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Determinants

Question
CBSEENMA12034910

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


Solution

The given equations are
2x + y – z = 1
x – y + z = 2
3x + y – 2 z = – 1
These equations can be written as
                      open square brackets table row 2 cell space space space space space 1 end cell cell space space space minus 1 end cell row 1 cell space minus 1 end cell cell space space space space space space 1 end cell row 3 cell space space space space space 1 end cell cell space space space minus 2 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 1 end cell row cell space space 2 end cell row cell negative 1 end cell end table close square brackets
or space space space space space space space space space space space space 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 1 end cell row 1 cell space space minus 1 end cell cell space space space space space space 1 end cell row 3 cell space space space space 1 end cell cell space space space space minus 2 end cell end table close square brackets comma 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 cell space space space space space 1 end cell row cell space space space space space 2 end cell row cell negative 1 end cell end table close square brackets space space space
space space space space space space space space space 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 1 end cell cell space space space minus 1 end cell row 1 cell space minus 1 end cell cell space space space space space space space 1 end cell row 3 cell space space space space space 1 end cell cell space space space minus 2 end cell end table close vertical bar space equals space 2 open vertical bar table row cell negative 1 end cell cell space space space space space space space space 1 end cell row 1 cell space space space space minus 2 end cell end table close vertical bar minus 1 space open vertical bar table row 1 cell space space space space space space 1 end cell row 3 cell space space minus 2 end cell end table close vertical bar space plus space left parenthesis negative 1 right parenthesis space open vertical bar table row 1 cell space space minus 1 end cell row 3 cell space space space space space 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 space space equals space 2 left parenthesis 2 minus 1 right parenthesis space minus space 1 left parenthesis negative 2 minus 3 right parenthesis space minus space 1 left parenthesis 1 plus 3 right parenthesis space equals space 2 left parenthesis 1 right parenthesis space minus space 1 left parenthesis negative 5 right parenthesis space minus space 1 left parenthesis 4 right parenthesis
space space space space space space space space space space space space space space space space space space space equals 2 plus 5 plus 4 space equals space 3 space not equal to space 0
therefore space space space space space straight A to the power of negative 1 end exponent space exists.
space space space space space space space space space space space space space space space space space space space space space space space space space space space space
Co-factors of the elements of first row of | A | are
       open vertical bar table row cell negative 1 end cell cell space space space space space space space 1 end cell row cell space 1 end cell cell space 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 1 end cell row 3 cell space space minus 2 end cell end table close vertical bar comma space space space space open vertical bar table row 1 cell space space space space minus 1 end cell row 3 cell space space space space space space space 1 end cell end table close vertical bar space space space or space space space 2 comma space space minus 1 comma space space minus left parenthesis negative 2 minus 3 right parenthesis comma space space space space 1 plus 3 space
i.e.     1, 5, 4 respectively.
Co-factors of the elements of second row of open vertical bar straight A close vertical bar are 
 negative open vertical bar table row 1 cell space space space minus 1 end cell row 1 cell space space space minus 2 end cell end table close vertical bar comma space space space space open vertical bar table row 2 cell space space space minus 1 end cell row 3 cell space space minus 2 end cell end table close vertical bar comma space space space space minus open vertical bar table row 2 cell space space space space 1 end cell row 3 cell space space space 1 end cell end table close vertical bar space space or space space space minus left parenthesis negative 2 plus 1 right parenthesis comma space space space space minus 4 plus 3 comma space space space minus left parenthesis 2 minus 3 right parenthesis
i.e. 1. – 1, 1 respectively.
Co-factors of the elements of third row of | A | are
open vertical bar table row cell space space 1 end cell cell space space minus 1 end cell row cell negative 1 end cell cell space space space space 1 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 1 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space space space space space space space space 1 space end cell row 1 cell space space space space minus 1 end cell end table close vertical bar space space space or space space 1 comma space minus 1 comma space minus left parenthesis 2 plus 1 right parenthesis comma space space minus 2 minus 1
i.e. 0, – 3, – 3 respectively.
 therefore space space space space space adj space straight A space equals space open square brackets table row 1 cell space space space space space space 5 end cell cell space space space space space 4 end cell row 1 cell space space minus 1 end cell cell space space space space 1 end cell row 0 cell space space minus 3 end cell cell space space minus 3 end cell end table close square brackets to the power of apostrophe space space equals space open square brackets table row 1 cell space space space space space space 1 end cell cell space space space space space space space 0 end cell row 5 cell space space minus 1 end cell cell space space space minus 3 end cell row 4 cell space space space space space 1 end cell cell space space space minus 3 end cell end table close square brackets
therefore 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 third open square brackets table row 1 cell space space space space space 1 end cell cell space space space space space space space 0 end cell row 5 cell space minus 1 end cell cell space space space minus 3 end cell row 4 cell space space space space space 1 end cell cell space space space minus 3 end cell end table close square brackets
Now comma space space space AX space equals space straight B space 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
therefore space space space 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 third open square brackets table row 1 cell space space space space space 1 end cell cell space space space space space 0 end cell row 5 cell space minus 1 end cell cell space space space minus 3 end cell row 4 cell space space space space space 1 end cell cell space space minus 3 end cell end table close square brackets space open square brackets table row cell space space 1 end cell row cell space space 2 end cell row cell negative 1 end cell end table close square brackets space 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 third open square brackets table row cell 1 plus 2 plus 0 end cell row cell 5 minus 2 plus 3 end cell row cell 4 plus 2 plus 3 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 third open square brackets table row 3 row 6 row 9 end table close square brackets space 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 1 row 2 row 3 end table close square brackets
therefore space space space space space space straight x space equals space 1 comma space space space space space straight y space equals space 2 comma space space straight z space equals space 3

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