Mixing
How to bring a Matrix to an advanced Coefficient matrix in row form?
$begingroup$
I just found this website on the Internet and you are really my last chance to help me with one task.
For $ a in mathbb{R}$, let the following linear equation system be given:
$$
begin{array}{rcrcrcc}
(a+1)x_{1} & + & (-a^2+6a-9)x_{2} & + & (a-2)x_{3} & = & 1\
(a+1)(a-3)x_{1} & + & (a^2-6a+9)x_{2} & + & 3x_{3} & = & a-3\
(a+1)x_{1} & + & (-a^2+6a-9)x_{2} & + & (a+1)x_{3} & = & 1\
end{array}
$$
Now I should bring that in the Matrix form $mathbf{A}x=b$, which isn't a problem.
$$
left(begin{array}{ccc}
(a+1) & (-a^2+6a-9) & (a-2) \
(a+1)(a-3) & (a^2-6a+9) & 3 \
(a+1) & (-a^2+6a-9) & (a+1)
end{array}right)
begin{pmatrix}
x_{1} \
x_{2} \
x_{3}
end{pmatrix}
=
begin{pmatrix}
1 \
a-3 \
1
end{pmatrix}
$$
But now I should bring it to the advanced Coefficient matrix in row form.
Can someone explain to me how to do it?
Thank you very much!
linear-algebra matrices
edited Jan 31 at 19:23
dantopa
6,65442245
asked Jan 30 at 20:14
AliceAlice
11
$endgroup$
add a comment |
$begingroup$
I just found this website on the Internet and you are really my last chance to help me with one task.
For $ a in mathbb{R}$, let the following linear equation system be given:
$$
begin{array}{rcrcrcc}
(a+1)x_{1} & + & (-a^2+6a-9)x_{2} & + & (a-2)x_{3} & = & 1\
(a+1)(a-3)x_{1} & + & (a^2-6a+9)x_{2} & + & 3x_{3} & = & a-3\
(a+1)x_{1} & + & (-a^2+6a-9)x_{2} & + & (a+1)x_{3} & = & 1\
end{array}
$$
Now I should bring that in the Matrix form $mathbf{A}x=b$, which isn't a problem.
$$
left(begin{array}{ccc}
(a+1) & (-a^2+6a-9) & (a-2) \
(a+1)(a-3) & (a^2-6a+9) & 3 \
(a+1) & (-a^2+6a-9) & (a+1)
end{array}right)
begin{pmatrix}
x_{1} \
x_{2} \
x_{3}
end{pmatrix}
=
begin{pmatrix}
1 \
a-3 \
1
end{pmatrix}
$$
But now I should bring it to the advanced Coefficient matrix in row form.
Can someone explain to me how to do it?
Thank you very much!
linear-algebra matrices
edited Jan 31 at 19:23
dantopa
6,65442245
asked Jan 30 at 20:14
AliceAlice
11
$endgroup$
$begingroup$
Thank you, I saw that, i just need a hint how to bring it to the coefficient matrix
$endgroup$
– Alice
Jan 31 at 8:36
add a comment |
$begingroup$
I just found this website on the Internet and you are really my last chance to help me with one task.
For $ a in mathbb{R}$, let the following linear equation system be given:
$$
begin{array}{rcrcrcc}
(a+1)x_{1} & + & (-a^2+6a-9)x_{2} & + & (a-2)x_{3} & = & 1\
(a+1)(a-3)x_{1} & + & (a^2-6a+9)x_{2} & + & 3x_{3} & = & a-3\
(a+1)x_{1} & + & (-a^2+6a-9)x_{2} & + & (a+1)x_{3} & = & 1\
end{array}
$$
Now I should bring that in the Matrix form $mathbf{A}x=b$, which isn't a problem.
$$
left(begin{array}{ccc}
(a+1) & (-a^2+6a-9) & (a-2) \
(a+1)(a-3) & (a^2-6a+9) & 3 \
(a+1) & (-a^2+6a-9) & (a+1)
end{array}right)
begin{pmatrix}
x_{1} \
x_{2} \
x_{3}
end{pmatrix}
=
begin{pmatrix}
1 \
a-3 \
1
end{pmatrix}
$$
But now I should bring it to the advanced Coefficient matrix in row form.
Can someone explain to me how to do it?
Thank you very much!
linear-algebra matrices
edited Jan 31 at 19:23
dantopa
6,65442245
asked Jan 30 at 20:14
AliceAlice
11
$endgroup$
I just found this website on the Internet and you are really my last chance to help me with one task.
For $ a in mathbb{R}$, let the following linear equation system be given:
$$
begin{array}{rcrcrcc}
(a+1)x_{1} & + & (-a^2+6a-9)x_{2} & + & (a-2)x_{3} & = & 1\
(a+1)(a-3)x_{1} & + & (a^2-6a+9)x_{2} & + & 3x_{3} & = & a-3\
(a+1)x_{1} & + & (-a^2+6a-9)x_{2} & + & (a+1)x_{3} & = & 1\
end{array}
$$
Now I should bring that in the Matrix form $mathbf{A}x=b$, which isn't a problem.
$$
left(begin{array}{ccc}
(a+1) & (-a^2+6a-9) & (a-2) \
(a+1)(a-3) & (a^2-6a+9) & 3 \
(a+1) & (-a^2+6a-9) & (a+1)
end{array}right)
begin{pmatrix}
x_{1} \
x_{2} \
x_{3}
end{pmatrix}
=
begin{pmatrix}
1 \
a-3 \
1
end{pmatrix}
$$
But now I should bring it to the advanced Coefficient matrix in row form.
Can someone explain to me how to do it?
Thank you very much!
linear-algebra matrices
linear-algebra matrices
edited Jan 31 at 19:23
dantopa
6,65442245
asked Jan 30 at 20:14
AliceAlice
11
edited Jan 31 at 19:23
dantopa
6,65442245
asked Jan 30 at 20:14
AliceAlice
11
edited Jan 31 at 19:23
dantopa
6,65442245
edited Jan 31 at 19:23
dantopa
6,65442245
edited Jan 31 at 19:23
dantopa
6,65442245
6,65442245
asked Jan 30 at 20:14
AliceAlice
11
asked Jan 30 at 20:14
AliceAlice
11
asked Jan 30 at 20:14
AliceAlice
11
11
$begingroup$
Thank you, I saw that, i just need a hint how to bring it to the coefficient matrix
$endgroup$
– Alice
Jan 31 at 8:36
add a comment |
$begingroup$
Thank you, I saw that, i just need a hint how to bring it to the coefficient matrix
$endgroup$
– Alice
Jan 31 at 8:36
$begingroup$
Thank you, I saw that, i just need a hint how to bring it to the coefficient matrix
$endgroup$
– Alice
Jan 31 at 8:36
$begingroup$
Thank you, I saw that, i just need a hint how to bring it to the coefficient matrix
$endgroup$
– Alice
Jan 31 at 8:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is a situation where Cramer's Rule applies quite nicely.
First, note that
$$
det(A)
= 3 , {left(a + 1right)} {left(a - 2right)} {left(a - 3right)}^{2}
$$
So, $A$ is invertible as long as $anotin{-1, 2, 3}$.
Next, define $A_i$ as the matrix obtained by replacing the $i$th column of $A$ with the vector $vec{b}=leftlangle1,,a - 3,,1rightrangle$. This gives
$$
begin{align*}
det(A_1) &= 3 , {left(a - 2right)} {left(a - 3right)}^{2} &
det(A_2) &= 0 &
det(A_3) &= 0
end{align*}
$$
By Cramer's Rule, we have
$$
begin{align*}
x_1 &= frac{det(A_1)}{det(A)} = frac{1}{a + 1} &
x_2 &= frac{det(A_2)}{det(A)} = 0 &
x_3 &= frac{det(A_3)}{det(A)} = 0
end{align*}
$$
Of course, we should also account for what happens if $ain{-1, 2, 3}$.
For $a=-1$, we can row-reduce the system to obtain
$$
operatorname{rref}left[begin{array}{rrr|r}
0 & -16 & -3 & 1 \
0 & 16 & 3 & -4 \
0 & -16 & 0 & 1
end{array}right]
=left[begin{array}{rrr|r}
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{array}right]
$$
For $a=2$, we have
$$
operatorname{rref}left[begin{array}{rrr|r}
3 & -1 & 0 & 1 \
-3 & 1 & 3 & -1 \
3 & -1 & 3 & 1
end{array}right]
=left[begin{array}{rrr|r}
1 & -frac{1}{3} & 0 & frac{1}{3} \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0
end{array}right]
$$
For $a=3$, we have
$$
operatorname{rref}left[begin{array}{rrr|r}
4 & 0 & 1 & 1 \
0 & 0 & 3 & 0 \
4 & 0 & 4 & 1
end{array}right]
=left[begin{array}{rrr|r}
1 & 0 & 0 & frac{1}{4} \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0
end{array}right]
$$
answered Jan 31 at 22:39
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
$endgroup$
add a comment |
$begingroup$
Problem statement
Solve the linear system $mathbf{A}x=b$:
$$
left[begin{array}{crc}
a+1 & -(a-3)^2 & a-2 \
(a-3) (a+1) & (a-3)^2 & 3 \
a+1 & -(a-3)^2 & a+1
end{array}right]
left[begin{array}{c}
x_{1} \ x_{2} \x_{3}
end{array}right]
=
left[begin{array}{c}
1 \ a-3 \ 1
end{array}right].
$$
Gaussian Elimination
Two reduction steps are used resulting in
$$
mathbf{E}_{2}mathbf{E}_{1}mathbf{A}x=mathbf{E}_{2}mathbf{E}_{1}b
$$
where
$$
mathbf{E}_{1} =
left[
begin{array}{ccc}
frac{1}{a+1} & 0 & 0 \
0 & frac{1}{a+1} & 0 \
0 & 0 & frac{1}{a+1} \
end{array}
right], qquad
mathbf{E}_{2} =
left[
begin{array}{rcc}
1 & 0 & 0 \
-1 & frac{1}{a-3} & 0 \
-1 & 0 & 1 \
end{array}
right].
$$
Solution via back substitution
The final linear system is
$$
left[
begin{array}{ccc}
1 & -frac{(a-3)^2}{a+1} &
frac{a-2}{a+1} \
0 & frac{a^2-5 a+6}{a+1} &
frac{a^2-5 a+3}{-a^2+2 a+3} \
0 & 0 & frac{3}{a+1} \
end{array}
right]
left[begin{array}{c}
x_{1} \ x_{2} \x_{3}
end{array}right]
=
left[begin{array}{c}
frac{1}{a+1} \ 0 \ 0
end{array}right].
$$
Solution vis back substitution produces
$$
left[begin{array}{c}
x_{1} \ x_{2} \ x_{3}
end{array}right]
=
left[begin{array}{c}
frac{1}{a+1} \ 0 \ 0
end{array}right],
qquad ane1.
$$
answered Jan 31 at 19:11
dantopadantopa
6,65442245
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a situation where Cramer's Rule applies quite nicely.
First, note that
$$
det(A)
= 3 , {left(a + 1right)} {left(a - 2right)} {left(a - 3right)}^{2}
$$
So, $A$ is invertible as long as $anotin{-1, 2, 3}$.
Next, define $A_i$ as the matrix obtained by replacing the $i$th column of $A$ with the vector $vec{b}=leftlangle1,,a - 3,,1rightrangle$. This gives
$$
begin{align*}
det(A_1) &= 3 , {left(a - 2right)} {left(a - 3right)}^{2} &
det(A_2) &= 0 &
det(A_3) &= 0
end{align*}
$$
By Cramer's Rule, we have
$$
begin{align*}
x_1 &= frac{det(A_1)}{det(A)} = frac{1}{a + 1} &
x_2 &= frac{det(A_2)}{det(A)} = 0 &
x_3 &= frac{det(A_3)}{det(A)} = 0
end{align*}
$$
Of course, we should also account for what happens if $ain{-1, 2, 3}$.
For $a=-1$, we can row-reduce the system to obtain
$$
operatorname{rref}left[begin{array}{rrr|r}
0 & -16 & -3 & 1 \
0 & 16 & 3 & -4 \
0 & -16 & 0 & 1
end{array}right]
=left[begin{array}{rrr|r}
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{array}right]
$$
For $a=2$, we have
$$
operatorname{rref}left[begin{array}{rrr|r}
3 & -1 & 0 & 1 \
-3 & 1 & 3 & -1 \
3 & -1 & 3 & 1
end{array}right]
=left[begin{array}{rrr|r}
1 & -frac{1}{3} & 0 & frac{1}{3} \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0
end{array}right]
$$
For $a=3$, we have
$$
operatorname{rref}left[begin{array}{rrr|r}
4 & 0 & 1 & 1 \
0 & 0 & 3 & 0 \
4 & 0 & 4 & 1
end{array}right]
=left[begin{array}{rrr|r}
1 & 0 & 0 & frac{1}{4} \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0
end{array}right]
$$
answered Jan 31 at 22:39
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
$endgroup$
add a comment |
$begingroup$
This is a situation where Cramer's Rule applies quite nicely.
First, note that
$$
det(A)
= 3 , {left(a + 1right)} {left(a - 2right)} {left(a - 3right)}^{2}
$$
So, $A$ is invertible as long as $anotin{-1, 2, 3}$.
Next, define $A_i$ as the matrix obtained by replacing the $i$th column of $A$ with the vector $vec{b}=leftlangle1,,a - 3,,1rightrangle$. This gives
$$
begin{align*}
det(A_1) &= 3 , {left(a - 2right)} {left(a - 3right)}^{2} &
det(A_2) &= 0 &
det(A_3) &= 0
end{align*}
$$
By Cramer's Rule, we have
$$
begin{align*}
x_1 &= frac{det(A_1)}{det(A)} = frac{1}{a + 1} &
x_2 &= frac{det(A_2)}{det(A)} = 0 &
x_3 &= frac{det(A_3)}{det(A)} = 0
end{align*}
$$
Of course, we should also account for what happens if $ain{-1, 2, 3}$.
For $a=-1$, we can row-reduce the system to obtain
$$
operatorname{rref}left[begin{array}{rrr|r}
0 & -16 & -3 & 1 \
0 & 16 & 3 & -4 \
0 & -16 & 0 & 1
end{array}right]
=left[begin{array}{rrr|r}
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{array}right]
$$
For $a=2$, we have
$$
operatorname{rref}left[begin{array}{rrr|r}
3 & -1 & 0 & 1 \
-3 & 1 & 3 & -1 \
3 & -1 & 3 & 1
end{array}right]
=left[begin{array}{rrr|r}
1 & -frac{1}{3} & 0 & frac{1}{3} \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0
end{array}right]
$$
For $a=3$, we have
$$
operatorname{rref}left[begin{array}{rrr|r}
4 & 0 & 1 & 1 \
0 & 0 & 3 & 0 \
4 & 0 & 4 & 1
end{array}right]
=left[begin{array}{rrr|r}
1 & 0 & 0 & frac{1}{4} \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0
end{array}right]
$$
answered Jan 31 at 22:39
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
$endgroup$
add a comment |
$begingroup$
This is a situation where Cramer's Rule applies quite nicely.
First, note that
$$
det(A)
= 3 , {left(a + 1right)} {left(a - 2right)} {left(a - 3right)}^{2}
$$
So, $A$ is invertible as long as $anotin{-1, 2, 3}$.
Next, define $A_i$ as the matrix obtained by replacing the $i$th column of $A$ with the vector $vec{b}=leftlangle1,,a - 3,,1rightrangle$. This gives
$$
begin{align*}
det(A_1) &= 3 , {left(a - 2right)} {left(a - 3right)}^{2} &
det(A_2) &= 0 &
det(A_3) &= 0
end{align*}
$$
By Cramer's Rule, we have
$$
begin{align*}
x_1 &= frac{det(A_1)}{det(A)} = frac{1}{a + 1} &
x_2 &= frac{det(A_2)}{det(A)} = 0 &
x_3 &= frac{det(A_3)}{det(A)} = 0
end{align*}
$$
Of course, we should also account for what happens if $ain{-1, 2, 3}$.
For $a=-1$, we can row-reduce the system to obtain
$$
operatorname{rref}left[begin{array}{rrr|r}
0 & -16 & -3 & 1 \
0 & 16 & 3 & -4 \
0 & -16 & 0 & 1
end{array}right]
=left[begin{array}{rrr|r}
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{array}right]
$$
For $a=2$, we have
$$
operatorname{rref}left[begin{array}{rrr|r}
3 & -1 & 0 & 1 \
-3 & 1 & 3 & -1 \
3 & -1 & 3 & 1
end{array}right]
=left[begin{array}{rrr|r}
1 & -frac{1}{3} & 0 & frac{1}{3} \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0
end{array}right]
$$
For $a=3$, we have
$$
operatorname{rref}left[begin{array}{rrr|r}
4 & 0 & 1 & 1 \
0 & 0 & 3 & 0 \
4 & 0 & 4 & 1
end{array}right]
=left[begin{array}{rrr|r}
1 & 0 & 0 & frac{1}{4} \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0
end{array}right]
$$
answered Jan 31 at 22:39
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
$endgroup$
This is a situation where Cramer's Rule applies quite nicely.
First, note that
$$
det(A)
= 3 , {left(a + 1right)} {left(a - 2right)} {left(a - 3right)}^{2}
$$
So, $A$ is invertible as long as $anotin{-1, 2, 3}$.
Next, define $A_i$ as the matrix obtained by replacing the $i$th column of $A$ with the vector $vec{b}=leftlangle1,,a - 3,,1rightrangle$. This gives
$$
begin{align*}
det(A_1) &= 3 , {left(a - 2right)} {left(a - 3right)}^{2} &
det(A_2) &= 0 &
det(A_3) &= 0
end{align*}
$$
By Cramer's Rule, we have
$$
begin{align*}
x_1 &= frac{det(A_1)}{det(A)} = frac{1}{a + 1} &
x_2 &= frac{det(A_2)}{det(A)} = 0 &
x_3 &= frac{det(A_3)}{det(A)} = 0
end{align*}
$$
Of course, we should also account for what happens if $ain{-1, 2, 3}$.
For $a=-1$, we can row-reduce the system to obtain
$$
operatorname{rref}left[begin{array}{rrr|r}
0 & -16 & -3 & 1 \
0 & 16 & 3 & -4 \
0 & -16 & 0 & 1
end{array}right]
=left[begin{array}{rrr|r}
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{array}right]
$$
For $a=2$, we have
$$
operatorname{rref}left[begin{array}{rrr|r}
3 & -1 & 0 & 1 \
-3 & 1 & 3 & -1 \
3 & -1 & 3 & 1
end{array}right]
=left[begin{array}{rrr|r}
1 & -frac{1}{3} & 0 & frac{1}{3} \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0
end{array}right]
$$
For $a=3$, we have
$$
operatorname{rref}left[begin{array}{rrr|r}
4 & 0 & 1 & 1 \
0 & 0 & 3 & 0 \
4 & 0 & 4 & 1
end{array}right]
=left[begin{array}{rrr|r}
1 & 0 & 0 & frac{1}{4} \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 0
end{array}right]
$$
answered Jan 31 at 22:39
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
answered Jan 31 at 22:39
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
answered Jan 31 at 22:39
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
answered Jan 31 at 22:39
Brian FitzpatrickBrian Fitzpatrick
21.8k42959
21.8k42959
add a comment |
add a comment |
$begingroup$
Problem statement
Solve the linear system $mathbf{A}x=b$:
$$
left[begin{array}{crc}
a+1 & -(a-3)^2 & a-2 \
(a-3) (a+1) & (a-3)^2 & 3 \
a+1 & -(a-3)^2 & a+1
end{array}right]
left[begin{array}{c}
x_{1} \ x_{2} \x_{3}
end{array}right]
=
left[begin{array}{c}
1 \ a-3 \ 1
end{array}right].
$$
Gaussian Elimination
Two reduction steps are used resulting in
$$
mathbf{E}_{2}mathbf{E}_{1}mathbf{A}x=mathbf{E}_{2}mathbf{E}_{1}b
$$
where
$$
mathbf{E}_{1} =
left[
begin{array}{ccc}
frac{1}{a+1} & 0 & 0 \
0 & frac{1}{a+1} & 0 \
0 & 0 & frac{1}{a+1} \
end{array}
right], qquad
mathbf{E}_{2} =
left[
begin{array}{rcc}
1 & 0 & 0 \
-1 & frac{1}{a-3} & 0 \
-1 & 0 & 1 \
end{array}
right].
$$
Solution via back substitution
The final linear system is
$$
left[
begin{array}{ccc}
1 & -frac{(a-3)^2}{a+1} &
frac{a-2}{a+1} \
0 & frac{a^2-5 a+6}{a+1} &
frac{a^2-5 a+3}{-a^2+2 a+3} \
0 & 0 & frac{3}{a+1} \
end{array}
right]
left[begin{array}{c}
x_{1} \ x_{2} \x_{3}
end{array}right]
=
left[begin{array}{c}
frac{1}{a+1} \ 0 \ 0
end{array}right].
$$
Solution vis back substitution produces
$$
left[begin{array}{c}
x_{1} \ x_{2} \ x_{3}
end{array}right]
=
left[begin{array}{c}
frac{1}{a+1} \ 0 \ 0
end{array}right],
qquad ane1.
$$
answered Jan 31 at 19:11
dantopadantopa
6,65442245
$endgroup$
add a comment |
$begingroup$
Problem statement
Solve the linear system $mathbf{A}x=b$:
$$
left[begin{array}{crc}
a+1 & -(a-3)^2 & a-2 \
(a-3) (a+1) & (a-3)^2 & 3 \
a+1 & -(a-3)^2 & a+1
end{array}right]
left[begin{array}{c}
x_{1} \ x_{2} \x_{3}
end{array}right]
=
left[begin{array}{c}
1 \ a-3 \ 1
end{array}right].
$$
Gaussian Elimination
Two reduction steps are used resulting in
$$
mathbf{E}_{2}mathbf{E}_{1}mathbf{A}x=mathbf{E}_{2}mathbf{E}_{1}b
$$
where
$$
mathbf{E}_{1} =
left[
begin{array}{ccc}
frac{1}{a+1} & 0 & 0 \
0 & frac{1}{a+1} & 0 \
0 & 0 & frac{1}{a+1} \
end{array}
right], qquad
mathbf{E}_{2} =
left[
begin{array}{rcc}
1 & 0 & 0 \
-1 & frac{1}{a-3} & 0 \
-1 & 0 & 1 \
end{array}
right].
$$
Solution via back substitution
The final linear system is
$$
left[
begin{array}{ccc}
1 & -frac{(a-3)^2}{a+1} &
frac{a-2}{a+1} \
0 & frac{a^2-5 a+6}{a+1} &
frac{a^2-5 a+3}{-a^2+2 a+3} \
0 & 0 & frac{3}{a+1} \
end{array}
right]
left[begin{array}{c}
x_{1} \ x_{2} \x_{3}
end{array}right]
=
left[begin{array}{c}
frac{1}{a+1} \ 0 \ 0
end{array}right].
$$
Solution vis back substitution produces
$$
left[begin{array}{c}
x_{1} \ x_{2} \ x_{3}
end{array}right]
=
left[begin{array}{c}
frac{1}{a+1} \ 0 \ 0
end{array}right],
qquad ane1.
$$
answered Jan 31 at 19:11
dantopadantopa
6,65442245
$endgroup$
add a comment |
$begingroup$
Problem statement
Solve the linear system $mathbf{A}x=b$:
$$
left[begin{array}{crc}
a+1 & -(a-3)^2 & a-2 \
(a-3) (a+1) & (a-3)^2 & 3 \
a+1 & -(a-3)^2 & a+1
end{array}right]
left[begin{array}{c}
x_{1} \ x_{2} \x_{3}
end{array}right]
=
left[begin{array}{c}
1 \ a-3 \ 1
end{array}right].
$$
Gaussian Elimination
Two reduction steps are used resulting in
$$
mathbf{E}_{2}mathbf{E}_{1}mathbf{A}x=mathbf{E}_{2}mathbf{E}_{1}b
$$
where
$$
mathbf{E}_{1} =
left[
begin{array}{ccc}
frac{1}{a+1} & 0 & 0 \
0 & frac{1}{a+1} & 0 \
0 & 0 & frac{1}{a+1} \
end{array}
right], qquad
mathbf{E}_{2} =
left[
begin{array}{rcc}
1 & 0 & 0 \
-1 & frac{1}{a-3} & 0 \
-1 & 0 & 1 \
end{array}
right].
$$
Solution via back substitution
The final linear system is
$$
left[
begin{array}{ccc}
1 & -frac{(a-3)^2}{a+1} &
frac{a-2}{a+1} \
0 & frac{a^2-5 a+6}{a+1} &
frac{a^2-5 a+3}{-a^2+2 a+3} \
0 & 0 & frac{3}{a+1} \
end{array}
right]
left[begin{array}{c}
x_{1} \ x_{2} \x_{3}
end{array}right]
=
left[begin{array}{c}
frac{1}{a+1} \ 0 \ 0
end{array}right].
$$
Solution vis back substitution produces
$$
left[begin{array}{c}
x_{1} \ x_{2} \ x_{3}
end{array}right]
=
left[begin{array}{c}
frac{1}{a+1} \ 0 \ 0
end{array}right],
qquad ane1.
$$
answered Jan 31 at 19:11
dantopadantopa
6,65442245
$endgroup$
Problem statement
Solve the linear system $mathbf{A}x=b$:
$$
left[begin{array}{crc}
a+1 & -(a-3)^2 & a-2 \
(a-3) (a+1) & (a-3)^2 & 3 \
a+1 & -(a-3)^2 & a+1
end{array}right]
left[begin{array}{c}
x_{1} \ x_{2} \x_{3}
end{array}right]
=
left[begin{array}{c}
1 \ a-3 \ 1
end{array}right].
$$
Gaussian Elimination
Two reduction steps are used resulting in
$$
mathbf{E}_{2}mathbf{E}_{1}mathbf{A}x=mathbf{E}_{2}mathbf{E}_{1}b
$$
where
$$
mathbf{E}_{1} =
left[
begin{array}{ccc}
frac{1}{a+1} & 0 & 0 \
0 & frac{1}{a+1} & 0 \
0 & 0 & frac{1}{a+1} \
end{array}
right], qquad
mathbf{E}_{2} =
left[
begin{array}{rcc}
1 & 0 & 0 \
-1 & frac{1}{a-3} & 0 \
-1 & 0 & 1 \
end{array}
right].
$$
Solution via back substitution
The final linear system is
$$
left[
begin{array}{ccc}
1 & -frac{(a-3)^2}{a+1} &
frac{a-2}{a+1} \
0 & frac{a^2-5 a+6}{a+1} &
frac{a^2-5 a+3}{-a^2+2 a+3} \
0 & 0 & frac{3}{a+1} \
end{array}
right]
left[begin{array}{c}
x_{1} \ x_{2} \x_{3}
end{array}right]
=
left[begin{array}{c}
frac{1}{a+1} \ 0 \ 0
end{array}right].
$$
Solution vis back substitution produces
$$
left[begin{array}{c}
x_{1} \ x_{2} \ x_{3}
end{array}right]
=
left[begin{array}{c}
frac{1}{a+1} \ 0 \ 0
end{array}right],
qquad ane1.
$$
answered Jan 31 at 19:11
dantopadantopa
6,65442245
answered Jan 31 at 19:11
dantopadantopa
6,65442245
answered Jan 31 at 19:11
dantopadantopa
6,65442245
answered Jan 31 at 19:11
dantopadantopa
6,65442245
6,65442245
add a comment |
add a comment |
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$begingroup$
Thank you, I saw that, i just need a hint how to bring it to the coefficient matrix
$endgroup$
– Alice
Jan 31 at 8:36