Given,

A =

A ^{- 1} =

The determinant of matrix A is

|A| =

= 2( 2 × - 2 – ( - 4)×1) + 3(3× - 2 – ( - 4)×1) + 5(3×1 – 2×1)

= 2( - 4 + 4 ) + 3( - 6 + 4 ) + 5( 3 – 2 )

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

= - 6 + 5

= - 1

|A| ≠ 0

∴ A ^{- 1} is possible.

A^T =

Adj(A) =

A ^{- 1} =

A ^{- 1} =

A ^{- 1} =

Given set of lines are : -

2x – 3y + 5z = 11

3x + 2y – 4z = - 5

x + y – 2z = - 3

Converting 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 = 2 , z = 3