Factorizing determinant

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  1. music lover's Avatar
    1 14-07-2012  10:40
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    Factorizing determinant
    Hi, I'm a bit stuck on factorizing the following:
    \begin{bmatrix} 1+a^2 & a & 1\\ 1+b^2 & b & 1\\ 1+c^2 & c & 1 \end{bmatrix}. I can't really see how to start i tried subtracting column 3 from column 1 to give \begin{bmatrix} a^2 & a & 1\\b^2 & b & 1\\ c^2 & c & 1\end{bmatrix}
  2. WarriorInAWig's Avatar
    2 14-07-2012  10:57
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    Re: Factorizing determinant
    I'd just do it the long way.

    With a 2x2 matrix, the determinant of
    \begin{bmatrix} a & b \\c & d \end{bmatrix}
    is ad-bc. For a 3x3 matrix
    \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}

    you need to split it into 3 2x2 determinants
    a is associated with \begin{vmatrix} e & f \\h & i \end{vmatrix}
    For the rest, look at the rows underneath and omit the column of the letter in top row.
    Now you add each of these BUT the sign of them alternate + - + etc.

    Wikipedia will explain it better than me about why this is the case.
    Last edited by WarriorInAWig; 14-07-2012 at 11:04.
  3. notnek's Avatar
    3 14-07-2012  10:57
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    Re: Factorizing determinant
    The trick with these is normally to make some of the entries 0.

    Try subtracting the first row from the second and third row. Does that help you?
  4. music lover's Avatar
    4 14-07-2012  12:23
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    Re: Factorizing determinant
    (Original post by notnek)
    The trick with these is normally to make some of the entries 0.

    Try subtracting the first row from the second and third row. Does that help you?
    hmm: i ended up with \begin{bmatrix} 1+a^2 & a & 1\\ 1+b^2 - 1+a^2 & b-a & 0\\ 1+c^2-1+a^2 & c-a & 0\end{bmatrix}. I then tried to take the determinant and got: c(a^2+b^2)a(-a^2-b^2). Not really too sure if this is the right approach.
  5. notnek's Avatar
    5 14-07-2012  12:31
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    Re: Factorizing determinant
    (Original post by music lover)
    hmm: i ended up with \begin{bmatrix} 1+a^2 & a & 1\\ 1+b^2 - 1+a^2 & b-a & 0\\ 1+c^2-1+a^2 & c-a & 0\end{bmatrix}. I then tried to take the determinant and got: c(a^2+b^2)a(-a^2-b^2). Not really too sure if this is the right approach.
    (1+b^2) - (1+a^2) \neq b^2+a^2

    You made a sign error.

    Once you've made the correction, use difference of two squares to factorise all the quadratics.
    Last edited by notnek; 14-07-2012 at 12:32.
  6. Slumpy's Avatar
    6 14-07-2012  12:32
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    Re: Factorizing determinant
    Off the top of my head; two rows being equal mean the determinant is 0. So (a-b) is a factor of the determinant. There are two other pretty obvious factors, then I think you can just equate coefficients.
  7. music lover's Avatar
    7 14-07-2012  12:57
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    Re: Factorizing determinant
    (Original post by Slumpy)
    Off the top of my head; two rows being equal mean the determinant is 0. So (a-b) is a factor of the determinant. There are two other pretty obvious factors, then I think you can just equate coefficients.
    Where do you get this from?
  8. music lover's Avatar
    8 14-07-2012  13:08
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    Re: Factorizing determinant
    (Original post by notnek)
    (1+b^2) - (1+a^2) \neq b^2+a^2

    You made a sign error.

    Once you've made the correction, use difference of two squares to factorise all the quadratics.
    hmm ok so now i have: \begin{bmatrix} 1+b^2-1+a^2 & b-a\\1+c^2-1+a^2 & c-a \end{bmatrix}=(1+b^2-1+a^2)(c-a)-[(b-a)(1+c^2-1+a^2)]=b^2c+a^2c-b^2a-c^2b-a^2b+c^2a

    On a side note is [latex]b^2c=c^2b{/latex]? I'm guessing not I tested it with a few numbers.
  9. Slumpy's Avatar
    9 14-07-2012  13:14
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    Re: Factorizing determinant
    (Original post by music lover)
    Where do you get this from?
    If a=b, we get:
    \begin{bmatrix} a^2 & a & 1\\b^2 & b & 1\\ c^2 & c & 1\end{bmatrix}=\begin{bmatrix} a^2 & a & 1\\a^2 & a & 1\\ c^2 & c & 1\end{bmatrix}

    This has a determinant of 0, so we know that if a=b, the determinant is 0, thus a-b is a factor of it. This is often a useful trick.
  10. notnek's Avatar
    10 14-07-2012  13:23
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    Re: Factorizing determinant
    (Original post by music lover)
    hmm ok so now i have: \begin{bmatrix} 1+b^2-1+a^2 & b-a\\1+c^2-1+a^2 & c-a \end{bmatrix}=(1+b^2-1+a^2)(c-a)-[(b-a)(1+c^2-1+a^2)]=b^2c+a^2c-b^2a-c^2b-a^2b+c^2a
    You're still making sign errors and not simplifying properly.

    If you do the row operation then the left middle entry becomes:

    (1+b^2)-(1+a^2) = 1+b^2-1-a^2

    And since 1-1=0 this simplifies to

    b^2-a^2

    which can be factorised:

    (b-a)(b+a)


    Now do the same with the other entries and then find the determinant. Let me know if there's any part that you're not sure about.

    On a side note is b^2c=c^2b? I'm guessing not I tested it with a few numbers.
    No, it is not true in general.
    Last edited by notnek; 14-07-2012 at 13:26.
  11. music lover's Avatar
    11 14-07-2012  13:56
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    Re: Factorizing determinant
    (Original post by notnek)
    You're still making sign errors and not simplifying properly.

    If you do the row operation then the left middle entry becomes:

    (1+b^2)-(1+a^2) = 1+b^2-1-a^2

    And since 1-1=0 this simplifies to

    b^2-a^2

    which can be factorised:

    (b-a)(b+a)


    Now do the same with the other entries and then find the determinant. Let me know if there's any part that you're not sure about.


    No, it is not true in general.
    done it thanks very much! Now onto this one. \begin{bmatrix} 1 & 1 & 1\\ cos^2 & cos^4 & sec^2\\ sin^2 & sin^4 & tan^2\end{bmatrix}. I need to show it's equal to 2sin^4cos^2

    Is this right so far: cos^4tan^2-sec^2sin^4-cos^2tan^2+sec^2sin^2+cos^2sin^4-cos^4sin^2 which is equal to cos^2sin^2-sin^4-\frac{sin^6}{cos^2}-sin^2+sin^2+\frac{sin^4}{cos^2}+ cos^2sin^4-cos^4sin^2. Sorry for all the questions!
    Last edited by music lover; 14-07-2012 at 14:04.
  12. notnek's Avatar
    12 14-07-2012  14:34
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    Re: Factorizing determinant
    (Original post by music lover)
    done it thanks very much! Now onto this one. \begin{bmatrix} 1 & 1 & 1\\ cos^2 & cos^4 & sec^2\\ sin^2 & sin^4 & tan^2\end{bmatrix}. I need to show it's equal to 2sin^4cos^2

    Is this right so far: cos^4tan^2-sec^2sin^4-cos^2tan^2+sec^2sin^2+cos^2sin^4-cos^4sin^2 which is equal to cos^2sin^2-sin^4-\frac{sin^6}{cos^2}-sin^2+sin^2+\frac{sin^4}{cos^2}+ cos^2sin^4-cos^4sin^2. Sorry for all the questions!
    Again, you should be doing some row/column operations to make some of the entries zero before expanding the determinant.

    So subtract the first column from the other two columns.
  13. music lover's Avatar
    13 14-07-2012  15:30
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    Re: Factorizing determinant
    (Original post by notnek)
    Again, you should be doing some row/column operations to make some of the entries zero before expanding the determinant.

    So subtract the first column from the other two columns.
    Ah ok so I have gotten to \begin{vmatrix}1 & 0 & 0\\cos^2 & cos^2 & sec^2-cos^2\\sin^2 & sin^2 & tan^2-sin^2\end{vmatrix}. Taking the determinant of this gives; \begin{vmatrix}cos^2 & sec^2-cos^2\\sin^2 & tan^2-sin^2\end{vmatrix} but how would i then simplify it further? I did difference of two squares to get (sec+cos)(sec-cos) and (tan+sin)(tan-sin).
    Last edited by music lover; 14-07-2012 at 15:34.
  14. notnek's Avatar
    14 14-07-2012  15:54
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    Re: Factorizing determinant
    (Original post by music lover)
    Ah ok so I have gotten to \begin{vmatrix}1 & 0 & 0\\cos^2 & cos^2 & sec^2-cos^2\\sin^2 & sin^2 & tan^2-sin^2\end{vmatrix}. Taking the determinant of this gives; \begin{vmatrix}cos^2 & sec^2-cos^2\\sin^2 & tan^2-sin^2\end{vmatrix} but how would i then simplify it further? I did difference of two squares to get (sec+cos)(sec-cos) and (tan+sin)(tan-sin).
    The (2,2) entry is wrong. You should have cos^4 - cos^2 instead of cos^2. I think you may have divided instead of subtracted. The (3,2) entry is also wrong for a similar reason.

    From here, I always like to change sin, cos and tan to s,c and t so that I concentrate more on the algebra instead of the trig. Using sec^2 x = 1+tan^2 x the determinant becomes:

    \begin{vmatrix}c^4-c^2 & 1+t^2-c^2\\s^4-s^2 & t^2-s^2\end{vmatrix}

    Now using 1-c^2=s^2, \ \ c^4-c^2=-c^2(1-c^2)=-c^2s^2

    and s^4-s^2=-s^2(1-s^2)=-s^2c^2, the determinant simplifies to


    \begin{vmatrix}-c^2s^2 & t^2+s^2\\-c^2s^2 & t^2-s^2\end{vmatrix}

    Can you finish it off?
    Last edited by notnek; 14-07-2012 at 15:58.
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  15. music lover's Avatar
    15 14-07-2012  16:24
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    Re: Factorizing determinant
    (Original post by notnek)
    The (2,2) entry is wrong. You should have cos^4 - cos^2 instead of cos^2. I think you may have divided instead of subtracted. The (3,2) entry is also wrong for a similar reason.

    From here, I always like to change sin, cos and tan to s,c and t so that I concentrate more on the algebra instead of the trig. Using sec^2 x = 1+tan^2 x the determinant becomes:

    \begin{vmatrix}c^4-c^2 & 1+t^2-c^2\\s^4-s^2 & t^2-s^2\end{vmatrix}

    Now using 1-c^2=s^2, \ \ c^4-c^2=-c^2(1-c^2)=-c^2s^2

    and s^4-s^2=-s^2(1-s^2)=-s^2c^2, the determinant simplifies to





    \begin{vmatrix}-c^2s^2 & t^2+s^2\\-c^2s^2 & t^2-s^2\end{vmatrix}

    Can you finish it off?


    Thanks very much , yeah I divided, silly mistake. What happened to the 1 with the sec^2 identity? Worked it out :P.
    Last edited by music lover; 14-07-2012 at 16:47.
  16. music lover's Avatar
    16 14-07-2012  17:03
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    Re: Factorizing determinant
    (Original post by notnek)
    The (2,2) entry is wrong. You should have cos^4 - cos^2 instead of cos^2. I think you may have divided instead of subtracted. The (3,2) entry is also wrong for a similar reason.

    From here, I always like to change sin, cos and tan to s,c and t so that I concentrate more on the algebra instead of the trig. Using sec^2 x = 1+tan^2 x the determinant becomes:

    \begin{vmatrix}c^4-c^2 & 1+t^2-c^2\\s^4-s^2 & t^2-s^2\end{vmatrix}

    Now using 1-c^2=s^2, \ \ c^4-c^2=-c^2(1-c^2)=-c^2s^2

    and s^4-s^2=-s^2(1-s^2)=-s^2c^2, the determinant simplifies to


    \begin{vmatrix}-c^2s^2 & t^2+s^2\\-c^2s^2 & t^2-s^2\end{vmatrix}

    Can you finish it off?
    Thought I had solved it but I get up to the very last part...

    \begin{vmatrix}-cos^2sin^2 & tan^2+sin^2\\-sin^2cos^2 & tan^2-sin^2\end{vmatrix}. This becomes (-cos^2sin^2)(tan^2-sin^2)-[(tan^2+sin^2)(-sin^2cos^2)]=-cos^2tan^2+cos^2sin^2-sin^2tan^2+sin^4-[-tan^2sin^2-tan^2cos^2-sin^4-cos^2sin^2]

    2sin^4+2cos^2sin^2+tan^2cos^2-cos^4tan^2
    2sin^2(sin^2+cos^2)+tan^2cos^2-cos^4tan^2=2sin^2+tan^2cos^2(1-cos^2) Simplifying this last bit i'm not too sure on.
  17. notnek's Avatar
    17 14-07-2012  17:08
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    Re: Factorizing determinant
    (Original post by music lover)
    Thought I had solved it but I get up to the very last part...

    \begin{vmatrix}-cos^2sin^2 & tan^2+sin^2\\-sin^2cos^2 & tan^2-sin^2\end{vmatrix}. This becomes (-cos^2sin^2)(tan^2-sin^2)-[(tan^2+sin^2)(-sin^2cos^2)]
    Notice here that you have a factor of \cos^2\sin^2 in both terms so you can take it outside the expression to get:

    (\cos^2\sin^2)[ \ \ \ \ \ ... \ \ \ ]

    See if you can fill in the blank and simplify.

    A general tip in algebra is to always try to factorise fully before expanding.
    Last edited by notnek; 14-07-2012 at 17:12.
  18. music lover's Avatar
    18 14-07-2012  17:12
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    Re: Factorizing determinant
    (Original post by notnek)
    Notice here that you have a factor of \cos^2\sin^2 in both terms so you can take it outside the expression to get:

    (-cos^2sin^2)[ \ \ \ \ \ ... \ \ \ ]

    See if you can fill in the blank and simplify.

    A general tip in algebra is to always try to factorise fully before expanding.
    done it!!!! Thank you so much
  19. notnek's Avatar
    19 14-07-2012  17:17
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    Re: Factorizing determinant
    (Original post by music lover)
    Thought I had solved it but I get up to the very last part...

    \begin{vmatrix}-cos^2sin^2 & tan^2+sin^2\\-sin^2cos^2 & tan^2-sin^2\end{vmatrix}. This becomes (-cos^2sin^2)(tan^2-sin^2)-[(tan^2+sin^2)(-sin^2cos^2)]=-cos^2tan^2+cos^2sin^2-sin^2tan^2+sin^4-[-tan^2sin^2-tan^2cos^2-sin^4-cos^2sin^2]

    2sin^4+2cos^2sin^2+tan^2cos^2-cos^4tan^2
    2sin^2(sin^2+cos^2)+tan^2cos^2-cos^4tan^2=2sin^2+tan^2cos^2(1-cos^2) Simplifying this last bit i'm not too sure on.
    On a separate note: you should try to be more careful when expanding. There are many errors in this working. You have said:

    (-\cos^2\sin^2)(\tan^2-\sin^2)=-\cos^2\tan^2+\cos^2\sin^2-\sin^2\tan^2+\sin^4

    Every term in your expansion is wrong. E.g. the first term should be -\cos^2\sin^2\tan^2 instead of -\cos^2\tan^2

    Or maybe I'm missing something? I'm very confused how you arrived at your expansion.


    Edit Now I see what you did. You expanded

    (-\cos^2-\sin^2)(\tan^2-\sin^2) instead of expanding (-\cos^2\sin^2)(\tan^2-\sin^2)
    Last edited by notnek; 14-07-2012 at 17:24.
  20. music lover's Avatar
    20 14-07-2012  17:21
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    Re: Factorizing determinant
    (Original post by notnek)
    On a separate note: you should try to be more careful when expanding. There are many errors in this working. You have said:

    (-\cos^2\sin^2)(\tan^2-\sin^2)=-\cos^2\tan^2+\cos^2\sin^2-\sin^2\tan^2+\sin^4

    Every term in your expansion is wrong. E.g. the first term should be -\cos^2\sin^2\tan^2 instead of -\cos^2\tan^2

    Or maybe I'm missing something? I'm very confused how you arrived at your expansion.
    Yeah i realized after. Epic fail...
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