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    I need help on this question:
    Solve the equations
    3x-y-2z =14
    2x+y-z=7
    4-y+3z=7
    I need to solve these simultaneous equations, but I don't need to solve to get an exact value for x and y, rather the way a system of equations is solved in Further Pure A level. The textbook doesn't give an explanation or an appropriate example. Please can you tell me how to solve...thanks
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    (Original post by Sibghy)
    I need help on this question:
    ...
    Call them eqn 1 ,2 ,and 3. Eliminate the z variable from 1) and 2) by doing the same thing you'd do with 1) and 2) if it was a normal two variable simultaneous pair.
    You will be left with a new equation involving only x and y. call that 4)

    Eliminate the z from 2) and 3) Call this new one 5).

    You now have 4) and 5) which are two equations both in x and y.

    Finish this off yourself maybe?
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    (Original post by Sibghy)
    I need help on this question:
    Solve the equations
    3x-y-2z =14
    2x+y-z=7
    4-y+3z=7
    I need to solve these simultaneous equations, but I don't need to solve to get an exact value for x and y, rather the way a system of equations is solved in Further Pure A level. The textbook doesn't give an explanation or an appropriate example. Please can you tell me how to solve...thanks
    Express the system of equations in matrix form: \begin{pmatrix}3x-y-2z\\2x+y-z\\4-y+3z\end{pmatrix} = \begin{pmatrix}14\\7\\7 \end{pmatrix}, that is, \begin{pmatrix}3&-1&2\\2&1&-1\\4&-1&3\end{pmatrix}\begin{pmatrix}x  \\y\\z\end{pmatrix} = \begin{pmatrix}14\\7\\7 \end{pmatrix}.

    Then pre-multiply both sides by \begin{pmatrix}3&-1&2\\2&1&-1\\4&-1&3\end{pmatrix}^{-1} in order to find the solution matrix \begin{pmatrix}x\\y\\z \end{pmatrix}.
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    Do you know how to use 3x3 matrices yet? That may be what your exam board wants you to use.
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    (Original post by Sibghy)
    I need help on this question:
    Solve the equations
    3x-y-2z =14
    2x+y-z=7
    4-y+3z=7
    I need to solve these simultaneous equations, but I don't need to solve to get an exact value for x and y, rather the way a system of equations is solved in Further Pure A level. The textbook doesn't give an explanation or an appropriate example. Please can you tell me how to solve...thanks
    Either use 3x3 matrices and inverses to solve, or rearrange one of them for one variable (like rearrange 2nd equation for z) and substitute it into the other 2. You'll then have 2 equations in terms of 2 unknowns which you can do. Solve this then come back to z.
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    (Original post by marioman)
    Express the system of equations in matrix form:..
    I doubt inverting 3x3 matrices is in FP1
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    No, only two by two, the exam board is ocr, but I don't have the actual textbook, just an incomplete textbook photocopy.
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    (Original post by Arcane1729)
    Call them eqn 1 ,2 ,and 3. Eliminate the z variable from 1) and 2) by doing the same thing you'd do with 1) and 2) if it was a normal two variable simultaneous pair.
    You will be left with a new equation involving only x and y. call that 4)


    Eliminate the z from 2) and 3) Call this new one 5).

    You now have 4) and 5) which are two equations both in x and y.

    Finish this off yourself maybe?
    I tried doing this, but then the multiplication of the top one would go wrong.
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    (Original post by marioman)
    Express the system of equations in matrix form: \begin{pmatrix}3x-y-2z\\2x+y-z\\4-y+3z\end{pmatrix} = \begin{pmatrix}14\\7\\7 \end{pmatrix}, that is, \begin{pmatrix}3&-1&2\\2&1&-1\\4&-1&3\end{pmatrix}\begin{pmatrix}x  \\y\\z\end{pmatrix} = \begin{pmatrix}14\\7\\7 \end{pmatrix}.

    Then pre-multiply both sides by \begin{pmatrix}3&-1&2\\2&1&-1\\4&-1&3\end{pmatrix}^{-1} in order to find the solution matrix \begin{pmatrix}x\\y\\z \end{pmatrix}.
    So do I multiply the original matrix with itself but to the power of -1, why is that? As in what is the explanation behind this.
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    (Original post by Sibghy)
    So do I multiply the original matrix with itself but to the power of -1, why is that? As in what is the explanation behind this.
    If you haven't covered inverse matrices, or 3x3 matrices for that matter, ignore this as inverse of a 3x3 matrix by hand will prove to be confusing for you. Unless you use a calculator. Just stick to the substitution method for now.

    To answer your question, it's not to the power of -1. Just like \sin^{-1}x represents inverse of sine, this represents the inverse matrix.
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    (Original post by Sibghy)
    I tried doing this, but then the multiplication of the top one would go wrong.
    Ok this makes sense now thanks.
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    (Original post by RDKGames)
    If you haven't covered inverse matrices, or 3x3 matrices for that matter, ignore this as inverse of a 3x3 matrix by hand will prove to be confusing for you. Unless you use a calculator. Just stick to the substitution method for now.

    To answer your question, it's not to the power of -1. Just like \sin^{-1}x represents inverse of sine, this represents the inverse matrix.
    Ok thanks I'll do it by hand, hopefully I haven't missed this.
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    (Original post by Arcane1729)
    I doubt inverting 3x3 matrices is in FP1
    It is for OCR (not MEI).
 
 
 
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