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Math Help - Find the value of 'q' in the matrix

  1. #1
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    Red face Find the value of 'q' in the matrix

    Question : If P= \begin{bmatrix}1 & -1 \\ 2 & -1\end{bmatrix} and Q = \begin{bmatrix} 1 & 1 \\ q & -1\end{bmatrix} and (P+Q)^2 = P^2 + Q^2 , determine the value of q
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    Quote Originally Posted by zorro View Post
    Question : If P= \begin{bmatrix}1 & -1 \\ 2 & -1\end{bmatrix} and Q = \begin{bmatrix} 1 & 1 \\ q & -1\end{bmatrix} and (P+Q)^2 = P^2 + Q^2 , determine the value of q
    What you have boils down to finding the value of q that satisfies PQ + QP = 0 \Rightarrow PQ = -QP. So do the necessary multiplications (it's assumed you can do this without trouble) and then equate the appropriate matrix elements to get an equation in q.
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    Please let me know if the value of q = 0

    Quote Originally Posted by mr fantastic View Post
    What you have boils down to finding the value of q that satisfies PQ + QP = 0 \Rightarrow PQ = -QP. So do the necessary multiplications (it's assumed you can do this without trouble) and then equate the appropriate matrix elements to get an equation in q.


    Please let me know if the value of q = 0 or not
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  4. #4
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    Quote Originally Posted by zorro View Post
    Please let me know if the value of q = 0 or not
    Have you substituted the value of q and checked that it works? If you post your work I will review it.
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    (P+Q)^2 = P^2 + Q^2\Leftrightarrow 2PQ=0\Leftrightarrow P=0 or Q=0
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    Quote Originally Posted by Raoh View Post
    (P+Q)^2 = P^2 + Q^2\Leftrightarrow 2PQ=0\Leftrightarrow P=0 or Q=0
    NO! matrices do not satisfy the cancellation law. It is quite possible to have PQ= 0 without either P or Q equal to 0.

    For example, if P= \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} and Q= \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix} then neither is the 0 matrix but PQ= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}.
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  7. #7
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    Matrices aren't always commutative, as well... So  PQ + QP = 0 \Leftrightarrow 2PQ = 0 isn't neccesarily correct. Just do what Mr.F said and you should be fine.
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    Quote Originally Posted by HallsofIvy View Post
    NO! matrices do not satisfy the cancellation law. It is quite possible to have PQ= 0 without either P or Q equal to 0.

    For example, if P= \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} and Q= \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix} then neither is the 0 matrix but PQ= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}.
    thank you
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  9. #9
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    zorro, just go ahead and do the computation.

    What is P+ Q? What is (P+Q)^2?

    What are P^2 and Q^2? What is P^2+ Q^2?

    Set (P+ Q)^2 equal to P^2+ Q^2 and set terms equal component by component. You will have four equations in q. Do they all give the same value for q?
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    Is this correct?

    Quote Originally Posted by mr fantastic View Post
    Have you substituted the value of q and checked that it works? If you post your work I will review it.


    P \ = \ \begin{bmatrix}1 & -1 \\ 2 & -1 \end{bmatrix}

    Q \ = \ \begin{bmatrix}1 & 1 \\ q & -1 \end{bmatrix}

     \therefore (P + Q) \ = \ \begin{bmatrix} 1 & 0 \\ (2 + q) & -2 \end{bmatrix}

    (P + Q)^2 \ = \ \begin{bmatrix} 1 & 0 \\ (q + 2)^2 & 4 \end{bmatrix} \ = \ 4 - (q + 2)^2 \ = \ [- q^2 - 4q] \longmapsto  eq\ 1

    P^2 \ = \ \begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix}

    Q^2 \ = \ \begin{bmatrix} 1 & 1 \\ q^2 & 1 \end{bmatrix}

    P^2 + Q^2 \ = \ \begin{bmatrix} 2 & 2 \\ (4 + q^2) & 2 \end{bmatrix} \ = \ 4 - 2(4 + q^2) \ = \ [-4 - 2q^2] \longmapsto eq \ 2





    Since

    (P + Q)^2 \ = \ (P^2 + Q^2) From eq 1 and eq 2

    -q^2 - 4q \ = \ -4 - 2q^2

    -q^2 + 2q^2 - 4q \ = \ -4

    q^2 - 4q + 4 \ = \ 0

    (q - 2)(q - 2)

    \therefore q = 2


    Is this correct ????????
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  11. #11
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    Quote Originally Posted by zorro View Post
    P \ = \ \begin{bmatrix}1 & -1 \\ 2 & -1 \end{bmatrix}

    Q \ = \ \begin{bmatrix}1 & 1 \\ q & -1 \end{bmatrix}

     \therefore (P + Q) \ = \ \begin{bmatrix} {\color{red}1} & 0 \\ (2 + q) & -2 \end{bmatrix} Mr F says: That red 1 should be a 2.

    (P + Q)^2 \ = \ \begin{bmatrix} 1 & 0 \\ (q + 2)^2 & 4 \end{bmatrix} Mr F says: This is completely wrong. Please go back and review how to multiply two matrices (which is what you're doing when you take the square of a matrix).

     \ = \ 4 - (q + 2)^2 \ = \ [- q^2 - 4q] \longmapsto eq\ 1 Mr F says: What is this meant to be? What is its relevance?

    P^2 \ = \ \begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix}

    Q^2 \ = \ \begin{bmatrix} 1 & 1 \\ q^2 & 1 \end{bmatrix}

    Mr F says: The above is completely wrong. Please go back and review how to multiply two matrices (which is what you're doing when you take the square of a matrix).

    P^2 + Q^2 \ = \ \begin{bmatrix} 2 & 2 \\ (4 + q^2) & 2 \end{bmatrix} \ = \ 4 - 2(4 + q^2) \ = \ [-4 - 2q^2] \longmapsto eq \ 2


    Since

    (P + Q)^2 \ = \ (P^2 + Q^2) From eq 1 and eq 2

    -q^2 - 4q \ = \ -4 - 2q^2

    -q^2 + 2q^2 - 4q \ = \ -4

    q^2 - 4q + 4 \ = \ 0

    (q - 2)(q - 2)

    \therefore q = 2


    Is this correct ????????
    There are many mistakes you need to fix. Because of these mistakes, all working that follows from them will be wrong.
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  12. #12
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    I have reworked the problem could u please check

    Quote Originally Posted by mr fantastic View Post
    There are many mistakes you need to fix. Because of these mistakes, all working that follows from them will be wrong.

    Is it right now

    (P + Q)^2 = \begin{bmatrix} 2 & 0 \\ (2 +q) & -2 \end{bmatrix} . \begin{bmatrix} 2 & 0 \\ (2 +q) & -2 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}

    P^2 = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} . \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}  ; \qquad \qquad  Q^2 = \begin{bmatrix} 1 & 1 \\ q & -1 \end{bmatrix} . \begin{bmatrix} 1 & 1 \\ q & -1 \end{bmatrix} = \begin{bmatrix} (1-q) & 0 \\ 0 & (1+q) \end{bmatrix}

    (P^2 + Q^2) = \begin{bmatrix} -q & 0 \\ 0 & q \end{bmatrix}

    (P^2 + Q^2) = P^2 + Q^2

    16 = 1+1-q^2

    16 = 2 - q^2

    q^2 = 2 - 16

    q^2 = -14

    q = - \sqrt{14}

    Is this correct ????
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  13. #13
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    Quote Originally Posted by zorro View Post
    Is it right now

    (P + Q)^2 = \begin{bmatrix} 2 & 0 \\ (2 +q) & -2 \end{bmatrix} . \begin{bmatrix} 2 & 0 \\ (2 +q) & -2 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}

    P^2 = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} . \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}  ; \qquad \qquad Q^2 = \begin{bmatrix} 1 & 1 \\ q & -1 \end{bmatrix} . \begin{bmatrix} 1 & 1 \\ q & -1 \end{bmatrix} = \begin{bmatrix} (1 {\color{red}+}q) & 0 \\ 0 & (1+q) \end{bmatrix}

    (P^2 + Q^2) = \begin{bmatrix} -q & 0 \\ 0 & q \end{bmatrix}

    (P^2 + Q^2) = P^2 + Q^2

    16 = 1+1-q^2

    16 = 2 - q^2

    q^2 = 2 - 16

    q^2 = -14

    q = - \sqrt{14}

    Is this correct ????
    Look carefully - there is a correction in red to your expression for Q^2. That will change everything that follows.
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