Q. 295.0( 2 Votes )

Show that each of

Answer :

Given: - Three equation


2x + y – 2z = 4


x – 2y + z = – 2


5x – 5y + z = – 2


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 solution.


(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,


2x + y – 2z = 4


x – 2y + z = – 2


5x – 5y + z = – 2


Lets find D



Expanding along 1st row


D = 2[ – 2 – ( – 5)(1)] – (1)[(1)1 – 5(1)] + ( – 2)[ – 5 – 5( – 2)]


D = 2[3] – 1[ – 4] – 2[5]


D = 0


Again, D1 by replacing 1st column by B


Here




D1 = 4[ – 2 – ( – 5)(1)] – (1)[( – 2)1 – ( – 2)(1)] + ( – 2)[( – 2)( – 5) – ( – 2)( – 2)]


D1 = 4[ – 2 + 5] – [ – 2 + 2] – 2[6]


D1 = 12 + 0 – 12


D1 = 0


Also, D2 by replacing 2nd column by B


Here




D2 = 2[ – 2 – ( – 2)(1)] – (4)[(1)1 – (5)] + ( – 2)[ – 2 – 5( – 2)]


D2 = 2[ – 2 + 2] – 4[ – 4] + ( – 2)[8]


D2 = 0 + 16 – 16


D2 = 0


Again, D3 by replacing 3rd column by B


Here




D3 = 2[4 – ( – 2)( – 5)] – (1)[( – 2)1 – 5( – 2)] + 4[1( – 5) – 5( – 2)]


D3 = 2[ – 6] – [8] + 4[ – 5 + 10]


D3 = – 12 – 8 + 20


D3 = 0


So, here we can see that


D = D1 = D2 = D3 = 0


Thus,


Either the system is consistent with infinitely many solutions or it is inconsistent.


Now, by 1st two equations, written as


x – 2y = – 2 – z


5x – 5y = – 2 – z


Now by applying Cramer’s rule to solve them,


New D and D1, D2



D = – 5 + 10


D = 5


Again, D1 by replacing 1st column with




D1 = 10 + 5z – ( – 2)( – 2 – z)


D1 = 6 + 3z


Again, D2 by replacing 2nd column with




D2 = – 2 – z – 5 ( – 2 – z)


D2 = 8 + 4z


Hence, using Cramer’s rule




again,




Let, z = k


Then




And z = k


By changing value of k you may get infinite solutions


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