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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