Results 1 to 6 of 6

Math Help - Determinant simplification

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    58

    Determinant simplification

    Hi,

    Can anyone see a way of getting from A to B?

    A
    <br /> <br />
\begin{bmatrix}<br />
1 & z & z+1 \\<br />
z+1   & 1 & z \\<br />
z       & z+1 & 1<br />
 \end{bmatrix}<br />

    B
    <br />
2*<br />
\begin{bmatrix}<br />
1 & 0 & z \\<br />
z   & 1 & 0 \\<br />
0       & z & 1<br />
 \end{bmatrix}<br />



    So far the best I can do is:
    <br />
2*<br />
 \begin{bmatrix}<br />
1 & 0 & z \\<br />
z   & 1 & z-1 \\<br />
0       & z & 1<br />
  \end{bmatrix}<br />

    Thanks guys
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2010
    From
    Bratislava
    Posts
    116
    Thanks
    1
    Quote Originally Posted by aceband View Post
    Hi,

    Can anyone see a way of getting from A to B?

    A
    <br /> <br />
\begin{bmatrix}<br />
1 & z & z+1 \\<br />
z+1   & 1 & z \\<br />
z       & z+1 & 1<br />
 \end{bmatrix}<br />

    B
    <br />
2*<br />
\begin{bmatrix}<br />
1 & 0 & z \\<br />
z   & 1 & 0 \\<br />
0       & z & 1<br />
 \end{bmatrix}<br />



    So far the best I can do is:
    <br />
2*<br />
 \begin{bmatrix}<br />
1 & 0 & z \\<br />
z   & 1 & z-1 \\<br />
0       & z & 1<br />
  \end{bmatrix}<br />

    Thanks guys

    <br />
\begin{bmatrix}<br />
1 & z & z+1 \\<br />
z+1   & 1 & z \\<br />
z       & z+1 & 1<br />
 \end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
1 & 0 & z \\<br />
z+1   & 1 & z \\<br />
z       & z+1 & 1<br />
 \end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 & z & 1 \\<br />
z+1   & 1 & z \\<br />
z       & z+1 & 1<br />
 \end{bmatrix}<br />
=<br />
    <br />
\begin{bmatrix}<br />
 1 & 0 & z \\<br />
 z   & 1 & 0 \\<br />
 z       & z+1 & 1<br />
  \end{bmatrix}<br />
 +<br />
\begin{bmatrix}<br />
 1 & 0 & z \\<br />
1   & 0 & z \\<br />
 z       & z+1 & 1<br />
  \end{bmatrix}<br />
 +<br />
\begin{bmatrix}<br />
0 & z & 1 \\<br />
1   & 0 & z \\<br />
 z       & z+1 & 1<br />
  \end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 & z & 1 \\<br />
 z   & 1 & 0 \\<br />
 z       & z+1 & 1<br />
  \end{bmatrix}<br />
    <br />
=<br />
\begin{bmatrix}<br />
  1 & 0 & z \\<br />
  z   & 1 & 0 \\<br />
  z       & z+1 & 1<br />
   \end{bmatrix}<br />
  +<br />
\begin{bmatrix}<br />
 0 & z & 1 \\<br />
1   & 0 & z \\<br />
  z       & z+1 & 1<br />
   \end{bmatrix}<br />
    <br />
 =<br />
 \begin{bmatrix}<br />
   1 & 0 & z \\<br />
   z   & 1 & 0 \\<br />
0       & z & 1<br />
    \end{bmatrix}<br />
   +<br />
 \begin{bmatrix}<br />
  0 & z & 1 \\<br />
 1   & 0 & z \\<br />
   z       & 1 & 0<br />
    \end{bmatrix}<br />
    <br />
 =<br />
 \begin{bmatrix}<br />
   1 & 0 & z \\<br />
   z   & 1 & 0 \\<br />
0       & z & 1<br />
    \end{bmatrix}<br />
   +<br />
 \begin{bmatrix}<br />
1   & 0 & z \\<br />
   z       & 1 & 0 \\<br />
  0 & z & 1 <br />
    \end{bmatrix}<br />

    In the first step I used the property, that if two matrices have all lines with the exception of one of them equal, then the sum of their determinants is the determinant of the matrix that has the sum of these two lines instead of this line (and the remaining lines are the same).

    Except for this property, all the remaining things I have used seems to be standard - determinant doesn't change when I subtract one line from another, switching two lines change the sign, if one of the lines is linear combination of the remaining ones, then the value of determinant is zero.

    I hope there is not a typo somewhere.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2010
    From
    Bratislava
    Posts
    116
    Thanks
    1
    Perhaps here Determinant as Sum of Determinants - ProofWiki you can find a better explanation of the result I have used in the first step.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,908
    Thanks
    766
    Hello, aceband!

    Can anyone see a way of getting from A to B?

    A \;=\;\begin{bmatrix} 1 & z & z+1 \\ z+1   & 1 & z \\ z & z+1 & 1 \end{bmatrix}

    B \;=\;2\cdot \begin{bmatrix} 1 & 0 & z \\ z & 1 & 0 \\ 0 & z & 1 \end{bmatrix}
    I used standard row operations . . .


    \text{Given: }\;\begin{bmatrix} 1 & z & z+1 \\ z+1   & 1 & z \\ z & z+1 & 1 \end{bmatrix}


    \begin{array}{c} \\ R_2-R_1 \\ \\ \end{array}\begin{bmatrix}1&z&z+1 \\ z&1-z&\text{-}1 \\ z&z+1& 1 \end{bmatrix}


    \begin{array}{c}\\ \\ R_3-R_2\end{array}\begin{bmatrix}1 & z&z+1 \\ z&1-z&\text{-}1 \\ 0 & 2z & 2 \end{bmatrix}


    \text{Factor: }\;2\!\cdot\!\begin{bmatrix}1&z&z+1 \\ z&1-z&\text{-}1 \\ 0&z&1\end{bmatrix}


    \begin{array}{c}R_1-R_3 \\ \\ \\ \end{array}\;2\!\cdot\!\begin{bmatrix}1&0&z \\ z & 1-z & \text{-}1 \\ 0&z&1\end{bmatrix}


    \begin{array}{c} \\ R_2+R_3 \\ \\ \end{array}\quad2\!\cdot\! \begin{bmatrix}1&0&z \\ z&1&0 \\ 0&z&1\end{bmatrix}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2009
    Posts
    58
    Wow, did not know you could do that! I think that'll be one of those techniques i never forget now! Thank you so much.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Mar 2010
    From
    Bratislava
    Posts
    116
    Thanks
    1
    Quote Originally Posted by aceband View Post
    Wow, did not know you could do that! I think that'll be one of those techniques i never forget now! Thank you so much.
    If you mean the result about the sum of determinants, it is often used in the proofs of results about effect of elementary row operations on the value of determinant, like here Multiple of Row Added to Row of Determinant - ProofWiki
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Determinant
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: September 26th 2011, 03:48 PM
  2. determinant 1
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 6th 2010, 12:56 AM
  3. Determinant help
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: January 16th 2010, 01:55 AM
  4. determinant
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 20th 2008, 01:30 AM
  5. Determinant
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 1st 2008, 07:52 AM

Search Tags


/mathhelpforum @mathhelpforum