Q. 25.0( 1 Vote )

# Solve the followi

Answer :

The system can be written as

A X = 0

Now, |A| = 2(– 15 + 5) + 1(– 25 + 1) + 2(25 – 3)

|A| = – 20 – 24 + 44

|A| = 0

Hence, the system has infinite solutions

Let z = k

2x – y = – 2k

5x + 3y = k

A X = B

|A| = 6 + 5 = 11≠0 So, A ^{– 1} exist

Now adj A = =

X = A ^{– 1} B =

X =

Hence, x = , y = and z = k

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