Q. 164.2( 32 Votes )

# Solve the s

Answer :

Given System of equations are Let Given system of equation becomes

2p + 3q + 10r = 4

4p – 6q + 5r = 1

6p + 9q – 20r = 2

The given System of Equations can be written in the form of AX = B

Here A = and B = , X = Now we need to find |A|

|A| = = 2 × (120 – 45) – 3 × (- 80 – 30) + 10 × (36 + 36)

= 150 + 330 + 720

= 1200

Since, |A| ≠ 0

A is non-singular. So, it’s inverse exists

Cofactor of an element aij = Aij

A11 = (-1)1+1× 75 = 1 × 75 = 75

A12 = (-1)1+2× (-110) = (-1) × (-110) = 110

A13 = (-1)1+3× 72 = 1 × (72) = 72

A21 = (-1)2+1× (-150) = (-1) × (-150) = 150

A22 = (-1)2+2 × (-100) = 1 × (-100) = -100

A23 = (-1)2+3 × 0 = (-1) × 0 = 0

A31 = (-1)3+1 × 75 = 1 × 75 = 75

A32 = (-1)3+2 × (-30) = (-1) × (-30) = 30

A33 = (-1)3+3 × (-24) = 1 × (-24) = -24

Adj A = = A-1 = (Adj A)/|A|

A-1 = Now,

Since, AX = B

X = A-1B      p = �, q = 1/3, r = 1/5

x = 1/p = 2; y = 1/q = 3; z = 1/r = 5

So, x = 2; y = 3; z = 5

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