Q. 3 D

# Show that each of

This can be written as:

|A| = 1(2) – 1(4) + 1(2)

= 2 – 4 + 2

|A| = 0

So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many solution according to as:

Cofactors of A are:

C11 = (– 1)1 + 1 14 – 12 = 2

C21 = (– 1)2 + 1 7 – 4 = – 3

C31 = (– 1)3 + 1 3 – 2 = 1

C12 = (– 1)1 + 2 7 – 3 = – 4

C22 = (– 1)2 + 1 7 – 1 = 6

C32 = (– 1)3 + 1 3 – 1 = 2

C13 = (– 1)1 + 2 4 – 2 = 2

C23 = (– 1)2 + 1 4 – 1 = – 3

C33 = (– 1)3 + 1 2 – 1 = 1

=

Now, AX = B has infinite many solution

Let z = k

Then, x + y = 6 – k

X + 2y = 14 – 3k

This can be written as:

|A| = 1

Now, X = A – 1B =

=

=

There values of x,y,z satisfy the third equation

Hence, x = k – 2, y = 8 – 2k, z = k

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