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Determinants

Question
CBSEENMA12034710
Wired Faculty App

Examine the consistency of the system of equations:
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = – 1

Solution

The given equations are
5x – y + 4z = 5
2x + 3y + 5z = 2
5 x – 2 y + 6 z = – 1
These equations can be written as
open square brackets table row 5 cell space space minus 1 end cell cell space space space 4 end cell row 2 cell space space space space space 3 end cell cell space space space 5 end cell row 5 cell space space minus 2 end cell cell space space 6 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 cell space 5 end cell row cell space space 2 end cell row cell negative 1 end cell end table close square brackets
therefore space space space space space space AX space space equals space straight B space where space straight A space equals space open square brackets table row 5 cell space space minus 1 end cell cell space space space 4 end cell row 2 cell space space space space space space 3 end cell cell space space space 5 end cell row 5 cell space space minus 2 end cell cell space space space 6 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 cell space space space 5 end cell row cell space space space space 2 end cell row cell negative 1 end cell end table close square brackets space space space space space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 5 cell space space minus 1 end cell cell space space 4 end cell row 2 cell space space space space space 3 end cell cell space space 5 end cell row 5 cell space space minus 2 end cell cell space space 6 end cell end table close vertical bar space equals space 5 open vertical bar table row cell space space 3 end cell cell space space space 5 end cell row cell negative 2 end cell cell space space 6 end cell end table close vertical bar minus left parenthesis negative 1 right parenthesis open vertical bar table row 2 cell space space 5 end cell row 5 cell space space 6 end cell end table close vertical bar plus 4 open vertical bar table row 2 cell space space space space space 3 end cell row 5 cell space space minus 2 end cell end table close vertical bar space space space space space space space space space space space space space space space space equals 5 space left parenthesis 18 plus 10 right parenthesis plus 1 space left parenthesis 12 minus 25 right parenthesis plus 4 space left parenthesis negative 4 minus 15 right parenthesis space space space space space space space space space space space space space space space space space equals 5 left parenthesis 28 right parenthesis plus 1 left parenthesis negative 13 right parenthesis plus 4 left parenthesis negative 19 right parenthesis space equals 140 minus 13 minus 76 space equals space 51 not equal to 0 therefore space space space space straight A to the power of negative 1 end exponent space exists.
 ∴  given system of equations "has a unique solution and so system of equations is consistent.

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