sol1 (1)

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11.2MH1 LINEAR ALGEBRA
EXAMPLES 1: LINEAR SYSTEMS – SOLUTIONS
1. The system has the same solutions as the systems corresponding to the arrays:
3 1 4 5
1 1 1 2
2 0 1 3
1 1 2 5
r1 r2
1 1 1 2
3 1 4 5
2 0 1 3
1 1 2 5
r2 r2 3r1
r3 r3 2r1
r4 r4 r1
1 1 1 2
0 4 7 1
0 2 3 1
0 0 3 3
r3 2r3 r2
1 1 1 2
0 4 7 1
0 0 1 1
0 0 3 3
r4 r4 3r3
1 1 1 2
0 4 7 1
0 0 1 1
0 0 0 0
i.e.
x1 x2 x3 2
4x2 7x3 1
x3 1
So x3 1, x2 1 7x3 4 2, x1 2 x2 x3 1. The system
is consistent and has the unique solution x1 1 x2 2 x3 1.
2. The system has the same solutions as the systems corresponding to
1 1 1 1 1 0
2 1 2 1 2 0
4 3 0 1 0 0
5 3 3 1 3 0
1 1 7 5 7 0
r2 r2 2r1
r3 r3 4r1
r4 r4 5r1
r5 r5 r1
1 1 1 1 1 0
0 1 4 3 4 0
0 1 4 3 4 0
0 2 8 6 8 0
0 2 8 6 8 0
r3 r3 r2
r4 r4 2r2
r5 r5 2r2
1 1 1 1 1 0
0 1 4 3 4 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
i e x y z u v 0y 4z 3u 4v 0
Thus we obtain the solution v , u , z , y 4 3 4 , x y z u v
3 2 3 for any constants .
3. The system has the same solutions as the systems corresponding to
1 2 1 2
2 1 1 3
1 1 2 2
r2 r2 2r1
r3 r3 r1
1 2 1 2
0 3 3 1
0 3 3 0
r3 r3 r2
1 2 1 2
0 3 3 1
0 0 0 1
The last row corresponds to the equation 0x 0y 0z 1 which has no solution. Hence the
system is inconsistent.
4. The system has the same solutions as the systems corresponding to the arrays:
2 1 3 a
3 1 5 b
5 5 21 c
r2 2r2 3r1
r3 2r3 5r1
2 1 3 a
0 5 19 2b 3a
0 15 57 2c 5a
r3 r3 3r2
1
2 1 3 a
0 5 19 2b 3a
0 0 0 4a 6b 2c
Row 3 corresponds to the equation 0x1 0x2 0x3 4a 6b 2c so the system is incon-
sistent if and only if 4a 6b 2c 0 i.e. 2a 3b c 0.
5. The system has the same solutions as the systems corresponding to the following arrays:
1 1 1 2
1 2 1 3
1 1 a2 5 a
r2 r2 r1
r3 r3 r1
1 1 1 2
0 1 2 1
0 0 a2 4 a 2
The last row corresponds to the equation a2 4 x3 a 2. If a2 4 0, i.e. a 2 the
system has a unique solution. If a 2 the last row becomes zero and the system has infinitely
many solutions. If a 2 the last row corresponds to 0x1 0x2 0x3 4 so the system is
inconsistent.
6. Consider the equation ax1 bx2 cx3 dx4 0. If x1 1 x2 1 x3 1 x4 2 is a
solution we must have a b c 2d 0 1 .
If x1 2 x2 0 x3 3 x4 1 is a solution we must have 2a 3c d 0 2 .
(1) and (2) have the same solutions as the systems represented by
1 1 1 2 0
2 0 3 1 0 r2 r2 2r1
1 1 1 2 0
0 2 1 5 0
i e a b c 2d 02b c 5d 0 Solutions are given by d , c , b
1
2 5 , a
1
2 3 .
Each choice of yields an equation with the desired property. Choosing 0 1 we
obtain a 32 , b
1
2 c 1 d 0. Choosing 1 0 we obtain a
1
2 b
5
2 c
0 d 1. So an appropriate system of equations is
3
2x1
1
2x2 x3 0
1
2x1
5
2x2 x4 0
7. The system has the same solutions as systems corresponding to the arrays
1 1 1 9
1 5 10 44 r2 r2 r1
1 1 1 9
0 4 9 35
i e x y z 9 14y 9z 35 2
Since x y z are positive integers (2) can not be satisfied unless z 4. Trying z 1 2 3 it is
found that z 3, y 2, x 4 is the only positive integer solution.
8. Examples of inconsistent linear systems with more unknowns than equations:
a) 0x 0y 1
b) x1 x2 x3 12x1 2x2 2x3 3
There are many such systems, but none of them can be homogeneous.
9. A non-trivial solution is any solution other than the trivial solution x 0.
2
(a) The system is homogeneous with more unknowns than equations, so has (infinitely
many) non-trivial solutions.
(b) x3 0 x2 0 x1 0 so this system has no non-trivial solutions.
(c) The system has general solutions x2 , x1 where is an arbitrary constant so
the system has (infinitely many) non-trivial solutions.
10. A x1 x2 Ax1 Ax2 (by properties of matrices)
0 0 since x1 and x2 are solutions of Ax 0
0
Hence x1 x2 is a solution of Ax 0.
Also A x1 Ax1 (properties of matrices)
0
0
So x1 is a solution of Ax 0.
This is false for inhomogeneous systems. For example consider the system
x y 1. i.e. 1 1 xy 1
1
0 and
0
1 are solutions, but
1
0
0
1
1
1 is not.
3