Q. 2 E4.3( 3 Votes )

# Solve the f

The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 5

= 5(4 – 6) – 3(8 – 3) + 1(4 – 2)

= – 10 – 15 + 3

= – 22

So, the above system has a unique solution, given by

X = A – 1B

Cofactors of A are:

C11 = (– 1)1 + 1 (4 – 6) = – 2

C21 = (– 1)2 + 1(12 – 2) = – 10

C31 = (– 1)3 + 1(9 – 1) = 8

C12 = (– 1)1 + 2 (8 – 3) = – 5

C22 = (– 1)2 + 1 20 – 1 = 19

C32 = (– 1)3 + 1 15 – 2 = – 13

C13 = (– 1)1 + 2 (4 – 2) = 2

C23 = (– 1)2 + 1 10 – 3 = – 7

C33 = (– 1)3 + 1 5 – 6 = – 1

=

A – 1 =

Now, X = A – 1B =

X =

X =

X =

Hence, X = 1,Y = 2 and Z = 5

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