Question

2. In each of the following, find conditions on a, b, and c (if any) such that the system has (i) no solution, (ii) a unique solution, and (iii) infinitely many solutions. (b) Ix-2y = 4 | 2x +y – z = a 2x +az= 6 2y +3z= b | 3x – 4y + 5z = b IX – cz = 1 [3]

Answer 1

2.(a). The augmented matrix of the given linear system is A (say) =

1	-2	0	4
2	0	a	6
3	-4	5	b

To determine the possible solutions to the given linear system, we will carry out the following row operations:

Add -2 times row 1 to row 2

Add -3 times row 1 to row 3

Multiply row 2 by 1/4

Add 2 times row 2 to row 1

Add -2 times row 2 to row 3

Multiply row 3 by 2/(10-a) ( if a ≠ 10).

Add -a/2 times row 3 to row 1

Add -a/4 times row 3 to row 2

Then A changes to

1	0	0	[a(b-11)+3a-30]/(a-10)
0	1	0	[a(b-11)-a+10]/2(a-10)
0	0	1	2(11-b)/(a-10)

It implies that the given linear system has a unique solution x = [a(b-11)+3a-30]/(a-10), y = [a(b-11)-a+10]/2(a-10) and z = 2(11-b)/(a-10) if a ≠ 10.

If a = 10, thenn the augmented matrix of the given linear system has the RREF

1	0	5	3
0	1	5/2	-1/2
0	0	0	b-11

Thus if a = 10 and b =11, then the given linear system has infinite solutions . However, if a = 10 and b ≠ 11, then the given linear system has no solution.

(b). The augmented matrix of the given linear system is M (say) =

2	1	-1	a
0	2	3	b
1	0	-c	1

To determine the possible solutions to the given linear system, we will carry out the following row operations:

Multiply row 1 by 1/2

Add -1 times row 1 to row 3

Multiply row 2 by 1/2

Add -1/2 times row 2 to row 3

Multiply row 3 by 4/(5c-4) ( if c ≠4/5)

Add 5/4 times row 3 to row 1

Add -3/2 times row 3 to row 2

Then M changes to

1	0	0	(2ac-bc-5)/(4c-5)
0	1	0	(-3a+2bc-b+6)/(4c-5)
0	0	1	(2a-b-4)/(4c-5).

It implies that the given linear system has a unique solution x = (2ac-bc-5)/(4c-5), y = (-3a+2bc-b+6)/(4c-5) and z = (2a-b-4)/(4c-5) if c ≠4/5

If c = 5/4, then M has the RREF

1	0	-5/4	(2a-b)/4
0	1	3/2	b/2
0	0	0	(-2a+b+4)/4

In this case, the given linear system has no solution if -2a+b+4 ≠ 0, i.e. if b ≠ 2a-4.

However, if c = 5/4 and b = 2a-4, then the RREF of M is

1	0	-5/4	1
0	1	3/2	a-2
0	0	0	0

In this case, the given linear system has infinite solutions.