+91-8076753736[email protected]
logo
logo
-->

Determinants

Question
CBSEENMA12034864
Wired Faculty App

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

Solution

The given equations are
2x + y + 2z = 3
x + y + 2z = 2
2x + 3y – z = – 2
These equations can be written as
  open square brackets table row 2 cell space space space space 1 end cell cell space space space space space 2 end cell row 1 cell space space space space 1 end cell cell space space space space space 2 end cell row 2 cell space space space 3 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 cell space space space 3 end cell row cell space space space space 2 end cell row cell negative 2 end cell end table close square brackets or space 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 space space space 1 end cell cell space space space space space 2 end cell row 1 cell space space space space space 1 end cell cell space space space space space 2 end cell row 2 cell space space space space space 3 end cell cell space space minus 1 end cell end table close square brackets comma space 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 cell space space space space 3 end cell row cell space space space space 2 end cell row cell negative 2 end cell end table close square brackets open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space space space 1 end cell cell space space space space space 2 end cell row 1 cell space space space space space 1 end cell cell space space space space space 2 end cell row 2 cell space space space space 3 end cell cell space space minus 1 end cell end table close vertical bar space equals space 2 space open vertical bar table row 1 cell space space space space space space 2 end cell row 3 cell space space minus 1 end cell end table close vertical bar space minus space 1 open vertical bar table row 1 cell space space space space 2 end cell row 2 cell space space minus 1 end cell end table close vertical bar space plus space 2 open vertical bar table row 1 cell space space space 1 end cell row 2 cell space space space 3 end cell end table close vertical bar space space space space equals space 2 left parenthesis negative 1 minus 6 right parenthesis space minus space 1 left parenthesis negative 1 minus 4 right parenthesis space plus space 2 left parenthesis 3 minus 2 right parenthesis space equals space minus 14 plus 5 plus 2 space equals space minus 7 space not equal to space 0 therefore space space space space straight A to the power of negative 1 end exponent space exists.
Co-factors of the elements of first row of | A | are
open vertical bar table row 1 cell space space space space space space 2 end cell row 3 cell space space minus 1 end cell end table close vertical bar comma space space space space minus open vertical bar table row 1 cell space space space space 2 end cell row 2 cell space space minus 1 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space space space space 1 end cell row 2 cell space space space space space 3 end cell end table close vertical bar
i.e.   -7,  5, 1 respectively.
Co-factors of the elements of second row of | A | are
negative open vertical bar table row 1 cell space space space space space space 2 end cell row 3 cell space space minus 1 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 2 end cell row 2 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 1 end cell row 2 cell space space space 3 end cell end table close vertical bar
i.e.  7, – 6, – 4 respectively
Co-factors of the elements of third row of | A | are
open vertical bar table row 1 cell space space space space 1 end cell row 2 cell space space space space 2 end cell end table close vertical bar comma space space minus open vertical bar table row 2 cell space space space 2 end cell row 1 cell space space space 2 end cell end table close vertical bar comma space space space space open vertical bar table row 2 cell space space space space 1 end cell row 1 cell space space space space 1 end cell end table close vertical bar
i.e. 0. – 2, 1 respectively
therefore space space space space adj space straight A space equals space open square brackets table row cell negative 7 end cell cell space space space space 5 end cell cell space space space space space 1 end cell row 7 cell space minus 6 end cell cell space minus 4 end cell row 0 cell space minus 2 end cell cell space space space space 1 end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row cell negative 7 end cell cell space space space space 7 end cell cell space space space space space 0 end cell row 5 cell space minus 6 end cell cell space space minus 2 end cell row 1 cell space minus 4 end cell cell space space space space space 1 end cell end table close square brackets
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 7 open square brackets table row cell negative 7 end cell cell space space space space space space 7 end cell cell space space space space space space 0 end cell row 5 cell space minus 6 end cell cell space space space minus 2 end cell row 1 cell space minus 4 end cell cell space space space space space 1 end cell end table close square brackets
Now     AX space equals space straight B
rightwards double arrow space space space space space straight X equals space straight A to the power of negative 1 end exponent straight B
rightwards double arrow 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 7 open square brackets table row cell negative 7 end cell cell space space space space space 7 end cell cell space space space 0 end cell row 5 cell space space minus 6 end cell cell space minus 2 end cell row 1 cell space space space minus 4 end cell cell space space space space 1 end cell end table close square brackets space open square brackets table row cell space space 3 end cell row cell space space 2 end cell row cell negative 2 end cell end table close square brackets 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 minus 1 over 7 open square brackets table row cell negative 21 plus 14 plus 0 end cell row cell 15 minus 12 plus 4 end cell row cell 3 minus 8 minus 2 end cell end table close square brackets 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 minus 1 over 7 open square brackets table row cell negative 7 end cell row cell space space space 7 end cell row cell negative 7 end cell end table close square brackets space space space space space rightwards double arrow 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 open square brackets table row cell space space space space 1 end cell row cell negative 1 space end cell row cell space space space 1 end cell end table close square brackets therefore space space space space solution space is space straight x space equals space 1 comma space space space space straight y space equals space minus 1 comma space space space straight z space equals space 1

Some More Questions From Determinants Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation