Det of matrix $4times 4$ [duplicate]

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This question already has an answer here:

• 
  Can we use Sarrus' method for finding the determinant of matrix greater than $3times3$?
  
  1 answer
  
  

Is the method of calculating determinant of $3times 3$ matrix by diagonals,
apply also on $4times 4$ matrix?

for example:

$$beginmatrix2&2&1&3|\1&4&4&5|\5&1&1&6|\7&1&4&5|endmatrixbeginmatrix2&2&1\1&4&4\5&1&1\7&1&4endmatrix$$

$det = 2cdot4cdot1cdot5+dotsb+3cdot1cdot1cdot4 - 7cdot1cdot4cdot3-dotsb-5cdot5cdot4cdot1 = 171$

Is this valid?

matrices determinant-functions

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edited Aug 1 at 15:40

Daniel Buck

2,2641623

asked Aug 1 at 15:17

Naor Yehuda

1

marked as duplicate by Hans Lundmark, Adrian Keister, Leucippus, amWhy, Xander Henderson Aug 2 at 1:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to Maths SX! I don't understand what your ‘method’ is, nor what the matrix is?
  – Bernard
  Aug 1 at 15:21
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

This question already has an answer here:

• 
  Can we use Sarrus' method for finding the determinant of matrix greater than $3times3$?
  
  1 answer
  
  

Is the method of calculating determinant of $3times 3$ matrix by diagonals,
apply also on $4times 4$ matrix?

for example:

$$beginmatrix2&2&1&3|\1&4&4&5|\5&1&1&6|\7&1&4&5|endmatrixbeginmatrix2&2&1\1&4&4\5&1&1\7&1&4endmatrix$$

$det = 2cdot4cdot1cdot5+dotsb+3cdot1cdot1cdot4 - 7cdot1cdot4cdot3-dotsb-5cdot5cdot4cdot1 = 171$

Is this valid?

matrices determinant-functions

share|cite|improve this question

edited Aug 1 at 15:40

Daniel Buck

2,2641623

asked Aug 1 at 15:17

Naor Yehuda

1

marked as duplicate by Hans Lundmark, Adrian Keister, Leucippus, amWhy, Xander Henderson Aug 2 at 1:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to Maths SX! I don't understand what your ‘method’ is, nor what the matrix is?
  – Bernard
  Aug 1 at 15:21
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

This question already has an answer here:

• 
  Can we use Sarrus' method for finding the determinant of matrix greater than $3times3$?
  
  1 answer
  
  

Is the method of calculating determinant of $3times 3$ matrix by diagonals,
apply also on $4times 4$ matrix?

for example:

$$beginmatrix2&2&1&3|\1&4&4&5|\5&1&1&6|\7&1&4&5|endmatrixbeginmatrix2&2&1\1&4&4\5&1&1\7&1&4endmatrix$$

$det = 2cdot4cdot1cdot5+dotsb+3cdot1cdot1cdot4 - 7cdot1cdot4cdot3-dotsb-5cdot5cdot4cdot1 = 171$

Is this valid?

matrices determinant-functions

share|cite|improve this question

edited Aug 1 at 15:40

Daniel Buck

2,2641623

asked Aug 1 at 15:17

Naor Yehuda

1

This question already has an answer here:

• 
  Can we use Sarrus' method for finding the determinant of matrix greater than $3times3$?
  
  1 answer
  
  

Is the method of calculating determinant of $3times 3$ matrix by diagonals,
apply also on $4times 4$ matrix?

for example:

$$beginmatrix2&2&1&3|\1&4&4&5|\5&1&1&6|\7&1&4&5|endmatrixbeginmatrix2&2&1\1&4&4\5&1&1\7&1&4endmatrix$$

$det = 2cdot4cdot1cdot5+dotsb+3cdot1cdot1cdot4 - 7cdot1cdot4cdot3-dotsb-5cdot5cdot4cdot1 = 171$

Is this valid?

This question already has an answer here:

• 
  Can we use Sarrus' method for finding the determinant of matrix greater than $3times3$?
  
  1 answer
  
  

matrices determinant-functions

share|cite|improve this question

edited Aug 1 at 15:40

Daniel Buck

2,2641623

asked Aug 1 at 15:17

Naor Yehuda

1

share|cite|improve this question

share|cite|improve this question

edited Aug 1 at 15:40

Daniel Buck

2,2641623

edited Aug 1 at 15:40

Daniel Buck

2,2641623

edited Aug 1 at 15:40

Daniel Buck

2,2641623

2,2641623

asked Aug 1 at 15:17

Naor Yehuda

1

asked Aug 1 at 15:17

Naor Yehuda

1

asked Aug 1 at 15:17

Naor Yehuda

1

1

marked as duplicate by Hans Lundmark, Adrian Keister, Leucippus, amWhy, Xander Henderson Aug 2 at 1:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Hans Lundmark, Adrian Keister, Leucippus, amWhy, Xander Henderson Aug 2 at 1:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to Maths SX! I don't understand what your ‘method’ is, nor what the matrix is?
  – Bernard
  Aug 1 at 15:21
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to Maths SX! I don't understand what your ‘method’ is, nor what the matrix is?
  – Bernard
  Aug 1 at 15:21
  
  

  
  
  
  

Welcome to Maths SX! I don't understand what your ‘method’ is, nor what the matrix is?
– Bernard
Aug 1 at 15:21

Welcome to Maths SX! I don't understand what your ‘method’ is, nor what the matrix is?
– Bernard
Aug 1 at 15:21

add a comment | 

4 Answers
4

active

oldest

votes

up vote
2
down vote

No the Rule of Sarrus is only valid for 3-by -3 matrices, in general for n-by -n matrices we can refer to Laplace expansion method, that is by the first row

$$beginvmatrix
2& 2& 1& 3\
1& 4& 4& 5\
5& 1& 1& 6\
7& 1& 4& 5
endvmatrix=2beginvmatrix
4& 4& 5\
1& 1& 6\
1& 4& 5
endvmatrix-2beginvmatrix
1& 4& 5\
5& 1& 6\
7& 4& 5
endvmatrix+1beginvmatrix
1& 4& 5\
5& 1& 6\
7& 1& 5
endvmatrix-3beginvmatrix
1& 4& 4 \
5& 1& 1 \
7& 1& 4
endvmatrix$$

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edited Aug 1 at 15:29

answered Aug 1 at 15:20

gimusi

63.9k73480

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up vote
1
down vote

No, it is not valid. My computation using Octave shows that the determinant is negative.

octave:1> A = [ 2 2 1 3; 1 4 4 5; 5 1 1 6; 7 1 4 5]
A =

 2 2 1 3
 1 4 4 5
 5 1 1 6
 7 1 4 5

octave:2> det(A)
ans = -114.00

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answered Aug 1 at 15:24

Siong Thye Goh

76.7k134794

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up vote
1
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The fastest method is Gaussian elimination to obtain a triangular matrix – in this case the determinant is the product of the diagonal elements:
beginalign
&beginvmatrix
2&2&1&3\1&4&4&5\5&1&1&6\7&1&4&5
endvmatrix
=-beginvmatrix
1&4&4&5\2&2&1&3\5&1&1&6\7&1&4&5
endvmatrix
=-beginvmatrixbeginarrayrrrr
1&4&4&5\0&-6&-7&-7\0&-19&-19&-19\0&-27&-24&-30
endarrayendvmatrix=19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&6&7&7\0&1&1&1\0&9&8&10
endarrayendvmatrix\[1ex]
=&-19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&1&1&1\0&6&7&7\0&9&8&10
endarrayendvmatrix=-19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&1&1&1\0&0&1&1\0&0&-1&1
endarrayendvmatrix
=-19times 3beginvmatrix
1&4&4&5\0&1&1&1\0&0&1&1\0&0&0&2
endvmatrix=-114\
&
endalign

However, in the case of $4times4$ matrices, you have two other possibilities, for a computation by blocks:

• If $M=beginpmatrixA&B\C&D endpmatrix$ is a $4×4$ matrix consisting of blocks of size $2$, and if $CD=DC$, then we can compute a $2×2$ determinant:
  $$det M=det(AD-BC).$$


• We can use Laplace's expansion along the first two columns. To explain the computation, we introduce some notations:

beginarrayll
p_ij&textis the $2×2$ determinant of rows $i$ and $j$ of the first two columns,\
q_ij&textis the $2×2$ determinant of rows $i$ and $j$ of the last two columns.
endarray
With these notations, one has
$$det M=p_12q_34+p_13q_42+p_14q_23+q_12p_34+q_13p_42+q_14p_23.$$

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answered Aug 1 at 16:45

Bernard

110k635102

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0
down vote

No, it isn't valid. When you calculate the determinant of an $ntimes n$ matrix by minors, you compute $n!$ products. When $n=3, n!=6$ and there are $6$ broken diagonals, and the "spaghetti rule" still involves $6$ products. But when $n=4, n!=24,$ and there are only $8$ broken diagonals, so there isn't much hope.

You can easily come up with examples to prove it doesn't work. (In fact, I imagine it's harder to come up with examples where it does work.)

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edited Aug 1 at 15:34

answered Aug 1 at 15:26

saulspatz

10.5k21324

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

No the Rule of Sarrus is only valid for 3-by -3 matrices, in general for n-by -n matrices we can refer to Laplace expansion method, that is by the first row

$$beginvmatrix
2& 2& 1& 3\
1& 4& 4& 5\
5& 1& 1& 6\
7& 1& 4& 5
endvmatrix=2beginvmatrix
4& 4& 5\
1& 1& 6\
1& 4& 5
endvmatrix-2beginvmatrix
1& 4& 5\
5& 1& 6\
7& 4& 5
endvmatrix+1beginvmatrix
1& 4& 5\
5& 1& 6\
7& 1& 5
endvmatrix-3beginvmatrix
1& 4& 4 \
5& 1& 1 \
7& 1& 4
endvmatrix$$

share|cite|improve this answer

edited Aug 1 at 15:29

answered Aug 1 at 15:20

gimusi

63.9k73480

add a comment | 

up vote
2
down vote

No the Rule of Sarrus is only valid for 3-by -3 matrices, in general for n-by -n matrices we can refer to Laplace expansion method, that is by the first row

$$beginvmatrix
2& 2& 1& 3\
1& 4& 4& 5\
5& 1& 1& 6\
7& 1& 4& 5
endvmatrix=2beginvmatrix
4& 4& 5\
1& 1& 6\
1& 4& 5
endvmatrix-2beginvmatrix
1& 4& 5\
5& 1& 6\
7& 4& 5
endvmatrix+1beginvmatrix
1& 4& 5\
5& 1& 6\
7& 1& 5
endvmatrix-3beginvmatrix
1& 4& 4 \
5& 1& 1 \
7& 1& 4
endvmatrix$$

share|cite|improve this answer

edited Aug 1 at 15:29

answered Aug 1 at 15:20

gimusi

63.9k73480

add a comment | 

up vote
2
down vote

up vote
2
down vote

No the Rule of Sarrus is only valid for 3-by -3 matrices, in general for n-by -n matrices we can refer to Laplace expansion method, that is by the first row

$$beginvmatrix
2& 2& 1& 3\
1& 4& 4& 5\
5& 1& 1& 6\
7& 1& 4& 5
endvmatrix=2beginvmatrix
4& 4& 5\
1& 1& 6\
1& 4& 5
endvmatrix-2beginvmatrix
1& 4& 5\
5& 1& 6\
7& 4& 5
endvmatrix+1beginvmatrix
1& 4& 5\
5& 1& 6\
7& 1& 5
endvmatrix-3beginvmatrix
1& 4& 4 \
5& 1& 1 \
7& 1& 4
endvmatrix$$

share|cite|improve this answer

edited Aug 1 at 15:29

answered Aug 1 at 15:20

gimusi

63.9k73480

No the Rule of Sarrus is only valid for 3-by -3 matrices, in general for n-by -n matrices we can refer to Laplace expansion method, that is by the first row

$$beginvmatrix
2& 2& 1& 3\
1& 4& 4& 5\
5& 1& 1& 6\
7& 1& 4& 5
endvmatrix=2beginvmatrix
4& 4& 5\
1& 1& 6\
1& 4& 5
endvmatrix-2beginvmatrix
1& 4& 5\
5& 1& 6\
7& 4& 5
endvmatrix+1beginvmatrix
1& 4& 5\
5& 1& 6\
7& 1& 5
endvmatrix-3beginvmatrix
1& 4& 4 \
5& 1& 1 \
7& 1& 4
endvmatrix$$

share|cite|improve this answer

edited Aug 1 at 15:29

answered Aug 1 at 15:20

gimusi

63.9k73480

share|cite|improve this answer

share|cite|improve this answer

edited Aug 1 at 15:29

edited Aug 1 at 15:29

edited Aug 1 at 15:29

answered Aug 1 at 15:20

gimusi

63.9k73480

answered Aug 1 at 15:20

gimusi

63.9k73480

answered Aug 1 at 15:20

gimusi

63.9k73480

63.9k73480

add a comment | 

add a comment | 

up vote
1
down vote

No, it is not valid. My computation using Octave shows that the determinant is negative.

octave:1> A = [ 2 2 1 3; 1 4 4 5; 5 1 1 6; 7 1 4 5]
A =

 2 2 1 3
 1 4 4 5
 5 1 1 6
 7 1 4 5

octave:2> det(A)
ans = -114.00

share|cite|improve this answer

answered Aug 1 at 15:24

Siong Thye Goh

76.7k134794

add a comment | 

up vote
1
down vote

No, it is not valid. My computation using Octave shows that the determinant is negative.

octave:1> A = [ 2 2 1 3; 1 4 4 5; 5 1 1 6; 7 1 4 5]
A =

 2 2 1 3
 1 4 4 5
 5 1 1 6
 7 1 4 5

octave:2> det(A)
ans = -114.00

share|cite|improve this answer

answered Aug 1 at 15:24

Siong Thye Goh

76.7k134794

add a comment | 

up vote
1
down vote

up vote
1
down vote

No, it is not valid. My computation using Octave shows that the determinant is negative.

octave:1> A = [ 2 2 1 3; 1 4 4 5; 5 1 1 6; 7 1 4 5]
A =

 2 2 1 3
 1 4 4 5
 5 1 1 6
 7 1 4 5

octave:2> det(A)
ans = -114.00

share|cite|improve this answer

answered Aug 1 at 15:24

Siong Thye Goh

76.7k134794

No, it is not valid. My computation using Octave shows that the determinant is negative.

octave:1> A = [ 2 2 1 3; 1 4 4 5; 5 1 1 6; 7 1 4 5]
A =

 2 2 1 3
 1 4 4 5
 5 1 1 6
 7 1 4 5

octave:2> det(A)
ans = -114.00

share|cite|improve this answer

answered Aug 1 at 15:24

Siong Thye Goh

76.7k134794

share|cite|improve this answer

share|cite|improve this answer

answered Aug 1 at 15:24

Siong Thye Goh

76.7k134794

answered Aug 1 at 15:24

Siong Thye Goh

76.7k134794

answered Aug 1 at 15:24

Siong Thye Goh

76.7k134794

76.7k134794

add a comment | 

add a comment | 

up vote
1
down vote

The fastest method is Gaussian elimination to obtain a triangular matrix – in this case the determinant is the product of the diagonal elements:
beginalign
&beginvmatrix
2&2&1&3\1&4&4&5\5&1&1&6\7&1&4&5
endvmatrix
=-beginvmatrix
1&4&4&5\2&2&1&3\5&1&1&6\7&1&4&5
endvmatrix
=-beginvmatrixbeginarrayrrrr
1&4&4&5\0&-6&-7&-7\0&-19&-19&-19\0&-27&-24&-30
endarrayendvmatrix=19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&6&7&7\0&1&1&1\0&9&8&10
endarrayendvmatrix\[1ex]
=&-19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&1&1&1\0&6&7&7\0&9&8&10
endarrayendvmatrix=-19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&1&1&1\0&0&1&1\0&0&-1&1
endarrayendvmatrix
=-19times 3beginvmatrix
1&4&4&5\0&1&1&1\0&0&1&1\0&0&0&2
endvmatrix=-114\
&
endalign

However, in the case of $4times4$ matrices, you have two other possibilities, for a computation by blocks:

• If $M=beginpmatrixA&B\C&D endpmatrix$ is a $4×4$ matrix consisting of blocks of size $2$, and if $CD=DC$, then we can compute a $2×2$ determinant:
  $$det M=det(AD-BC).$$


• We can use Laplace's expansion along the first two columns. To explain the computation, we introduce some notations:

beginarrayll
p_ij&textis the $2×2$ determinant of rows $i$ and $j$ of the first two columns,\
q_ij&textis the $2×2$ determinant of rows $i$ and $j$ of the last two columns.
endarray
With these notations, one has
$$det M=p_12q_34+p_13q_42+p_14q_23+q_12p_34+q_13p_42+q_14p_23.$$

share|cite|improve this answer

answered Aug 1 at 16:45

Bernard

110k635102

add a comment | 

up vote
1
down vote

The fastest method is Gaussian elimination to obtain a triangular matrix – in this case the determinant is the product of the diagonal elements:
beginalign
&beginvmatrix
2&2&1&3\1&4&4&5\5&1&1&6\7&1&4&5
endvmatrix
=-beginvmatrix
1&4&4&5\2&2&1&3\5&1&1&6\7&1&4&5
endvmatrix
=-beginvmatrixbeginarrayrrrr
1&4&4&5\0&-6&-7&-7\0&-19&-19&-19\0&-27&-24&-30
endarrayendvmatrix=19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&6&7&7\0&1&1&1\0&9&8&10
endarrayendvmatrix\[1ex]
=&-19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&1&1&1\0&6&7&7\0&9&8&10
endarrayendvmatrix=-19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&1&1&1\0&0&1&1\0&0&-1&1
endarrayendvmatrix
=-19times 3beginvmatrix
1&4&4&5\0&1&1&1\0&0&1&1\0&0&0&2
endvmatrix=-114\
&
endalign

However, in the case of $4times4$ matrices, you have two other possibilities, for a computation by blocks:

• If $M=beginpmatrixA&B\C&D endpmatrix$ is a $4×4$ matrix consisting of blocks of size $2$, and if $CD=DC$, then we can compute a $2×2$ determinant:
  $$det M=det(AD-BC).$$


• We can use Laplace's expansion along the first two columns. To explain the computation, we introduce some notations:

beginarrayll
p_ij&textis the $2×2$ determinant of rows $i$ and $j$ of the first two columns,\
q_ij&textis the $2×2$ determinant of rows $i$ and $j$ of the last two columns.
endarray
With these notations, one has
$$det M=p_12q_34+p_13q_42+p_14q_23+q_12p_34+q_13p_42+q_14p_23.$$

share|cite|improve this answer

answered Aug 1 at 16:45

Bernard

110k635102

add a comment | 

up vote
1
down vote

up vote
1
down vote

The fastest method is Gaussian elimination to obtain a triangular matrix – in this case the determinant is the product of the diagonal elements:
beginalign
&beginvmatrix
2&2&1&3\1&4&4&5\5&1&1&6\7&1&4&5
endvmatrix
=-beginvmatrix
1&4&4&5\2&2&1&3\5&1&1&6\7&1&4&5
endvmatrix
=-beginvmatrixbeginarrayrrrr
1&4&4&5\0&-6&-7&-7\0&-19&-19&-19\0&-27&-24&-30
endarrayendvmatrix=19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&6&7&7\0&1&1&1\0&9&8&10
endarrayendvmatrix\[1ex]
=&-19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&1&1&1\0&6&7&7\0&9&8&10
endarrayendvmatrix=-19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&1&1&1\0&0&1&1\0&0&-1&1
endarrayendvmatrix
=-19times 3beginvmatrix
1&4&4&5\0&1&1&1\0&0&1&1\0&0&0&2
endvmatrix=-114\
&
endalign

However, in the case of $4times4$ matrices, you have two other possibilities, for a computation by blocks:

• If $M=beginpmatrixA&B\C&D endpmatrix$ is a $4×4$ matrix consisting of blocks of size $2$, and if $CD=DC$, then we can compute a $2×2$ determinant:
  $$det M=det(AD-BC).$$


• We can use Laplace's expansion along the first two columns. To explain the computation, we introduce some notations:

beginarrayll
p_ij&textis the $2×2$ determinant of rows $i$ and $j$ of the first two columns,\
q_ij&textis the $2×2$ determinant of rows $i$ and $j$ of the last two columns.
endarray
With these notations, one has
$$det M=p_12q_34+p_13q_42+p_14q_23+q_12p_34+q_13p_42+q_14p_23.$$

share|cite|improve this answer

answered Aug 1 at 16:45

Bernard

110k635102

The fastest method is Gaussian elimination to obtain a triangular matrix – in this case the determinant is the product of the diagonal elements:
beginalign
&beginvmatrix
2&2&1&3\1&4&4&5\5&1&1&6\7&1&4&5
endvmatrix
=-beginvmatrix
1&4&4&5\2&2&1&3\5&1&1&6\7&1&4&5
endvmatrix
=-beginvmatrixbeginarrayrrrr
1&4&4&5\0&-6&-7&-7\0&-19&-19&-19\0&-27&-24&-30
endarrayendvmatrix=19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&6&7&7\0&1&1&1\0&9&8&10
endarrayendvmatrix\[1ex]
=&-19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&1&1&1\0&6&7&7\0&9&8&10
endarrayendvmatrix=-19times 3beginvmatrixbeginarrayrrrr
1&4&4&5\0&1&1&1\0&0&1&1\0&0&-1&1
endarrayendvmatrix
=-19times 3beginvmatrix
1&4&4&5\0&1&1&1\0&0&1&1\0&0&0&2
endvmatrix=-114\
&
endalign

However, in the case of $4times4$ matrices, you have two other possibilities, for a computation by blocks:

• If $M=beginpmatrixA&B\C&D endpmatrix$ is a $4×4$ matrix consisting of blocks of size $2$, and if $CD=DC$, then we can compute a $2×2$ determinant:
  $$det M=det(AD-BC).$$


• We can use Laplace's expansion along the first two columns. To explain the computation, we introduce some notations:

beginarrayll
p_ij&textis the $2×2$ determinant of rows $i$ and $j$ of the first two columns,\
q_ij&textis the $2×2$ determinant of rows $i$ and $j$ of the last two columns.
endarray
With these notations, one has
$$det M=p_12q_34+p_13q_42+p_14q_23+q_12p_34+q_13p_42+q_14p_23.$$

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answered Aug 1 at 16:45

Bernard

110k635102

share|cite|improve this answer

share|cite|improve this answer

answered Aug 1 at 16:45

Bernard

110k635102

answered Aug 1 at 16:45

Bernard

110k635102

answered Aug 1 at 16:45

Bernard

110k635102

110k635102

add a comment | 

add a comment | 

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No, it isn't valid. When you calculate the determinant of an $ntimes n$ matrix by minors, you compute $n!$ products. When $n=3, n!=6$ and there are $6$ broken diagonals, and the "spaghetti rule" still involves $6$ products. But when $n=4, n!=24,$ and there are only $8$ broken diagonals, so there isn't much hope.

You can easily come up with examples to prove it doesn't work. (In fact, I imagine it's harder to come up with examples where it does work.)

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edited Aug 1 at 15:34

answered Aug 1 at 15:26

saulspatz

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up vote
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down vote

No, it isn't valid. When you calculate the determinant of an $ntimes n$ matrix by minors, you compute $n!$ products. When $n=3, n!=6$ and there are $6$ broken diagonals, and the "spaghetti rule" still involves $6$ products. But when $n=4, n!=24,$ and there are only $8$ broken diagonals, so there isn't much hope.

You can easily come up with examples to prove it doesn't work. (In fact, I imagine it's harder to come up with examples where it does work.)

share|cite|improve this answer

edited Aug 1 at 15:34

answered Aug 1 at 15:26

saulspatz

10.5k21324

add a comment | 

up vote
0
down vote

up vote
0
down vote

No, it isn't valid. When you calculate the determinant of an $ntimes n$ matrix by minors, you compute $n!$ products. When $n=3, n!=6$ and there are $6$ broken diagonals, and the "spaghetti rule" still involves $6$ products. But when $n=4, n!=24,$ and there are only $8$ broken diagonals, so there isn't much hope.

You can easily come up with examples to prove it doesn't work. (In fact, I imagine it's harder to come up with examples where it does work.)

share|cite|improve this answer

edited Aug 1 at 15:34

answered Aug 1 at 15:26

saulspatz

10.5k21324

No, it isn't valid. When you calculate the determinant of an $ntimes n$ matrix by minors, you compute $n!$ products. When $n=3, n!=6$ and there are $6$ broken diagonals, and the "spaghetti rule" still involves $6$ products. But when $n=4, n!=24,$ and there are only $8$ broken diagonals, so there isn't much hope.

You can easily come up with examples to prove it doesn't work. (In fact, I imagine it's harder to come up with examples where it does work.)

share|cite|improve this answer

edited Aug 1 at 15:34

answered Aug 1 at 15:26

saulspatz

10.5k21324

share|cite|improve this answer

share|cite|improve this answer

edited Aug 1 at 15:34

edited Aug 1 at 15:34

edited Aug 1 at 15:34

answered Aug 1 at 15:26

saulspatz

10.5k21324

answered Aug 1 at 15:26

saulspatz

10.5k21324

answered Aug 1 at 15:26

saulspatz

10.5k21324

10.5k21324

add a comment | 

add a comment |