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Determinants

Question
CBSEENMA12034702
Wired Faculty App

Examine the consistency of the system of equations:
x+y+z = 1
2x + 3y + 2z = 2
ax+ay+2az = 4

Solution
The given equations are
straight x plus straight y plus straight z space equals space 1 2 straight x plus 3 straight y plus 2 straight z space equals space 2 straight x plus straight y plus 2 straight z space equals space 4 over straight a
These equations can be written as
space space space space space space space space open square brackets table row 1 cell space space 1 end cell cell space space space 1 end cell row 2 cell space space 3 end cell cell space space 2 end cell row 1 cell space space 1 end cell cell space space 2 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 1 row 2 row cell 4 over straight a end cell end table close square brackets
therefore space space space 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 space 1 end cell row 2 cell space space 3 end cell cell space space space 2 end cell row 1 cell space space 1 end cell cell space space space 2 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 1 row 2 row cell 4 over straight a end cell end table close square brackets space space space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 1 cell space space 1 end cell cell space space 1 end cell row 2 cell space space 3 end cell cell space space 2 end cell row 1 cell space space 1 end cell cell space 2 end cell end table close vertical bar space equals space open square brackets table row 1 cell space space 0 end cell cell space space 0 end cell row 2 cell space 1 end cell cell space space 0 end cell row 1 cell space space 0 end cell cell space space 1 end cell end table close square brackets comma space space by space straight C subscript 2 space rightwards arrow space straight C subscript 2 minus 2 straight C subscript 1 comma space space straight C subscript 3 rightwards arrow space space straight C subscript 3 space minus space straight C subscript 1 space space space space space space space space space space space space space space equals space 1 open vertical bar table row 1 cell space space 0 end cell row 0 cell space 1 end cell end table close vertical bar space equals space 1 left parenthesis 1 minus 0 right parenthesis space equals space 1 space not equal to space 0 therefore space space space space straight A to the power of negative 1 end exponent space exists.
∴    given equations has a unique solution and so system of equations is consistent.

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