Q. 8 B5.0( 1 Vote )

# Solve the followi

A =

|A| = 3(3 – 0) + 4(2 – 5) + 2(0 – 3)

= 9 – 12 – 6

= – 9

Now, the cofactors of A

C11 = (– 1)1 + 1 3 – 0 = 3

C21 = (– 1)2 + 1 – 4 – 0 = 4

C31 = (– 1)3 + 1 – 20 – 6 = – 26

C12 = (– 1)1 + 2 2 – 5 = 3

C22 = (– 1)2 + 1 3 – 2 = 1

C32 = (– 1)3 + 1 15 – 4 = – 11

C13 = (– 1)1 + 2 0 – 3 = – 3

C23 = (– 1)2 + 1 0 + 4 = – 4

C33 = (– 1)3 + 1 9 + 8 = 17

A – 1 =

A – 1 =

Now the given equation can be written as:

A X B

Or, X = A – 1B

=

X =

X =

Hence, x = 3,y = 2 and z = – 1

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