Simplify the determinant of a $4 times 4$ matrix.












1












$begingroup$


I have to find the determinant of the following 4x4 matrix:
$quad A=begin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}$



So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$detbegin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}=begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}xrightarrow{3R_3-R_1}begin{vmatrix}3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$$



Since I know from the solutions that the determinant is -64, I suppose that I need to simplify the third row in the reduced form to $quad 0 quad 0 quad 2 quad 0 quad$ and then multiply the elements in the upper-left-to-bottom-right diagonal, which is indeed -64. But this doesn't make much sense since there's also a $-4$ that we can simplify. Can someone explain me the actual rules we need to follow?






linear-algebra matrices determinant







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$endgroup$












  • $begingroup$
    Expansion by minors along the second row or column and the last row or column makes this easy.
    $endgroup$
    – Robert Israel
    Dec 19 '18 at 18:19










  • $begingroup$
    The rule is that you can add a multiple of a row to another row. You cannot replace a row by a multiple of itself and add another row as well.
    $endgroup$
    – Klaas van Aarsen
    Dec 19 '18 at 18:57










  • $begingroup$
    @IlikeSerena That’s not entirely right. The latter isn’t an elementary row operation per se, but it can be broken down into two elementary operations, one of which changes the determinant in a known way.
    $endgroup$
    – amd
    Dec 19 '18 at 20:26












  • $begingroup$
    That's a different rule @amd. We can multiply a row by a constant if we divide the whole determinant by the same constant. Mixing these rules up is causing the confusion.
    $endgroup$
    – Klaas van Aarsen
    Dec 19 '18 at 20:39










  • $begingroup$
    @IlikeSerena My point is that, contrary to what you wrote, one can replace a row by a multiple of itself and then add another to it—it produces a perfectly valid result—but it’s not an elementary row operation and doing so changes the value of the determinant.
    $endgroup$
    – amd
    Dec 19 '18 at 20:43
















1












$begingroup$


I have to find the determinant of the following 4x4 matrix:
$quad A=begin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}$



So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$detbegin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}=begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}xrightarrow{3R_3-R_1}begin{vmatrix}3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$$



Since I know from the solutions that the determinant is -64, I suppose that I need to simplify the third row in the reduced form to $quad 0 quad 0 quad 2 quad 0 quad$ and then multiply the elements in the upper-left-to-bottom-right diagonal, which is indeed -64. But this doesn't make much sense since there's also a $-4$ that we can simplify. Can someone explain me the actual rules we need to follow?






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$












  • $begingroup$
    Expansion by minors along the second row or column and the last row or column makes this easy.
    $endgroup$
    – Robert Israel
    Dec 19 '18 at 18:19










  • $begingroup$
    The rule is that you can add a multiple of a row to another row. You cannot replace a row by a multiple of itself and add another row as well.
    $endgroup$
    – Klaas van Aarsen
    Dec 19 '18 at 18:57










  • $begingroup$
    @IlikeSerena That’s not entirely right. The latter isn’t an elementary row operation per se, but it can be broken down into two elementary operations, one of which changes the determinant in a known way.
    $endgroup$
    – amd
    Dec 19 '18 at 20:26












  • $begingroup$
    That's a different rule @amd. We can multiply a row by a constant if we divide the whole determinant by the same constant. Mixing these rules up is causing the confusion.
    $endgroup$
    – Klaas van Aarsen
    Dec 19 '18 at 20:39










  • $begingroup$
    @IlikeSerena My point is that, contrary to what you wrote, one can replace a row by a multiple of itself and then add another to it—it produces a perfectly valid result—but it’s not an elementary row operation and doing so changes the value of the determinant.
    $endgroup$
    – amd
    Dec 19 '18 at 20:43














1












1








1


1



$begingroup$


I have to find the determinant of the following 4x4 matrix:
$quad A=begin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}$



So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$detbegin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}=begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}xrightarrow{3R_3-R_1}begin{vmatrix}3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$$



Since I know from the solutions that the determinant is -64, I suppose that I need to simplify the third row in the reduced form to $quad 0 quad 0 quad 2 quad 0 quad$ and then multiply the elements in the upper-left-to-bottom-right diagonal, which is indeed -64. But this doesn't make much sense since there's also a $-4$ that we can simplify. Can someone explain me the actual rules we need to follow?






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$




I have to find the determinant of the following 4x4 matrix:
$quad A=begin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}$



So I apply the Gaussian elimination to obtain an upper-triangle matrix:
$$detbegin{bmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{bmatrix}=begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}xrightarrow{3R_3-R_1}begin{vmatrix}3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$$



Since I know from the solutions that the determinant is -64, I suppose that I need to simplify the third row in the reduced form to $quad 0 quad 0 quad 2 quad 0 quad$ and then multiply the elements in the upper-left-to-bottom-right diagonal, which is indeed -64. But this doesn't make much sense since there's also a $-4$ that we can simplify. Can someone explain me the actual rules we need to follow?






linear-algebra matrices determinant




linear-algebra matrices determinant






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share|cite|improve this question













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share|cite|improve this question








edited Dec 19 '18 at 18:59









user1101010

9011830




9011830










asked Dec 19 '18 at 18:12









Kevin MooreKevin Moore

14818




14818












  • $begingroup$
    Expansion by minors along the second row or column and the last row or column makes this easy.
    $endgroup$
    – Robert Israel
    Dec 19 '18 at 18:19










  • $begingroup$
    The rule is that you can add a multiple of a row to another row. You cannot replace a row by a multiple of itself and add another row as well.
    $endgroup$
    – Klaas van Aarsen
    Dec 19 '18 at 18:57










  • $begingroup$
    @IlikeSerena That’s not entirely right. The latter isn’t an elementary row operation per se, but it can be broken down into two elementary operations, one of which changes the determinant in a known way.
    $endgroup$
    – amd
    Dec 19 '18 at 20:26












  • $begingroup$
    That's a different rule @amd. We can multiply a row by a constant if we divide the whole determinant by the same constant. Mixing these rules up is causing the confusion.
    $endgroup$
    – Klaas van Aarsen
    Dec 19 '18 at 20:39










  • $begingroup$
    @IlikeSerena My point is that, contrary to what you wrote, one can replace a row by a multiple of itself and then add another to it—it produces a perfectly valid result—but it’s not an elementary row operation and doing so changes the value of the determinant.
    $endgroup$
    – amd
    Dec 19 '18 at 20:43


















  • $begingroup$
    Expansion by minors along the second row or column and the last row or column makes this easy.
    $endgroup$
    – Robert Israel
    Dec 19 '18 at 18:19










  • $begingroup$
    The rule is that you can add a multiple of a row to another row. You cannot replace a row by a multiple of itself and add another row as well.
    $endgroup$
    – Klaas van Aarsen
    Dec 19 '18 at 18:57










  • $begingroup$
    @IlikeSerena That’s not entirely right. The latter isn’t an elementary row operation per se, but it can be broken down into two elementary operations, one of which changes the determinant in a known way.
    $endgroup$
    – amd
    Dec 19 '18 at 20:26












  • $begingroup$
    That's a different rule @amd. We can multiply a row by a constant if we divide the whole determinant by the same constant. Mixing these rules up is causing the confusion.
    $endgroup$
    – Klaas van Aarsen
    Dec 19 '18 at 20:39










  • $begingroup$
    @IlikeSerena My point is that, contrary to what you wrote, one can replace a row by a multiple of itself and then add another to it—it produces a perfectly valid result—but it’s not an elementary row operation and doing so changes the value of the determinant.
    $endgroup$
    – amd
    Dec 19 '18 at 20:43
















$begingroup$
Expansion by minors along the second row or column and the last row or column makes this easy.
$endgroup$
– Robert Israel
Dec 19 '18 at 18:19




$begingroup$
Expansion by minors along the second row or column and the last row or column makes this easy.
$endgroup$
– Robert Israel
Dec 19 '18 at 18:19












$begingroup$
The rule is that you can add a multiple of a row to another row. You cannot replace a row by a multiple of itself and add another row as well.
$endgroup$
– Klaas van Aarsen
Dec 19 '18 at 18:57




$begingroup$
The rule is that you can add a multiple of a row to another row. You cannot replace a row by a multiple of itself and add another row as well.
$endgroup$
– Klaas van Aarsen
Dec 19 '18 at 18:57












$begingroup$
@IlikeSerena That’s not entirely right. The latter isn’t an elementary row operation per se, but it can be broken down into two elementary operations, one of which changes the determinant in a known way.
$endgroup$
– amd
Dec 19 '18 at 20:26






$begingroup$
@IlikeSerena That’s not entirely right. The latter isn’t an elementary row operation per se, but it can be broken down into two elementary operations, one of which changes the determinant in a known way.
$endgroup$
– amd
Dec 19 '18 at 20:26














$begingroup$
That's a different rule @amd. We can multiply a row by a constant if we divide the whole determinant by the same constant. Mixing these rules up is causing the confusion.
$endgroup$
– Klaas van Aarsen
Dec 19 '18 at 20:39




$begingroup$
That's a different rule @amd. We can multiply a row by a constant if we divide the whole determinant by the same constant. Mixing these rules up is causing the confusion.
$endgroup$
– Klaas van Aarsen
Dec 19 '18 at 20:39












$begingroup$
@IlikeSerena My point is that, contrary to what you wrote, one can replace a row by a multiple of itself and then add another to it—it produces a perfectly valid result—but it’s not an elementary row operation and doing so changes the value of the determinant.
$endgroup$
– amd
Dec 19 '18 at 20:43




$begingroup$
@IlikeSerena My point is that, contrary to what you wrote, one can replace a row by a multiple of itself and then add another to it—it produces a perfectly valid result—but it’s not an elementary row operation and doing so changes the value of the determinant.
$endgroup$
– amd
Dec 19 '18 at 20:43










4 Answers
4






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2












$begingroup$

It should be: $$begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}rightarrow begin{vmatrix}3&0&1&0\0&2&0&0\0&0&color{red}{3-{1over 3}}&0\0&0&0&-4end{vmatrix}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why not merely compute the determinant and find that it is $-64$?
    $endgroup$
    – David G. Stork
    Dec 19 '18 at 18:18










  • $begingroup$
    I did $3R_3-R_1$, shouldn't it be fine?
    $endgroup$
    – Kevin Moore
    Dec 19 '18 at 18:22






  • 2




    $begingroup$
    No, it shoul be $R_3-R_1/3$
    $endgroup$
    – Maria Mazur
    Dec 19 '18 at 18:24










  • $begingroup$
    Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
    $endgroup$
    – Kevin Moore
    Dec 19 '18 at 18:27






  • 2




    $begingroup$
    $R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
    $endgroup$
    – Shubham Johri
    Dec 19 '18 at 18:30



















1












$begingroup$

Rather than applying row operations, expand the minors.



$det A = -4begin{vmatrix} 3&0&1\0&2&0\1&0&3end{vmatrix} = (-4)(18-2) = -64$



If you apply row operations, you don't want your row operations to change the determinant.



I think of multiplying by an elementary matrix.



$begin{vmatrix} 1\&1\-frac 13&&1\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&frac {8}{3}&0\0&0&0&-4end{vmatrix}$



keeps the determinant unchanged.



while



$begin{vmatrix} 1\&1\-1&&3\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$



will change the determinant by a factor of $3.$






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    0












    $begingroup$

    Just expand by the first row:
    begin{align}
    begin{vmatrix}3&0&1&0 \
    0&2&0&0 \
    1&0&3&0 \
    0&0&0&-4
    end{vmatrix}&=
    3,begin{vmatrix}
    2&0&0 \
    0&3&0 \
    0&0&-4
    end{vmatrix}+
    1,begin{vmatrix}
    0&2&0 \
    1&0&0 \
    0&0&-4
    end{vmatrix}
    =3(2cdot3cdot(-4))+1(-1)begin{vmatrix}
    2&0 \
    0&-4
    end{vmatrix}\
    &=-72+8=-64.
    end{align}






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      The common mistake you made is that replacing $R_3$ with $3R_3-R_1$ is not an elementary row operation. You can add a multiple of $R_1$ to $R_3$ without changing the determinant, but not the other way around. What you did is a combination of two elementary row operations: $R_3to3R_3$ and $R_3to R_3+R_1$. The second operation doesn’t affect the determinant, but the first one multiplies it by $3$, so you have to divide by $3$ at the end when you combine the diagonal entries. Once you’ve accounted for this factor of $3$ that you introduced, you get the correct value.



      Alternatively, you could’ve performed $R_3to R_3-frac13R_1$ to clear the first column without changing the value of the determinant. Note the difference between this and what you did: the latter operation multiplies by a scalar a different row from the one being replaced.



      As others have noted, a much easier way to compute the determinant of this matrix is to take advantage of all of those zeros in the first and second row/column and expand by minors along them.






      share|cite|improve this answer









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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        It should be: $$begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}rightarrow begin{vmatrix}3&0&1&0\0&2&0&0\0&0&color{red}{3-{1over 3}}&0\0&0&0&-4end{vmatrix}$$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Why not merely compute the determinant and find that it is $-64$?
          $endgroup$
          – David G. Stork
          Dec 19 '18 at 18:18










        • $begingroup$
          I did $3R_3-R_1$, shouldn't it be fine?
          $endgroup$
          – Kevin Moore
          Dec 19 '18 at 18:22






        • 2




          $begingroup$
          No, it shoul be $R_3-R_1/3$
          $endgroup$
          – Maria Mazur
          Dec 19 '18 at 18:24










        • $begingroup$
          Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
          $endgroup$
          – Kevin Moore
          Dec 19 '18 at 18:27






        • 2




          $begingroup$
          $R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
          $endgroup$
          – Shubham Johri
          Dec 19 '18 at 18:30
















        2












        $begingroup$

        It should be: $$begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}rightarrow begin{vmatrix}3&0&1&0\0&2&0&0\0&0&color{red}{3-{1over 3}}&0\0&0&0&-4end{vmatrix}$$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Why not merely compute the determinant and find that it is $-64$?
          $endgroup$
          – David G. Stork
          Dec 19 '18 at 18:18










        • $begingroup$
          I did $3R_3-R_1$, shouldn't it be fine?
          $endgroup$
          – Kevin Moore
          Dec 19 '18 at 18:22






        • 2




          $begingroup$
          No, it shoul be $R_3-R_1/3$
          $endgroup$
          – Maria Mazur
          Dec 19 '18 at 18:24










        • $begingroup$
          Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
          $endgroup$
          – Kevin Moore
          Dec 19 '18 at 18:27






        • 2




          $begingroup$
          $R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
          $endgroup$
          – Shubham Johri
          Dec 19 '18 at 18:30














        2












        2








        2





        $begingroup$

        It should be: $$begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}rightarrow begin{vmatrix}3&0&1&0\0&2&0&0\0&0&color{red}{3-{1over 3}}&0\0&0&0&-4end{vmatrix}$$






        share|cite|improve this answer











        $endgroup$



        It should be: $$begin{vmatrix}3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix}rightarrow begin{vmatrix}3&0&1&0\0&2&0&0\0&0&color{red}{3-{1over 3}}&0\0&0&0&-4end{vmatrix}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 19 '18 at 18:18

























        answered Dec 19 '18 at 18:15









        Maria MazurMaria Mazur

        49.6k1361124




        49.6k1361124












        • $begingroup$
          Why not merely compute the determinant and find that it is $-64$?
          $endgroup$
          – David G. Stork
          Dec 19 '18 at 18:18










        • $begingroup$
          I did $3R_3-R_1$, shouldn't it be fine?
          $endgroup$
          – Kevin Moore
          Dec 19 '18 at 18:22






        • 2




          $begingroup$
          No, it shoul be $R_3-R_1/3$
          $endgroup$
          – Maria Mazur
          Dec 19 '18 at 18:24










        • $begingroup$
          Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
          $endgroup$
          – Kevin Moore
          Dec 19 '18 at 18:27






        • 2




          $begingroup$
          $R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
          $endgroup$
          – Shubham Johri
          Dec 19 '18 at 18:30


















        • $begingroup$
          Why not merely compute the determinant and find that it is $-64$?
          $endgroup$
          – David G. Stork
          Dec 19 '18 at 18:18










        • $begingroup$
          I did $3R_3-R_1$, shouldn't it be fine?
          $endgroup$
          – Kevin Moore
          Dec 19 '18 at 18:22






        • 2




          $begingroup$
          No, it shoul be $R_3-R_1/3$
          $endgroup$
          – Maria Mazur
          Dec 19 '18 at 18:24










        • $begingroup$
          Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
          $endgroup$
          – Kevin Moore
          Dec 19 '18 at 18:27






        • 2




          $begingroup$
          $R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
          $endgroup$
          – Shubham Johri
          Dec 19 '18 at 18:30
















        $begingroup$
        Why not merely compute the determinant and find that it is $-64$?
        $endgroup$
        – David G. Stork
        Dec 19 '18 at 18:18




        $begingroup$
        Why not merely compute the determinant and find that it is $-64$?
        $endgroup$
        – David G. Stork
        Dec 19 '18 at 18:18












        $begingroup$
        I did $3R_3-R_1$, shouldn't it be fine?
        $endgroup$
        – Kevin Moore
        Dec 19 '18 at 18:22




        $begingroup$
        I did $3R_3-R_1$, shouldn't it be fine?
        $endgroup$
        – Kevin Moore
        Dec 19 '18 at 18:22




        2




        2




        $begingroup$
        No, it shoul be $R_3-R_1/3$
        $endgroup$
        – Maria Mazur
        Dec 19 '18 at 18:24




        $begingroup$
        No, it shoul be $R_3-R_1/3$
        $endgroup$
        – Maria Mazur
        Dec 19 '18 at 18:24












        $begingroup$
        Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
        $endgroup$
        – Kevin Moore
        Dec 19 '18 at 18:27




        $begingroup$
        Sorry for my ignorance, is there any specific reason for that? I mean, for the Gaussian-elimination process in matrices shouldn't it be valid?
        $endgroup$
        – Kevin Moore
        Dec 19 '18 at 18:27




        2




        2




        $begingroup$
        $R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
        $endgroup$
        – Shubham Johri
        Dec 19 '18 at 18:30




        $begingroup$
        $R_3to3R_3-R_1$ is a compound row transformation composed of $R_3to3R_3,R_3to R_3-R_1$. Recall that multiplying any row or column of a square matrix by $k$ will multiply its determinant by $k$. When you multiply $R_3$ by $3$, the determinant of the resulting matrix is thrice that of the original matrix
        $endgroup$
        – Shubham Johri
        Dec 19 '18 at 18:30











        1












        $begingroup$

        Rather than applying row operations, expand the minors.



        $det A = -4begin{vmatrix} 3&0&1\0&2&0\1&0&3end{vmatrix} = (-4)(18-2) = -64$



        If you apply row operations, you don't want your row operations to change the determinant.



        I think of multiplying by an elementary matrix.



        $begin{vmatrix} 1\&1\-frac 13&&1\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&frac {8}{3}&0\0&0&0&-4end{vmatrix}$



        keeps the determinant unchanged.



        while



        $begin{vmatrix} 1\&1\-1&&3\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$



        will change the determinant by a factor of $3.$






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

          Rather than applying row operations, expand the minors.



          $det A = -4begin{vmatrix} 3&0&1\0&2&0\1&0&3end{vmatrix} = (-4)(18-2) = -64$



          If you apply row operations, you don't want your row operations to change the determinant.



          I think of multiplying by an elementary matrix.



          $begin{vmatrix} 1\&1\-frac 13&&1\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&frac {8}{3}&0\0&0&0&-4end{vmatrix}$



          keeps the determinant unchanged.



          while



          $begin{vmatrix} 1\&1\-1&&3\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$



          will change the determinant by a factor of $3.$






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            Rather than applying row operations, expand the minors.



            $det A = -4begin{vmatrix} 3&0&1\0&2&0\1&0&3end{vmatrix} = (-4)(18-2) = -64$



            If you apply row operations, you don't want your row operations to change the determinant.



            I think of multiplying by an elementary matrix.



            $begin{vmatrix} 1\&1\-frac 13&&1\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&frac {8}{3}&0\0&0&0&-4end{vmatrix}$



            keeps the determinant unchanged.



            while



            $begin{vmatrix} 1\&1\-1&&3\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$



            will change the determinant by a factor of $3.$






            share|cite|improve this answer











            $endgroup$



            Rather than applying row operations, expand the minors.



            $det A = -4begin{vmatrix} 3&0&1\0&2&0\1&0&3end{vmatrix} = (-4)(18-2) = -64$



            If you apply row operations, you don't want your row operations to change the determinant.



            I think of multiplying by an elementary matrix.



            $begin{vmatrix} 1\&1\-frac 13&&1\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&frac {8}{3}&0\0&0&0&-4end{vmatrix}$



            keeps the determinant unchanged.



            while



            $begin{vmatrix} 1\&1\-1&&3\&&&1end{vmatrix}begin{vmatrix} 3&0&1&0\0&2&0&0\1&0&3&0\0&0&0&-4end{vmatrix} = begin{vmatrix} 3&0&1&0\0&2&0&0\0&0&8&0\0&0&0&-4end{vmatrix}$



            will change the determinant by a factor of $3.$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 19 '18 at 18:39









            Bernard

            123k741117




            123k741117










            answered Dec 19 '18 at 18:26









            Doug MDoug M

            45.4k31954




            45.4k31954























                0












                $begingroup$

                Just expand by the first row:
                begin{align}
                begin{vmatrix}3&0&1&0 \
                0&2&0&0 \
                1&0&3&0 \
                0&0&0&-4
                end{vmatrix}&=
                3,begin{vmatrix}
                2&0&0 \
                0&3&0 \
                0&0&-4
                end{vmatrix}+
                1,begin{vmatrix}
                0&2&0 \
                1&0&0 \
                0&0&-4
                end{vmatrix}
                =3(2cdot3cdot(-4))+1(-1)begin{vmatrix}
                2&0 \
                0&-4
                end{vmatrix}\
                &=-72+8=-64.
                end{align}






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Just expand by the first row:
                  begin{align}
                  begin{vmatrix}3&0&1&0 \
                  0&2&0&0 \
                  1&0&3&0 \
                  0&0&0&-4
                  end{vmatrix}&=
                  3,begin{vmatrix}
                  2&0&0 \
                  0&3&0 \
                  0&0&-4
                  end{vmatrix}+
                  1,begin{vmatrix}
                  0&2&0 \
                  1&0&0 \
                  0&0&-4
                  end{vmatrix}
                  =3(2cdot3cdot(-4))+1(-1)begin{vmatrix}
                  2&0 \
                  0&-4
                  end{vmatrix}\
                  &=-72+8=-64.
                  end{align}






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Just expand by the first row:
                    begin{align}
                    begin{vmatrix}3&0&1&0 \
                    0&2&0&0 \
                    1&0&3&0 \
                    0&0&0&-4
                    end{vmatrix}&=
                    3,begin{vmatrix}
                    2&0&0 \
                    0&3&0 \
                    0&0&-4
                    end{vmatrix}+
                    1,begin{vmatrix}
                    0&2&0 \
                    1&0&0 \
                    0&0&-4
                    end{vmatrix}
                    =3(2cdot3cdot(-4))+1(-1)begin{vmatrix}
                    2&0 \
                    0&-4
                    end{vmatrix}\
                    &=-72+8=-64.
                    end{align}






                    share|cite|improve this answer









                    $endgroup$



                    Just expand by the first row:
                    begin{align}
                    begin{vmatrix}3&0&1&0 \
                    0&2&0&0 \
                    1&0&3&0 \
                    0&0&0&-4
                    end{vmatrix}&=
                    3,begin{vmatrix}
                    2&0&0 \
                    0&3&0 \
                    0&0&-4
                    end{vmatrix}+
                    1,begin{vmatrix}
                    0&2&0 \
                    1&0&0 \
                    0&0&-4
                    end{vmatrix}
                    =3(2cdot3cdot(-4))+1(-1)begin{vmatrix}
                    2&0 \
                    0&-4
                    end{vmatrix}\
                    &=-72+8=-64.
                    end{align}







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 19 '18 at 18:53









                    BernardBernard

                    123k741117




                    123k741117























                        0












                        $begingroup$

                        The common mistake you made is that replacing $R_3$ with $3R_3-R_1$ is not an elementary row operation. You can add a multiple of $R_1$ to $R_3$ without changing the determinant, but not the other way around. What you did is a combination of two elementary row operations: $R_3to3R_3$ and $R_3to R_3+R_1$. The second operation doesn’t affect the determinant, but the first one multiplies it by $3$, so you have to divide by $3$ at the end when you combine the diagonal entries. Once you’ve accounted for this factor of $3$ that you introduced, you get the correct value.



                        Alternatively, you could’ve performed $R_3to R_3-frac13R_1$ to clear the first column without changing the value of the determinant. Note the difference between this and what you did: the latter operation multiplies by a scalar a different row from the one being replaced.



                        As others have noted, a much easier way to compute the determinant of this matrix is to take advantage of all of those zeros in the first and second row/column and expand by minors along them.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The common mistake you made is that replacing $R_3$ with $3R_3-R_1$ is not an elementary row operation. You can add a multiple of $R_1$ to $R_3$ without changing the determinant, but not the other way around. What you did is a combination of two elementary row operations: $R_3to3R_3$ and $R_3to R_3+R_1$. The second operation doesn’t affect the determinant, but the first one multiplies it by $3$, so you have to divide by $3$ at the end when you combine the diagonal entries. Once you’ve accounted for this factor of $3$ that you introduced, you get the correct value.



                          Alternatively, you could’ve performed $R_3to R_3-frac13R_1$ to clear the first column without changing the value of the determinant. Note the difference between this and what you did: the latter operation multiplies by a scalar a different row from the one being replaced.



                          As others have noted, a much easier way to compute the determinant of this matrix is to take advantage of all of those zeros in the first and second row/column and expand by minors along them.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The common mistake you made is that replacing $R_3$ with $3R_3-R_1$ is not an elementary row operation. You can add a multiple of $R_1$ to $R_3$ without changing the determinant, but not the other way around. What you did is a combination of two elementary row operations: $R_3to3R_3$ and $R_3to R_3+R_1$. The second operation doesn’t affect the determinant, but the first one multiplies it by $3$, so you have to divide by $3$ at the end when you combine the diagonal entries. Once you’ve accounted for this factor of $3$ that you introduced, you get the correct value.



                            Alternatively, you could’ve performed $R_3to R_3-frac13R_1$ to clear the first column without changing the value of the determinant. Note the difference between this and what you did: the latter operation multiplies by a scalar a different row from the one being replaced.



                            As others have noted, a much easier way to compute the determinant of this matrix is to take advantage of all of those zeros in the first and second row/column and expand by minors along them.






                            share|cite|improve this answer









                            $endgroup$



                            The common mistake you made is that replacing $R_3$ with $3R_3-R_1$ is not an elementary row operation. You can add a multiple of $R_1$ to $R_3$ without changing the determinant, but not the other way around. What you did is a combination of two elementary row operations: $R_3to3R_3$ and $R_3to R_3+R_1$. The second operation doesn’t affect the determinant, but the first one multiplies it by $3$, so you have to divide by $3$ at the end when you combine the diagonal entries. Once you’ve accounted for this factor of $3$ that you introduced, you get the correct value.



                            Alternatively, you could’ve performed $R_3to R_3-frac13R_1$ to clear the first column without changing the value of the determinant. Note the difference between this and what you did: the latter operation multiplies by a scalar a different row from the one being replaced.



                            As others have noted, a much easier way to compute the determinant of this matrix is to take advantage of all of those zeros in the first and second row/column and expand by minors along them.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 19 '18 at 20:32









                            amdamd

                            31.5k21052




                            31.5k21052






























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