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:



enter image description here



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




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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














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:



enter image description here



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




share|cite|improve this question















  • 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












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:



enter image description here



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




share|cite|improve this question











I'm reading this text and I'm a bit confused about how they determined the determinant of this matrix:



enter image description here



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





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









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












  • 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










1 Answer
1






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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



$$beginvmatrix1&0\1&1endvmatrix=1$$






share|cite|improve this answer























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    1 Answer
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    active

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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



    $$beginvmatrix1&0\1&1endvmatrix=1$$






    share|cite|improve this answer



























      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



      $$beginvmatrix1&0\1&1endvmatrix=1$$






      share|cite|improve this answer

























        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



        $$beginvmatrix1&0\1&1endvmatrix=1$$






        share|cite|improve this answer















        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



        $$beginvmatrix1&0\1&1endvmatrix=1$$







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 6 at 21:15


























        answered Aug 6 at 21:09









        gimusi

        65.5k73684




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