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Determinants

Question
CBSEENMA12034895

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

Solution

The given equations are
2 x – y – z = 1
x + y + 2z = 1
3x – 2y – 2z = 1
These equations can be written as
                         open square brackets table row 2 cell space space space minus 1 end cell cell space space minus 1 end cell row 1 cell space space space space space space 1 end cell cell space space space space 2 end cell row 3 cell space space space space minus 2 end cell cell space minus 2 end cell end table close square brackets 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 1 row 1 row 1 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 2 cell space space minus 1 end cell cell space space space minus 1 end cell row 1 cell space space space space 1 end cell cell space space space space space space space 2 end cell row 3 cell space space minus 2 end cell cell space space minus 2 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 space straight B space equals space open square brackets table row 1 row 1 row 1 end table close square brackets
open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space minus 1 end cell cell space space space minus 1 end cell row 1 cell space space space space 1 end cell cell space space space space space space space 2 end cell row 3 cell space minus 2 end cell cell space space minus 2 end cell end table close vertical bar space equals space 2 open vertical bar table row cell space 1 end cell cell space space space space 2 end cell row cell negative 2 end cell cell space space minus 2 end cell end table close vertical bar space minus left parenthesis negative 1 right parenthesis space open vertical bar table row 1 cell space space space space 2 end cell row 3 cell space space minus 2 end cell end table close vertical bar plus left parenthesis negative 1 right parenthesis space open vertical bar table row 1 cell space space space space 1 end cell row 3 cell space space minus 2 end cell end table close vertical bar
space space space space space equals 2 left parenthesis negative 2 plus 4 right parenthesis space plus 1 left parenthesis negative 2 minus 6 right parenthesis minus 1 left parenthesis negative 2 minus 3 right parenthesis space equals space 4 minus 8 plus 5 space equals space 1 space not equal to space 0
therefore 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 space space 2 end cell row cell negative 2 end cell cell space space minus 2 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space space space space space 2 end cell row 3 cell space space minus 2 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space space space 1 end cell row 3 cell space space minus 2 end cell end table close vertical bar
i.e.  2, 8, – 5 respectively
Co-factors of the elements of second row of | A | are
negative open vertical bar table row cell negative 1 end cell cell space space space space space minus 1 end cell row cell negative 2 end cell cell space space minus 2 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space space minus 1 end cell row 3 cell space space minus 2 end cell end table close vertical bar comma space space minus space open vertical bar table row 2 cell space space space space space space minus 1 end cell row 3 cell space space space space minus 2 end cell end table close vertical bar

i.e. 0, – 1, 1 respectively
Co-factors of the elements of third row of | A | are
open vertical bar table row cell negative 1 end cell cell space space space minus 1 end cell row 1 cell space space space space space 2 end cell end table close vertical bar comma space space space space minus open vertical bar table row 2 cell space space space minus 1 end cell row 1 cell space space space space space space 2 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space space minus 1 end cell row 1 cell space space space space space space 1 end cell end table close vertical bar comma space space straight i. straight e. space space space space minus 1 comma space minus 5 comma space space 3 space respectively
therefore space space space space adj space space straight A space equals space open square brackets table row 2 cell space space space space 8 end cell cell space space space minus 5 end cell row 0 cell space space minus 1 end cell cell space space space space space 1 end cell row cell negative 1 end cell cell space minus 5 end cell cell space space space space space 3 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 2 cell space space space space space 0 end cell cell space space minus 1 end cell row 8 cell space minus 1 end cell cell space space minus 5 end cell row cell negative 5 end cell cell space space space space 1 end cell cell space space space space space 3 end cell end table close square brackets
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 space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space open square brackets table row 2 cell space space space space 0 end cell cell space space space minus 1 end cell row cell space 8 end cell cell space minus 1 end cell cell space space minus 5 end cell row cell negative 5 end cell cell space space space space space 1 end cell cell space space space space space 3 end cell end table close square brackets
Now space space space AX space equals space straight B space space rightwards double arrow space space space straight X space equals space straight A to the power of negative 1 end exponent straight B
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 open square brackets table row cell space 2 end cell cell space space space space 0 end cell cell space space space space minus 1 end cell row cell space 8 end cell cell space space minus 1 end cell cell space space minus 5 end cell row cell negative 5 end cell cell space space space space space 1 end cell cell space space space space space space 3 end cell end table close square brackets space open square brackets table row 1 row 1 row 1 end table close square brackets 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 open square brackets table row cell 2 plus 0 minus 1 end cell row cell 8 minus 1 minus 5 end cell row cell negative 5 plus 1 plus 3 end cell end table close square brackets 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 open square brackets table row cell space space 1 end cell row cell space space space space 2 end cell row cell negative 1 end cell end table close square brackets.
therefore space space space space space solution space is space straight x space equals space 1 comma space space space straight y space equals space 2 comma space space space straight z space equals space 1.

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