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









=

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