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Determinants

Question
CBSEENMA12034790
Wired Faculty App

Solve system of linear equations, using matrix method:
x – y + z = 4
2x + y – 3z = 0
x + y + z = 2 

Solution

The given equations are
x– y + z = 4
2x + y – 3z = 0
x + y + z = 2
These equations can be written as
                      open square brackets table row 1 cell space space space minus 1 end cell cell space space 1 end cell row 2 cell space space space space minus 3 end cell cell space space 1 end cell row 1 cell space space space space space 1 end cell cell space 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 4 row 0 row 2 end table close square brackets
or      AX space equals space straight B space where space space space straight A space equals space open square brackets table row 1 cell space space minus 1 end cell cell space space minus 1 end cell row 2 cell space space space space 1 end cell cell space space space minus 3 end cell row 1 cell space space space 1 end cell cell space space space space 1 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 straight B space equals space open square brackets table row 4 row 0 row 2 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 minus 1 end cell cell space space minus 1 end cell row 2 cell space space space space 1 end cell cell space space minus 3 end cell row 1 cell space space space space 1 end cell cell space space space space 1 end cell end table close vertical bar space equals space 1 open vertical bar table row 1 cell space space space minus 3 end cell row 1 cell space space space space space 1 end cell end table close vertical bar space minus left parenthesis negative 1 right parenthesis space open vertical bar table row 2 cell space space space minus 3 end cell row 1 cell space space space space space space space 1 space end cell end table close vertical bar plus 1 open vertical bar table row 2 cell space space space 1 end cell row 1 cell space space 1 end cell end table close vertical bar space space space space space space equals 1 thin space left parenthesis 1 plus 3 right parenthesis space plus space 1 thin space left parenthesis 2 plus 3 right parenthesis space space plus space 1 left parenthesis 2 minus 1 right parenthesis space equals space 4 plus 5 plus 1 space equals space 10 not equal to 0 therefore 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 3 end cell row 1 cell space space space space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space space minus 3 end cell row 1 cell space space space space space 1 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space space space 1 end cell row 1 cell space space space space space 1 end cell end table close vertical bar
i.e. 4,  –5,  1 respectively.
Co-factors of the elements of second row of | A | are
negative open vertical bar table row cell negative 1 end cell cell space space space space 1 end cell row 1 cell space space space 1 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space 1 end cell row 1 cell space space space space 1 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space space minus 1 end cell row 1 cell space space space space space space 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 cell negative 1 end cell cell space space space space space space 1 end cell row 1 cell space space minus 3 end cell end table close vertical bar comma space 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 3 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
i.e.  2, 5, 3 respectively
therefore space space space space adj. space straight A space equals space open square brackets table row 4 cell space space minus 5 end cell cell space space space space space 1 end cell row 2 cell space space space 0 end cell cell space minus 2 end cell row 2 cell space space 5 end cell cell space space space 3 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell space space space 4 end cell cell space space space 2 end cell cell space space 2 end cell row cell negative 5 end cell cell space space space 0 end cell cell space space 5 end cell row cell space space 1 end cell cell space minus 2 end cell cell space space 3 end cell end table close square brackets space Now 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 10 open square brackets table row 4 cell space space space space 2 end cell cell space space space space 2 end cell row cell negative 5 end cell cell space space space 0 end cell cell space space space 5 end cell row 1 cell negative 2 end cell cell space space space 3 end cell end table close square brackets Now space space space straight A space straight X 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 1 over 10 open square brackets table row cell space space 4 end cell cell space space space 2 end cell cell space space 2 end cell row cell negative 5 end cell cell space space 0 end cell cell space space 5 end cell row 1 cell negative 2 end cell cell space space 3 end cell end table close square brackets space open square brackets table row 4 row 0 row 2 end table close square brackets 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 1 over 10 open square brackets table row cell 16 plus 0 plus 4 end cell row cell negative 20 plus 0 plus 10 end cell row cell 4 plus 0 plus 6 end cell 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 1 over 10 open square brackets table row cell space space space space 20 end cell row cell negative 10 end cell row cell space space space 10 end cell end table close square brackets space 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 open square brackets table row cell space space space space 2 end cell row cell negative 1 end cell row cell space space 1 end cell end table close square brackets rightwards double arrow space space space space space straight x space equals space 2 comma space space space space straight y space space equals negative 1 comma space space space space straight z space equals space 1 space is space the space required space solution. space

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