Q. 245.0( 3 Votes )

Show that each of

Answer :

Given: - Three equation


3x – y + 2z = 3


2x + y + 3z = 5


x – 2y – z = 1


Tip: - We know that


For a system of 3 simultaneous linear equation with 3 unknowns


(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by



(ii) If D = 0 and D1 = D2 = D3 = 0, then the given system of equation may or may not be consistent. However if consistent, then it has infinitely many solutions.


(iii) If D = 0 and at least one of the determinants D1, D2 and D3 is non – zero, then the system is inconsistent.


Now,


We have,


3x – y + 2z = 3


2x + y + 3z = 5


x – 2y – z = 1


Lets find D



Expanding along 1st row


D = 3[ – 1 – 3( – 2)] – ( – 1)[( – 1)2 – 3] + 2[ – 4 – 1]


D = 3[5] + 1[ – 5] + 2[ – 5]


D = 0


Again, D1 by replacing 1st column by B


Here




D1 = 3[ – 1 – 3( – 2)] – ( – 1)[( – 1)5 – 3] + 2[ – 10 – 1]


D1 = 3[5] + [ – 8] + 2[ – 11]


D1 = 15 – 8 – 22


D1 = – 15


D1 ≠ 0


So, here we can see that


D = 0 and D1 is non – zero


Hence the given system of equation is inconsistent.


Hence Proved


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