If A = , find A ^{- 1}.

Using A ^{- 1}, solve the following

system of linear equations:

2x + y + z = 1;

X – 2y – z = ;

3y – 5z = 9.

HINT: Here A = ,

X = and B = .

Given,

A =

A ^{- 1} =

The determinant of matrix A is

|A| =

= 2( - 2× - 5 - ( - 1)×3) – (1× - 5 - ( - 1)×0) + (1×3 – ( - 2)×0)

= 2(10 + 3) – ( - 5) + (3)

= 26 + 5 + 3

= 34

|A| ≠ 0

∴ A ^{- 1} is possible.

A^T =

Adj(A) =

A ^{- 1} =

A ^{- 1} =

Given set of lines are : -

2x + y + z = 1

X – 2y – z =

3y – 5z = 9

Converting the following equations in matrix form,

AX = B

Where A = , X = , B =

Pre - multiplying by A ^{- 1}

A ^{- 1}AX = A ^{- 1}B

IX = A ^{- 1}B

X = A ^{- 1}B

=

=

= =

∴ x = 1 , y = , z = -