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Determinants

Question
CBSEENMA12034862

Use matrix method to solve the following system of equations:
x + y + z = 1
x – 2y + 3z = 2
x – 3y + 5z = 3

Solution

The given equations are
x + y + z =1
x – 2y + 3z = 2
x – 3y + 5z = 3
These equations can be written as
             open square brackets table row 1 cell space space space space space space space space space 1 end cell cell space space space space space space 1 end cell row 1 cell space space space space space minus 2 end cell cell space space space space space space 3 end cell row 1 cell space space space space space minus 3 end cell cell space space space space space space 5 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 space space 3 end cell 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 1 end cell cell space space 1 end cell row 1 cell space minus 2 end cell cell space space 3 end cell row 1 cell space minus 3 end cell cell space space 5 end cell end table close square brackets space space space space 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 space space space straight B space equals space open square brackets table row 1 row 2 row 3 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 1 end cell row 1 cell space minus 2 end cell cell space space space 3 end cell row 1 cell negative 3 end cell cell space space space 5 end cell end table close vertical bar space equals space 1 space open vertical bar table row cell negative 2 end cell cell space space space 3 end cell row cell negative 3 end cell cell space space space 5 end cell end table close vertical bar space minus space 1 space open vertical bar table row 1 cell space space space 3 end cell row 1 cell space space space 5 end cell end table close vertical bar plus 1 space open vertical bar table row 1 cell space space space minus 2 end cell row 1 cell space space space minus 3 end cell end table close vertical bar
space space space space space space space equals 1 left parenthesis negative 10 plus 9 right parenthesis space minus space 1 left parenthesis 5 minus 3 right parenthesis space plus space 1 left parenthesis negative 3 plus 2 right parenthesis space equals space minus 1 minus 2 minus 1 space equals space minus 4 space not equal to space 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 cell negative 2 end cell cell space space space 3 end cell row cell negative 3 end cell cell space space space 5 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space space 3 end cell row 1 cell space space space 5 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space minus 2 end cell row 1 cell space space minus 3 end cell end table close vertical bar
i.e. – 1, – 2, – 1 respectively
Co-factors of the elements of second row of | A | are
negative open vertical bar table row cell space space space 1 end cell cell space space space 1 end cell row cell negative 3 end cell cell space space space 5 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 1 cell space space space space 5 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 space 1 end cell row 1 cell space space space minus 3 end cell end table close vertical bar
i.e. – 8, 4, 4 respectively
Co-factors of the elements of third row of | A | are
open vertical bar table row 1 cell space space space 1 end cell row cell negative 2 end cell cell space space space 3 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space space 1 end cell row 1 cell space space space space 3 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 1 cell space space minus 2 end cell end table close vertical bar
i.e. 5, – 2, – 3 respectively
therefore space space space adj. space straight A space equals space open square brackets table row cell negative 1 end cell cell space space minus 2 end cell cell space space minus 1 end cell row cell negative 8 end cell cell space space space space space 4 end cell cell space space space space space space 4 end cell row cell space 5 end cell cell space space minus 2 end cell cell space space minus 3 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 1 end cell cell space space space minus 8 end cell cell space space space space space 5 end cell row cell negative 2 end cell cell space space space space space 4 end cell cell space minus 2 end cell row cell negative 1 end cell cell space space space space 4 end cell cell negative 3 end cell end table close square brackets
therefore space 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 minus 1 fourth open square brackets table row cell negative 1 end cell cell space space space minus 8 end cell cell space space space 5 end cell row cell negative 2 end cell cell space space space space 4 end cell cell space minus 2 end cell row cell negative 1 end cell cell space space space space 4 end cell cell negative 3 end cell end table close square brackets
Now space space space AX space equals space straight B
rightwards double arrow space space space 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 open square brackets table row straight x row straight y row straight z end table close square brackets space equals space minus 1 fourth open square brackets table row cell negative 1 end cell cell space space space minus 8 end cell cell space space space space space 5 end cell row cell negative 2 end cell cell space space space space 4 end cell cell space minus 2 end cell row cell negative 1 end cell cell space space space 4 end cell cell space minus 3 end cell end table close square brackets space space open square brackets table row 1 row 2 row 3 end table close square brackets
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 minus 1 fourth open square brackets table row cell negative 1 minus 16 plus 15 end cell row cell negative 2 plus 8 minus 6 end cell row cell negative 1 plus 8 minus 9 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 cell 1 half end cell row 0 row cell 1 half end cell end table close square brackets
therefore space space space space space soluiton space is space straight x space equals space 1 half comma space space space straight y space equals space 0 comma space space space space straight z space equals space 1 half

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