Q. 255.0( 3 Votes )

# <span lang="EN-US

Answer :

A = |A| = 2( – 4 + 4) + 3( – 6 + 4) + 5(3 – 2) = 0 – 6 + 5 = – 1 ≠ 0

Now,     …… (i)

Now the given system of equations can be written in the form of AX = B, where

A = , X = , B = The solution of the system of equations is given by X = X =  [Using (i)]  Hence, x = 1, y = 2, z = 3.

OR

A = |A| = 1( – 25 + 28) – 2( – 10 + 14) + 3( – 8 + 100)

= 3 – 2(4) + 3(2)

= 9 – 8

= 1 ≠ 0

A-1 exists.

A.A-1 = I        = Rate this question :

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