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Determinants

Question
CBSEENMA12034887

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

Solution

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

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