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Determinants

Question
CBSEENMA12034716
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Examine the consistencies of the system of equations:
x - y+ z = 3
2x - y – z = 2
– x – 2y + 2z = 1

Solution

The given equations are
x – y + z = 3    ...(1)
2 x – y – z = 2    ...(2)
– x –2y + 2z = 1    ...(3)
These equations can be written as
open square brackets table row 1 cell space space space minus 1 end cell cell space space space space space 1 end cell row cell space 2 end cell cell space space space space 1 end cell cell space space minus 1 end cell row cell negative 1 end cell cell space minus 2 end cell cell space space space space space 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 3 row 2 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 1 cell space space space minus 1 end cell cell space space space space space 1 end cell row 2 cell space space space space space space 1 end cell cell space minus 1 end cell row cell negative 1 end cell cell space space space minus 2 end cell cell space space space space 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 space straight B space equals space open square brackets table row 3 row 2 row 1 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 space minus 1 end cell cell space space space space space 1 end cell row 2 cell space 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 minus 2 end cell cell space space space space 2 end cell end table close vertical bar space equals space open vertical bar table row cell space 1 end cell cell space space space 0 end cell cell space space space space 0 end cell row cell space space 2 end cell cell space space space 3 end cell cell space space minus 3 end cell row cell negative 1 end cell cell space space minus 3 end cell cell space space space space space 3 end cell end table close vertical bar space space space space space space space space equals space open vertical bar table row 3 cell space space space minus 3 end cell row cell negative 3 end cell cell space space space space space space space 3 end cell end table close vertical bar space equals space 9 minus 9 space equals space 0
Co-factors of the elements of first row of | A | are
open vertical bar table row 1 cell space space space space minus 1 end cell row cell negative 2 end cell cell space space space space space space 2 end cell end table close vertical bar comma space space space space minus open vertical bar table row cell space 2 end cell cell space space space minus 1 end cell row cell negative 1 end cell cell space space space space space 2 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space space space space 1 end cell row cell negative 1 end cell cell space space space minus 2 end cell end table close vertical bar

or  0, – 3, – 3 respectively.
Co-factors of the elements of 2nd row of | A | are
negative open vertical bar table row cell negative 1 end cell cell space space space space space 1 end cell row cell negative 2 end cell cell space space space space space 2 end cell end table close vertical bar comma space space open vertical bar table row 1 cell space space space space 1 end cell row cell negative 1 end cell cell space space space space 2 end cell end table close vertical bar comma space space space minus 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 minus 2 end cell end table close vertical bar

or 0, 3, 3 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 space space space 1 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 1 cell space space space space space space 1 end cell row 2 cell space space minus 1 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space minus 1 end cell row 2 cell space space space space space space 1 end cell end table close vertical bar space space space or space space 0 comma space 3 comma space 3 space respectively.
therefore space space space adj. space straight A space equals space open square brackets table row 0 cell space space minus 3 end cell cell space space space minus 3 end cell row 0 cell space space space space space 3 end cell cell space space space space space space space 3 space end cell row 0 cell space space space space 3 end cell cell space space space space space space space 3 space end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 0 cell space space space 0 end cell cell space space 0 end cell row cell negative 3 end cell cell space space space 3 end cell cell space space space 3 end cell row cell negative 3 end cell cell space space space 3 end cell cell space space space 3 end cell end table close square brackets left parenthesis adj. space straight A right parenthesis space straight B space equals space open square brackets table row 0 cell space space space 0 end cell cell space space 0 end cell row cell negative 3 end cell cell space space space 3 end cell cell space space space 3 end cell row cell negative 3 end cell cell space space space 3 end cell cell space space space 3 end cell end table close square brackets space open square brackets table row 3 row 2 row 1 end table close square brackets space equals space open square brackets table row cell 0 plus 0 plus 0 end cell row cell negative 9 plus 6 plus 3 end cell row cell negative 9 plus 6 plus 3 end cell end table close square brackets space equals space open square brackets table row 0 row 0 row 0 end table close square brackets space equals space straight O therefore space space space given space equations space have space infinite space solutions. space Adding space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space get comma space space 3 straight x space equals space 5 space space space space space space rightwards double arrow space space space space straight x space equals space 5 over 3 Adding space left parenthesis 1 right parenthesis space and space left parenthesis 3 right parenthesis comma space we space get comma space space minus 3 straight y plus 3 straight z space equals space 4 comma space space Put space straight z space equals space straight k therefore space space space space space 3 space straight y space equals space 3 space straight k space minus 4 space space space space space space space space space rightwards double arrow space space space space straight y space equals space straight k minus 4 over 3 therefore space space space solutions space are space straight x space equals space 5 over 3 comma space space straight y space equals space straight k minus 4 over 3 comma space space straight z space equals space straight k where space straight k space is space straight a space parameter.

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