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

Determinants

Question
CBSEENMA12034816
Wired Faculty App

Use matrix method to solve the following system of equations:
x – y + 2z = 7  
3x + 4 y – 5 z = – 5
2x – y + 3z = 12

Solution

The given equations are
x – y + 2z  = 7
3x + 4y – 5z= – 5
2 x – y + 3z = 12
These equations can be written as
                       open square brackets table row 1 cell space space space minus 1 end cell cell space space space space 2 end cell row 3 cell space space space space space 4 end cell cell space minus 5 end cell row 2 cell space space minus 1 end cell cell space space space 3 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 7 end cell row cell negative 5 end cell row cell space 12 end cell end table close square brackets
or space space AX space equals space straight B space space space space where space space space straight A space equals space open square brackets table row 1 cell space space minus 1 end cell cell space space space space space space 2 end cell row 3 cell space space space space space space 4 end cell cell space space minus 5 end cell row 2 cell space space space minus 1 end cell cell space space space space 3 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 space straight B equals space open square brackets table row cell space space 7 end cell row cell negative 5 end cell row cell space space 12 end cell end table close square brackets space space 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 space minus 1 end cell cell space space space 2 end cell row 3 cell space space space space space 4 end cell cell space minus 5 end cell row 2 cell space space minus 1 end cell cell space space space space 3 end cell end table close vertical bar space equals space 1 left parenthesis 12 minus 5 right parenthesis space minus space left parenthesis negative 1 right parenthesis thin space left parenthesis 9 plus 10 right parenthesis space plus space 2 left parenthesis negative 3 minus 8 right parenthesis space space space space space space space space space space space space space space space space space equals space 7 plus 19 minus 22 space equals space 4 space not equal to space 0 space space space space space space space rightwards double arrow space space space space straight A to the power of negative 1 end exponent space exists. space space space space space space space space space space space space space space space space space space       
  
 Co-factors of the elements of first row of | A | are
             open vertical bar table row 4 cell space space space minus 5 end cell row cell negative 1 end cell cell space space space space space 3 end cell end table close vertical bar comma space space minus open vertical bar table row 3 cell space space space minus 5 end cell row 2 cell space space space space space space 3 end cell end table close vertical bar comma space space space space space open vertical bar table row 3 cell space space space space space space 4 end cell row 2 cell space space minus 1 end cell end table close vertical bar
or   7, 19, – 11 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 2 end cell row cell negative 1 end cell cell space space 3 end cell end table close vertical bar comma space space space space space open vertical bar table row 1 cell space space space space 2 end cell row 2 cell space space space space 3 end cell end table close vertical bar comma space space minus open vertical bar table row 1 cell space space space minus 1 end cell row 2 cell space space minus 1 end cell end table close vertical bar
or    1, – 1, – 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 2 end cell row 4 cell space minus 5 end cell end table close vertical bar comma space space space space space minus open vertical bar table row 1 cell space space space space space space 2 end cell row 3 cell space space minus 5 end cell end table close vertical bar comma space space space open vertical bar table row 1 cell space space minus 1 end cell row 3 cell space space space space space 4 end cell end table close vertical bar
or    – 3, 11, 7 respectively.
therefore space space space adj. space straight A space equals space open square brackets table row 7 cell space space space minus 19 end cell cell space space space space minus 11 end cell row 1 cell space space minus 1 end cell cell space space space space minus 1 end cell row cell negative 3 end cell cell space space space 11 end cell cell space space space space space 7 end cell end table close square brackets space equals space open square brackets table row cell space space space 7 end cell cell space space space space space 1 end cell cell space space space minus 3 end cell row cell negative 19 end cell cell space minus 1 end cell cell space space space space space 11 end cell row cell negative 11 end cell cell space space space minus 1 end cell cell space space space space 7 end cell end table close square brackets space space 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 1 fourth open square brackets table row 7 cell space space space space space 1 end cell cell space space space minus 3 end cell row cell negative 19 end cell cell space space minus 1 end cell cell space space space space 11 end cell row cell negative 11 end cell cell space minus 1 end cell cell space space space space 7 end cell end table close square brackets Now space space space space AX space equals space straight B space space space space space rightwards double arrow space space space space straight X equals space straight A to the power of negative 1 end exponent straight B space 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 1 fourth open square brackets table row 7 cell space space space space 1 end cell cell space space minus 3 end cell row cell negative 19 end cell cell space space space minus 1 end cell cell space space space 11 end cell row cell negative 11 end cell cell space space minus 1 end cell cell space space space 7 end cell end table close square brackets space open square brackets table row cell space 7 end cell row cell negative 5 end cell row 12 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 1 fourth open square brackets table row cell 49 minus 5 minus 36 end cell row cell negative 133 plus 5 plus 132 end cell row cell negative 77 plus 5 plus 84 end cell end table close square brackets space space space space space space space space space space space 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 2 row 1 row 3 end table close square brackets rightwards double arrow 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

Some More Questions From Determinants Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation