Page 97 - DMTH502_LINEAR_ALGEBRA

Unit 5: Summary of Row-Equivalence

x + 2x = 0 or x = –2x Notes
4 5 4 5
So we may assign any values to x , x , and x , say x = a, x = b, x = c, and obtain the solution
1 3 5 1 3 5
1
(a, 3b – , c b, – 2c, c).
2
Let us observe one thing more in connection with the system of equations RX = 0. If the number
r of non-zero rows in R is less than n, then the system RX = 0 has a non-trivial solution, that is,
a solution (x ,...,x ) in which not every x in 0. For, since r < n, we can choose some x which is not
1 n j j
among the r unknowns x ,...,x , and we can then construct a solution as above in which this x
k1 kr j
is 1. This observation leads us to one of the most fundamental facts concerning systems of
homogeneous linear equations.
Theorem 3: If A is an m × n matrix and m < n, then the homogeneous system of linear equations
AX = 0 has a non-trivial solution.
Proof: Let R be a row-reduced echelon matrix which is row-equivalent to A. Then the systems
AX = 0 and RX = 0 have the same solutions by Theorem 3. If r is the number of non-zero rows in
R, then certainly r  m, and since m < n, we have r < n. It follows immediately from our remarks
above that AX = 0 has a non-trivial solution.
Theorem 4: If A is an n × n (square) matrix, then A is row-equivalent to the n × n identity matrix
if and only if the system of equations AX = 0 has only the trivial solution.
Proof: If A is row-equivalent to I, then AX = 0 and IX = 0 have the same solutions. Conversely,
suppose AX = 0 has only the trivial solution X = 0. Let R be an n × n row-reduced echelon matrix
which is row-equivalent to A, and let r be the number of non-zero rows of R. Then RX = 0 has no
non-trivial solution. Thus r  n. But since R has n rows, certainly r  n, and we have r = n. Since
this means that R actually has a leading non-zero entry of 1 in each of its n rows, and since these
1’s occur each in a different one of the n columns, R must be the n × n identity matrix.

Let us now ask what elementary row operations do toward solving a system of linear equations
AX = Y which is not homogeneous. At the outset, one must observe one basic difference between
this and the homogeneous case, namely, that while the homogeneous system always has the
trivial solution x =  = x = 0, an inhomogeneous system need have no solution at all.
1 n
We form the augmented matrix A’ of the system AX = Y. This is the m × (n + 1) matrix whose first
n columns are the columns of A and whose last column is Y. More precisely,

'
A  A if j  n
, ij
ij
A '  y
( i n 1) i
Suppose we perform a sequence of elementary row operations on A arriving at a row-reduced
echelon matrix R. If we perform this same sequence of row operations on the augmented matrix
A’, we will arrive at a matrix R’ whose first n columns are the columns of R and whose last
column contains certain scalars z ,...,z . The scalars z are the entries of the m × 1 matrix
1 m i
z  1 
 
Z  
 
z  
 m
which results from applying the sequence of row operations to the matrix Y. It should be clear
to the reader that, just as in the proof of Theorem 3 the systems AX = Y and RX = Z are equivalent
and hence have the same solutions. It is very easy to determine whether the system RX = Z has
any solutions and to determine all the solutions if any exist. For, if R has r non-zero rows, with
the leading non-zero entry of row i occurring in column k , i = 1,...,r, then the first r equations of
i

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