Question

If $straight A equals space open square brackets table row 3 cell space space space minus 2 end cell cell space space space space space 1 end cell row 2 cell space space space space 1 end cell cell space minus 3 end cell row cell negative 1 end cell cell space space space 2 end cell cell space space space 1 end cell end table close square brackets comma$ Find A^–1.

Using A solve the following systems of linear equations:
3x – 2y + z = 2
2y + y – 3z = – 5
– x + 2y + z = 6.

Solution

Here,
   $straight A space equals space open square brackets table row 3 cell space space space minus 2 end cell cell space space space space space 1 end cell row 2 cell space space space space space 1 end cell cell space minus 3 end cell row 1 cell space space space space 2 end cell cell space space space space space 1 space end cell end table close square brackets open vertical bar straight A close vertical bar space equals space open vertical bar table row 3 cell space space space space minus 2 end cell cell space space space 1 end cell row 2 cell space space space space space 1 end cell cell negative 3 end cell row cell negative 1 end cell cell space space space space space 2 end cell cell space space space 1 end cell end table close vertical bar space equals space 3 space open vertical bar table row 1 cell space space space minus 3 end cell row 2 cell space space space space space space 1 end cell end table close vertical bar minus left parenthesis negative 2 right parenthesis space open vertical bar table row cell space space 2 end cell cell space space minus 3 end cell row cell negative 1 end cell cell space space space space space 1 end cell end table close vertical bar plus 1 space open vertical bar table row cell space space 2 end cell cell space space space space 1 end cell row cell negative 1 end cell cell space space space space space 2 end cell end table close vertical bar space space space space space space equals space 3 space left parenthesis 1 plus 6 right parenthesis space plus space 2 left parenthesis 2 minus 3 right parenthesis space plus space 1 left parenthesis 4 plus 1 right parenthesis space equals space 21 minus 2 plus 5 space equals space 24 space not equal to 0 therefore space space space space space space space straight A to the power of negative 1 end exponent space exists.$
Co-factors of the elements of first row of | A | are
$open vertical bar table row 1 cell space space space minus 3 end cell row 2 cell space space space space space 1 end cell end table close vertical bar comma space space space minus open vertical bar table row cell space space space 2 end cell cell space space space minus 3 end cell row cell negative 1 end cell cell space space space space space 1 end cell end table close vertical bar comma space space space space space open vertical bar table row cell space space 2 end cell cell space space space 1 end cell row cell negative 1 end cell cell space space space 2 end cell end table close vertical bar$
i.e., 7, 1, 5 respectively
Co-factors of the elements of second row of | A | are
$negative open vertical bar table row cell negative 2 end cell cell space space space 1 end cell row cell space 2 end cell cell space space space 1 end cell end table close vertical bar comma space space space space open vertical bar table row cell space 3 end cell cell space space space 1 end cell row cell negative 1 end cell cell space space space 1 end cell end table close vertical bar comma space space space minus open vertical bar table row cell space space 3 end cell cell space space space minus 2 end cell row cell negative 1 end cell cell space space space space space 2 end cell end table close vertical bar$
i.e.   4, 4, – 4 respectively
Co-factors of the elements of third row of | A | are
$open vertical bar table row cell negative 2 end cell cell space space space space space space space space 1 end cell row 1 cell space space space space minus 3 end cell end table close vertical bar comma space space space space space minus open vertical bar table row 3 cell space space space space space 1 end cell row 2 cell space space minus 3 end cell end table close vertical bar comma space space space open vertical bar table row 3 cell space space space minus 2 end cell row 2 cell space space space space space 1 end cell end table close vertical bar$
i.e. 5, 11, 7 respectively
$therefore space space space adj. space straight A space equals space open square brackets table row 7 cell space space space space 1 end cell cell space space space 5 end cell row 4 cell space space space 4 end cell cell negative 4 end cell row 5 cell space space space space 11 end cell cell space space space 7 end cell end table close square brackets to the power of apostrophe space space equals space open square brackets table row 7 cell space space space space 4 end cell cell space 5 end cell row 1 cell space space space space 4 end cell cell space 11 end cell row 5 cell space minus 4 end cell 7 end table close square brackets$
$straight A to the power of negative 1 end exponent space equals space fraction numerator adj space straight A over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 over 24 open square brackets table row 7 cell space space space space 4 end cell cell space space 5 end cell row 1 cell space space space space space 4 space end cell cell space space 11 end cell row 5 cell negative 4 end cell cell space space 7 end cell end table close square brackets$

The given equations are
3x – 2y + z = 2
2y + y – 3 z = – 5
– x + 2y + z = 6
These equations can be written as
$open square brackets table row 3 cell space space space space minus 2 end cell cell space space space space space 1 end cell row 2 cell space space space space space 1 end cell cell space minus 3 end cell row cell negative 1 end cell cell space space space space 2 end cell cell space space space space space 1 end cell end table close square brackets space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row cell space space space 2 end cell row cell negative 5 end cell row cell space space 6 end cell end table close square brackets space$
or $AX space equals space straight B space where space straight A space equals space open square brackets table row cell space space 3 end cell cell space space space minus 2 end cell cell space space space 1 end cell row cell space space space 2 end cell cell space space space space 1 end cell cell negative 3 end cell row cell negative 1 end cell cell space space space 2 end cell cell space space 1 end cell end table close square brackets comma space space space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets comma space space straight B space equals space open square brackets table row cell space space 2 end cell row cell negative 5 end cell row cell space space 6 end cell end table close square brackets$
Now $Now space space space AX space equals space straight B space space space space rightwards double arrow space space space space straight X space equals space straight A to the power of negative 1 end exponent straight B$
$therefore space space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 24 open square brackets table row 7 cell space space space 4 end cell cell space space space 5 end cell row 1 cell space space space space 4 end cell cell space space 11 end cell row 5 cell space minus 4 end cell cell space space 7 end cell end table close square brackets space open square brackets table row cell space space space 2 end cell row cell negative 5 end cell row cell space space space 6 end cell end table close square brackets space space space rightwards double arrow space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 24 open square brackets table row cell 14 minus 20 plus 30 end cell row cell 2 minus 20 plus 66 end cell row cell 10 plus 20 plus 42 end cell end table close square brackets rightwards double arrow space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space 1 over 24 open square brackets table row 24 row 48 row 72 end table close square brackets space space space space space space rightwards double arrow space space space space space space open square brackets table row straight x row straight y row straight z end table close square brackets space equals space open square brackets table row 1 row 2 row 3 end table close square brackets therefore space space space space space straight x space equals space 1 comma space space space space straight y equals space 2 comma space space straight z space equals space 3.$

For Daily Free Study Material Join wiredfaculty

Determinants

Some More Questions From Determinants Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction