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Determinants

Question
CBSEENMA12034907

Use matrix method to solve the following system of equations:
9x – 5y - 11z = 12 
x – 3y + z = 1
2x + 3y – 7z = 2

Solution

The given equations are
9x – 5y – 11z = 12
x – 3y + z = 1
2x + 3y – 7z = 2
These equations can be written as
                         open square brackets table row 9 cell space space space space minus 5 end cell cell space space space minus 11 end cell row 1 cell space space space space minus 3 end cell cell space space space space space space space 1 end cell row 2 cell space space space space space space 3 end cell cell space space space minus 7 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 12 row 1 row 2 end table close square brackets
or              AX space equals space straight B space where space straight A space equals space open square brackets table row 9 cell space space space minus 5 end cell cell space space space minus 11 end cell row 1 cell space space space minus 3 end cell cell space space space space space space 1 end cell row 2 cell space space space space 3 end cell cell space space space space space space 7 end cell end table close square brackets comma space 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 12 row 1 row 2 end table close square brackets
                   open vertical bar straight A close vertical bar space equals space open vertical bar table row 9 cell space space space minus 5 end cell cell space space minus 11 end cell row 1 cell space space minus 3 end cell cell space space space space space space 1 end cell row 2 cell space space space space space 3 end cell cell space space space space minus 7 end cell end table close vertical bar space equals space 9 open vertical bar table row cell negative 3 end cell cell space space space space space space 1 end cell row cell space 3 end cell cell space minus 7 end cell end table close vertical bar minus left parenthesis negative 5 right parenthesis 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 7 end cell end table close vertical bar space minus space 11 space open vertical bar table row 1 cell space space space minus 3 end cell row 2 cell space space space space space space space 3 end cell end table close vertical bar
space space space space space space space equals space 9 space left parenthesis 21 minus 3 right parenthesis space plus space 5 left parenthesis negative 7 minus 2 right parenthesis space minus 11 space left parenthesis 3 plus 6 right parenthesis space equals space 9 space left parenthesis 18 right parenthesis space plus space 5 thin space left parenthesis negative 9 right parenthesis space minus space 11 left parenthesis 9 right parenthesis
space space space space space space space space equals space 162 space minus 45 space minus 99 space equals space 18 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 cell negative 3 end cell cell space space space space space space 1 end cell row 3 cell space minus 7 end cell end table close vertical bar comma space space space space space space minus open vertical bar table row 1 cell space space space space space space space 1 end cell row 2 cell space space minus 7 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space minus 3 end cell row 2 cell space space space space space space 3 end cell end table close vertical bar space space space or space space 21 minus 3 comma space space space space minus left parenthesis negative 7 minus 2 right parenthesis comma space space 3 space plus space 6
straight i. straight e. space space 18 comma space 9 comma space 9 space space respectively.
Co-factors of the elements of second row of | A | are
negative open vertical bar table row cell negative 5 end cell cell space space space minus 11 end cell row 3 cell space space minus 7 end cell end table close vertical bar comma space space space space open vertical bar table row 9 cell space space space minus 11 end cell row 2 cell space space space minus 7 end cell end table close vertical bar comma space space minus open vertical bar table row 9 cell space space space minus 5 end cell row 2 cell space space space space space 3 end cell end table close vertical bar space space or space space space minus left parenthesis 35 plus 33 right parenthesis comma space space minus 63 plus 22 comma space space minus left parenthesis 27 plus 10 right parenthesis
i.e. – 68 – 41, – 37 respectively.
Co-factors of the elements of third row of | A | are
open vertical bar table row cell negative 5 end cell cell space space space space minus 11 end cell row cell negative 3 end cell cell space space space space space space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 9 cell space space space minus 11 end cell row 1 cell space space space space space space space 1 end cell end table close vertical bar comma space space space space open vertical bar table row 9 cell space space space minus 5 end cell row 1 cell space space space minus 3 end cell end table close vertical bar space space or space space minus 5 minus 33 comma space space space space minus left parenthesis 9 plus 11 right parenthesis comma space space minus 27 plus 5
i.e. – 38, – 20, – 22 respectively.
therefore space space space space space space space space adj space straight A space equals space open square brackets table row 18 cell space space space space space space 9 end cell cell space space space space 9 end cell row cell negative 68 end cell cell space space minus 41 end cell cell negative 37 end cell row cell negative 38 end cell cell space minus 20 end cell cell negative 22 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 18 cell space space space space minus 68 end cell cell space space space minus 38 end cell row 9 cell space space space minus 41 end cell cell space minus 20 end cell row 9 cell space minus 37 end cell cell space minus 22 end cell end table close square brackets
therefore space space 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 18 open square brackets table row 18 cell space space minus 68 end cell cell space space minus 38 end cell row 9 cell space space minus 41 end cell cell space minus 20 end cell row 9 cell space space minus 37 end cell cell space minus 22 end cell end table close square brackets
Now,     
      AX space equals space straight B space space space space space space rightwards double arrow 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 space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 18 open square brackets table row 18 cell space space space minus 68 end cell cell space space minus 38 end cell row 9 cell negative 41 end cell cell negative 20 end cell row 9 cell negative 37 end cell cell negative 22 end cell end table close square brackets space open square brackets table row 12 row 1 row 2 end table close square brackets space space space space rightwards double arrow 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 18 open square brackets table row cell 216 minus 68 minus 76 end cell row cell 108 minus 41 minus 40 end cell row cell 108 minus 37 minus 44 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 18 open square brackets table row 72 row 27 row 27 end table close square brackets space space space space space 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 4 row cell 3 over 2 end cell row cell 3 over 2 end cell end table close square brackets
therefore space space space space space space straight x space equals space 4 comma space space space space straight y space equals space 3 over 2 comma space space space space straight z space equals space 3 over 2.

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