Answer :


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.


AT =


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 - 1AX = A - 1B


IX = A - 1B


X = A - 1B



=


=


=


x = 1 , y = 2 , z = 3


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