Use matrix method to solve the following system of equations:x �
    • Zigya Logo

    Previous Year Papers

    Download Solved Question Papers Free for Offline Practice and view Solutions Online.

    Test Series

    Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

    Test Yourself

    Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
    Advertisement

    Determinants

    Multiple Choice QuestionsLong Answer Type

    271.

    Use matrix method to solve the following system of equations:
    2x – 3y + 3z = 1
    2x+ 2y+ 3z = 2
    3x – 2y + 2z = 3

    85 Views
    Answer
    272.

    Use matrix method to solve the following system of equations:
    2x – 3y + 3z = 1
    2x+ 2y+ 3z = 2
    3x – 2y + 2z = 3

    75 Views
    Answer
    273.

    Use matrix method to solve the following system of equations:
    2x – 3y + 3z = 1
    2x+ 2y+ 3z = 2
    3x – 2y + 2z = 3

    75 Views
    Answer
    274. Use matrix method to solve the following system of equations:
    y + 2z = 4
    2z + x = 5
    x + 2y = 7
    78 Views
    Answer
    Advertisement
    275. Use matrix method to solve the following system of equations:
    3x – 4y + 2z = – 1
    2x + 3y + 5z = 7
    x + z = 2 
    75 Views
    Answer
    Advertisement
    Zigya App

    276. Use matrix method to solve the following system of equations:
    x – y = 5
    y + z = 3
    z + x = 4

    The given equations are
    x + y + 0z = 5
    0x + y + z = 3
    x + 0y + z = 4
    These equations can be written as
    open square brackets table row 1 cell space space 1 end cell cell space space 0 end cell row 0 cell space space 1 end cell cell space space 1 end cell row 1 cell space space 0 end cell cell space space 1 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 5 row 3 row 4 end table close square brackets
    or     AX space equals straight B space where space straight A space equals space open square brackets table row 1 cell space space space 1 end cell cell space space space 0 end cell row 0 cell space space space 1 end cell cell space space space 1 end cell row 1 cell space space space 0 end cell cell space space space 1 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 5 row 3 row 4 end table close square brackets

                open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space space 1 end cell cell space space 0 end cell row 0 cell space space 1 end cell cell space space 1 end cell row 1 cell space space 0 end cell cell space space 1 end cell end table close vertical bar space equals space 1 open vertical bar table row 1 cell space space 1 end cell row 0 cell space space 1 end cell end table close vertical bar minus 1 open vertical bar table row 0 cell space space space space 1 end cell row 1 cell space space space space 1 end cell end table close vertical bar plus 0 open vertical bar table row 0 cell space space 1 end cell row 1 cell space space 0 end cell end table close vertical bar space space space space space space equals space 1 left parenthesis 1 minus 0 right parenthesis space minus space 1 space left parenthesis 0 minus 1 right parenthesis space plus space 0 left parenthesis 0 minus 1 right parenthesis space equals space 1 plus 1 plus 0 space equals space 2
    Co-factors of the elements of first row of |A| are
    open vertical bar table row 1 cell space space space 1 end cell row 0 cell space space space 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 0 cell space space space space 1 end cell row 1 cell space space space space 1 end cell end table close vertical bar comma space space space space open vertical bar table row 0 cell space space space space 1 end cell row 1 cell space space space space 0 end cell end table close vertical bar

    i.e., 1, 1, – 1 respectively
    Co-factors of the elements of second row of |A| are
    negative open vertical bar table row 1 cell space space space space space 0 end cell row 0 cell space space space space 1 end cell end table close vertical bar comma space space space space space open vertical bar table row 1 cell space space space 0 end cell row 1 cell space space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space 1 end cell row 1 cell space space space 0 end cell end table close vertical bar
    i.e.    – 1, 1, 1 respectively
    Co-factors of the elements of third row of | A | are
    open vertical bar table row 1 cell space space space 0 end cell row 1 cell space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space space 0 end cell row 0 cell space space space space 1 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space 1 end cell row 0 cell space space 1 end cell end table close vertical bar
    i.e., 1, – 1, 1 respectively
    therefore space space space adj space straight A space equals space open square brackets table row 1 cell space space space 1 end cell cell space space minus 1 end cell row cell negative 1 end cell cell space space space space space 1 end cell cell space space space space space 1 end cell row 1 cell space minus 1 end cell cell space space space space space 1 end cell end table close square brackets to the power of apostrophe space space equals space open square brackets table row 1 cell space minus 1 end cell cell space space space space space 1 end cell row 1 cell space space space space 1 end cell cell space minus 1 end cell row cell negative 1 end cell cell space space space 1 end cell cell space space space space 1 end cell end table close square brackets 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 1 half open square brackets table row 1 cell space space space minus 1 end cell cell space space space space space 1 end cell row 1 cell space space space space space 1 end cell cell space minus 1 end cell row cell negative 1 end cell cell space space space space space space 1 end cell cell space space space space 1 end cell end table close square brackets Now comma space space AX space equals space straight B space space space space rightwards double arrow space space space space space space straight X space equals space straight A to the power of negative 1 end exponent straight B space space space space 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 half open square brackets table row 1 cell space space minus 1 end cell cell space space space space 1 end cell row cell space 1 end cell cell space space space space space space 1 end cell cell space space minus 1 end cell row cell negative 1 end cell cell space space space space space space 1 end cell cell space space space space 1 end cell end table close square brackets space open square brackets table row 5 row 3 row 4 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 space equals space 1 half open square brackets table row cell 5 minus 3 plus 4 end cell row cell 5 plus 3 minus 4 end cell row cell negative 5 plus 3 plus 4 end cell end table close square brackets space space rightwards double arrow 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 half open square brackets table row 6 row 4 row 2 end table close square brackets space space 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 open square brackets table row 3 row 2 row 1 end table close square brackets therefore space space space space space space space space straight x space equals space 3 comma space space space space straight y space equals space 2 comma space space space straight z space equals space 1

     

    85 Views
  • Switch Icon Switch
    • Advertisement
    277. Use matrix method to solve the following system of equations:
    x + y + z = 3
    x + 2y + 3z = 4
    x + 4y + 9z = 6
    76 Views
    Answer
    278. Use matrix method to solve the following system of equations:
    3x + y + z = 3
    2x – y –z = 2
    –x –y + z = 1
    77 Views
    Answer
    Advertisement
    279. Use matrix method to solve the following system of equations:
    x + 2y – 3z = 6
    3x + 2y – 2z = 3
    2x – y + z = 2
    78 Views
    Answer
    280. Use matrix method to solve the following system of equations:
    x + y + z = 5         
    2 x + y – z = 2       
    2 x – y + z = 2 
    78 Views
    Answer
    Advertisement
    Flag Popup Icon
    Flag it
    Submit
    Flag Popup Icon
    Switch to
    flexipad
    Share It Icon
    Share it
    Share It
    Share it