Answer :

Let the numbers are x, y, z


3x + 5y – 4 z = 6000 …… (i)


Also,


2x – 3y + z = 5000 …… (ii)


Again,


– x + 4y + 6z = 13000 ……(iii)



A X = B


|A| = 3(– 18 – 4) – 2(30 + 16) – 1(5 – 12)


= 3(– 22) – 2(46) + 7


= – 66 – 92 + 7


= – 151


Hence, the unique solution given by x = A – 1B


C11 = (– 1)1 + 1 (– 18 – 4) = – 22


C12 = (– 1)1 + 2 (12 + 1) = – 13


C13 = (– 1)1 + 3 (8 – 3) = 5


C21 = (– 1)2 + 1 (30 + 16) = – 46


C22 = (– 1)2 + 2 (18 – 4) = 14


C23 = (– 1)2 + 3 (12 + 5 ) = – 17


C31 = (– 1)3 + 1 (5 – 12) = – 7


C32 = (– 1)3 + 2 (3 + 8) = – 11


C33 = (– 1)3 + 3 (– 9 – 10) = – 19


Adj A =


X = A – 1 B =


X =


X =


=


Hence, x = 3000, y = 1000 and z = 2000


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