Posted at 5:58 pm by Miss Loi in A-Maths Questions
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Miss Loi avatar

Matrices - Enter The Matrix

(10)

In order to put Miss Loi’s new found freedom into good use, it’s about time the Matrix is loaded onto this website … hohoho.

Do take particular note on the second question of Part 1. Recently Miss Loi’s been seeing that more and more school papers are testing students’ understanding of concepts like these (which is the right thing to do).

From experience, many students don’t pay enough attention to mundane little things like the conditions for inverses to exist. Even Miss Loi is sometimes guilty of that in her uni days :P.

  1. Given that A = delim{[}{matrix{2}{3}{{-2} 2 3 1 {-1} 0}}{]} and B = delim{[}{matrix{3}{2}{3 {-4} 1 2 {-1} 1}}{]}, find AB.
    State, with reason, whether A-1 and (AB)-1 exist.
  2. Given that M = delim{[}{matrix{2}{2}{4 5 2 3}}{]}, find the matrix N such that MN = NM.

So what would you do Neo?

がんばって!!!
Miss Loi

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10 Comments

  1. 1
    日
    曜
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    2007
    Sun
    30
    Sep
    1:30PM
    123 says,

    How to prove for part 1 ar?
    Can i just say …. “A is a non-singular matrix, hence there is an inverse matrix for A”?

  2. 2
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    2007
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    12:49AM
    Miss Loi avatar Miss Loi says,

    123, Miss Loi shall reproduce this from an obscure corner of your text-book

    A matrix whose determinant is zero is singular and has no inverse.

    On the other hand, a matrix whose determinant is not zero is non-singular and has an inverse.

    But the tricky part here is that A is a non-square matrix. So can you even calculate its determinant? So can A ever be non-singular? :)

  3. 3
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    10:07PM
    Li-sa says,


    1. AB = delim{[}{matrix{2}{2}{{-7} 15 2 {-6}}}{]}
    A-1 does not exist because it isn’t a square matrix.
    (AB)-1 exists because it’s a square matrix.

    OK to make things clearer, a matrix is invertible if it is a square matrix i.e. n by n AND if its determinant ≠ 0.

    So A-1 does not exist since A is not even a square matrix to begin with.

    For (AB)-1, MAKE SURE you do a quick calculation of its determinant i.e.

    1/{(-7)(-6) - (15)(2)} = 1/12

    just to check that it’s not zero before stating that the inverse exists!

    P.S. your -1 should appear properly now ;)

    2. Let N = delim{[}{matrix{2}{2}{a c b d}}{]}.
    MN=delim{[}{matrix{2}{2}{4 5 2 3}}{]} delim{[}{matrix{2}{2}{a c b d}}{]}
    MN=delim{[}{matrix{2}{2}{{4a+5b} {4c+5d} {2a+3b} {2c+3d}}}{]}
    NM=delim{[}{matrix{2}{2}{a c b d}}{]} delim{[}{matrix{2}{2}{4 5 2 3}}{]}
    NM=delim{[}{matrix{2}{2}{{4a+2c} {5a+3c} {4b+2d} {5b+3d}}}{]}

    By equality of matrices,
    4a+5b=4a+2c –>5b=2c
    4c+5d=5a+3c –> c+5d=5a –(1)
    2a+3b=4b+2d –> 2a=b+2d–(2)
    2c+3d=5b+3d –> 2c=5b
    .’. b:c = 2:5

    Let’s take b=2k and c=5k where k is any number.

    (1): 5k+5d=5a
    k+d=a
    a-d=k=b/2=c/5

    (2):2a=2k+2d
    a-d=k=b/2=c/5

    So we need to find values such that a-d=b/2=c/5.
    If we take b=2 and c=5, then a-d=1
    a and d can be 1 and 0 respectively.
    Checking…MN=delim{[}{matrix{2}{2}{14 20 8 10}}{]}; NM=delim{[}{matrix{2}{2}{14 20 8 10}}{]} .’. N can bedelim{[}{matrix{2}{2}{1 5 2 0}}{]}.

    Of course there are many other solutions to this question.你认为矩阵(matrices)麻烦吗？之前的ratios问题(http://www.exampaper.com.sg/questions/e-maths/similarity-ratios-fetish)中，其实ratios叫「比」。

    Please see Miss Loi’s comment #6 below.

    Why hide part of your comments? It’s so interesting for 新加坡個小朋友 to know that Matrices=「矩阵」 and Ratios=「比」!

  4. 4
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    Jul
    10:08PM
    Li-sa says,

    I wonder why my -1 does not go superscript.

  5. 5
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    2008
    Sun
    6
    Jul
    6:32PM
    Li-sa says,

    Thanks!

  6. 6
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    6
    Jul
    11:58PM
    Miss Loi avatar Miss Loi says,

    Wait Li-sa Miss Loi hasn’t dismissed you yet!

    Yes there can be many answers to N in Part 2 but one thing that wasn’t stated is that Part 2 only carries 2 marks in the actual question.

    So in the interest of grabbing this puny 2 marks in the quickest of time under stressful exam conditions, it’s not really advisable to go through that lengthy 長氣workings of yours (even though your final answer is correct).

    A simple recall of the Identity Matrix (which is [{matrix{2}{2}{1 0 0 1}}] in this case) that satisfies the conditions of N ought to do the trick ;)

  7. 7
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    12:34AM
    Li-sa says,

    Maybe I should tell you that I have hidden some information in the source code.

  8. 8
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    12:36AM
    Li-sa says,

    *無奈*

  9. 9
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    12:44AM
    Li-sa says,

    I also tried the “This is A-Maths not Physics”.

  10. 10
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    Mon
    7
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    5:46PM
    Li-sa says,

    Oh, and “Thanks!” is not equal to fleeeeeeing and somehow I wonder how you linked it to unintended dismissal. It is what it it is.

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