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Determinants

Question
CBSEENMA12034906

Use matrix method to solve the following system of equations:
x + y + z = 6
x – y + z = 2
2x + y – z = 1

Solution

The given equations are
x + y + z = 6
x – y + z = 2
2x + y – z = 1
These equations can be written as
                          open square brackets table row 1 cell space space space space space 1 end cell cell space space space space space 1 end cell row 1 cell space minus 1 end cell cell space space space space space 1 end cell row 2 cell space space space space space 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 6 row 2 row 1 end table close square brackets
or         AX space space equals space straight B space where space straight A space equals space open square brackets table row 1 cell space space space space space 1 end cell cell space space space space space space 1 end cell row 1 cell space minus 1 end cell cell space space space space space space 1 end cell row 2 cell space space space 1 end cell cell space space minus 1 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 6 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 space 1 end cell cell space space space space space 1 end cell row 1 cell space minus 1 end cell cell space space space space space 1 end cell row 2 cell space space space 1 end cell cell space minus 1 end cell end table close vertical bar space equals space 1 open vertical bar table row cell negative 1 end cell cell space space space space space 1 end cell row 1 cell space space minus 1 end cell end table close vertical bar space minus 1 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 plus 1 open vertical bar table row 1 cell space space space space minus 1 end cell row 2 cell space space space space space space space 1 end cell end table close vertical bar
space space space space space space equals 1 left parenthesis 1 minus 1 right parenthesis minus 1 left parenthesis negative 1 minus 2 right parenthesis plus 1 left parenthesis 1 plus 2 right parenthesis space equals space 1 left parenthesis 0 right parenthesis minus 1 left parenthesis negative 3 right parenthesis plus 1 left parenthesis 3 right parenthesis
space space space space space space space equals 0 plus 3 plus 3 space equals space 6 space not equal to space 0
therefore 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 1 end cell cell space space space space space 1 end cell row cell space space 1 end cell cell space minus 1 end cell end table close vertical bar comma space 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 space minus 1 end cell row 2 cell space space space space space space 1 end cell end table close vertical bar space space or space space space space space 1 comma space space space minus 1 space comma space space minus left parenthesis negative 1 minus 2 right parenthesis comma space space space space 1 plus 2
straight i. straight e.. space space space 0 comma space 3 comma space 3 space respectively.
Co-factors of the elements of second row of | A | are
negative open vertical bar table row 1 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 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 minus open vertical bar table row 1 cell space space space space space 1 end cell row 2 cell space space space space space 1 end cell end table close vertical bar space space space or space space minus left parenthesis negative 1 minus 1 right parenthesis comma space space minus 1 minus 2 comma space space minus left parenthesis 1 minus 2 right parenthesis
i.e. 2, – 3,   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 1 end cell row cell negative 1 end cell cell space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space 1 end cell row 1 cell space space space 1 end cell end table close vertical bar comma space space space space open vertical bar table row 1 cell space space space space space space 1 end cell row 1 cell space space minus 1 end cell end table close vertical bar space space or space space space 1 plus 1 comma space space minus left parenthesis 1 minus 1 right parenthesis comma space space space space minus 1 minus 1
i.e. 2, 0, – 2 respectively.
therefore space space adj space straight A space equals space open square brackets table row 0 cell space space space space space 3 end cell cell space space space space 3 end cell row 2 cell space minus 3 end cell cell space space space space 1 end cell row 2 cell space space space space 0 end cell cell space minus 2 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 space space space 2 end cell cell space space space space space 2 end cell row 3 cell space space minus 3 end cell cell space space space space space 0 end cell row 3 cell space space space space space space 1 end cell cell space minus 2 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 over 6 open square brackets table row 0 cell space space space space space 2 end cell cell space space space space space 2 end cell row 3 cell space minus 3 end cell cell space space space space space 0 end cell row 3 cell space space space space 1 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 rightwards double arrow space space space straight X space equals space straight A to the power of negative 1 end exponent straight B
therefore 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 6 open square brackets table row 0 cell space space space space space 2 end cell cell space space space space space space 2 end cell row 3 cell space minus 3 end cell cell space space space space space 0 end cell row 3 cell space space space space 1 end cell cell space space space minus 2 end cell end table close square brackets space open square brackets table row 6 row 2 row 1 end table close square brackets space 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 space equals space 1 over 6 open square brackets table row cell 0 plus 4 plus 2 end cell row cell 18 minus 6 plus 0 end cell row cell 18 plus 2 minus 2 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 6 open square brackets table row 6 row 12 row 18 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 open square brackets table row 1 row 2 row 3 end table close square brackets
therefore space space space space space straight x space equals space 1 comma space space space straight y space equals space 2 comma space space space straight z equals space 3.

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