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Math Help - Matrix help + binomial

  1. #1
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    Matrix help + binomial

    Q1) Given that A =  \left(\begin{array}{cc}1&-1\\0&3\end{array}\right) and B =  \left(\begin{array}{cc}1&3\\3&4\end{array}\right) , find the matrix  X such that  AX = B + 2I , where  I is a 2 X 2 identity matrix


    and

    Q2) The coefficient of X in the expansion of  (2 + \frac{x}{3})^n is 3 times the coefficient of  X^2 . find the value of n.

    i somehow worked my way through and got r = -2. is there something wrong?

    Q3) The equation of a circle is X^2 + Y^2 - 10X - 4Y + 25 = 0
    Find the set of values of K if the circle touches the line Y = K.
    I subbed in the value of K into the equation but i have 2 unknowns. so how?

    Hope someone could help me with this! thanks! (:
    Last edited by teddybear67; October 28th 2009 at 07:10 PM.
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  2. #2
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    Hello, teddybear67!

    1) Given that: . A \:=\: \begin{pmatrix}1&\text{-}1\\0&3\end{pmatrix}\:\text{ and }\:B \:=\:\begin{pmatrix}1&3\\3&4\end{pmatrix},

    find the matrix X such that AX \:=\: B + 2I, where I is the 22 identity matrix
    B + 2I \;=\;\begin{pmatrix}1&3\\3&4\end{pmatrix} + \begin{pmatrix}2&0\\0&2\end{pmatrix} \;=\;\begin{pmatrix}3&3\\3&6\end{pmatrix}

    Let X \:=\:\begin{pmatrix}a&b\\c&d\end{pmatrix}

    Then: . AX \;=\;B + 2I\quad\Rightarrow\quad \begin{pmatrix}1&\text{-}1\\0&3\end{pmatrix} \begin{pmatrix}a&b\\c&d\end{pmatrix} \;=\;\begin{pmatrix}3&3\\3&6\end{pmatrix}

    . . \begin{pmatrix}a-c & b-d \\ 3c & 3d\end{pmatrix} \;=\;\begin{pmatrix}3&3\\3&6\end{pmatrix} \quad\Rightarrow\quad \begin{bmatrix}a-c\:=\:3 & b-d \:=\:3 \\ 3c \:=\:3 & 3d \:=\:6\end{bmatrix}


    And we have: . X \;=\;\begin{pmatrix}a&b\\c&d\end{pmatrix} \;=\;\begin{pmatrix}4&5\\1&2\end{pmatrix}




    2) The coefficient of x in the expansion of \left(2 + \frac{x}{3}\right)^n is 3 times the coefficient of  x^2.
    Find the value of n.
    The expansion begins: . 2^n + n\!\cdot\!2^{n-1}\left(\frac{x}{3}\right) + \frac{n(n-1)}{2}\!\cdot\!2^{n-2}\left(\frac{x}{3}\right)^2 + \hdots

    The coefficient of x is: . \frac{n\!\cdot\!2^{n-2}}{3}

    The coefficient of x^2 is: . \frac{n(n-1)\!\cdot\!2^{n-2}}{2\cdot3^2} \:=\:\frac{n(n-1)\!\cdot\!2^{n-3}}{9}

    We are told that: . \frac{n\!\cdot\!2^{n-1}}{3} \;=\;3\bigg[\frac{n(n-1)\!\cdot\!2^{n-3}}{9}\bigg] \quad\Rightarrow\quad \frac{n\!\cdot\!2^{n-1}}{3} \;=\;\frac{n(n-1)\!\cdot\!2^{n-3}}{3}

    Assuming n \neq 0, multiply by \frac{3}{n}\!:\quad 2^{n-1} \:=\:(n-1)\!\cdot\!2^{n-3}

    Divide by 2^{n-3}\!:\quad 2^2 \:=\:n-1 \quad\Rightarrow\quad\boxed{ n \:=\:5}

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, teddybear67!

    B + 2I \;=\;\begin{pmatrix}1&3\\3&4\end{pmatrix} + \begin{pmatrix}2&0\\0&2\end{pmatrix} \;=\;\begin{pmatrix}3&3\\3&6\end{pmatrix}

    Let X \:=\:\begin{pmatrix}a&b\\c&d\end{pmatrix}

    Then: . AX \;=\;B + 2I\quad\Rightarrow\quad \begin{pmatrix}1&\text{-}1\\0&3\end{pmatrix} \begin{pmatrix}a&b\\c&d\end{pmatrix} \;=\;\begin{pmatrix}3&3\\3&6\end{pmatrix}

    . . \begin{pmatrix}a-c & b-d \\ 3c & 3d\end{pmatrix} \;=\;\begin{pmatrix}3&3\\3&6\end{pmatrix} \quad\Rightarrow\quad \begin{bmatrix}a-c\:=\:3 & b-d \:=\:3 \\ 3c \:=\:3 & 3d \:=\:6\end{bmatrix}


    And we have: . X \;=\;\begin{pmatrix}a&b\\c&d\end{pmatrix} \;=\;\begin{pmatrix}4&5\\1&2\end{pmatrix}




    The expansion begins: . 2^n + n\!\cdot\!2^{n-1}\left(\frac{x}{3}\right) + \frac{n(n-1)}{2}\!\cdot\!2^{n-2}\left(\frac{x}{3}\right)^2 + \hdots

    The coefficient of x is: . \frac{n\!\cdot\!2^{n-2}}{3}

    The coefficient of x^2 is: . \frac{n(n-1)\!\cdot\!2^{n-2}}{2\cdot3^2} \:=\:\frac{n(n-1)\!\cdot\!2^{n-3}}{9}

    We are told that: . \frac{n\!\cdot\!2^{n-1}}{3} \;=\;3\bigg[\frac{n(n-1)\!\cdot\!2^{n-3}}{9}\bigg] \quad\Rightarrow\quad \frac{n\!\cdot\!2^{n-1}}{3} \;=\;\frac{n(n-1)\!\cdot\!2^{n-3}}{3}

    Assuming n \neq 0, multiply by n-1)\!\cdot\!2^{n-3}" alt="\frac{3}{n}\!:\quad 2^{n-1} \:=\n-1)\!\cdot\!2^{n-3}" />

    Divide by 2^{n-3}\!:\quad 2^2 \:=\:n-1 \quad\Rightarrow\quad\boxed{ n \:=\:5}

    thanks! i think ive gotten it! (:
    but just one question. Identity matrix =  \left(\begin{array}{cc}1&0\\0&1\end{array}\right)
    then what is singular matrix?
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  4. #4
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    You have a LaTex error. If you mean "identity matrix= \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}", then you are right.

    A "singular matrix" is a matrix that does not have a multiplicative inverse- which is the same as saying its determinant is 0.

    The matrix, A, in your first problem does have an inverse (which is why you were able to find a unique solution to this problem) and you could have done this as X= A^{-1}AX= A^{-1}(B+ 2I).


    It is not difficult to calculate that A^{-1}= \begin{pmatrix}1 & \frac{1}{3} \\ 0 & \frac{1}{3}\end{pmatrix} and, as Soroban said, B+ 2I= \begin{pmatrix}3 & 3 \\ 3 & 6\end{pmatrix}

    So X= \begin{pmatrix}1 & \frac{1}{3} \\ 0 & \frac{1}{3}\end{pmatrix} \begin{pmatrix}3 & 3 \\ 3 & 6\end{pmatrix}

    My method is a little simpler if you know how to find the inverse matrix, Soroban's method has the advantage that you don't have to know how to find the inverse.
    Last edited by HallsofIvy; October 29th 2009 at 05:22 AM.
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