Simplify the determinant of a $4 times 4$ matrix.
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I have to find the determinant of the following 4x4 matrix:
$quad A=begin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}$
So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$detbegin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}=begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}xrightarrow{3R_3-R_1}begin{vmatrix}3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$$
Since I know from the solutions that the determinant is -64, I suppose that I need to simplify the third row in the reduced form to $quad 0 quad 0 quad 2 quad 0 quad$ and then multiply the elements in the upper-left-to-bottom-right diagonal, which is indeed -64. But this doesn't make much sense since there's also a $-4$ that we can simplify. Can someone explain me the actual rules we need to follow?
linear-algebra matrices determinant
edited Dec 19 '18 at 18:59
user1101010
8021730
asked Dec 19 '18 at 18:12
Kevin MooreKevin Moore
16918
$endgroup$
add a comment |
$begingroup$
I have to find the determinant of the following 4x4 matrix:
$quad A=begin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}$
So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$detbegin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}=begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}xrightarrow{3R_3-R_1}begin{vmatrix}3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$$
Since I know from the solutions that the determinant is -64, I suppose that I need to simplify the third row in the reduced form to $quad 0 quad 0 quad 2 quad 0 quad$ and then multiply the elements in the upper-left-to-bottom-right diagonal, which is indeed -64. But this doesn't make much sense since there's also a $-4$ that we can simplify. Can someone explain me the actual rules we need to follow?
linear-algebra matrices determinant
edited Dec 19 '18 at 18:59
user1101010
8021730
asked Dec 19 '18 at 18:12
Kevin MooreKevin Moore
16918
$endgroup$
$begingroup$
Expansion by minors along the second row or column and the last row or column makes this easy.
$endgroup$
– Robert Israel
Dec 19 '18 at 18:19
$begingroup$
The rule is that you can add a multiple of a row to another row. You cannot replace a row by a multiple of itself and add another row as well.
$endgroup$
– I like Serena
Dec 19 '18 at 18:57
$begingroup$
@IlikeSerena That’s not entirely right. The latter isn’t an elementary row operation per se, but it can be broken down into two elementary operations, one of which changes the determinant in a known way.
$endgroup$
– amd
Dec 19 '18 at 20:26
$begingroup$
That's a different rule @amd. We can multiply a row by a constant if we divide the whole determinant by the same constant. Mixing these rules up is causing the confusion.
$endgroup$
– I like Serena
Dec 19 '18 at 20:39
$begingroup$
@IlikeSerena My point is that, contrary to what you wrote, one can replace a row by a multiple of itself and then add another to it—it produces a perfectly valid result—but it’s not an elementary row operation and doing so changes the value of the determinant.
$endgroup$
– amd
Dec 19 '18 at 20:43
add a comment |
$begingroup$
I have to find the determinant of the following 4x4 matrix:
$quad A=begin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}$
So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$detbegin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}=begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}xrightarrow{3R_3-R_1}begin{vmatrix}3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$$
Since I know from the solutions that the determinant is -64, I suppose that I need to simplify the third row in the reduced form to $quad 0 quad 0 quad 2 quad 0 quad$ and then multiply the elements in the upper-left-to-bottom-right diagonal, which is indeed -64. But this doesn't make much sense since there's also a $-4$ that we can simplify. Can someone explain me the actual rules we need to follow?
linear-algebra matrices determinant
edited Dec 19 '18 at 18:59
user1101010
8021730
asked Dec 19 '18 at 18:12
Kevin MooreKevin Moore
16918
$endgroup$
I have to find the determinant of the following 4x4 matrix:
$quad A=begin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}$
So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$detbegin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}=begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}xrightarrow{3R_3-R_1}begin{vmatrix}3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$$
Since I know from the solutions that the determinant is -64, I suppose that I need to simplify the third row in the reduced form to $quad 0 quad 0 quad 2 quad 0 quad$ and then multiply the elements in the upper-left-to-bottom-right diagonal, which is indeed -64. But this doesn't make much sense since there's also a $-4$ that we can simplify. Can someone explain me the actual rules we need to follow?
linear-algebra matrices determinant
linear-algebra matrices determinant
edited Dec 19 '18 at 18:59
user1101010
8021730
asked Dec 19 '18 at 18:12
Kevin MooreKevin Moore
16918
edited Dec 19 '18 at 18:59
user1101010
8021730
asked Dec 19 '18 at 18:12
Kevin MooreKevin Moore
16918
edited Dec 19 '18 at 18:59
user1101010
8021730
edited Dec 19 '18 at 18:59
user1101010
8021730
edited Dec 19 '18 at 18:59
user1101010
8021730
8021730
asked Dec 19 '18 at 18:12
Kevin MooreKevin Moore
16918
asked Dec 19 '18 at 18:12
Kevin MooreKevin Moore
16918
asked Dec 19 '18 at 18:12
Kevin MooreKevin Moore
16918
16918
$begingroup$
Expansion by minors along the second row or column and the last row or column makes this easy.
$endgroup$
– Robert Israel
Dec 19 '18 at 18:19
$begingroup$
The rule is that you can add a multiple of a row to another row. You cannot replace a row by a multiple of itself and add another row as well.
$endgroup$
– I like Serena
Dec 19 '18 at 18:57
$begingroup$
@IlikeSerena That’s not entirely right. The latter isn’t an elementary row operation per se, but it can be broken down into two elementary operations, one of which changes the determinant in a known way.
$endgroup$
– amd
Dec 19 '18 at 20:26
$begingroup$
That's a different rule @amd. We can multiply a row by a constant if we divide the whole determinant by the same constant. Mixing these rules up is causing the confusion.
$endgroup$
– I like Serena
Dec 19 '18 at 20:39
$begingroup$
@IlikeSerena My point is that, contrary to what you wrote, one can replace a row by a multiple of itself and then add another to it—it produces a perfectly valid result—but it’s not an elementary row operation and doing so changes the value of the determinant.
$endgroup$
– amd
Dec 19 '18 at 20:43
add a comment |
$begingroup$
Expansion by minors along the second row or column and the last row or column makes this easy.
$endgroup$
– Robert Israel
Dec 19 '18 at 18:19
$begingroup$
The rule is that you can add a multiple of a row to another row. You cannot replace a row by a multiple of itself and add another row as well.
$endgroup$
– I like Serena
Dec 19 '18 at 18:57
$begingroup$
@IlikeSerena That’s not entirely right. The latter isn’t an elementary row operation per se, but it can be broken down into two elementary operations, one of which changes the determinant in a known way.
$endgroup$
– amd
Dec 19 '18 at 20:26
$begingroup$
That's a different rule @amd. We can multiply a row by a constant if we divide the whole determinant by the same constant. Mixing these rules up is causing the confusion.
$endgroup$
– I like Serena
Dec 19 '18 at 20:39
$begingroup$
@IlikeSerena My point is that, contrary to what you wrote, one can replace a row by a multiple of itself and then add another to it—it produces a perfectly valid result—but it’s not an elementary row operation and doing so changes the value of the determinant.
$endgroup$
– amd
Dec 19 '18 at 20:43
$begingroup$
Expansion by minors along the second row or column and the last row or column makes this easy.
$endgroup$
– Robert Israel
Dec 19 '18 at 18:19
$begingroup$
Expansion by minors along the second row or column and the last row or column makes this easy.
$endgroup$
– Robert Israel
Dec 19 '18 at 18:19
$begingroup$
The rule is that you can add a multiple of a row to another row. You cannot replace a row by a multiple of itself and add another row as well.
$endgroup$
– I like Serena
Dec 19 '18 at 18:57
$begingroup$
The rule is that you can add a multiple of a row to another row. You cannot replace a row by a multiple of itself and add another row as well.
$endgroup$
– I like Serena
Dec 19 '18 at 18:57
$begingroup$
@IlikeSerena That’s not entirely right. The latter isn’t an elementary row operation per se, but it can be broken down into two elementary operations, one of which changes the determinant in a known way.
$endgroup$
– amd
Dec 19 '18 at 20:26
$begingroup$
@IlikeSerena That’s not entirely right. The latter isn’t an elementary row operation per se, but it can be broken down into two elementary operations, one of which changes the determinant in a known way.
$endgroup$
– amd
Dec 19 '18 at 20:26
$begingroup$
That's a different rule @amd. We can multiply a row by a constant if we divide the whole determinant by the same constant. Mixing these rules up is causing the confusion.
$endgroup$
– I like Serena
Dec 19 '18 at 20:39
$begingroup$
That's a different rule @amd. We can multiply a row by a constant if we divide the whole determinant by the same constant. Mixing these rules up is causing the confusion.
$endgroup$
– I like Serena
Dec 19 '18 at 20:39
$begingroup$
@IlikeSerena My point is that, contrary to what you wrote, one can replace a row by a multiple of itself and then add another to it—it produces a perfectly valid result—but it’s not an elementary row operation and doing so changes the value of the determinant.
$endgroup$
– amd
Dec 19 '18 at 20:43
$begingroup$
@IlikeSerena My point is that, contrary to what you wrote, one can replace a row by a multiple of itself and then add another to it—it produces a perfectly valid result—but it’s not an elementary row operation and doing so changes the value of the determinant.
$endgroup$
– amd
Dec 19 '18 at 20:43
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
It should be: $$begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}rightarrow begin{vmatrix}3&0&1&0\0&2&0&0\0&0&color{red}{3-{1over 3}}&0\0&0&0&-4end{vmatrix}$$
answered Dec 19 '18 at 18:15
greedoidgreedoid
42.7k1153105
$endgroup$
$begingroup$
Why not merely compute the determinant and find that it is $-64$?
$endgroup$
– David G. Stork
Dec 19 '18 at 18:18
$begingroup$
I did $3R_3-R_1$, shouldn't it be fine?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:22
2
$begingroup$
No, it shoul be $R_3-R_1/3$
$endgroup$
– greedoid
Dec 19 '18 at 18:24
$begingroup$
Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:27
2
$begingroup$
$R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
$endgroup$
– Shubham Johri
Dec 19 '18 at 18:30
add a comment |
$begingroup$
Rather than applying row operations, expand the minors.
$det A = -4begin{vmatrix} 3&0&1\0&2&0\1&0&3end{vmatrix} = (-4)(18-2) = -64$
If you apply row operations, you don't want your row operations to change the determinant.
I think of multiplying by an elementary matrix.
$begin{vmatrix} 1\&1\-frac 13&&1\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&frac {8}{3}&0\0&0&0&-4end{vmatrix}$
keeps the determinant unchanged.
while
$begin{vmatrix} 1\&1\-1&&3\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$
will change the determinant by a factor of $3.$
edited Dec 19 '18 at 18:39
Bernard
121k740116
answered Dec 19 '18 at 18:26
Doug MDoug M
45.2k31854
$endgroup$
add a comment |
$begingroup$
Just expand by the first row:
begin{align}
begin{vmatrix}3&0&1&0 \
0&2&0&0 \
1&0&3&0 \
0&0&0&-4
end{vmatrix}&=
3,begin{vmatrix}
2&0&0 \
0&3&0 \
0&0&-4
end{vmatrix}+
1,begin{vmatrix}
0&2&0 \
1&0&0 \
0&0&-4
end{vmatrix}
=3(2cdot3cdot(-4))+1(-1)begin{vmatrix}
2&0 \
0&-4
end{vmatrix}\
&=-72+8=-64.
end{align}
answered Dec 19 '18 at 18:53
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
The common mistake you made is that replacing $R_3$ with $3R_3-R_1$ is not an elementary row operation. You can add a multiple of $R_1$ to $R_3$ without changing the determinant, but not the other way around. What you did is a combination of two elementary row operations: $R_3to3R_3$ and $R_3to R_3+R_1$. The second operation doesn’t affect the determinant, but the first one multiplies it by $3$, so you have to divide by $3$ at the end when you combine the diagonal entries. Once you’ve accounted for this factor of $3$ that you introduced, you get the correct value.
Alternatively, you could’ve performed $R_3to R_3-frac13R_1$ to clear the first column without changing the value of the determinant. Note the difference between this and what you did: the latter operation multiplies by a scalar a different row from the one being replaced.
As others have noted, a much easier way to compute the determinant of this matrix is to take advantage of all of those zeros in the first and second row/column and expand by minors along them.
answered Dec 19 '18 at 20:32
amdamd
30.4k21050
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It should be: $$begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}rightarrow begin{vmatrix}3&0&1&0\0&2&0&0\0&0&color{red}{3-{1over 3}}&0\0&0&0&-4end{vmatrix}$$
answered Dec 19 '18 at 18:15
greedoidgreedoid
42.7k1153105
$endgroup$
$begingroup$
Why not merely compute the determinant and find that it is $-64$?
$endgroup$
– David G. Stork
Dec 19 '18 at 18:18
$begingroup$
I did $3R_3-R_1$, shouldn't it be fine?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:22
2
$begingroup$
No, it shoul be $R_3-R_1/3$
$endgroup$
– greedoid
Dec 19 '18 at 18:24
$begingroup$
Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:27
2
$begingroup$
$R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
$endgroup$
– Shubham Johri
Dec 19 '18 at 18:30
add a comment |
$begingroup$
It should be: $$begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}rightarrow begin{vmatrix}3&0&1&0\0&2&0&0\0&0&color{red}{3-{1over 3}}&0\0&0&0&-4end{vmatrix}$$
answered Dec 19 '18 at 18:15
greedoidgreedoid
42.7k1153105
$endgroup$
$begingroup$
Why not merely compute the determinant and find that it is $-64$?
$endgroup$
– David G. Stork
Dec 19 '18 at 18:18
$begingroup$
I did $3R_3-R_1$, shouldn't it be fine?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:22
2
$begingroup$
No, it shoul be $R_3-R_1/3$
$endgroup$
– greedoid
Dec 19 '18 at 18:24
$begingroup$
Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:27
2
$begingroup$
$R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
$endgroup$
– Shubham Johri
Dec 19 '18 at 18:30
add a comment |
$begingroup$
It should be: $$begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}rightarrow begin{vmatrix}3&0&1&0\0&2&0&0\0&0&color{red}{3-{1over 3}}&0\0&0&0&-4end{vmatrix}$$
answered Dec 19 '18 at 18:15
greedoidgreedoid
42.7k1153105
$endgroup$
It should be: $$begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}rightarrow begin{vmatrix}3&0&1&0\0&2&0&0\0&0&color{red}{3-{1over 3}}&0\0&0&0&-4end{vmatrix}$$
answered Dec 19 '18 at 18:15
greedoidgreedoid
42.7k1153105
edited Dec 19 '18 at 18:18
answered Dec 19 '18 at 18:15
greedoidgreedoid
42.7k1153105
answered Dec 19 '18 at 18:15
greedoidgreedoid
42.7k1153105
answered Dec 19 '18 at 18:15
greedoidgreedoid
42.7k1153105
42.7k1153105
$begingroup$
Why not merely compute the determinant and find that it is $-64$?
$endgroup$
– David G. Stork
Dec 19 '18 at 18:18
$begingroup$
I did $3R_3-R_1$, shouldn't it be fine?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:22
2
$begingroup$
No, it shoul be $R_3-R_1/3$
$endgroup$
– greedoid
Dec 19 '18 at 18:24
$begingroup$
Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:27
2
$begingroup$
$R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
$endgroup$
– Shubham Johri
Dec 19 '18 at 18:30
add a comment |
$begingroup$
Why not merely compute the determinant and find that it is $-64$?
$endgroup$
– David G. Stork
Dec 19 '18 at 18:18
$begingroup$
I did $3R_3-R_1$, shouldn't it be fine?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:22
2
$begingroup$
No, it shoul be $R_3-R_1/3$
$endgroup$
– greedoid
Dec 19 '18 at 18:24
$begingroup$
Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:27
2
$begingroup$
$R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
$endgroup$
– Shubham Johri
Dec 19 '18 at 18:30
$begingroup$
Why not merely compute the determinant and find that it is $-64$?
$endgroup$
– David G. Stork
Dec 19 '18 at 18:18
$begingroup$
Why not merely compute the determinant and find that it is $-64$?
$endgroup$
– David G. Stork
Dec 19 '18 at 18:18
$begingroup$
I did $3R_3-R_1$, shouldn't it be fine?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:22
$begingroup$
I did $3R_3-R_1$, shouldn't it be fine?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:22
2
2
$begingroup$
No, it shoul be $R_3-R_1/3$
$endgroup$
– greedoid
Dec 19 '18 at 18:24
$begingroup$
No, it shoul be $R_3-R_1/3$
$endgroup$
– greedoid
Dec 19 '18 at 18:24
$begingroup$
Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:27
$begingroup$
Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
$endgroup$
– Kevin Moore
Dec 19 '18 at 18:27
2
2
$begingroup$
$R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
$endgroup$
– Shubham Johri
Dec 19 '18 at 18:30
$begingroup$
$R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
$endgroup$
– Shubham Johri
Dec 19 '18 at 18:30
add a comment |
$begingroup$
Rather than applying row operations, expand the minors.
$det A = -4begin{vmatrix} 3&0&1\0&2&0\1&0&3end{vmatrix} = (-4)(18-2) = -64$
If you apply row operations, you don't want your row operations to change the determinant.
I think of multiplying by an elementary matrix.
$begin{vmatrix} 1\&1\-frac 13&&1\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&frac {8}{3}&0\0&0&0&-4end{vmatrix}$
keeps the determinant unchanged.
while
$begin{vmatrix} 1\&1\-1&&3\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$
will change the determinant by a factor of $3.$
edited Dec 19 '18 at 18:39
Bernard
121k740116
answered Dec 19 '18 at 18:26
Doug MDoug M
45.2k31854
$endgroup$
add a comment |
$begingroup$
Rather than applying row operations, expand the minors.
$det A = -4begin{vmatrix} 3&0&1\0&2&0\1&0&3end{vmatrix} = (-4)(18-2) = -64$
If you apply row operations, you don't want your row operations to change the determinant.
I think of multiplying by an elementary matrix.
$begin{vmatrix} 1\&1\-frac 13&&1\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&frac {8}{3}&0\0&0&0&-4end{vmatrix}$
keeps the determinant unchanged.
while
$begin{vmatrix} 1\&1\-1&&3\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$
will change the determinant by a factor of $3.$
edited Dec 19 '18 at 18:39
Bernard
121k740116
answered Dec 19 '18 at 18:26
Doug MDoug M
45.2k31854
$endgroup$
add a comment |
$begingroup$
Rather than applying row operations, expand the minors.
$det A = -4begin{vmatrix} 3&0&1\0&2&0\1&0&3end{vmatrix} = (-4)(18-2) = -64$
If you apply row operations, you don't want your row operations to change the determinant.
I think of multiplying by an elementary matrix.
$begin{vmatrix} 1\&1\-frac 13&&1\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&frac {8}{3}&0\0&0&0&-4end{vmatrix}$
keeps the determinant unchanged.
while
$begin{vmatrix} 1\&1\-1&&3\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$
will change the determinant by a factor of $3.$
edited Dec 19 '18 at 18:39
Bernard
121k740116
answered Dec 19 '18 at 18:26
Doug MDoug M
45.2k31854
$endgroup$
Rather than applying row operations, expand the minors.
$det A = -4begin{vmatrix} 3&0&1\0&2&0\1&0&3end{vmatrix} = (-4)(18-2) = -64$
If you apply row operations, you don't want your row operations to change the determinant.
I think of multiplying by an elementary matrix.
$begin{vmatrix} 1\&1\-frac 13&&1\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&frac {8}{3}&0\0&0&0&-4end{vmatrix}$
keeps the determinant unchanged.
while
$begin{vmatrix} 1\&1\-1&&3\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$
will change the determinant by a factor of $3.$
edited Dec 19 '18 at 18:39
Bernard
121k740116
answered Dec 19 '18 at 18:26
Doug MDoug M
45.2k31854
edited Dec 19 '18 at 18:39
Bernard
121k740116
edited Dec 19 '18 at 18:39
Bernard
121k740116
edited Dec 19 '18 at 18:39
Bernard
121k740116
121k740116
answered Dec 19 '18 at 18:26
Doug MDoug M
45.2k31854
answered Dec 19 '18 at 18:26
Doug MDoug M
45.2k31854
answered Dec 19 '18 at 18:26
Doug MDoug M
45.2k31854
45.2k31854
add a comment |
add a comment |
$begingroup$
Just expand by the first row:
begin{align}
begin{vmatrix}3&0&1&0 \
0&2&0&0 \
1&0&3&0 \
0&0&0&-4
end{vmatrix}&=
3,begin{vmatrix}
2&0&0 \
0&3&0 \
0&0&-4
end{vmatrix}+
1,begin{vmatrix}
0&2&0 \
1&0&0 \
0&0&-4
end{vmatrix}
=3(2cdot3cdot(-4))+1(-1)begin{vmatrix}
2&0 \
0&-4
end{vmatrix}\
&=-72+8=-64.
end{align}
answered Dec 19 '18 at 18:53
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Just expand by the first row:
begin{align}
begin{vmatrix}3&0&1&0 \
0&2&0&0 \
1&0&3&0 \
0&0&0&-4
end{vmatrix}&=
3,begin{vmatrix}
2&0&0 \
0&3&0 \
0&0&-4
end{vmatrix}+
1,begin{vmatrix}
0&2&0 \
1&0&0 \
0&0&-4
end{vmatrix}
=3(2cdot3cdot(-4))+1(-1)begin{vmatrix}
2&0 \
0&-4
end{vmatrix}\
&=-72+8=-64.
end{align}
answered Dec 19 '18 at 18:53
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Just expand by the first row:
begin{align}
begin{vmatrix}3&0&1&0 \
0&2&0&0 \
1&0&3&0 \
0&0&0&-4
end{vmatrix}&=
3,begin{vmatrix}
2&0&0 \
0&3&0 \
0&0&-4
end{vmatrix}+
1,begin{vmatrix}
0&2&0 \
1&0&0 \
0&0&-4
end{vmatrix}
=3(2cdot3cdot(-4))+1(-1)begin{vmatrix}
2&0 \
0&-4
end{vmatrix}\
&=-72+8=-64.
end{align}
answered Dec 19 '18 at 18:53
BernardBernard
121k740116
$endgroup$
Just expand by the first row:
begin{align}
begin{vmatrix}3&0&1&0 \
0&2&0&0 \
1&0&3&0 \
0&0&0&-4
end{vmatrix}&=
3,begin{vmatrix}
2&0&0 \
0&3&0 \
0&0&-4
end{vmatrix}+
1,begin{vmatrix}
0&2&0 \
1&0&0 \
0&0&-4
end{vmatrix}
=3(2cdot3cdot(-4))+1(-1)begin{vmatrix}
2&0 \
0&-4
end{vmatrix}\
&=-72+8=-64.
end{align}
answered Dec 19 '18 at 18:53
BernardBernard
121k740116
answered Dec 19 '18 at 18:53
BernardBernard
121k740116
answered Dec 19 '18 at 18:53
BernardBernard
121k740116
answered Dec 19 '18 at 18:53
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
$begingroup$
The common mistake you made is that replacing $R_3$ with $3R_3-R_1$ is not an elementary row operation. You can add a multiple of $R_1$ to $R_3$ without changing the determinant, but not the other way around. What you did is a combination of two elementary row operations: $R_3to3R_3$ and $R_3to R_3+R_1$. The second operation doesn’t affect the determinant, but the first one multiplies it by $3$, so you have to divide by $3$ at the end when you combine the diagonal entries. Once you’ve accounted for this factor of $3$ that you introduced, you get the correct value.
Alternatively, you could’ve performed $R_3to R_3-frac13R_1$ to clear the first column without changing the value of the determinant. Note the difference between this and what you did: the latter operation multiplies by a scalar a different row from the one being replaced.
As others have noted, a much easier way to compute the determinant of this matrix is to take advantage of all of those zeros in the first and second row/column and expand by minors along them.
answered Dec 19 '18 at 20:32
amdamd
30.4k21050
$endgroup$
add a comment |
$begingroup$
The common mistake you made is that replacing $R_3$ with $3R_3-R_1$ is not an elementary row operation. You can add a multiple of $R_1$ to $R_3$ without changing the determinant, but not the other way around. What you did is a combination of two elementary row operations: $R_3to3R_3$ and $R_3to R_3+R_1$. The second operation doesn’t affect the determinant, but the first one multiplies it by $3$, so you have to divide by $3$ at the end when you combine the diagonal entries. Once you’ve accounted for this factor of $3$ that you introduced, you get the correct value.
Alternatively, you could’ve performed $R_3to R_3-frac13R_1$ to clear the first column without changing the value of the determinant. Note the difference between this and what you did: the latter operation multiplies by a scalar a different row from the one being replaced.
As others have noted, a much easier way to compute the determinant of this matrix is to take advantage of all of those zeros in the first and second row/column and expand by minors along them.
answered Dec 19 '18 at 20:32
amdamd
30.4k21050
$endgroup$
add a comment |
$begingroup$
The common mistake you made is that replacing $R_3$ with $3R_3-R_1$ is not an elementary row operation. You can add a multiple of $R_1$ to $R_3$ without changing the determinant, but not the other way around. What you did is a combination of two elementary row operations: $R_3to3R_3$ and $R_3to R_3+R_1$. The second operation doesn’t affect the determinant, but the first one multiplies it by $3$, so you have to divide by $3$ at the end when you combine the diagonal entries. Once you’ve accounted for this factor of $3$ that you introduced, you get the correct value.
Alternatively, you could’ve performed $R_3to R_3-frac13R_1$ to clear the first column without changing the value of the determinant. Note the difference between this and what you did: the latter operation multiplies by a scalar a different row from the one being replaced.
As others have noted, a much easier way to compute the determinant of this matrix is to take advantage of all of those zeros in the first and second row/column and expand by minors along them.
answered Dec 19 '18 at 20:32
amdamd
30.4k21050
$endgroup$
The common mistake you made is that replacing $R_3$ with $3R_3-R_1$ is not an elementary row operation. You can add a multiple of $R_1$ to $R_3$ without changing the determinant, but not the other way around. What you did is a combination of two elementary row operations: $R_3to3R_3$ and $R_3to R_3+R_1$. The second operation doesn’t affect the determinant, but the first one multiplies it by $3$, so you have to divide by $3$ at the end when you combine the diagonal entries. Once you’ve accounted for this factor of $3$ that you introduced, you get the correct value.
Alternatively, you could’ve performed $R_3to R_3-frac13R_1$ to clear the first column without changing the value of the determinant. Note the difference between this and what you did: the latter operation multiplies by a scalar a different row from the one being replaced.
As others have noted, a much easier way to compute the determinant of this matrix is to take advantage of all of those zeros in the first and second row/column and expand by minors along them.
answered Dec 19 '18 at 20:32
amdamd
30.4k21050
answered Dec 19 '18 at 20:32
amdamd
30.4k21050
answered Dec 19 '18 at 20:32
amdamd
30.4k21050
answered Dec 19 '18 at 20:32
amdamd
30.4k21050
30.4k21050
add a comment |
add a comment |
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$begingroup$
Expansion by minors along the second row or column and the last row or column makes this easy.
$endgroup$
– Robert Israel
Dec 19 '18 at 18:19
$begingroup$
The rule is that you can add a multiple of a row to another row. You cannot replace a row by a multiple of itself and add another row as well.
$endgroup$
– I like Serena
Dec 19 '18 at 18:57
$begingroup$
@IlikeSerena That’s not entirely right. The latter isn’t an elementary row operation per se, but it can be broken down into two elementary operations, one of which changes the determinant in a known way.
$endgroup$
– amd
Dec 19 '18 at 20:26
$begingroup$
That's a different rule @amd. We can multiply a row by a constant if we divide the whole determinant by the same constant. Mixing these rules up is causing the confusion.
$endgroup$
– I like Serena
Dec 19 '18 at 20:39
$begingroup$
@IlikeSerena My point is that, contrary to what you wrote, one can replace a row by a multiple of itself and then add another to it—it produces a perfectly valid result—but it’s not an elementary row operation and doing so changes the value of the determinant.
$endgroup$
– amd
Dec 19 '18 at 20:43