Using matrices, solve the following system of equations:2x + y �
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    Determinants

    Multiple Choice QuestionsLong Answer Type

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    251.

    Using matrices, solve the following system of equations:
    2x + y – 3r = 13
    x + y – z = 6
    2x – y + 4z = – 12

    The given equations are
    2x + y – 3 z = 13
    x + y – z – 6
    2x – y + 4 z = – 12
    These equations can be written as
                         open square brackets table row 2 cell space space space 1 end cell cell space space space minus 3 end cell row 1 cell space space space space 1 end cell cell space space minus 1 end cell row 2 cell space minus 1 end cell cell space space space space 4 end cell end table close square brackets space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row cell space space 13 end cell row cell space space 6 end cell row cell negative 12 end cell end table close square brackets        

         or space space AX space equals space straight B space where space straight A space equals space open square brackets table row 2 cell space space space space space space 1 end cell cell space space space minus 3 end cell row 1 cell space space space space space 1 end cell cell space space minus 1 end cell row 2 cell space minus 1 end cell cell space space space space space 4 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space straight B space equals space open square brackets table row cell space space space 13 end cell row cell space space 6 end cell row cell negative 12 end cell end table close square brackets Now space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space space space 1 end cell cell space space space space minus 3 end cell row 1 cell space space space space space space 1 end cell cell space space space minus 1 end cell row 2 cell space minus 1 end cell cell space space space space space 4 end cell end table close vertical bar space equals space 2 open vertical bar table row cell space space 1 end cell cell space space space minus 1 end cell row cell negative 1 end cell cell space space space space space space 4 end cell end table close vertical bar space minus 1 space open vertical bar table row 1 cell space space minus 1 end cell row 2 cell space space space space space space 4 end cell end table close vertical bar plus left parenthesis negative 3 right parenthesis space open vertical bar table row 1 cell space space space space space 1 end cell row 2 cell space minus 1 end cell end table close vertical bar space space space space space space space space space space space space space space space space space space equals space 2 left parenthesis 4 minus 1 right parenthesis minus 1 left parenthesis 4 plus 2 right parenthesis minus 3 left parenthesis negative 1 minus 2 right parenthesis space equals space 6 minus 6 plus 9 space equals space 9 space not equal to space 0 therefore space space space space space space space space straight A to the power of negative 1 end exponent space exists.

    Co-factors of the elements of first row of | A | are
    open vertical bar table row 1 cell space space space minus 1 end cell row cell negative 1 end cell cell space space space space 4 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space space minus 1 end cell row 2 cell space space space space space 4 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space space space space 1 end cell row 2 cell space space space minus 1 end cell end table close vertical bar

    i.e. 3, – 6, – 3 respectively.
    Co-factors of the elements of second row of | A | are
    negative open vertical bar table row cell space space 1 end cell cell space space minus 3 end cell row cell negative 1 end cell cell space space space space 4 end cell end table close vertical bar comma space space space space open vertical bar table row 2 cell space space space minus 3 end cell row 2 cell space space space space space 4 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space space space 1 end cell row 2 cell space minus 1 end cell end table close vertical bar
    i.e. – 1,  14,  4 respectively.
    Co-factors of the elements of third row of | A | are
    open vertical bar table row 1 cell space space minus 3 end cell row 1 cell space space minus 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space minus 3 end cell row 1 cell space space space minus 1 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space 1 end cell row 1 cell space space space 1 end cell end table close vertical bar
    i.e.   2, – 1, 1 respectively.
    therefore space space space space space adj. space straight A space equals space open square brackets table row 3 cell space space space minus 6 end cell cell space space minus 3 end cell row cell negative 1 end cell cell space space space space 14 end cell cell space space space 4 end cell row 2 cell space minus 1 end cell cell space space 1 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell space space 3 end cell cell space space minus 1 end cell cell space space space space space 2 end cell row cell negative 6 end cell cell space space space 14 end cell cell space space minus 1 end cell row cell negative 3 end cell cell space space 4 end cell cell space space space space 1 end cell end table close square brackets therefore space space space space space space space straight A to the power of negative 1 end exponent space equals space fraction numerator adj space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 over 9 open square brackets table row 3 cell space space minus 1 end cell cell space space space space 2 end cell row cell negative 6 end cell cell space space space 14 end cell cell space minus 1 end cell row cell negative 3 end cell cell space space 4 end cell cell space 1 end cell end table close square brackets Now space space space AX space equals space straight B space space space space rightwards double arrow space space space space space straight X space equals space straight A to the power of negative 1 end exponent straight B therefore space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 9 open square brackets table row cell space space space space 3 end cell cell space space space space space space minus 1 end cell cell space space space space space 2 end cell row cell negative 6 end cell cell space space space space 14 end cell cell space space minus 1 end cell row cell negative 3 end cell cell space space space space 4 end cell cell space space space space 1 end cell end table close square brackets space space open square brackets table row cell space space 13 end cell row cell space space 6 end cell row cell negative 12 end cell end table close square brackets rightwards double arrow space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 9 open square brackets table row cell 39 minus 6 minus 24 end cell row cell negative 78 plus 84 plus 12 end cell row cell negative 39 plus 24 minus 12 end cell end table close square brackets space space space space space space space space space space space space space space space space rightwards double arrow space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 9 open square brackets table row 9 row 18 row 27 end table close square brackets rightwards double arrow space space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row cell space space space space 1 end cell row cell space space space 2 end cell row cell negative 3 end cell end table close square brackets therefore space space space space space space straight x space equals space 1 comma space space straight y space equals space 2 comma space space space straight z space equals space minus 3 space is space the space required space solution. space

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    252.

    Using matrices, solve the following system of linear equations.
    3x + 4y + 2z = 8
    2y –3z = 3
    x – 2y + 6z = –2  

    87 Views
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    253. Using matrices, solve the following system of linear equations:
    x – y + z = 1
    2x –  yz = 2
    x – 2y – z
    81 Views
    Answer
    254.

    Using matrices, solve the following system of linear equations:
    x – y = 3
    2x + 3y + 4z = 17
    y + 2 z = 7

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    Answer
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    255.

    Solve by matrix method:
    y + 2z = – 8
    x + 2y + 3z = – 14
    3x + y + z = – 8

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    256.

    Solve the following system of equations by matrix method:
    3x – 2y + 3z = 8
    2x + y – z = 1
    4x – 3y + 2z = 4

    89 Views
    Answer
    257.

    Solve the following equation by matrix method:
    2 straight x plus straight y plus straight z space equals space 1 straight x minus 2 straight y minus straight z space equals 3 over 2 space space 3 straight y minus 5 straight z space space equals 9

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    Answer
    258. Solve system of linear equations, using matrix method:
    x – y + z = 4
    2x + y – 3z = 0
    x + y + z = 2 
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    Answer
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    259.

    Solve the following system of equations by matrix method:
    x – y + z = 2
    2x – y = 0
    2y – z = 1 

    86 Views
    Answer
    260.

    Solve the following system of equations by matrix method:
    2x – y + z = – 3
    3 x – z = – 8
    2x + 6y    = 2

    74 Views
    Answer
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