Mixing
Factoring out multiples to determine determinants. Linear algebra
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I'm reading this text and I'm a bit confused about how they determined the determinant of this matrix:
The factoring out of the -7 and the -3 is because of rule 2 in theoreum 3.3 right?
What is the intuition behind these rules? Aren't row operations supposed to keep matrices equal? Aren't matrices that result from the elementary row operations supposed to be equivalent? So why would that change the determinants if you say interchange two rows?
linear-algebra
asked Aug 6 at 21:06
Jwan622
1,61211224
add a comment |Â
up vote
0
down vote
favorite
I'm reading this text and I'm a bit confused about how they determined the determinant of this matrix:
The factoring out of the -7 and the -3 is because of rule 2 in theoreum 3.3 right?
What is the intuition behind these rules? Aren't row operations supposed to keep matrices equal? Aren't matrices that result from the elementary row operations supposed to be equivalent? So why would that change the determinants if you say interchange two rows?
linear-algebra
asked Aug 6 at 21:06
Jwan622
1,61211224
2
Row operations do not keep matrices "equal" but they do keep them "equivalent." There are many similarities between two matrices which are equivalent, such as dimension of the row space, dimension of the column space, dimension of the kernel, invertability, etc... but of course equality is not one of them.
– JMoravitz
Aug 6 at 21:09
1
Perhaps a clearer explanation of what is happening is in pointing out the property that for two square matrices $A$ and $B$ of the same shape, one has that $det(AB)=det(A)cdot det(B)$. Then remember that performing elementary row operations is like "factoring out elementary matrices." (note: to be fair, the properties mentioned in the image are usually used to prove the property I mention here in my comment)
– JMoravitz
Aug 6 at 21:11
Indeed, in some treatments those rules are taken as defining properties of the determinant.
– amd
Aug 6 at 22:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm reading this text and I'm a bit confused about how they determined the determinant of this matrix:
The factoring out of the -7 and the -3 is because of rule 2 in theoreum 3.3 right?
What is the intuition behind these rules? Aren't row operations supposed to keep matrices equal? Aren't matrices that result from the elementary row operations supposed to be equivalent? So why would that change the determinants if you say interchange two rows?
linear-algebra
asked Aug 6 at 21:06
Jwan622
1,61211224
I'm reading this text and I'm a bit confused about how they determined the determinant of this matrix:
The factoring out of the -7 and the -3 is because of rule 2 in theoreum 3.3 right?
What is the intuition behind these rules? Aren't row operations supposed to keep matrices equal? Aren't matrices that result from the elementary row operations supposed to be equivalent? So why would that change the determinants if you say interchange two rows?
linear-algebra
asked Aug 6 at 21:06
Jwan622
1,61211224
asked Aug 6 at 21:06
Jwan622
1,61211224
asked Aug 6 at 21:06
Jwan622
1,61211224
asked Aug 6 at 21:06
Jwan622
1,61211224
1,61211224
2
Row operations do not keep matrices "equal" but they do keep them "equivalent." There are many similarities between two matrices which are equivalent, such as dimension of the row space, dimension of the column space, dimension of the kernel, invertability, etc... but of course equality is not one of them.
– JMoravitz
Aug 6 at 21:09
1
Perhaps a clearer explanation of what is happening is in pointing out the property that for two square matrices $A$ and $B$ of the same shape, one has that $det(AB)=det(A)cdot det(B)$. Then remember that performing elementary row operations is like "factoring out elementary matrices." (note: to be fair, the properties mentioned in the image are usually used to prove the property I mention here in my comment)
– JMoravitz
Aug 6 at 21:11
Indeed, in some treatments those rules are taken as defining properties of the determinant.
– amd
Aug 6 at 22:07
add a comment |Â
2
Row operations do not keep matrices "equal" but they do keep them "equivalent." There are many similarities between two matrices which are equivalent, such as dimension of the row space, dimension of the column space, dimension of the kernel, invertability, etc... but of course equality is not one of them.
– JMoravitz
Aug 6 at 21:09
1
Perhaps a clearer explanation of what is happening is in pointing out the property that for two square matrices $A$ and $B$ of the same shape, one has that $det(AB)=det(A)cdot det(B)$. Then remember that performing elementary row operations is like "factoring out elementary matrices." (note: to be fair, the properties mentioned in the image are usually used to prove the property I mention here in my comment)
– JMoravitz
Aug 6 at 21:11
Indeed, in some treatments those rules are taken as defining properties of the determinant.
– amd
Aug 6 at 22:07
2
2
Row operations do not keep matrices "equal" but they do keep them "equivalent." There are many similarities between two matrices which are equivalent, such as dimension of the row space, dimension of the column space, dimension of the kernel, invertability, etc... but of course equality is not one of them.
– JMoravitz
Aug 6 at 21:09
Row operations do not keep matrices "equal" but they do keep them "equivalent." There are many similarities between two matrices which are equivalent, such as dimension of the row space, dimension of the column space, dimension of the kernel, invertability, etc... but of course equality is not one of them.
– JMoravitz
Aug 6 at 21:09
1
1
Perhaps a clearer explanation of what is happening is in pointing out the property that for two square matrices $A$ and $B$ of the same shape, one has that $det(AB)=det(A)cdot det(B)$. Then remember that performing elementary row operations is like "factoring out elementary matrices." (note: to be fair, the properties mentioned in the image are usually used to prove the property I mention here in my comment)
– JMoravitz
Aug 6 at 21:11
Perhaps a clearer explanation of what is happening is in pointing out the property that for two square matrices $A$ and $B$ of the same shape, one has that $det(AB)=det(A)cdot det(B)$. Then remember that performing elementary row operations is like "factoring out elementary matrices." (note: to be fair, the properties mentioned in the image are usually used to prove the property I mention here in my comment)
– JMoravitz
Aug 6 at 21:11
Indeed, in some treatments those rules are taken as defining properties of the determinant.
– amd
Aug 6 at 22:07
Indeed, in some treatments those rules are taken as defining properties of the determinant.
– amd
Aug 6 at 22:07
add a comment |Â
1 Answer
1
active
oldest
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up vote
2
down vote
For a simple intuition and check of the Properties of the determinant let consider
$$beginvmatrix1&0\0&1endvmatrix=1$$
but for interchanging rows/colums
$$beginvmatrix0&1\1&0endvmatrix=-1$$
multiply a row/column by a scalar
$$beginvmatrix2&0\0&1endvmatrix=2$$
add first row to second row
$$beginvmatrix1&0\1&1endvmatrix=1$$
answered Aug 6 at 21:09
gimusi
65.5k73684
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
For a simple intuition and check of the Properties of the determinant let consider
$$beginvmatrix1&0\0&1endvmatrix=1$$
but for interchanging rows/colums
$$beginvmatrix0&1\1&0endvmatrix=-1$$
multiply a row/column by a scalar
$$beginvmatrix2&0\0&1endvmatrix=2$$
add first row to second row
$$beginvmatrix1&0\1&1endvmatrix=1$$
answered Aug 6 at 21:09
gimusi
65.5k73684
add a comment |Â
up vote
2
down vote
For a simple intuition and check of the Properties of the determinant let consider
$$beginvmatrix1&0\0&1endvmatrix=1$$
but for interchanging rows/colums
$$beginvmatrix0&1\1&0endvmatrix=-1$$
multiply a row/column by a scalar
$$beginvmatrix2&0\0&1endvmatrix=2$$
add first row to second row
$$beginvmatrix1&0\1&1endvmatrix=1$$
answered Aug 6 at 21:09
gimusi
65.5k73684
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For a simple intuition and check of the Properties of the determinant let consider
$$beginvmatrix1&0\0&1endvmatrix=1$$
but for interchanging rows/colums
$$beginvmatrix0&1\1&0endvmatrix=-1$$
multiply a row/column by a scalar
$$beginvmatrix2&0\0&1endvmatrix=2$$
add first row to second row
$$beginvmatrix1&0\1&1endvmatrix=1$$
answered Aug 6 at 21:09
gimusi
65.5k73684
For a simple intuition and check of the Properties of the determinant let consider
$$beginvmatrix1&0\0&1endvmatrix=1$$
but for interchanging rows/colums
$$beginvmatrix0&1\1&0endvmatrix=-1$$
multiply a row/column by a scalar
$$beginvmatrix2&0\0&1endvmatrix=2$$
add first row to second row
$$beginvmatrix1&0\1&1endvmatrix=1$$
answered Aug 6 at 21:09
gimusi
65.5k73684
edited Aug 6 at 21:15
answered Aug 6 at 21:09
gimusi
65.5k73684
answered Aug 6 at 21:09
gimusi
65.5k73684
answered Aug 6 at 21:09
gimusi
65.5k73684
65.5k73684
add a comment |Â
add a comment |Â
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2
Row operations do not keep matrices "equal" but they do keep them "equivalent." There are many similarities between two matrices which are equivalent, such as dimension of the row space, dimension of the column space, dimension of the kernel, invertability, etc... but of course equality is not one of them.
– JMoravitz
Aug 6 at 21:09
1
Perhaps a clearer explanation of what is happening is in pointing out the property that for two square matrices $A$ and $B$ of the same shape, one has that $det(AB)=det(A)cdot det(B)$. Then remember that performing elementary row operations is like "factoring out elementary matrices." (note: to be fair, the properties mentioned in the image are usually used to prove the property I mention here in my comment)
– JMoravitz
Aug 6 at 21:11
Indeed, in some treatments those rules are taken as defining properties of the determinant.
– amd
Aug 6 at 22:07