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    Hi I am doing my GCSEs (just finished astually) I just had year 12 Induction week were you seewhat A Level subjects are like. I am doing: Maths, Phsyics and Further Maths. I done some rearch on some topics in Further Maths and came across 'matrices'. This is the problem;

    The Matrix M is given by M = a 2 -1
    2 3 -1
    2 -1 1 where a is a constant.

    i) show that the determinat of M is 2a. (2marks)

    ii) given that a ≠ 0, find the inverse matrix M^-1 (4marks)


    I don't get matrices? :confused:

    Can anyone in year 12/13 please help me
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    Sorry that didn't come out well, the matrix, M is;

    a 2 -1
    2 3 -1
    2 -1 1

    Pretent its in a big bracket. [/B]
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    This would get a better response in the maths academic help forum. xx
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    I think using the formular:
    (M) = a b c
    d e f
    g h i

    det (M) = a(ei -hf) -b(di - gf) + c(dh - eg)

    You can easily find out det (M) = 2a
    I am not sure in the 12 GCSE has used this ... That comes from A Level P6
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    (Original post by 00sdhun)
    Hi I am doing my GCSEs (just finished astually) I just had year 12 Induction week were you seewhat A Level subjects are like. I am doing: Maths, Phsyics and Further Maths. I done some rearch on some topics in Further Maths and came across 'matrices'. This is the problem;

    The Matrix M is given by M = a 2 -1
    2 3 -1
    2 -1 1 where a is a constant.

    i) show that the determinat of M is 2a. (2marks)

    ii) given that a ≠ 0, find the inverse matrix M^-1 (4marks)


    I don't get matrices? :confused:

    Can anyone in year 12/13 please help me
    i. The determinant of a 3x3 matrix written as
    a b c
    d e f
    g h i
    is given by a(ei-hf)-b(di-gf)+c(dh-ge). Apply this on your matrix and you'll get that its determinant is 2a.

    ii. First, you find the determinant--you have this, it is 2a. Then you need to find the matrix of minors. This is done by finding the determinant of the 2x2 matrix left by covering a given entry in the matrix. For example, the upper-left hand entry in the matrix of minors is 3x1-(-1x-1)=2. You then multiply this matrix by
    + - +
    - + -
    + - +
    so that you get the matrix of cofactors. Transpose this and multiply by 1/det(M), and you're done!
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    (Original post by J.F.N)
    ii. First, you find the determinant--you have this, it is 2a. Then you need to find the matrix of minors. This is done by finding the determinant of the 2x2 matrix left by covering a given entry in the matrix. For example, the upper-left hand entry in the matrix of minors is 3x1-(-1x-1)=2. You then multiply this matrix by
    + - +
    - + -
    + - +
    so that you get the matrix of cofactors. Transpose this and multiply by 1/det(M), and you're done!
    I've never understood why so many find the matrix of minors - I think it's easier just to find the matrix of cofactors straight away. Say you've got a transpose matrix of:
    a b c
    d e f
    g h i

    Write this out four times in a square (perhaps with lines to separate them):

    a b c | a b c
    d e f | d e f
    g h i | g h i
    a b c | a b c
    d e f | d e f
    g h i | g h i

    Then just find the determinant of each 2x2 matrix that lies south-east of a-i. So, for example, the entry in the bottom-right element of the matrix of co-factors is given by ae - bd; the entry in the middle element of the matrix of cofactors is given by ia - gc [etc.].

    So yeah. Find the matrix of cofactors, then divide by 2a, the determinant of the original matrix. I kept forgetting which matrix to take the determinant of, myself.
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    (Original post by JohnSPals)
    I've never understood why so many find the matrix of minors - I think it's easier just to find the matrix of cofactors straight away. Say you've got a transpose matrix of:
    a b c
    d e f
    g h i

    Write this out four times in a square (perhaps with lines to separate them):

    a b c | a b c
    d e f | d e f
    g h i | g h i
    a b c | a b c
    d e f | d e f
    g h i | g h i

    Then just find the determinant of each 2x2 matrix that lies south-east of a-i. So, for example, the entry in the bottom-right element of the matrix of co-factors is given by ae - bd; the entry in the middle element of the matrix of cofactors is given by ia - gc [etc.].

    So yeah. Find the matrix of cofactors, then divide by 2a, the determinant of the original matrix. I kept forgetting which matrix to take the determinant of, myself.
    Whatever works for you. I personally don't believe that this method is particularly quicker, especially since the matrix of cofactors can be found smiply once you find the matrix of minors. In a similar way, I don't use Sarrus' rule to compute the determinant of a 3x3 matrix. Old-fashioned, eh?
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    (Original post by J.F.N)
    Whatever works for you. I personally don't believe that this method is particularly quicker, especially since the matrix of cofactors can be found smiply once you find the matrix of minors. In a similar way, I don't use Sarrus' rule to compute the determinant of a 3x3 matrix. Old-fashioned, eh?
    I don't think it's much quicker, but it's far more fool-proof. I find myself making all kinds of errors when computing a matrix of minors, and it's easy to forget to change some signs. My method seems comparitively foolproof.

    I'd comment on Sarrus' rule, but I don't know what it is .
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    (Original post by JohnSPals)
    I'd comment on Sarrus' rule, but I don't know what it is .
    It works only for 3x3 matrices, but consider the matrix
    a b c
    d e f
    g h i
    If you write it as
    a b c | a b
    d e f | d e
    g h i | g h
    and add the product of the diagonals pointing to the right and subtract the product product of the diagonals pointing to the left, you get the determinant of the matrix, i.e. aei + bfg + cdh - ceg - afh - bdi
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    (Original post by J.F.N)
    It works only for 3x3 matrices, but consider the matrix
    a b c
    d e f
    g h i
    If you write it as
    a b c | a b
    d e f | d e
    g h i | g h
    and add the product of the diagonals pointing to the right and subtract the product product of the diagonals pointing to the left, you get the determinant of the matrix, i.e. aei + bfg + cdh - ceg - afh - bdi
    I like that method very much . Again, it's comparitively foolproof!
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    I remember doing matrices very long ago in year 9 to prepare for maths igcse. Wow never knew igcse maths overlaps with A-lvl further maths!
    :rolleyes:
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    (Original post by blah888)
    I remember doing matrices very long ago in year 9 to prepare for maths igcse. Wow never knew igcse maths overlaps with A-lvl further maths!
    :rolleyes:
    Well you study matrices further at university - all the way till your final year if you wish - but I don't really think it implies "overlap" :rolleyes:
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    (Original post by RichE)
    Well you study matrices further at university - all the way till your final year if you wish - but I don't really think it implies "overlap" :rolleyes:
    like WHATEVER
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    (Original post by blah888)
    like WHATEVER
    aww shucks and I was hoping your next question was going to be on the Cayley-Hamilton Theorem :eek:
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    That's not the way we did matrices at all... If you call the first column of the matrix the vector a, the second b and the third c, then isn't the determinant a.(bXc) ? I find that easier to remember... And we never learnt any names like matrix of minors or cofactors! I can't quite remember how we did it... I think you call the rows a, b and c then make them into column vectors. Work out bXc, cXa and aXb, and these are the columns that make up the new matrix. Then divide by the determinant.
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    (Original post by blah888)
    I remember doing matrices very long ago in year 9 to prepare for maths igcse. Wow never knew igcse maths overlaps with A-lvl further maths!
    :rolleyes:
    What they do in Further Maths is Matrices from scratch to a pretty advanced level whereas the year 9 stuff is to a limited level. So yes, the first bit of FM Matrices is Year 9 stuff but the majority of the whole topic (afterwards) is a lot more advanced.
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    (Original post by Showsni)
    That's not the way we did matrices at all... If you call the first column of the matrix the vector a, the second b and the third c, then isn't the determinant a.(bXc) ? I find that easier to remember... And we never learnt any names like matrix of minors or cofactors! I can't quite remember how we did it... I think you call the rows a, b and c then make them into column vectors. Work out bXc, cXa and aXb, and these are the columns that make up the new matrix. Then divide by the determinant.
    I have to say, I was disapointed in a way because the relationship between matrices and vectors wasn't really established enough for EdExcel P6. Either that or my tecaher didn't explain it enough (luckily I didn't need to know all the linear transformations stuff and base vector stuff for my exam, otherwise I would have been a bit stuck).

    Your method for finding the inverse of a 3x3 matrix sounds interesting.
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    Don't worry - Matrices are covered in FP1 and they're easy!
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    (Original post by Nima)
    the majority of the whole topic (afterwards) is a lot more advanced.
    It was? It all seemed quite algorithmic in nature.
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    (Original post by Gaz031)
    It was? It all seemed quite algorithmic in nature.
    I think there is quite a difference between knowing how to find eigenvalues and eigenvectors and understanding just what you're doing and why.

    If you can think about the matrix stuff from a geometric point of view then you really are on top of the subject.
 
 
 

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