-->

Determinants

Question
CBSEENMA12034760

Using matrices, solve the following system of equations:
2x – y + z = 0
x + y – z = 6
3x – y – 4 z = 7

Solution

The given equations are
2x – y + z = 0
x + y – z = 6
3x – y – 4z = 7
These equations can be written as
                     open square brackets table row 2 cell space space minus 1 end cell cell space space space space space space 1 end cell row 1 cell space space space space space space 1 end cell cell space space minus 1 end cell row 3 cell space space space minus 1 end cell cell space space space minus 4 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 0 row 6 row 7 end table close square brackets
or          space space AX space equals space straight B space where space straight A space equals space open square brackets table row 2 cell space space minus 1 end cell cell space space space 1 end cell row 1 cell space space space space space 1 end cell cell space minus 1 end cell row 3 cell space space minus 1 end cell cell space minus 4 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 space straight B space equals space open square brackets table row 0 row 6 row 7 end table close square brackets
Now                         straight A space equals space open square brackets table row 2 cell space space minus 1 end cell cell space space space 1 end cell row 1 cell space space space space 1 end cell cell space minus 1 end cell row 3 cell space minus 1 end cell cell space minus 4 end cell end table close square brackets
therefore space space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space minus 1 end cell cell space space space space space 1 end cell row 1 cell space space space space space 1 end cell cell space minus 1 end cell row 3 cell space space minus 1 end cell cell space space minus 4 end cell end table close vertical bar space space equals space 2 space open vertical bar table row 1 cell space space minus 1 end cell row cell negative 1 end cell cell space space space minus 4 end cell end table close vertical bar minus left parenthesis negative 1 right parenthesis space open vertical bar table row 1 cell space space minus 1 end cell row 3 cell space space minus 4 end cell end table close vertical bar plus open vertical bar table row 1 cell space space space space space 1 end cell row 3 cell space minus 1 end cell end table close vertical bar
space space space space space space space space space space space space space space space equals 2 left parenthesis negative 4 minus 1 right parenthesis space plus space 1 left parenthesis negative 4 plus 3 right parenthesis space plus space 1 left parenthesis negative 1 minus 3 right parenthesis
space space space space space space space space space space space space space space space space space equals 2 left parenthesis negative 5 right parenthesis space plus space 1 left parenthesis negative 1 right parenthesis space plus space 1 left parenthesis negative 4 right parenthesis space equals space minus 10 minus 1 minus 4 space equals space minus 15 not equal to space 0
therefore space space space space straight A to the power of negative 1 end exponent space space exists.
Co-factors of the elements of first row of | A | are
open vertical bar table row cell space space 1 end cell cell space space space minus 1 end cell row cell negative 1 end cell cell space space minus 4 end cell end table close vertical bar comma space space space minus open vertical bar table row 1 cell space space minus 1 end cell row 3 cell space space minus 4 end cell end table close vertical bar space space comma space space open vertical bar table row 1 cell space space space space space 1 end cell row 3 cell space minus 1 end cell end table close vertical bar
i.e. – 5, 1, – 4 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 1 end cell row cell negative 1 end cell cell space minus 4 end cell end table close vertical bar comma space space space open vertical bar table row 2 cell space space space space space space space 1 end cell row 3 cell space space minus 4 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space minus 1 end cell row 3 cell space space minus 1 end cell end table close vertical bar

i.e. – 5, – 11, – 1 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 1 end cell row 1 cell negative 1 end cell end table close vertical bar comma space space space minus open vertical bar table row 2 cell space space space space space 1 end cell row 1 cell space minus 1 end cell end table close vertical bar comma space space open vertical bar table row 2 cell space space minus 1 end cell row 1 cell space space space space 1 end cell end table close vertical bar
i.e. 0, 3, 3 respectively.
adj space straight A space equals space open square brackets table row cell negative 5 end cell cell space space space 1 end cell cell space space minus 4 end cell row cell negative 5 end cell cell space space minus 11 end cell cell space space space minus 1 end cell row 0 3 cell space space space 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 negative 5 end cell cell space space space minus 5 end cell cell space space space 0 end cell row 1 cell space minus 11 end cell cell space space 3 end cell row cell negative 4 end cell cell space minus 1 end cell cell space space 3 end cell end table close square brackets
therefore space space 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 minus 1 over 15 open square brackets table row cell negative 5 end cell cell space space minus 5 end cell cell space space space 0 end cell row cell space space 1 end cell cell space space minus 11 end cell cell space space 3 end cell row cell negative 4 end cell cell space space minus 1 end cell cell space space 3 end cell end table close square brackets
Now space space space space space space space AX space equals space straight B space space space space space rightwards double arrow space space space straight X space equals space straight A to the power of negative 1 end exponent straight B
therefore space space 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 minus 1 over 15 open square brackets table row cell negative 5 end cell cell space space space minus 5 end cell cell space space space 0 end cell row 1 cell space space minus 11 end cell cell space space 3 end cell row cell negative 4 end cell cell space minus 1 end cell cell space 3 end cell end table close square brackets space open square brackets table row 0 row 6 row 7 end table close square brackets
rightwards double arrow space space space 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 minus 1 over 15 open square brackets table row cell 0 minus 30 plus 0 end cell row cell 0 minus 66 plus 21 end cell row cell 0 minus 6 plus 21 end cell end table close square brackets space space space space rightwards double arrow 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 minus 1 over 15 open square brackets table row cell negative 30 end cell row cell negative 45 end cell row 15 end table close square brackets 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 2 end cell row cell space space space 3 end cell row cell negative 1 end cell end table close square brackets
therefore space space space space straight x space equals space 2 comma space space space straight y space equals space 3 comma space space space straight z space equals space minus 1 space space is space the space solution.

 

Some More Questions From Determinants Chapter