Using Cramer's Rule to solve a system of linear equation
$begingroup$
I have trouble solving the following question:
Question
How I approached it initially was as follow:
1) Since the questions asks for x3 only, I replaced the x3 column with the particular solution.
Step 1
2) Cramer's Rule asks of us to divide the det(Substituted Matrix in Step 1) by the det(Initial Matrix)
Step 2
3) To find the det of each 4x4 matrix I did C(1,1)(det(3x3 sub Matrix)). So 1(det(3x3 sub Matrix)). I evaluated the 3x3 sub Matrix as follow: 3x3 sub Matrix The red arrows are added and then subtracted to the addition of the blue arrows. Thus det(Substituted Matrix) = 3aei + 3bfg + 2cdh - 3ceg - 2afh - 3bdi.
4) I did step 3 for the initial Matrix as well and got det(Initial Matrix) = aei + bfg + cdh - ceg - afh - bdi
Now, I am unsure how to work around the determinants I just got. I should factor out something and then cancel the rest, but I don't see what. Did I miss something?
linear-algebra
asked Dec 13 '18 at 1:00
HeykanHeykan
11
$endgroup$
add a comment |
$begingroup$
I have trouble solving the following question:
Question
How I approached it initially was as follow:
1) Since the questions asks for x3 only, I replaced the x3 column with the particular solution.
Step 1
2) Cramer's Rule asks of us to divide the det(Substituted Matrix in Step 1) by the det(Initial Matrix)
Step 2
3) To find the det of each 4x4 matrix I did C(1,1)(det(3x3 sub Matrix)). So 1(det(3x3 sub Matrix)). I evaluated the 3x3 sub Matrix as follow: 3x3 sub Matrix The red arrows are added and then subtracted to the addition of the blue arrows. Thus det(Substituted Matrix) = 3aei + 3bfg + 2cdh - 3ceg - 2afh - 3bdi.
4) I did step 3 for the initial Matrix as well and got det(Initial Matrix) = aei + bfg + cdh - ceg - afh - bdi
Now, I am unsure how to work around the determinants I just got. I should factor out something and then cancel the rest, but I don't see what. Did I miss something?
linear-algebra
asked Dec 13 '18 at 1:00
HeykanHeykan
11
$endgroup$
add a comment |
$begingroup$
I have trouble solving the following question:
Question
How I approached it initially was as follow:
1) Since the questions asks for x3 only, I replaced the x3 column with the particular solution.
Step 1
2) Cramer's Rule asks of us to divide the det(Substituted Matrix in Step 1) by the det(Initial Matrix)
Step 2
3) To find the det of each 4x4 matrix I did C(1,1)(det(3x3 sub Matrix)). So 1(det(3x3 sub Matrix)). I evaluated the 3x3 sub Matrix as follow: 3x3 sub Matrix The red arrows are added and then subtracted to the addition of the blue arrows. Thus det(Substituted Matrix) = 3aei + 3bfg + 2cdh - 3ceg - 2afh - 3bdi.
4) I did step 3 for the initial Matrix as well and got det(Initial Matrix) = aei + bfg + cdh - ceg - afh - bdi
Now, I am unsure how to work around the determinants I just got. I should factor out something and then cancel the rest, but I don't see what. Did I miss something?
linear-algebra
asked Dec 13 '18 at 1:00
HeykanHeykan
11
$endgroup$
I have trouble solving the following question:
Question
How I approached it initially was as follow:
1) Since the questions asks for x3 only, I replaced the x3 column with the particular solution.
Step 1
2) Cramer's Rule asks of us to divide the det(Substituted Matrix in Step 1) by the det(Initial Matrix)
Step 2
3) To find the det of each 4x4 matrix I did C(1,1)(det(3x3 sub Matrix)). So 1(det(3x3 sub Matrix)). I evaluated the 3x3 sub Matrix as follow: 3x3 sub Matrix The red arrows are added and then subtracted to the addition of the blue arrows. Thus det(Substituted Matrix) = 3aei + 3bfg + 2cdh - 3ceg - 2afh - 3bdi.
4) I did step 3 for the initial Matrix as well and got det(Initial Matrix) = aei + bfg + cdh - ceg - afh - bdi
Now, I am unsure how to work around the determinants I just got. I should factor out something and then cancel the rest, but I don't see what. Did I miss something?
linear-algebra
linear-algebra
asked Dec 13 '18 at 1:00
HeykanHeykan
11
asked Dec 13 '18 at 1:00
HeykanHeykan
11
asked Dec 13 '18 at 1:00
HeykanHeykan
11
asked Dec 13 '18 at 1:00
HeykanHeykan
11
asked Dec 13 '18 at 1:00
HeykanHeykan
11
11
add a comment |
add a comment |
1 Answer
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$begingroup$
It is clear that $begin{bmatrix}x_1\x_2\x_3\x_4 end{bmatrix} = begin{bmatrix}0\0\3\2 end{bmatrix}$ without Cramer's rule...
What does Cramer say?
$x_1 = frac {detbegin{bmatrix}0&0&0&0\3b+2c&a&b&c\3e+2f&d&e&f\3h+2i&g&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
We know this because one row is all 0's
$x_2 = frac {detbegin{bmatrix}1&0&0&0\0&3b+2c&b&c\0&3e+2f&e&f\0&3h+2i&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
We know this because clearly the columns are not linearly independent.
$x_3 = frac {detbegin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
rather than calculating this out.
$begin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix} = begin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}begin{bmatrix}1&0&0&0\0&2&3&0\0&0&1&0\0&0&0&1 end{bmatrix}$
The determinant of the product equals the product of the determinants. And
$detbegin{bmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1 end{bmatrix} = 2$
And $x_4$ can be approached the same way.
answered Dec 13 '18 at 1:18
Doug MDoug M
44.8k31854
$endgroup$
$begingroup$
You should get $3$. The last matrix is off slightly.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:32
$begingroup$
I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:45
$begingroup$
@ChrisCuster thanks, I something came up and rushed the ending.
$endgroup$
– Doug M
Dec 13 '18 at 16:38
add a comment |
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1 Answer
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$begingroup$
It is clear that $begin{bmatrix}x_1\x_2\x_3\x_4 end{bmatrix} = begin{bmatrix}0\0\3\2 end{bmatrix}$ without Cramer's rule...
What does Cramer say?
$x_1 = frac {detbegin{bmatrix}0&0&0&0\3b+2c&a&b&c\3e+2f&d&e&f\3h+2i&g&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
We know this because one row is all 0's
$x_2 = frac {detbegin{bmatrix}1&0&0&0\0&3b+2c&b&c\0&3e+2f&e&f\0&3h+2i&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
We know this because clearly the columns are not linearly independent.
$x_3 = frac {detbegin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
rather than calculating this out.
$begin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix} = begin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}begin{bmatrix}1&0&0&0\0&2&3&0\0&0&1&0\0&0&0&1 end{bmatrix}$
The determinant of the product equals the product of the determinants. And
$detbegin{bmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1 end{bmatrix} = 2$
And $x_4$ can be approached the same way.
answered Dec 13 '18 at 1:18
Doug MDoug M
44.8k31854
$endgroup$
$begingroup$
You should get $3$. The last matrix is off slightly.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:32
$begingroup$
I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:45
$begingroup$
@ChrisCuster thanks, I something came up and rushed the ending.
$endgroup$
– Doug M
Dec 13 '18 at 16:38
add a comment |
$begingroup$
It is clear that $begin{bmatrix}x_1\x_2\x_3\x_4 end{bmatrix} = begin{bmatrix}0\0\3\2 end{bmatrix}$ without Cramer's rule...
What does Cramer say?
$x_1 = frac {detbegin{bmatrix}0&0&0&0\3b+2c&a&b&c\3e+2f&d&e&f\3h+2i&g&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
We know this because one row is all 0's
$x_2 = frac {detbegin{bmatrix}1&0&0&0\0&3b+2c&b&c\0&3e+2f&e&f\0&3h+2i&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
We know this because clearly the columns are not linearly independent.
$x_3 = frac {detbegin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
rather than calculating this out.
$begin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix} = begin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}begin{bmatrix}1&0&0&0\0&2&3&0\0&0&1&0\0&0&0&1 end{bmatrix}$
The determinant of the product equals the product of the determinants. And
$detbegin{bmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1 end{bmatrix} = 2$
And $x_4$ can be approached the same way.
answered Dec 13 '18 at 1:18
Doug MDoug M
44.8k31854
$endgroup$
$begingroup$
You should get $3$. The last matrix is off slightly.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:32
$begingroup$
I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:45
$begingroup$
@ChrisCuster thanks, I something came up and rushed the ending.
$endgroup$
– Doug M
Dec 13 '18 at 16:38
add a comment |
$begingroup$
It is clear that $begin{bmatrix}x_1\x_2\x_3\x_4 end{bmatrix} = begin{bmatrix}0\0\3\2 end{bmatrix}$ without Cramer's rule...
What does Cramer say?
$x_1 = frac {detbegin{bmatrix}0&0&0&0\3b+2c&a&b&c\3e+2f&d&e&f\3h+2i&g&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
We know this because one row is all 0's
$x_2 = frac {detbegin{bmatrix}1&0&0&0\0&3b+2c&b&c\0&3e+2f&e&f\0&3h+2i&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
We know this because clearly the columns are not linearly independent.
$x_3 = frac {detbegin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
rather than calculating this out.
$begin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix} = begin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}begin{bmatrix}1&0&0&0\0&2&3&0\0&0&1&0\0&0&0&1 end{bmatrix}$
The determinant of the product equals the product of the determinants. And
$detbegin{bmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1 end{bmatrix} = 2$
And $x_4$ can be approached the same way.
answered Dec 13 '18 at 1:18
Doug MDoug M
44.8k31854
$endgroup$
It is clear that $begin{bmatrix}x_1\x_2\x_3\x_4 end{bmatrix} = begin{bmatrix}0\0\3\2 end{bmatrix}$ without Cramer's rule...
What does Cramer say?
$x_1 = frac {detbegin{bmatrix}0&0&0&0\3b+2c&a&b&c\3e+2f&d&e&f\3h+2i&g&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
We know this because one row is all 0's
$x_2 = frac {detbegin{bmatrix}1&0&0&0\0&3b+2c&b&c\0&3e+2f&e&f\0&3h+2i&h&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
We know this because clearly the columns are not linearly independent.
$x_3 = frac {detbegin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix}}{detbegin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}} = 0$
rather than calculating this out.
$begin{bmatrix}1&0&0&0\0&a&3b+2c&c\0&d&3e+2f&f\0&g&3h+2i&i end{bmatrix} = begin{bmatrix}1&0&0&0\0&a&b&c\0&d&e&f\0&g&h&i end{bmatrix}begin{bmatrix}1&0&0&0\0&2&3&0\0&0&1&0\0&0&0&1 end{bmatrix}$
The determinant of the product equals the product of the determinants. And
$detbegin{bmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1 end{bmatrix} = 2$
And $x_4$ can be approached the same way.
answered Dec 13 '18 at 1:18
Doug MDoug M
44.8k31854
edited Dec 13 '18 at 16:37
answered Dec 13 '18 at 1:18
Doug MDoug M
44.8k31854
answered Dec 13 '18 at 1:18
Doug MDoug M
44.8k31854
answered Dec 13 '18 at 1:18
Doug MDoug M
44.8k31854
44.8k31854
$begingroup$
You should get $3$. The last matrix is off slightly.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:32
$begingroup$
I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:45
$begingroup$
@ChrisCuster thanks, I something came up and rushed the ending.
$endgroup$
– Doug M
Dec 13 '18 at 16:38
add a comment |
$begingroup$
You should get $3$. The last matrix is off slightly.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:32
$begingroup$
I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:45
$begingroup$
@ChrisCuster thanks, I something came up and rushed the ending.
$endgroup$
– Doug M
Dec 13 '18 at 16:38
$begingroup$
You should get $3$. The last matrix is off slightly.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:32
$begingroup$
You should get $3$. The last matrix is off slightly.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:32
$begingroup$
I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:45
$begingroup$
I think it's $begin{pmatrix}1&0&0&0\0&1&0&0\0&0&3&0\0&0&2&1end{pmatrix}$.
$endgroup$
– Chris Custer
Dec 13 '18 at 1:45
$begingroup$
@ChrisCuster thanks, I something came up and rushed the ending.
$endgroup$
– Doug M
Dec 13 '18 at 16:38
$begingroup$
@ChrisCuster thanks, I something came up and rushed the ending.
$endgroup$
– Doug M
Dec 13 '18 at 16:38
add a comment |
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