16
Homogeneous Systems
Any system of linear equations can be written in matrix form AX = B. The system
is called homogeneous, if the constant matrix B = 0. X = 0 is a solution for this system.
Any solution which is not equal to zero, is called a non-trivial solution.
Example 13. The following system is a homogeneous system.
x1 −2x2 +x3 +x4 = 0
−x1 +2x2
+x4 = 0
2x1 −4x2 +x3
= 0
Obviously, x1 = 0, x2 = 0, x3 = 0, x4 = 0 is a solution for this system. We can check that
x1 = 2, x2 = 1, x3 = 0, x4 = 0 is also a solution for this system. This solution is a non-trivial
solution.
Example 14. Solve the following homogeneous system.
2x + 4y + 6z = 0
4x + 5y + 6z = 0 .
3x + y − 2z = 0
Solution. The augmented matrix is

2 4 6 0
AG =  4 5 6 0  .
3 1 −2 0
By applying

2 4 6
 4 5 6
3 1 −2
elementary


0
1
R
2 1

0  −→
0

operations, we have to find its reduced row-echelon form.



1 2 3 9
1 2
3 0
R2 −4R1 ,R3 −3R1
 0 −3 −6 0 
4 5 6 24 
−→
3 1 −2 0
0 −5 −11 0





1 2
3 0
1 2 3 0
1 2 3 0
R3 +5R2
−R3
 0 1 2 0  −→
 0 1 2 0 .
2 0  −→
−→  0 1
0 −5 −11 0
0 0 −1 0
0 0 1 0

−1
R2
3
The corresponding system of equation is :
x + 2y + 3z = 0
y + 2z = 0 .
z = 0
 
x
Therefore, the only solution for this system is X =  y  = 0.
z
17
Remark 6. There are two possibilities for the set of solutions for a homogeneous system.
1. Unique solution X = 0
2. Infinitely many solutions
Basic Solutions
Consider the homogeneous system AX = 0. If this system has a nontrivial solutions,
then by Gaussian elimination we can find non-trivial solutions X1 , · · · , Xk such that any
other solution of the system is in the form
t1 X1 + t2 X2 + · · · + tk Xk ,
where t1 , · · · , tk are numbers. X1 , · · · , Xk are called basic solutions for the system and a
solution of the form t1 X1 + t2 X2 + · · · + tk Xk is called a linear combination of X1 , · · · , Xk .
Example 15. Solve the following homogeneous system of linear equations.
x + 2y
+ −z = 0
3x + −3y + 2z = 0 .
−x + −11y + 6z = 0
Solution. The augmented matrix is


1
2 −1 0
AG =  3 −3 2 0  .
−1 −11 6 0
By applying elementary operations, we have to find its reduced row-echelon form.






1
2 −1 0
1 2 −1 0 R +R , −1 R
1 2 −1 0
2 9
2
1 ,R3 +R1
 3 −3 2 0  R2 −3R
 0 −9 5 0  3 −→
 0 1 −5 0 
−→
9
−1 −11 6 0 
0 −9 5 0
0 0 0 0
1 0 19 0
R1 −2R2
0 .
−→  0 1 −5
9
0 0 0 0
The corresponding system of equation is :
x +
+
y +
1
z
9
−5
z
9
= 0
.
= 0
Take z = t. Therefore, the solutions are in the form
   −1 
 −1 
x
t
9
9
X =  y  =  59 t  = t  59  .
z
t
1
18
Here,

X1 = 
−1
9
5
9


1
is the basic solution.
Example 16. Find the basic solutions for the following homogeneous system.
x1
−2x1
3x1
2x1
−2x2 +4x3 −x4
+5x6
+4x2 −7x3 +x4 +2x5 −8x6
−6x2 +12x3 −3x4 +x5 +15x6
−4x2 +9x3 −3x4 +3x5 +12x6
Solutions:

1 −2 4 −1 0 5 0
 −2 4 −7 1 2 −8 0

 3 −6 12 −3 1 15 0
2 −4 9 −3 3 12 0
=
=
=
=
0
0
0
0


 R2 +2R1 ,R3 −3R1 ,R3 −2R1

−→


1 −2 4 −1 0 5 0
0 0 1 −1 2 2 0 
R4 −2R2 
3 ,R1 −4R2
 R4 −2R−→
−→ 
 0 0 0 0 1 0 0 
0 0 0 0 0 2 0

1 −2 4 −1
 0 0 1 −1

 0 0 0 0
0 0 1 −1
0
2
1
3
5
2
0
2

0
0 

0 
0

1 −2 0 3 −8 −3 0
 0 0 1 −1 2
2 0 


 0 0 0 0
1
0 0 
0 0 0 0
0
0 0


1 −2 0 3 0 −3 0
 0 0 1 −1 0 2 0 


 0 0 0 0 1 0 0 
0 0 0 0 0 0 0

R1 +8R3 ,R2 −2R3
−→
Assign non-leading variables as parameters. Then, x2 = t, x4 = s, x6 = u, x5 = 0, x3 =
t − 2u, x1 = 2t − 3s + 3u. The solutions are:




X=



x1
x2
x3
x4
x5
x6


 
 
 
=
 
 
 
2t − 3s + 3u
t
s − 2u
s
0
u


 
 
 
=
 
 
 
2t
t
0
0
0
0


 
 
 
+
 
 
 
−3s
0
s
s
0
0


 
 
 
+
 
 
 
3u
0
−2u
0
0
u




=



19



−3
3
 
 0 
 0 
 




 
 1 
 −2 





t
  + s 1  + u 0 .
 




 
 0 
 0 
0
1


 


2
−3
3
 0 
 0 
 1 


 


 1 
 −2 
 0 






Therefore, the basic solutions are X1 =   , X2 = 
 , X2 =  0  .
0
1


 


 0 
 0 
 0 
0
0
1

2
1
0
0
0
0


Remark 7. Consider a homogeneous system of linear equation with n variables. If the
augmented matrix of the system has rank r, then solutions consist of n − r parameters.
Thus, we can express any solution as a linear combination of n − r basic solutions.
Consider the linear system AX = B. The linear system AX = 0 is called the
associated homogeneous system for the system AX = B. Let Y be a solution for the
homogeneous system and X0 be a particular solution for the system AX = B. Then, since
A(X0 + Y ) = AX0 + AY = B + 0 = B
(X0 + Y ) also is a solution for the system. Conversely, if Z be a solution for the system
then
A(Z − X0 ) = B − B = 0.
Thus, Y = Z − X0 is a solution for the associated system. This shows any solution is in
the form X0 + Y .
Example 17.
Example 18. Solve the following system of equations.
x − y − z = 2
2x − y − 3z = 6 .
x
− 2z = 4
Solution. The augmented matrix is

1 −1 −1 2
AG =  2 −1 −3 6  .
1 0 −2 4

By applying elementary operations, we have to find its reduced row-echelon form.
20





1 −1 −1 2
1 −1 −1 2
1 −1 −1 2
1 ,R3 −R1
3 −R2
 2 −1 −3 6  R2 −2R
 0 1 −1 2  R−→
 0 1 −1 2 
−→
1 0 −2 4
0 1 −1 2
0 0
0 0


1 0 −2 4
R2 +R1
−→  0 1 −1 2 .
0 0 0 0

The corresponding system of equation is :
x
− 2z = 4
y − z = 2 .
0 = 0
z is non-leading variable. Therefore, we assign it as a parameter. If z = t, then the solution
has the parametric form
x = 2t + 4
y = t+2
z = t.


4
In this example the particular solution is X =  2  and the general solution for the
0
 
2
associated homogeneous system is Y = t  1 .
1
Example 19. Express the solutions of the following system as the sum of particular solution
and general solution of the associated homogeneous system.
x1 −2x2 −x3 +3x4 = 1
2x1 −4x2 +x3
= 5
x1 −2x2 +2x3 −3x4 = 4
The augmented matrix is:


1 −2 −1 3 1
0 5 .
AG =  2 −4 1
1 −2 2 −3 5
Apply elementary operations to get the row-echelon



1 −2 −1 3 1
1 −2 −1
R2 −2R1 ,R3 −R1
 2 −4 1
 0 0
0 5 
3
−→
1 −2 2 −3 5
0 0
3
form.



3 1 R −R , 1 R
1 −2 −1 3 1
3
2
2
−6 3  −→3  0 0
1 −2 1 
−6 3
0 0
0
0 0
21

1 −2 0
R1 +R2

0 0 1
−→
0 0 0
The corresponding

1 2
−2 1  .
0 0
system of equations is:
x1 −2x2
+x4 = 2
x3 −2x4 = 1
0
= 0
Let x2 = t, x4 = s. Then, the set of solutions are:

 
 
x1
2 + 2t − s
2
 x2  


t
 
  0
X=
 x3  =  1 + 2s  =  1
x4
s
0



2
−1

 1 
 0
 + t  + s

 0 
 2
0
1



.

In this example the particular solution is:


2
 0 
 
 1 
0
The general solutions for the associated homogeneous system is:


2
−1
 0
 1 


t
 0  + s 2
0
1



.

Matrix Inverses
Let A be a square matrix of size n. A matrix B is called inverse of A if
AB = BA = I.
If A has an inverse, then we say that A is invertible. The inverse of A is unique and we
denote it by A−1 .
·
¸
·
¸
1 1
1
−1
Example 20. Let A =
. Then A−1 =
.
0 1
0 1
¸
·
¸
·
a b
d −b
1
−1
Example 21. If A =
and ad − bc 6= 0 then A = ad−bc
.
c d
−c a
22



1 4 −1
−3 −3 11
1 −3 .
Example 22. A =  2 7 1 . Then, A−1 = 12  1
1 3 0
−1 1 −1

Theorem 6. If a square matrix A, has a zero row (zero column), then A is not invertible.
Proof. Suppose that the i-th row of A is zero. For any square matrix C, the (i, i)-entry of
AC is equal to dot product of the i-th row of A with i-th column of C. Therefore, (i, i)-entry
of AC is 0. But, the (i, i)-entry of identity matrix is equal to 1. Thus,
AC 6= I
for any C.
Therefore, A is not invertible.



1 12 13 3

 0 0 0 0 




Example 23. A = 
and
B
=

9 9 1 1 

0 9 3 2
1
2
1
4
4
0
0
0
0
0
5
9
8
2
6

9 11
5 1 

2 1 
 are not invertible.
1 1 
7 7
Theorem 7. If A and B are square matrices, then
• If A is invertible, then A−1 is invertible, and (A−1 )−1 = A.
• If A and B are invertible, then AB is invertible, and(AB)−1 = B −1 A−1 .
• If A is invertible, then AT is invertible, and(AT )−1 = (A−1 )T .
• If A is invertible, then 1c A is invertible for a number c 6= 0, and (cA)−1 = 1c A−1 .
Example 24. A,B, and C are invertible matrices. Simplify
C T B(AB)−1 [C −1 AT ]T .
Solution:
C T B(AB)−1 [C −1 AT ]T = C T B(B −1 A−1 )[(AT )T (C −1 )T ]
= C T IA−1 [A(C −1 )T ]
= C T I(C −1 )T
= I.
Example 25. Find the matrix A.
·
¸
£
¤T
−1
1 1
2
− 5A−1 = ( AT )−1 .
−2 3
4
23
Solution:
·
2
¸T
1 1
−2 3
Thus,
− 5(AT )−1 = −4(AT )−1
·
2
1 1
−2 3
Then,
1
A=
10
¸T
= (A−1 )T .
·
3 −1
2 1
¸
.