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Determinants

Question
CBSEENMA12034891
Wired Faculty App

Use matrix method to solve the following system of equations:
2x – 3y + 5z = 16
3x + 2y – 4z = – 4
x + y – 2z = – 3

Solution

2x – 3y + 5z = 16
3x + 2y – 4z = – 4
x + y – 2z = – 3
These equations can be written as
                  open square brackets table row 2 cell space space minus 3 end cell cell space space space space 5 end cell row 3 cell space space space 2 end cell cell negative 4 end cell row 1 cell space 1 end cell cell negative 2 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 space 16 end cell row cell negative 4 end cell row cell negative 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 2 cell space space minus 3 end cell cell space space space space 5 end cell row 3 cell space space space space space 2 end cell cell space space minus 4 end cell row 1 cell space space space space 1 end cell cell space minus 2 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 space space straight B space equals space open square brackets table row cell space space 16 end cell row cell negative 4 end cell row cell negative 3 end cell end table close square brackets
Now  AX = B      rightwards double arrow space space space space space straight X space equals space straight A to the power of negative 1 end exponent space straight B
therefore 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 open square brackets table row 0 cell space space space 1 end cell cell space space space space minus 2 end cell row cell negative 2 end cell cell space space space 9 end cell cell space space minus 23 end cell row cell negative 1 end cell cell space space space 5 end cell cell space minus 13 end cell end table close square brackets space open square brackets table row cell space space space 16 end cell row cell negative 4 end cell row cell negative 3 end cell end table close square brackets space space space space rightwards double arrow 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 0 minus 4 plus 6 end cell row cell negative 32 minus 36 plus 69 end cell row cell negative 16 minus 20 plus 39 end cell end table close square brackets space rightwards double arrow 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 2 row 1 row 3 end table close square brackets therefore space space space space space space space space space space straight x space equals space 2 comma space space space space straight y space equals space 1 comma space space space straight z space equals space 3 comma space space space which space is space required space solution. space

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