Page 1 of 2 12 LastLast
Results 1 to 15 of 23

Math Help - Weird Matrix Problem

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Weird Matrix Problem

    The Matrix A is  \left[ \begin{array}{cc}<br />
1 & 1 \\<br />
1 & 3 \end{array} \right]

    (i) Find the matrix B such that B=A^2 -2I where I is the identity matrix.

    (ii) Find the inverse matrix A^{-1}

    (iii) Use A^{-1} to solve the simultaneous equations
    [tex]

    2x+2y=6
    x+3y=8
    Last edited by Punch; January 15th 2010 at 04:00 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    Shouldn't this go to the Linear Algebra subforum ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Well, I am under the catergory "Pre University" and the description of Pre-Algebra and Algebra section is "Basic calculations, order of operations, basic matrices, solving for variables, functions, logs and exponents and more"
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    My mistake. Forgive me ...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1423
    Quote Originally Posted by Punch View Post
    The Matrix is  \left[ \begin{array}{cc}<br />
1 & 1 \\<br />
1 & 3 \end{array} \right]

    (i) Find the matrix B such that B=A^2 -2I where I is the identity matrix.

    (ii) Find the inverse matrix A^{-1}

    (iii) Use A^{-1} to solve the simultaneous equations
     <br /> <br />
2x+2y=6<br />
x+3y=8<br />
    Is this matrix you have posted matrix A?

    If so...

    B = A^2 - 2I

     = \left[\begin{matrix}1 & 1 \\ 1 & 3\end{matrix}\right]^2 - 2\left[\begin{matrix} 1 & 0\\ 0 & 1\end{matrix}\right]

     = \left[\begin{matrix}1 & 1\\ 1 & 3\end{matrix}\right] \left[\begin{matrix} 1 & 1\\ 1 & 3\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]

     = \left[\begin{matrix} 2 & 4\\ 4 & 10\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]

     = \left[\begin{matrix} 0 & 4\\ 4 & 8\end{matrix}\right].
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by Prove It View Post
    Is this matrix you have posted matrix A?

    If so...

    B = A^2 - 2I

     = \left[\begin{matrix}1 & 1 \\ 1 & 3\end{matrix}\right]^2 - 2\left[\begin{matrix} 1 & 0\\ 0 & 1\end{matrix}\right]

     = \left[\begin{matrix}1 & 1\\ 1 & 3\end{matrix}\right] \left[\begin{matrix} 1 & 1\\ 1 & 3\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]

     = \left[\begin{matrix} 2 & 4\\ 4 & 10\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]

     = \left[\begin{matrix} 0 & 4\\ 4 & 8\end{matrix}\right].
    Thanks but I don't understand what the identity matrix is. May I ask how I can derive the "identity matrix"?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1423
    Quote Originally Posted by Punch View Post
    The Matrix is  \left[ \begin{array}{cc}<br />
1 & 1 \\<br />
1 & 3 \end{array} \right]

    (i) Find the matrix B such that B=A^2 -2I where I is the identity matrix.

    (ii) Find the inverse matrix A^{-1}

    (iii) Use A^{-1} to solve the simultaneous equations
     <br /> <br />
2x+2y=6<br />
x+3y=8<br />
    ii) A = \left[\begin{matrix} 1 & 1\\ 1 & 3 \end{matrix}\right]


    A^{-1} = \frac{1}{1\cdot 3 - 1 \cdot 1}\left[\begin{matrix}3 & -1 \\ -1 & 1\end{matrix}\right]

     = \frac{1}{2}\left[\begin{matrix} 3 & -1\\ -1 & 1\end{matrix}\right]

     = \left[\begin{matrix} \frac{3}{2} & -\frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2}\end{matrix}\right].
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1423
    Quote Originally Posted by Punch View Post
    Thanks but I don't understand what the identity matrix is. May I ask how I can derive the "identity matrix"?
    The Identity Matrix I_n is a square n \times n matrix with 1's down the main diagonal and 0's everywhere else.

    It is usually abbreviated to I, and you have to remember what dimensions your matrices are. In your case, your matrices are 2 \times 2, so you will be using I_2 = \left[\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right].


    It is the matrix equivalent of the number 1, because when you premultiply or postmultiply any matrix by I, you get that original matrix.

    Also, if you multiply any nonsingular matrix by its inverse, you get the identity matrix. This gives us the ability to perform the matrix equivalent of division.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Dec 2009
    Posts
    755
    So basically an identity matrix is equals to one and I can use <br />
I_2 = \left[\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right]<br />
for all identity matrix?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Also, I haven't learnt or seen any summing of matrix or subtraction of matrix..

    How exactly do I solve <br />
= \left[\begin{matrix} 2 & 4\\ 4 & 10\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]<br />
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1423
    Quote Originally Posted by Punch View Post
    The Matrix A is  \left[ \begin{array}{cc}<br />
1 & 1 \\<br />
1 & 3 \end{array} \right]

    (i) Find the matrix B such that B=A^2 -2I where I is the identity matrix.

    (ii) Find the inverse matrix A^{-1}

    (iii) Use A^{-1} to solve the simultaneous equations
    [tex]

    2x+2y=6
    x+3y=8
    First, notice that you can rewrite the system of equations as

    x + y = 3
    x + 3y = 8

    Write the system in matrix form.

    \left[\begin{matrix}1 & 1 \\ 1 & 3\end{matrix}\right] \left[\begin{matrix} x \\ y\end{matrix}\right] = \left[\begin{matrix} 3 \\ 8\end{matrix}\right]

    I am going to write this equation as

    A\mathbf{x} = \mathbf{b}.


    Premultiply both sides of the equation by A^{-1}.

    Then you have

    A^{-1}A\mathbf{x} = A^{-1}\mathbf{b}.

    Recall that if you multiply any matrix by its inverse, you get the identity matrix, so

    I\mathbf{x} = A^{-1}\mathbf{b}

    And if you multiply any matrix by the identity matrix, you get back to original matrix.

    So \mathbf{x} = A^{-1}\mathbf{b}

    \left[\begin{matrix} x \\ y\end{matrix}\right] = \left[\begin{matrix} \frac{3}{2} & -\frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2} \end{matrix}\right] \left[\begin{matrix} 3 \\ 8 \end{matrix}\right]

    \left[\begin{matrix} x \\ y\end{matrix}\right] = \left[\begin{matrix} \frac{9}{2} - 4 \\ -\frac{3}{2} + 4\end{matrix} \right]

    \left[\begin{matrix} x \\ y\end{matrix}\right] = \left[\begin{matrix} \frac{1}{2} \\ \frac{5}{2}\end{matrix}\right]

    Therefore x = \frac{1}{2} and y = \frac{5}{2}.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1423
    Quote Originally Posted by Punch View Post
    Also, I haven't learnt or seen any summing of matrix or subtraction of matrix..

    How exactly do I solve <br />
= \left[\begin{matrix} 2 & 4\\ 4 & 10\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]<br />
    How can you be given questions on these things without having been taught them? Matrix addition, Matrix multiplication and the Identity Matrix are the most important things to know when dealing with matrices.


    Matrix addition is performed component-wise. So you add (or subtract) corresponding components.

    So in your case

    \left[\begin{matrix} 2 & 4\\ 4 & 10\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right] = \left[\begin{matrix} 2 - 2 & 4 - 0\\ 4 - 0 & 10 - 2\end{matrix}\right]
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,553
    Thanks
    1423
    Quote Originally Posted by Punch View Post
    So basically an identity matrix is equals to one and I can use <br />
I_2 = \left[\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right]<br />
for all identity matrix?
    No, an identity matrix is not equal to 1. It ACTS LIKE a 1 when you are dealing with matrices.

    And you can't ALWAYS use I_2 because the matrices you are working with might not be 2 \times 2. It depends on the dimensions of the matrices you are working with. In this case you use I_2 because the matrices you are working with are 2 \times 2.

    But what if you were dealing with 3 \times 3 matrices?

    Which I would you use then?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Well, I found this question in an assesment book and am not sure if this type of question is in the syllabus.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by Prove It View Post
    No, an identity matrix is not equal to 1. It ACTS LIKE a 1 when you are dealing with matrices.

    And you can't ALWAYS use I_2 because the matrices you are working with might not be 2 \times 2. It depends on the dimensions of the matrices you are working with. In this case you use I_2 because the matrices you are working with are 2 \times 2.

    But what if you were dealing with 3 \times 3 matrices?

    Which I would you use then?
    hmm, 3 I_2?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: January 2nd 2011, 08:20 PM
  2. Help with a weird problem
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: March 30th 2010, 11:45 PM
  3. Weird sum problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 21st 2009, 03:47 AM
  4. weird problem
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: July 13th 2009, 09:42 AM
  5. A weird problem.. Please take a look and help!
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: April 10th 2008, 09:59 AM

Search Tags


/mathhelpforum @mathhelpforum