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Determinants

Question
CBSEENMA12034713
Wired Faculty App

Examine the consistencies of the system of equations:
3x – y + 2z = 3
2x + y + 3z =5
x - 2y - z = 1

Solution

The given equations are
3x – y + 2z = 3
2x + y + 3z = 5
x – 2 – z = 1
These equations can be written as
                          open square brackets table row 3 cell space space minus 1 end cell cell space space space space space space 2 end cell row 2 cell space space space space space 1 end cell cell space space space space space space 3 end cell row 1 cell space space minus 2 end cell cell space 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 3 row 5 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 3 cell space space minus 1 end cell cell space space 2 end cell row 2 cell space space 1 end cell cell space 3 end cell row 1 cell space minus 2 end cell cell space 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 straight B space equals space open square brackets table row 3 row 5 row 1 end table close square brackets
open vertical bar straight A close vertical bar space equals space open vertical bar table row 3 cell space minus 1 end cell cell space space space space 2 end cell row 2 cell space space space 1 end cell cell space space space space space 3 space end cell row 1 cell negative 2 end cell cell space minus 1 end cell end table close vertical bar space equals space 3 left parenthesis negative 1 plus 6 right parenthesis minus left parenthesis negative 1 right parenthesis minus left parenthesis negative 2 minus 3 right parenthesis plus 2 left parenthesis negative 4 minus 1 right parenthesis space space space space space equals space 15 minus 5 minus 10 space equals space 0
Co-factors of the elements of the first row of | A | are
open vertical bar table row cell space space 1 end cell cell space space space space space space 3 end cell row cell negative 2 end cell cell space space minus 1 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space space space space 3 end cell row 1 cell space space space minus 1 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space space space space space 1 end cell row 1 cell space space space space minus 2 end cell end table close vertical bar

or – 1 + 6, – (– 2, – 3), – 4 – 1 or 5, 5 – 5 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 space space space 2 end cell row cell negative 2 end cell cell space space space minus 1 end cell end table close vertical bar comma space space space space open vertical bar table row 3 cell space space space space space space 2 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 3 cell space space space minus 1 end cell row 1 cell space space space minus 2 end cell end table close vertical bar or space space space minus 5 comma space space minus 5 comma space space 5 space respectively.
Co-factors of the elements of the third row of | A | are
open vertical bar table row cell negative 1 end cell cell space space space 2 end cell row 1 cell space space 3 end cell end table close vertical bar comma space space minus open vertical bar table row 3 cell space space space space space 2 end cell row 2 cell space space space space space 3 end cell end table close vertical bar comma space space open vertical bar table row 3 cell space space space space minus 1 end cell row 2 cell space space space space space space 1 end cell end table close vertical bar
or   – 5,   – 5,  5 respectively.
therefore space space adj. space straight A space equals space open square brackets table row cell space space space 5 end cell cell space space space space space space space 5 end cell cell space space space minus 5 end cell row cell negative 5 end cell cell space space space minus 5 end cell cell space space space space space space 5 end cell row cell negative 5 end cell cell space space space minus 5 end cell cell space space space space space 5 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row 5 cell space space space minus 5 end cell cell space space space space minus 5 end cell row 5 cell space space minus 5 end cell cell space space minus 5 end cell row cell negative 5 end cell cell space space space space space 5 end cell cell space space space space space 5 end cell end table close square brackets space space space space space space space left parenthesis adj. space straight A right parenthesis thin space straight B space equals space open square brackets table row cell space space 5 end cell cell space space space minus 5 end cell cell space space space space minus 5 end cell row cell space 5 end cell cell space space minus 5 end cell cell space space space minus 5 end cell row cell negative 5 space end cell cell space space space space space 5 end cell cell space space space space space 5 end cell end table close square brackets space open square brackets table row 3 row 5 row 1 end table close square brackets space space space space space space space space space space space space space space space space space space space space space space space equals space open square brackets table row cell 15 minus 25 minus 5 end cell row cell 15 minus 25 minus 5 end cell row cell negative 15 plus 25 plus 5 end cell end table close square brackets space equals space open square brackets table row cell negative 15 end cell row cell negative 15 end cell row 15 end table close square brackets not equal to space straight O
given equations have no solution.

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