{array}
right]
$$
It doesn't seem that we need to resolve the nullspace. But it is a byproduct of the augmented reduction producing $mathbf{E}_{mathbf{B}}$:
$$
mathcal{N} left( mathbf{B}^{mathrm{T}} right) =
text{span } left{ ,
%
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right], ,
%
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
%
, right}
$$
Solution
The solution to
$$
begin{align}
mathbf{B} x &= y \
%%
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
%
left[
begin{array}{c}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{array}
right]
%
&=
%
left[
begin{array}{r}
6 \
-4 \
2 \
-2
end{array}
right]
%
end{align}
$$
is
$$
%
left[
begin{array}{c}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{array}
right]
%
=
left[
begin{array}{r}
24 \
-14 \
0 \
0
end{array}
right]
%
+
%
s
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right]
%
+
%
t
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
$$
where the constants $s$ and $t$ are arbitrary.
To emphasize:
$$
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
left(
left[
begin{array}{r}
24 \
-14 \
0 \
0
end{array}
right]
%
+
%
s
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right]
%
+
%
t
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
right)
%
=
%
left[
begin{array}{r}
6 \
-4 \
2 \
-2
end{array}
right]
$$
answered Mar 19 '17 at 1:15
dantopadantopa
6,46942243
$endgroup$
$begingroup$
I'm confused what you are solving for because the question was asking for 3 vectors in $R^4$. I don't think I made it clear but it was asking for $c_1, c_2$ and $c_3$
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:23
$begingroup$
@JanoyCresva: Edited solution to be more explicit.
$endgroup$
– dantopa
Mar 19 '17 at 1:42
$begingroup$
It's saying these answers are incorrect, I don't know why
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:44
$begingroup$
So to get the solution you'd multiply the three vectors in the parentheses by the original matrix?
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:49
$begingroup$
Why didn’t you simply augment the matrix before row-reduction instead of going through all those extra steps? This would’ve given you the correct $alpha$ and $beta$ immediately.
$endgroup$
– amd
Mar 19 '17 at 2:38
|
show 2 more comments
$begingroup$
Start with the reduced row echelon form:
$$
begin{align}
mathbf{B} &mapsto mathbf{E}_{mathbf{B}} \
%
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
%
&mapsto
%
left[
begin{array}{rrrr}
color{blue}{1} & 0 & -1 & -8 \
0 & color{blue}{1} & 1 & 5 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
end{array}
right]
end{align}
$$
The message is that the fundamental columns are the first two, highlighted by blue coloring.
Your question is look for the values of $alpha$ and $beta$ which solve
$$
left[
begin{array}{r}
6 \ -4 \ 2 \ -2
end{array}
right]
%
=
%
alpha
left[
begin{array}{r}
2 \ 1 \ 3 \ 4
end{array}
right]
%
+
%
beta
left[
begin{array}{r}
3 \ 2 \ 5 \ 7
end{array}
right]
$$
We are look for two unknowns, so solve using two rows. Let's pick rows 1 and 3, liberating us from the chance of sign errors. The problem is
now
$$
%
left[
begin{array}{rr}
2 & 3 \
3 & 5
end{array}
right]
%
left[
begin{array}{c}
alpha \
beta
end{array}
right]
%
=
left[
begin{array}{c}
6 \
2
end{array}
right]
%%
quad Rightarrow quad
%
left[
begin{array}{c}
alpha \
beta
end{array}
right]
=
%
left[
begin{array}{rr}
5 & -3 \
-3 & 2
end{array}
right]
%
left[
begin{array}{c}
6 \
2
end{array}
right]
%
=
%
left[
begin{array}{r}
24 \
-14
end{array}
right]
$$
It doesn't seem that we need to resolve the nullspace. But it is a byproduct of the augmented reduction producing $mathbf{E}_{mathbf{B}}$:
$$
mathcal{N} left( mathbf{B}^{mathrm{T}} right) =
text{span } left{ ,
%
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right], ,
%
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
%
, right}
$$
Solution
The solution to
$$
begin{align}
mathbf{B} x &= y \
%%
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
%
left[
begin{array}{c}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{array}
right]
%
&=
%
left[
begin{array}{r}
6 \
-4 \
2 \
-2
end{array}
right]
%
end{align}
$$
is
$$
%
left[
begin{array}{c}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{array}
right]
%
=
left[
begin{array}{r}
24 \
-14 \
0 \
0
end{array}
right]
%
+
%
s
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right]
%
+
%
t
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
$$
where the constants $s$ and $t$ are arbitrary.
To emphasize:
$$
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
left(
left[
begin{array}{r}
24 \
-14 \
0 \
0
end{array}
right]
%
+
%
s
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right]
%
+
%
t
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
right)
%
=
%
left[
begin{array}{r}
6 \
-4 \
2 \
-2
end{array}
right]
$$
answered Mar 19 '17 at 1:15
dantopadantopa
6,46942243
$endgroup$
Start with the reduced row echelon form:
$$
begin{align}
mathbf{B} &mapsto mathbf{E}_{mathbf{B}} \
%
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
%
&mapsto
%
left[
begin{array}{rrrr}
color{blue}{1} & 0 & -1 & -8 \
0 & color{blue}{1} & 1 & 5 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
end{array}
right]
end{align}
$$
The message is that the fundamental columns are the first two, highlighted by blue coloring.
Your question is look for the values of $alpha$ and $beta$ which solve
$$
left[
begin{array}{r}
6 \ -4 \ 2 \ -2
end{array}
right]
%
=
%
alpha
left[
begin{array}{r}
2 \ 1 \ 3 \ 4
end{array}
right]
%
+
%
beta
left[
begin{array}{r}
3 \ 2 \ 5 \ 7
end{array}
right]
$$
We are look for two unknowns, so solve using two rows. Let's pick rows 1 and 3, liberating us from the chance of sign errors. The problem is
now
$$
%
left[
begin{array}{rr}
2 & 3 \
3 & 5
end{array}
right]
%
left[
begin{array}{c}
alpha \
beta
end{array}
right]
%
=
left[
begin{array}{c}
6 \
2
end{array}
right]
%%
quad Rightarrow quad
%
left[
begin{array}{c}
alpha \
beta
end{array}
right]
=
%
left[
begin{array}{rr}
5 & -3 \
-3 & 2
end{array}
right]
%
left[
begin{array}{c}
6 \
2
end{array}
right]
%
=
%
left[
begin{array}{r}
24 \
-14
end{array}
right]
$$
It doesn't seem that we need to resolve the nullspace. But it is a byproduct of the augmented reduction producing $mathbf{E}_{mathbf{B}}$:
$$
mathcal{N} left( mathbf{B}^{mathrm{T}} right) =
text{span } left{ ,
%
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right], ,
%
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
%
, right}
$$
Solution
The solution to
$$
begin{align}
mathbf{B} x &= y \
%%
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
%
left[
begin{array}{c}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{array}
right]
%
&=
%
left[
begin{array}{r}
6 \
-4 \
2 \
-2
end{array}
right]
%
end{align}
$$
is
$$
%
left[
begin{array}{c}
x_{1} \
x_{2} \
x_{3} \
x_{4}
end{array}
right]
%
=
left[
begin{array}{r}
24 \
-14 \
0 \
0
end{array}
right]
%
+
%
s
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right]
%
+
%
t
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
$$
where the constants $s$ and $t$ are arbitrary.
To emphasize:
$$
left[
begin{array}{rrrr}
2 & 3 & 1 & -1 \
1 & 2 & 1 & 2 \
3 & 5 & 2 & 1 \
4 & 7 & 3 & 3 \
end{array}
right]
left(
left[
begin{array}{r}
24 \
-14 \
0 \
0
end{array}
right]
%
+
%
s
left[
begin{array}{r}
8 \
-5 \
0 \
1
end{array}
right]
%
+
%
t
left[
begin{array}{r}
-1 \
-1 \
1 \
0
end{array}
right]
right)
%
=
%
left[
begin{array}{r}
6 \
-4 \
2 \
-2
end{array}
right]
$$
answered Mar 19 '17 at 1:15
dantopadantopa
6,46942243
edited Mar 19 '17 at 1:36
answered Mar 19 '17 at 1:15
dantopadantopa
6,46942243
answered Mar 19 '17 at 1:15
dantopadantopa
6,46942243
answered Mar 19 '17 at 1:15
dantopadantopa
6,46942243
6,46942243
$begingroup$
I'm confused what you are solving for because the question was asking for 3 vectors in $R^4$. I don't think I made it clear but it was asking for $c_1, c_2$ and $c_3$
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:23
$begingroup$
@JanoyCresva: Edited solution to be more explicit.
$endgroup$
– dantopa
Mar 19 '17 at 1:42
$begingroup$
It's saying these answers are incorrect, I don't know why
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:44
$begingroup$
So to get the solution you'd multiply the three vectors in the parentheses by the original matrix?
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:49
$begingroup$
Why didn’t you simply augment the matrix before row-reduction instead of going through all those extra steps? This would’ve given you the correct $alpha$ and $beta$ immediately.
$endgroup$
– amd
Mar 19 '17 at 2:38
|
show 2 more comments
$begingroup$
I'm confused what you are solving for because the question was asking for 3 vectors in $R^4$. I don't think I made it clear but it was asking for $c_1, c_2$ and $c_3$
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:23
$begingroup$
@JanoyCresva: Edited solution to be more explicit.
$endgroup$
– dantopa
Mar 19 '17 at 1:42
$begingroup$
It's saying these answers are incorrect, I don't know why
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:44
$begingroup$
So to get the solution you'd multiply the three vectors in the parentheses by the original matrix?
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:49
$begingroup$
Why didn’t you simply augment the matrix before row-reduction instead of going through all those extra steps? This would’ve given you the correct $alpha$ and $beta$ immediately.
$endgroup$
– amd
Mar 19 '17 at 2:38
$begingroup$
I'm confused what you are solving for because the question was asking for 3 vectors in $R^4$. I don't think I made it clear but it was asking for $c_1, c_2$ and $c_3$
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:23
$begingroup$
I'm confused what you are solving for because the question was asking for 3 vectors in $R^4$. I don't think I made it clear but it was asking for $c_1, c_2$ and $c_3$
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:23
$begingroup$
@JanoyCresva: Edited solution to be more explicit.
$endgroup$
– dantopa
Mar 19 '17 at 1:42
$begingroup$
@JanoyCresva: Edited solution to be more explicit.
$endgroup$
– dantopa
Mar 19 '17 at 1:42
$begingroup$
It's saying these answers are incorrect, I don't know why
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:44
$begingroup$
It's saying these answers are incorrect, I don't know why
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:44
$begingroup$
So to get the solution you'd multiply the three vectors in the parentheses by the original matrix?
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:49
$begingroup$
So to get the solution you'd multiply the three vectors in the parentheses by the original matrix?
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:49
$begingroup$
Why didn’t you simply augment the matrix before row-reduction instead of going through all those extra steps? This would’ve given you the correct $alpha$ and $beta$ immediately.
$endgroup$
– amd
Mar 19 '17 at 2:38
$begingroup$
Why didn’t you simply augment the matrix before row-reduction instead of going through all those extra steps? This would’ve given you the correct $alpha$ and $beta$ immediately.
$endgroup$
– amd
Mar 19 '17 at 2:38
|
show 2 more comments
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$begingroup$
That only works when the matrix is nonsingular. The idea is that if $c$ any solution to the equation, then so is $c+d$, where $d$ is any vector such that $Bd=0$, i.e., an element of the nullspace.
$endgroup$
– amd
Mar 19 '17 at 0:56
$begingroup$
Your method is for invertible matrices (i. e. Cramer systems of linear equations). In the general case, you have to perform a full row reduction on the augmented matrix.
$endgroup$
– Bernard
Mar 19 '17 at 0:57
$begingroup$
Since A is singular, $A^{-1}$ does not exist. So, how will you find a solution? Once you find 1 solution, call it $P$ and and find a basis for the kernel, then $P + $ any combination of vectors in the kernel is your solution set.
$endgroup$
– Doug M
Mar 19 '17 at 1:00
$begingroup$
So how would you find the first solution? I reduced the augmented matrix and got two free variables.
$endgroup$
– JanoyCresva
Mar 19 '17 at 1:04