Q. 2 J5.0( 1 Vote )

# Solve the following system of equations by matrix method:8x + 4y + 3z = 182x + y + z = 5X + 2y + z = 5

The given system can be written in matrix form as:

AX = B

Now, |A| = 8

= 8(– 1) – 4(1) + 3(3)

= – 8 – 4 + 9

= – 3

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

X = A – 1B

Cofactors of A are:

C11 = (– 1)1 + 1 1 – 2 = – 1

C21 = (– 1)2 + 1 4 – 6 = 2

C31 = (– 1)3 + 1 4 – 3 = 1

C12 = (– 1)1 + 2 2 – 1 = – 1

C22 = (– 1)2 + 1 8 – 3 = 5

C32 = (– 1)3 + 1 8 – 6 = – 2

C13 = (– 1)1 + 2 4 – 1 = 3

C23 = (– 1)2 + 1 16 – 4 = – 12

C33 = (– 1)3 + 1 8 – 8 = 0

=

A – 1 =

Now, X = A – 1B =

X =

X =

X =

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

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