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Thread: Matrices

  1. #1
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    Matrices

    I am going crazy. it is almost 3:00 in the morning and I have been trying to figure out a problem and I just can't get it.

    If you could show me how I would so appreciate it. Iam taking an online math class and the teacher has 7 other classes and never seems to answer the questions on time. Thank you so much!

    A =
    • 2 0 1
    • 2 -1 3
    • 4 1 2

    B =
    • -2 1 0
    • 4 3 -2
    • 1 2 -1

    I need to find (A + B)squared

    2 + -2 0 + 1 1 + 0 is 0 1 1

    and then squared is the same answer

    The book has the answer as

    Matrix

    • 11 5 2
    • 17 13 9
    • 23 14 9
    I know this is difficult to read but I hope you can understand it. i didn't know how to type any other way
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  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
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    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    $\displaystyle \displaystyle A+B=\begin{pmatrix}
    2 & 0 & 1\\
    2 & -1 & 3\\
    4 & 1 & 2\end{pmatrix}+
    \begin{pmatrix}
    -2 & 1 & 0\\
    4 & 3 & -2\\
    1 & 2 & -1\end{pmatrix}=
    \begin{pmatrix}
    0 & 1 & 1\\
    6 & 2 & 1\\
    5 & 3 & 1\end{pmatrix}$
    Then, $\displaystyle (A+B)^2=\begin{pmatrix}
    11 & 5 & 2\\
    17 & 13 & 9\\
    23 & 14 & 9\end{pmatrix}$
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  3. #3
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    $\displaystyle \left(\begin{bmatrix}
    {2}\;\; {0}\;\;{1}\\
    {2}\;\;{-1}\;\;{3} \\
    {4}\;\;{1}\;\;{2}

    \end{bmatrix} +\begin{bmatrix}
    {-2}\;\; {1}\;\;{0}\\
    {4}\;\;{3}\;\;{-2}\\
    {1}\;\;{2}\;\;{-1}\\

    \end{bmatrix}\right)^2$

    $\displaystyle =\left(\begin{bmatrix}
    {2-2\ }\;\; {\ 0+1\ }\;\;{\ 1+0}\\
    {2+4\ }\;\;{\ -1+3\ }\;\;{\ 3-2}\\
    {4+1\ }\;\;{\ 1+2\ }\;\;{\ 2-1}\\

    \end{bmatrix}\right)^2$

    $\displaystyle =\left(\begin{bmatrix}
    {0}\;\;{1}\;\;{1}\\
    {6}\;\;{2}\;\;{1}\\
    {5}\;\;{3}\;\;{1}\\
    \end{bmatrix}\right)^2$

    $\displaystyle =\begin{bmatrix}
    {0}\;\;{1}\;\;{1}\\
    {6}\;\;{2}\;\;{1}\\
    {5}\;\;{3}\;\;{1}\\
    \end{bmatrix} \cdot \begin{bmatrix}
    {0}\;\;{1}\;\;{1}\\
    {6}\;\;{2}\;\;{1}\\
    {5}\;\;{3}\;\;{1}\\
    \end{bmatrix}$

    $\displaystyle =\begin{bmatrix}
    {0\cdot 0 + 1 \cdot 6 +1\cdot 5\ }\;\;{\ 0\cdot 1+1\cdot 2 + 1 \cdot 3\ }\;\;{\ 0\cdot 1+1\cdot 1 + 1\cdot 1}\\
    {6\cdot 0 + 2 \cdot 6 + 1 \cdot 5\ }\;\;{\ 6\cdot 1+2\cdot 2 + 1 \cdot 3\ }\;\;{\ 6\cdot 1 +2 \cdot 1 + 1 \cdot 1}\\
    {5\cdot 0+ 3 \cdot 6 + 1 \cdot 5\ }\;\;{\ 5 \cdot 1 + 3 \cdot 2 + 1 \cdot 3\ }\;\;{\ 5 \cdot 1 + 3 \cdot 1 + 1 \cdot 1}\\
    \end{bmatrix}$

    $\displaystyle =\begin{bmatrix}
    {11}\;\;{5}\;\;{2}\\
    {17}\;\;{13}\;\;{9}\\
    {23}\;\;{14}\;\;{9}\\
    \end{bmatrix}$
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  4. #4
    Newbie
    Joined
    Dec 2007
    Posts
    9

    Thank you so very much Divideby0

    I knew there was an easy explaination. I just couldn't for the life of me figure it out. Next time I am just going to come here and post because I really do not like staying up till all hours of the night.

    Thank you again really!

    Lori
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