Using the concept of rank for a matrix, we can characterize the solvability of a system of linear equations. First a definition of the key concept:
The rank of a matrix is the number of non-zero rows in an echelon form of the matrix. We denote the rank of a matrix \(A\) by \(\text{rank}(A)\).
The number of non-zero rows in an echelon form of a matrix does not depend on the echelon form. This can be seen by observing that
- this number does not change when the echelon form is further reduced to a reduced echelon form;
- the reduced echelon form is unique, so the rank is unique in the case of a reduced echelon form.
The rank of a matrix is, by definition, not greater than the number of rows. But also the rank is not greater than the number of columns. After all, the number of rows having a leading element equal to #1# in the reduced echelon form is in fact no greater than the number of columns.
- A system of linear equations is inconsistent if and only if the rank of the coefficient matrix is smaller than the rank of the augmented matrix.
- If a system of #m# linear equations in #n# unknowns has a solution, then the solution is parameterized by \(n-r\) free parameters, where \(r\) is the rank of the associated coefficient matrix.
- If the coefficient matrix of a system of #m# linear equations in #n# unknowns has rank #n#, then the system has a unique solution.
In the definition of rank, and consequently in the above theorem, we deliberately used the echelon form and did not require that the matrix be in reduced echelon form. The latter form is not always necessary to determine solvability of a linear system of equations.
Bringing the augmented matrix to reduced echelon form does not change the rank of the augmented matrix, the rank of the coefficient matrix or the solution. We may therefore assume that the augmented matrix is in reduced echelon form. The corresponding coefficient matrix then has that form as well. The rank of a matrix in reduced echelon form is equal to the number of rows with a leading element #1#.
1. The rank of the coefficient matrix is smaller than the rank of the augmented matrix is and only if the augmented matrix has a row with a leading element #1# in the last column. In this case (and only then) will the equation corresponding to that row be equal to #0=1#, so that the system is inconsistent.
2. The role of the rank #r# is the same as in Solutions of systems of linear equations. The statement follows immediately from that theorem.
3. This statement follows from the previous two since by statement 1 there is a solution and by statement 2 there are no free parameters.
For homogeneous systems we already know that there is always a solution:
Each homogeneous system has a trivial solution in which all the values of the unknowns are equal to zero. A non-zero solution of a homogeneous system is called a non-trivial solution.
Each homogeneous system of linear equations having more unknowns than equations, possesses non-trivial solutions.
Apply the Gauss elimination method to such a system. The number of bound unknowns (the rank of the system) is at most equal to the number of equations. Since the number of unknowns is larger than the number of equations, there must be free parameters. By assigning one of these variables a non-zero value, we get a non-zero solution.
Determine the rank of the matrix \[A=\matrix{1 & 1 & 4 & -4 \\ 1 & 2 & 5 & -3 \\ 2 & 3 & 9 & -7 \\ }\]
#\text{rank}(A)=# #2#
With the aid of elementary operations on the rows we reduce the matrix to reduced echelon form: \[ \begin{array}{rcl}A = \matrix{1&1&4&-4\\1&2&5&-3\\2&3&9&-7\\}&\sim\matrix{1&1&4&-4\\0&1&1&1\\2&3&9&-7\\}&{\color{blue}{\begin{array}{c}\phantom{x} R_2-R_1\phantom{x}\end{array}}}\\&\sim\matrix{1&1&4&-4\\0&1&1&1\\0&1&1&1\\}&{\color{blue}{\begin{array}{c}\phantom{x} R_3-2R_1\end{array}}}\\&\sim\matrix{1&0&3&-5\\0&1&1&1\\0&1&1&1\\}&{\color{blue}{\begin{array}{c}R_1-R_2\phantom{x}\end{array}}}\\&\sim\matrix{1&0&3&-5\\0&1&1&1\\0&0&0&0\\}&{\color{blue}{\begin{array}{c}\phantom{x} R_3-R_2\end{array}}}\end{array}\]
Because the rank is the number of non-null rows of this matrix, we conclude that the rank of the matrix #A# equals #2#.
In the given solution, we have reduced the matrix #A# to reduced echelon form, although it is sufficient to reduce #A# to an echelon form.