Computing the determinant of a $4times4$ matrix

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up vote
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Compute the determinant of beginvmatrix 2 & -1 & 1 & 3 \ 1 & 2 & 0 & 1 \ 2 & 1 & 1 & -1 \ -1 & 1 & 2 & -2 endvmatrix

My try:
$$beginvmatrix 2 & -1 & 1 & 3 \ 1 & 2 & 0 & 1 \ 2 & 1 & 1 & -1 \ -1 & 1 & 2 & -2 endvmatrix_R_2rightarrow2R_2-R_1\R_3rightarrow R_3-R_1\R_4rightarrow2R_4+R_1$$
$$beginvmatrix 2 & -1 & 1 & 3 \ 0 & 5 & -1 & -1 \ 0 & 2 & 0 & -4 \ 0 & 1 & 5 & -1 endvmatrix$$
Now I took $$beginvmatrix 5 & -1 & -1 \ 2 & 0 & -4 \ 1 & 5 & -1 endvmatrix=92$$
But the answer is $46$. Where did I go wrong?

linear-algebra

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asked Aug 2 at 15:27

philip

938

add a comment | 

up vote
2
down vote

favorite

Compute the determinant of beginvmatrix 2 & -1 & 1 & 3 \ 1 & 2 & 0 & 1 \ 2 & 1 & 1 & -1 \ -1 & 1 & 2 & -2 endvmatrix

My try:
$$beginvmatrix 2 & -1 & 1 & 3 \ 1 & 2 & 0 & 1 \ 2 & 1 & 1 & -1 \ -1 & 1 & 2 & -2 endvmatrix_R_2rightarrow2R_2-R_1\R_3rightarrow R_3-R_1\R_4rightarrow2R_4+R_1$$
$$beginvmatrix 2 & -1 & 1 & 3 \ 0 & 5 & -1 & -1 \ 0 & 2 & 0 & -4 \ 0 & 1 & 5 & -1 endvmatrix$$
Now I took $$beginvmatrix 5 & -1 & -1 \ 2 & 0 & -4 \ 1 & 5 & -1 endvmatrix=92$$
But the answer is $46$. Where did I go wrong?

linear-algebra

share|cite|improve this question

asked Aug 2 at 15:27

philip

938

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Compute the determinant of beginvmatrix 2 & -1 & 1 & 3 \ 1 & 2 & 0 & 1 \ 2 & 1 & 1 & -1 \ -1 & 1 & 2 & -2 endvmatrix

My try:
$$beginvmatrix 2 & -1 & 1 & 3 \ 1 & 2 & 0 & 1 \ 2 & 1 & 1 & -1 \ -1 & 1 & 2 & -2 endvmatrix_R_2rightarrow2R_2-R_1\R_3rightarrow R_3-R_1\R_4rightarrow2R_4+R_1$$
$$beginvmatrix 2 & -1 & 1 & 3 \ 0 & 5 & -1 & -1 \ 0 & 2 & 0 & -4 \ 0 & 1 & 5 & -1 endvmatrix$$
Now I took $$beginvmatrix 5 & -1 & -1 \ 2 & 0 & -4 \ 1 & 5 & -1 endvmatrix=92$$
But the answer is $46$. Where did I go wrong?

linear-algebra

share|cite|improve this question

asked Aug 2 at 15:27

philip

938

Compute the determinant of beginvmatrix 2 & -1 & 1 & 3 \ 1 & 2 & 0 & 1 \ 2 & 1 & 1 & -1 \ -1 & 1 & 2 & -2 endvmatrix

My try:
$$beginvmatrix 2 & -1 & 1 & 3 \ 1 & 2 & 0 & 1 \ 2 & 1 & 1 & -1 \ -1 & 1 & 2 & -2 endvmatrix_R_2rightarrow2R_2-R_1\R_3rightarrow R_3-R_1\R_4rightarrow2R_4+R_1$$
$$beginvmatrix 2 & -1 & 1 & 3 \ 0 & 5 & -1 & -1 \ 0 & 2 & 0 & -4 \ 0 & 1 & 5 & -1 endvmatrix$$
Now I took $$beginvmatrix 5 & -1 & -1 \ 2 & 0 & -4 \ 1 & 5 & -1 endvmatrix=92$$
But the answer is $46$. Where did I go wrong?

linear-algebra

share|cite|improve this question

asked Aug 2 at 15:27

philip

938

share|cite|improve this question

share|cite|improve this question

asked Aug 2 at 15:27

philip

938

asked Aug 2 at 15:27

philip

938

asked Aug 2 at 15:27

philip

938

938

add a comment | 

add a comment | 

4 Answers
4

active

oldest

votes

up vote
3
down vote



accepted

When doing some operations on rows and columns of a matrix you have to be pretty careful: remember that the determinant is a multilinear map in the rows and the columns so it follows that

Multiplying or dividing a row or a column by a scalar $lambda$, multiplies or divides the determinant by the same scalar $lambda$

So if $A$ is your initial matrix and $B$ is the matrix obtained by multiplying a column of $A$ by a scalar $lambda$ you have that $$det(B) = lambdadet(A)$$
So in your case you've done $2$ multiplications by $2$ then the determinant of your final matrix will be $4$ times the determinant of the initial matrix. So you have that your determinant will be $$92/4times 2 = 46$$

Obviously the times two is there for the Laplace expansion.

share|cite|improve this answer

edited Aug 2 at 15:45

answered Aug 2 at 15:41

Davide Morgante

1,634220

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (96/4)*2 = 48 …
  – current_user
  Aug 2 at 15:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @current_user Yeah, just mistyped a LOT! ahah thanks
  – Davide Morgante
  Aug 2 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavideMorgante why did you multiply $dfrac924$ with $2$
  – philip
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As I said in the post is for the Laplace expansion: you're expanding your determinant on the first column so, with alternating signs, you multiply the determinant of the minors (that are found taking away the row and the column of a chosen element) by the element that identify the minor. In your case 3 out of 4 elements are zeros so the determinant of that minors is zero and the first one is 2 so the determinant has to be multiplied by 2
  – Davide Morgante
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Because you multiplied the first and the third rows by 2 while making the row operations, hence the determinant you found is 4 times bigger than the determinant of the original matrix. So the answer is $frac92*24$=46.

share|cite|improve this answer

answered Aug 2 at 15:39

Mark

5949

add a comment | 

up vote
1
down vote

I'm sure this question has been asked and answered many times, but I can't find it.

You can add or subtract a multiple of another row to a given row, without changing the value of the determinant, so $$R_1=R_1+4R_2$$
is perfectly OK. When you multiply a row by $k$, you multiply the determinant by $k$. When you do $$R_2=2R_2-R_1$$ you are really doing two operations. First, you multiply row $2$ by $2$, then you subtract row $1$ from row $2$. The effect of this is to multiply the determinant by $2$.

It's fine to do these operations, as long as you keep track of them, so you can adjust the value of the determinant at the end.

The second mistake you made is at the end. Because of the $2$ in the $(1,1)$ position, you should have multiplied the $3times3$ by $2$ giving $184$, which is $4$ times the true value. This is because you performed that "row-doubling" operation twice.

share|cite|improve this answer

answered Aug 2 at 15:40

saulspatz

10.5k21324

add a comment | 

up vote
1
down vote

You multiplied two rows by two, so your answer (184) is 4 times greater than the actual determinant. This video may help.

share|cite|improve this answer

answered Aug 2 at 15:41

Meeta Jo

1418

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

up vote
3
down vote



accepted

When doing some operations on rows and columns of a matrix you have to be pretty careful: remember that the determinant is a multilinear map in the rows and the columns so it follows that

Multiplying or dividing a row or a column by a scalar $lambda$, multiplies or divides the determinant by the same scalar $lambda$

So if $A$ is your initial matrix and $B$ is the matrix obtained by multiplying a column of $A$ by a scalar $lambda$ you have that $$det(B) = lambdadet(A)$$
So in your case you've done $2$ multiplications by $2$ then the determinant of your final matrix will be $4$ times the determinant of the initial matrix. So you have that your determinant will be $$92/4times 2 = 46$$

Obviously the times two is there for the Laplace expansion.

share|cite|improve this answer

edited Aug 2 at 15:45

answered Aug 2 at 15:41

Davide Morgante

1,634220

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (96/4)*2 = 48 …
  – current_user
  Aug 2 at 15:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @current_user Yeah, just mistyped a LOT! ahah thanks
  – Davide Morgante
  Aug 2 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavideMorgante why did you multiply $dfrac924$ with $2$
  – philip
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As I said in the post is for the Laplace expansion: you're expanding your determinant on the first column so, with alternating signs, you multiply the determinant of the minors (that are found taking away the row and the column of a chosen element) by the element that identify the minor. In your case 3 out of 4 elements are zeros so the determinant of that minors is zero and the first one is 2 so the determinant has to be multiplied by 2
  – Davide Morgante
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

When doing some operations on rows and columns of a matrix you have to be pretty careful: remember that the determinant is a multilinear map in the rows and the columns so it follows that

Multiplying or dividing a row or a column by a scalar $lambda$, multiplies or divides the determinant by the same scalar $lambda$

So if $A$ is your initial matrix and $B$ is the matrix obtained by multiplying a column of $A$ by a scalar $lambda$ you have that $$det(B) = lambdadet(A)$$
So in your case you've done $2$ multiplications by $2$ then the determinant of your final matrix will be $4$ times the determinant of the initial matrix. So you have that your determinant will be $$92/4times 2 = 46$$

Obviously the times two is there for the Laplace expansion.

share|cite|improve this answer

edited Aug 2 at 15:45

answered Aug 2 at 15:41

Davide Morgante

1,634220

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (96/4)*2 = 48 …
  – current_user
  Aug 2 at 15:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @current_user Yeah, just mistyped a LOT! ahah thanks
  – Davide Morgante
  Aug 2 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavideMorgante why did you multiply $dfrac924$ with $2$
  – philip
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As I said in the post is for the Laplace expansion: you're expanding your determinant on the first column so, with alternating signs, you multiply the determinant of the minors (that are found taking away the row and the column of a chosen element) by the element that identify the minor. In your case 3 out of 4 elements are zeros so the determinant of that minors is zero and the first one is 2 so the determinant has to be multiplied by 2
  – Davide Morgante
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

When doing some operations on rows and columns of a matrix you have to be pretty careful: remember that the determinant is a multilinear map in the rows and the columns so it follows that

Multiplying or dividing a row or a column by a scalar $lambda$, multiplies or divides the determinant by the same scalar $lambda$

So if $A$ is your initial matrix and $B$ is the matrix obtained by multiplying a column of $A$ by a scalar $lambda$ you have that $$det(B) = lambdadet(A)$$
So in your case you've done $2$ multiplications by $2$ then the determinant of your final matrix will be $4$ times the determinant of the initial matrix. So you have that your determinant will be $$92/4times 2 = 46$$

Obviously the times two is there for the Laplace expansion.

share|cite|improve this answer

edited Aug 2 at 15:45

answered Aug 2 at 15:41

Davide Morgante

1,634220

When doing some operations on rows and columns of a matrix you have to be pretty careful: remember that the determinant is a multilinear map in the rows and the columns so it follows that

Multiplying or dividing a row or a column by a scalar $lambda$, multiplies or divides the determinant by the same scalar $lambda$

So if $A$ is your initial matrix and $B$ is the matrix obtained by multiplying a column of $A$ by a scalar $lambda$ you have that $$det(B) = lambdadet(A)$$
So in your case you've done $2$ multiplications by $2$ then the determinant of your final matrix will be $4$ times the determinant of the initial matrix. So you have that your determinant will be $$92/4times 2 = 46$$

Obviously the times two is there for the Laplace expansion.

share|cite|improve this answer

edited Aug 2 at 15:45

answered Aug 2 at 15:41

Davide Morgante

1,634220

share|cite|improve this answer

share|cite|improve this answer

edited Aug 2 at 15:45

edited Aug 2 at 15:45

edited Aug 2 at 15:45

answered Aug 2 at 15:41

Davide Morgante

1,634220

answered Aug 2 at 15:41

Davide Morgante

1,634220

answered Aug 2 at 15:41

Davide Morgante

1,634220

1,634220

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (96/4)*2 = 48 …
  – current_user
  Aug 2 at 15:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @current_user Yeah, just mistyped a LOT! ahah thanks
  – Davide Morgante
  Aug 2 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavideMorgante why did you multiply $dfrac924$ with $2$
  – philip
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As I said in the post is for the Laplace expansion: you're expanding your determinant on the first column so, with alternating signs, you multiply the determinant of the minors (that are found taking away the row and the column of a chosen element) by the element that identify the minor. In your case 3 out of 4 elements are zeros so the determinant of that minors is zero and the first one is 2 so the determinant has to be multiplied by 2
  – Davide Morgante
  12 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (96/4)*2 = 48 …
  – current_user
  Aug 2 at 15:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @current_user Yeah, just mistyped a LOT! ahah thanks
  – Davide Morgante
  Aug 2 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavideMorgante why did you multiply $dfrac924$ with $2$
  – philip
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  As I said in the post is for the Laplace expansion: you're expanding your determinant on the first column so, with alternating signs, you multiply the determinant of the minors (that are found taking away the row and the column of a chosen element) by the element that identify the minor. In your case 3 out of 4 elements are zeros so the determinant of that minors is zero and the first one is 2 so the determinant has to be multiplied by 2
  – Davide Morgante
  12 hours ago
  

  
  
  
  

1

1

(96/4)*2 = 48 …
– current_user
Aug 2 at 15:44

(96/4)*2 = 48 …
– current_user
Aug 2 at 15:44

1

1

@current_user Yeah, just mistyped a LOT! ahah thanks
– Davide Morgante
Aug 2 at 15:45

@current_user Yeah, just mistyped a LOT! ahah thanks
– Davide Morgante
Aug 2 at 15:45

@DavideMorgante why did you multiply $dfrac924$ with $2$
– philip
12 hours ago

@DavideMorgante why did you multiply $dfrac924$ with $2$
– philip
12 hours ago

As I said in the post is for the Laplace expansion: you're expanding your determinant on the first column so, with alternating signs, you multiply the determinant of the minors (that are found taking away the row and the column of a chosen element) by the element that identify the minor. In your case 3 out of 4 elements are zeros so the determinant of that minors is zero and the first one is 2 so the determinant has to be multiplied by 2
– Davide Morgante
12 hours ago

As I said in the post is for the Laplace expansion: you're expanding your determinant on the first column so, with alternating signs, you multiply the determinant of the minors (that are found taking away the row and the column of a chosen element) by the element that identify the minor. In your case 3 out of 4 elements are zeros so the determinant of that minors is zero and the first one is 2 so the determinant has to be multiplied by 2
– Davide Morgante
12 hours ago

add a comment | 

up vote
2
down vote

Because you multiplied the first and the third rows by 2 while making the row operations, hence the determinant you found is 4 times bigger than the determinant of the original matrix. So the answer is $frac92*24$=46.

share|cite|improve this answer

answered Aug 2 at 15:39

Mark

5949

add a comment | 

up vote
2
down vote

Because you multiplied the first and the third rows by 2 while making the row operations, hence the determinant you found is 4 times bigger than the determinant of the original matrix. So the answer is $frac92*24$=46.

share|cite|improve this answer

answered Aug 2 at 15:39

Mark

5949

add a comment | 

up vote
2
down vote

up vote
2
down vote

Because you multiplied the first and the third rows by 2 while making the row operations, hence the determinant you found is 4 times bigger than the determinant of the original matrix. So the answer is $frac92*24$=46.

share|cite|improve this answer

answered Aug 2 at 15:39

Mark

5949

Because you multiplied the first and the third rows by 2 while making the row operations, hence the determinant you found is 4 times bigger than the determinant of the original matrix. So the answer is $frac92*24$=46.

share|cite|improve this answer

answered Aug 2 at 15:39

Mark

5949

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 15:39

Mark

5949

answered Aug 2 at 15:39

Mark

5949

answered Aug 2 at 15:39

Mark

5949

5949

add a comment | 

add a comment | 

up vote
1
down vote

I'm sure this question has been asked and answered many times, but I can't find it.

You can add or subtract a multiple of another row to a given row, without changing the value of the determinant, so $$R_1=R_1+4R_2$$
is perfectly OK. When you multiply a row by $k$, you multiply the determinant by $k$. When you do $$R_2=2R_2-R_1$$ you are really doing two operations. First, you multiply row $2$ by $2$, then you subtract row $1$ from row $2$. The effect of this is to multiply the determinant by $2$.

It's fine to do these operations, as long as you keep track of them, so you can adjust the value of the determinant at the end.

The second mistake you made is at the end. Because of the $2$ in the $(1,1)$ position, you should have multiplied the $3times3$ by $2$ giving $184$, which is $4$ times the true value. This is because you performed that "row-doubling" operation twice.

share|cite|improve this answer

answered Aug 2 at 15:40

saulspatz

10.5k21324

add a comment | 

up vote
1
down vote

I'm sure this question has been asked and answered many times, but I can't find it.

You can add or subtract a multiple of another row to a given row, without changing the value of the determinant, so $$R_1=R_1+4R_2$$
is perfectly OK. When you multiply a row by $k$, you multiply the determinant by $k$. When you do $$R_2=2R_2-R_1$$ you are really doing two operations. First, you multiply row $2$ by $2$, then you subtract row $1$ from row $2$. The effect of this is to multiply the determinant by $2$.

It's fine to do these operations, as long as you keep track of them, so you can adjust the value of the determinant at the end.

The second mistake you made is at the end. Because of the $2$ in the $(1,1)$ position, you should have multiplied the $3times3$ by $2$ giving $184$, which is $4$ times the true value. This is because you performed that "row-doubling" operation twice.

share|cite|improve this answer

answered Aug 2 at 15:40

saulspatz

10.5k21324

add a comment | 

up vote
1
down vote

up vote
1
down vote

I'm sure this question has been asked and answered many times, but I can't find it.

You can add or subtract a multiple of another row to a given row, without changing the value of the determinant, so $$R_1=R_1+4R_2$$
is perfectly OK. When you multiply a row by $k$, you multiply the determinant by $k$. When you do $$R_2=2R_2-R_1$$ you are really doing two operations. First, you multiply row $2$ by $2$, then you subtract row $1$ from row $2$. The effect of this is to multiply the determinant by $2$.

It's fine to do these operations, as long as you keep track of them, so you can adjust the value of the determinant at the end.

The second mistake you made is at the end. Because of the $2$ in the $(1,1)$ position, you should have multiplied the $3times3$ by $2$ giving $184$, which is $4$ times the true value. This is because you performed that "row-doubling" operation twice.

share|cite|improve this answer

answered Aug 2 at 15:40

saulspatz

10.5k21324

I'm sure this question has been asked and answered many times, but I can't find it.

You can add or subtract a multiple of another row to a given row, without changing the value of the determinant, so $$R_1=R_1+4R_2$$
is perfectly OK. When you multiply a row by $k$, you multiply the determinant by $k$. When you do $$R_2=2R_2-R_1$$ you are really doing two operations. First, you multiply row $2$ by $2$, then you subtract row $1$ from row $2$. The effect of this is to multiply the determinant by $2$.

It's fine to do these operations, as long as you keep track of them, so you can adjust the value of the determinant at the end.

The second mistake you made is at the end. Because of the $2$ in the $(1,1)$ position, you should have multiplied the $3times3$ by $2$ giving $184$, which is $4$ times the true value. This is because you performed that "row-doubling" operation twice.

share|cite|improve this answer

answered Aug 2 at 15:40

saulspatz

10.5k21324

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 15:40

saulspatz

10.5k21324

answered Aug 2 at 15:40

saulspatz

10.5k21324

answered Aug 2 at 15:40

saulspatz

10.5k21324

10.5k21324

add a comment | 

add a comment | 

up vote
1
down vote

You multiplied two rows by two, so your answer (184) is 4 times greater than the actual determinant. This video may help.

share|cite|improve this answer

answered Aug 2 at 15:41

Meeta Jo

1418

add a comment | 

up vote
1
down vote

You multiplied two rows by two, so your answer (184) is 4 times greater than the actual determinant. This video may help.

share|cite|improve this answer

answered Aug 2 at 15:41

Meeta Jo

1418

add a comment | 

up vote
1
down vote

up vote
1
down vote

You multiplied two rows by two, so your answer (184) is 4 times greater than the actual determinant. This video may help.

share|cite|improve this answer

answered Aug 2 at 15:41

Meeta Jo

1418

You multiplied two rows by two, so your answer (184) is 4 times greater than the actual determinant. This video may help.

share|cite|improve this answer

answered Aug 2 at 15:41

Meeta Jo

1418

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 15:41

Meeta Jo

1418

answered Aug 2 at 15:41

Meeta Jo

1418

answered Aug 2 at 15:41

Meeta Jo

1418

1418

add a comment | 

add a comment | 

 

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