Solution Check for Determinant

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Q1



Solution is 4.



Original matrix is simply [v1;v2;v3;v4]. It forms an identity matrix. Hence the only alteration of the determinant comes from row 1 operation where v1 is multiplied by 2. Then the determinant will also be multiplied by two so 2*2 =4.



We ignore the rest of row operations since they have no effect on the determinant.



Is this line of reasoning correct?




linear-algebra matrices determinant




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  • 2




    No. 2on row is multiplied by 4 and 4th row is multiplied by 2.
    – xarles
    2 days ago










  • math.stackexchange.com/questions/1802752/… From other sources, it appears that it does not have effect on determinant. Only row 1 has an effective change on the initial determinant A = 2
    – PERTURBATIONFLOW
    2 days ago






  • 1




    Adding a multiple of a row to another row does not affect the determinant. But you are adding a multiple of a row to a MULTIPLE of another row, so the determinant gets affected.
    – xarles
    2 days ago










  • I understand now. So in this case, we have three effects on the determinant. 1) first row is multiplied by 2. 2) Second row is multiplied by 4. 3) fourth row is multiplied by 2. Then in effect, our determinant is now 2*4*2*2? So 64?
    – PERTURBATIONFLOW
    2 days ago















up vote
0
down vote

favorite












Q1



Solution is 4.



Original matrix is simply [v1;v2;v3;v4]. It forms an identity matrix. Hence the only alteration of the determinant comes from row 1 operation where v1 is multiplied by 2. Then the determinant will also be multiplied by two so 2*2 =4.



We ignore the rest of row operations since they have no effect on the determinant.



Is this line of reasoning correct?




linear-algebra matrices determinant




share|cite|improve this question















  • 2




    No. 2on row is multiplied by 4 and 4th row is multiplied by 2.
    – xarles
    2 days ago










  • math.stackexchange.com/questions/1802752/… From other sources, it appears that it does not have effect on determinant. Only row 1 has an effective change on the initial determinant A = 2
    – PERTURBATIONFLOW
    2 days ago






  • 1




    Adding a multiple of a row to another row does not affect the determinant. But you are adding a multiple of a row to a MULTIPLE of another row, so the determinant gets affected.
    – xarles
    2 days ago










  • I understand now. So in this case, we have three effects on the determinant. 1) first row is multiplied by 2. 2) Second row is multiplied by 4. 3) fourth row is multiplied by 2. Then in effect, our determinant is now 2*4*2*2? So 64?
    – PERTURBATIONFLOW
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Q1



Solution is 4.



Original matrix is simply [v1;v2;v3;v4]. It forms an identity matrix. Hence the only alteration of the determinant comes from row 1 operation where v1 is multiplied by 2. Then the determinant will also be multiplied by two so 2*2 =4.



We ignore the rest of row operations since they have no effect on the determinant.



Is this line of reasoning correct?




linear-algebra matrices determinant




share|cite|improve this question











Q1



Solution is 4.



Original matrix is simply [v1;v2;v3;v4]. It forms an identity matrix. Hence the only alteration of the determinant comes from row 1 operation where v1 is multiplied by 2. Then the determinant will also be multiplied by two so 2*2 =4.



We ignore the rest of row operations since they have no effect on the determinant.



Is this line of reasoning correct?





linear-algebra matrices determinant





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked 2 days ago









PERTURBATIONFLOW

396




396







  • 2




    No. 2on row is multiplied by 4 and 4th row is multiplied by 2.
    – xarles
    2 days ago










  • math.stackexchange.com/questions/1802752/… From other sources, it appears that it does not have effect on determinant. Only row 1 has an effective change on the initial determinant A = 2
    – PERTURBATIONFLOW
    2 days ago






  • 1




    Adding a multiple of a row to another row does not affect the determinant. But you are adding a multiple of a row to a MULTIPLE of another row, so the determinant gets affected.
    – xarles
    2 days ago










  • I understand now. So in this case, we have three effects on the determinant. 1) first row is multiplied by 2. 2) Second row is multiplied by 4. 3) fourth row is multiplied by 2. Then in effect, our determinant is now 2*4*2*2? So 64?
    – PERTURBATIONFLOW
    2 days ago













  • 2




    No. 2on row is multiplied by 4 and 4th row is multiplied by 2.
    – xarles
    2 days ago










  • math.stackexchange.com/questions/1802752/… From other sources, it appears that it does not have effect on determinant. Only row 1 has an effective change on the initial determinant A = 2
    – PERTURBATIONFLOW
    2 days ago






  • 1




    Adding a multiple of a row to another row does not affect the determinant. But you are adding a multiple of a row to a MULTIPLE of another row, so the determinant gets affected.
    – xarles
    2 days ago










  • I understand now. So in this case, we have three effects on the determinant. 1) first row is multiplied by 2. 2) Second row is multiplied by 4. 3) fourth row is multiplied by 2. Then in effect, our determinant is now 2*4*2*2? So 64?
    – PERTURBATIONFLOW
    2 days ago








2




2




No. 2on row is multiplied by 4 and 4th row is multiplied by 2.
– xarles
2 days ago




No. 2on row is multiplied by 4 and 4th row is multiplied by 2.
– xarles
2 days ago












math.stackexchange.com/questions/1802752/… From other sources, it appears that it does not have effect on determinant. Only row 1 has an effective change on the initial determinant A = 2
– PERTURBATIONFLOW
2 days ago




math.stackexchange.com/questions/1802752/… From other sources, it appears that it does not have effect on determinant. Only row 1 has an effective change on the initial determinant A = 2
– PERTURBATIONFLOW
2 days ago




1




1




Adding a multiple of a row to another row does not affect the determinant. But you are adding a multiple of a row to a MULTIPLE of another row, so the determinant gets affected.
– xarles
2 days ago




Adding a multiple of a row to another row does not affect the determinant. But you are adding a multiple of a row to a MULTIPLE of another row, so the determinant gets affected.
– xarles
2 days ago












I understand now. So in this case, we have three effects on the determinant. 1) first row is multiplied by 2. 2) Second row is multiplied by 4. 3) fourth row is multiplied by 2. Then in effect, our determinant is now 2*4*2*2? So 64?
– PERTURBATIONFLOW
2 days ago





I understand now. So in this case, we have three effects on the determinant. 1) first row is multiplied by 2. 2) Second row is multiplied by 4. 3) fourth row is multiplied by 2. Then in effect, our determinant is now 2*4*2*2? So 64?
– PERTURBATIONFLOW
2 days ago











2 Answers
2






active

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up vote
0
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accepted










Note that:
$$beginalign|A|=beginvmatrixv_1\v_2\v_3\v_4endvmatrix&=2 Rightarrow \
beginvmatrix2v_1\4v_2\v_3\2v_4endvmatrix&=32 stackrelfrac32R_1+R_2to R_2\ frac14R_2+R_3to R_3\ frac12R_1+R_4to R_4Rightarrow \
beginvmatrix2v_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix&=32.endalign$$






share|cite|improve this answer























  • This was very concise. Thank you! Can you answer this question as well: math.stackexchange.com/questions/2871861/…
    – PERTURBATIONFLOW
    2 days ago











  • My pleasure, good luck.
    – farruhota
    2 days ago

















up vote
0
down vote













We have



$$beginvmatrix2v_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix=2beginvmatrixv_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix=2beginvmatrixv_1\4v_2\v_2+v_3\2v_4endvmatrix=16beginvmatrixv_1\v_2\v_2+v_3\v_4endvmatrix=16beginvmatrixv_1\v_2\v_3\v_4endvmatrix=32$$



To check we can try with



$$beginvmatrix2&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1endvmatrix=2$$



and



$$beginvmatrix2&0&0&0\3&4&0&0\0&1&1&0\1&0&0&2endvmatrix=32$$






share|cite|improve this answer























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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

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    up vote
    0
    down vote



    accepted










    Note that:
    $$beginalign|A|=beginvmatrixv_1\v_2\v_3\v_4endvmatrix&=2 Rightarrow \
    beginvmatrix2v_1\4v_2\v_3\2v_4endvmatrix&=32 stackrelfrac32R_1+R_2to R_2\ frac14R_2+R_3to R_3\ frac12R_1+R_4to R_4Rightarrow \
    beginvmatrix2v_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix&=32.endalign$$






    share|cite|improve this answer























    • This was very concise. Thank you! Can you answer this question as well: math.stackexchange.com/questions/2871861/…
      – PERTURBATIONFLOW
      2 days ago











    • My pleasure, good luck.
      – farruhota
      2 days ago














    up vote
    0
    down vote



    accepted










    Note that:
    $$beginalign|A|=beginvmatrixv_1\v_2\v_3\v_4endvmatrix&=2 Rightarrow \
    beginvmatrix2v_1\4v_2\v_3\2v_4endvmatrix&=32 stackrelfrac32R_1+R_2to R_2\ frac14R_2+R_3to R_3\ frac12R_1+R_4to R_4Rightarrow \
    beginvmatrix2v_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix&=32.endalign$$






    share|cite|improve this answer























    • This was very concise. Thank you! Can you answer this question as well: math.stackexchange.com/questions/2871861/…
      – PERTURBATIONFLOW
      2 days ago











    • My pleasure, good luck.
      – farruhota
      2 days ago












    up vote
    0
    down vote



    accepted







    up vote
    0
    down vote



    accepted






    Note that:
    $$beginalign|A|=beginvmatrixv_1\v_2\v_3\v_4endvmatrix&=2 Rightarrow \
    beginvmatrix2v_1\4v_2\v_3\2v_4endvmatrix&=32 stackrelfrac32R_1+R_2to R_2\ frac14R_2+R_3to R_3\ frac12R_1+R_4to R_4Rightarrow \
    beginvmatrix2v_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix&=32.endalign$$






    share|cite|improve this answer















    Note that:
    $$beginalign|A|=beginvmatrixv_1\v_2\v_3\v_4endvmatrix&=2 Rightarrow \
    beginvmatrix2v_1\4v_2\v_3\2v_4endvmatrix&=32 stackrelfrac32R_1+R_2to R_2\ frac14R_2+R_3to R_3\ frac12R_1+R_4to R_4Rightarrow \
    beginvmatrix2v_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix&=32.endalign$$







    share|cite|improve this answer















    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago


























    answered 2 days ago









    farruhota

    13.4k2632




    13.4k2632











    • This was very concise. Thank you! Can you answer this question as well: math.stackexchange.com/questions/2871861/…
      – PERTURBATIONFLOW
      2 days ago











    • My pleasure, good luck.
      – farruhota
      2 days ago
















    • This was very concise. Thank you! Can you answer this question as well: math.stackexchange.com/questions/2871861/…
      – PERTURBATIONFLOW
      2 days ago











    • My pleasure, good luck.
      – farruhota
      2 days ago















    This was very concise. Thank you! Can you answer this question as well: math.stackexchange.com/questions/2871861/…
    – PERTURBATIONFLOW
    2 days ago





    This was very concise. Thank you! Can you answer this question as well: math.stackexchange.com/questions/2871861/…
    – PERTURBATIONFLOW
    2 days ago













    My pleasure, good luck.
    – farruhota
    2 days ago




    My pleasure, good luck.
    – farruhota
    2 days ago










    up vote
    0
    down vote













    We have



    $$beginvmatrix2v_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix=2beginvmatrixv_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix=2beginvmatrixv_1\4v_2\v_2+v_3\2v_4endvmatrix=16beginvmatrixv_1\v_2\v_2+v_3\v_4endvmatrix=16beginvmatrixv_1\v_2\v_3\v_4endvmatrix=32$$



    To check we can try with



    $$beginvmatrix2&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1endvmatrix=2$$



    and



    $$beginvmatrix2&0&0&0\3&4&0&0\0&1&1&0\1&0&0&2endvmatrix=32$$






    share|cite|improve this answer



























      up vote
      0
      down vote













      We have



      $$beginvmatrix2v_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix=2beginvmatrixv_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix=2beginvmatrixv_1\4v_2\v_2+v_3\2v_4endvmatrix=16beginvmatrixv_1\v_2\v_2+v_3\v_4endvmatrix=16beginvmatrixv_1\v_2\v_3\v_4endvmatrix=32$$



      To check we can try with



      $$beginvmatrix2&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1endvmatrix=2$$



      and



      $$beginvmatrix2&0&0&0\3&4&0&0\0&1&1&0\1&0&0&2endvmatrix=32$$






      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        We have



        $$beginvmatrix2v_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix=2beginvmatrixv_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix=2beginvmatrixv_1\4v_2\v_2+v_3\2v_4endvmatrix=16beginvmatrixv_1\v_2\v_2+v_3\v_4endvmatrix=16beginvmatrixv_1\v_2\v_3\v_4endvmatrix=32$$



        To check we can try with



        $$beginvmatrix2&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1endvmatrix=2$$



        and



        $$beginvmatrix2&0&0&0\3&4&0&0\0&1&1&0\1&0&0&2endvmatrix=32$$






        share|cite|improve this answer















        We have



        $$beginvmatrix2v_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix=2beginvmatrixv_1\3v_1+4v_2\v_2+v_3\v_1+2v_4endvmatrix=2beginvmatrixv_1\4v_2\v_2+v_3\2v_4endvmatrix=16beginvmatrixv_1\v_2\v_2+v_3\v_4endvmatrix=16beginvmatrixv_1\v_2\v_3\v_4endvmatrix=32$$



        To check we can try with



        $$beginvmatrix2&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1endvmatrix=2$$



        and



        $$beginvmatrix2&0&0&0\3&4&0&0\0&1&1&0\1&0&0&2endvmatrix=32$$







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago


























        answered 2 days ago









        gimusi

        63.6k73480




        63.6k73480






















             

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