Q. 2 D4.3( 3 Votes )

# Solve the following system of equations by matrix method:3x + 4y + 7z = 142x – y + 3z = 4X + 2y – 3z = 0

The given system can be written in matrix form as:

or A X = B

A = , X = and B =

Now, |A| = 3

= 3(3 – 6) – 4(– 6 – 3) + 7(4 + 1)

= – 9 + 36 + 35

= 62

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

X = A – 1B

Cofactors of A are:

C11 = (– 1)1 + 1 3 – 6 = – 3

C21 = (– 1)2 + 1 – 12 – 14 = 26

C31 = (– 1)3 + 112 + 7 = 19

C12 = (– 1)1 + 2 – 6 – 3 = 9

C22 = (– 1)2 + 1 – 3 – 7 = – 10

C32 = (– 1)3 + 1 9 – 14 = 5

C13 = (– 1)1 + 2 4 + 1 = 5

C23 = (– 1)2 + 1 6 – 4 = – 2

C33 = (– 1)3 + 1 – 3 – 8 = – 11

=

A – 1 =

Now, X = A – 1B =

X =

X =

X =

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

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