Systems of linear equations: System of linear equations and matrices
Solvability of systems of linear equations
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:
Rank 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 \(\text{rank}(A)\).
Rank criteria for the existence of solutions of systems of linear equations
- 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 \(r\) the rank is of the associated coefficient matrix.
For homogeneous systems we already know that there is always a solution:
Nontrivial solutions of homogeneous systems Each homogeneous system has a trivial solution in which all the values of the unknowns are equal to zero. A nonzero solution of a homogeneous system is called a nontrivial solution.
Each homogeneous system of linear equations having more unknowns than equations, possesses nontrivial solutions.
#\text{rang}(A)=# #2#
Via elementary row operations we reduce the matrix to the reduced echelon form \[ \begin{array}{rcl}A = \matrix{1&1&4&-4\\1&2&5&-3\\2&3&9&-7\\}&\sim\matrix{1&1&4&-4\\0&1&1&1\\2&3&9&-7\\}&{\blue{\begin{array}{c}\phantom{x} R_2-R_1\phantom{x}\end{array}}}\\&\sim\matrix{1&1&4&-4\\0&1&1&1\\0&1&1&1\\}&{\blue{\begin{array}{c}\phantom{x} R_3-2R_1\end{array}}}\\&\sim\matrix{1&0&3&-5\\0&1&1&1\\0&1&1&1\\}&{\blue{\begin{array}{c}R_1-R_2\phantom{x}\end{array}}}\\&\sim\matrix{1&0&3&-5\\0&1&1&1\\0&0&0&0\\}&{\blue{\begin{array}{c}\phantom{x} R_3-R_2\end{array}}}\end{array}\]
Because the rank is the number of non-null rows of this matrix, we conclude that the rank of the matrix #A# equals #2#.
Via elementary row operations we reduce the matrix to the reduced echelon form \[ \begin{array}{rcl}A = \matrix{1&1&4&-4\\1&2&5&-3\\2&3&9&-7\\}&\sim\matrix{1&1&4&-4\\0&1&1&1\\2&3&9&-7\\}&{\blue{\begin{array}{c}\phantom{x} R_2-R_1\phantom{x}\end{array}}}\\&\sim\matrix{1&1&4&-4\\0&1&1&1\\0&1&1&1\\}&{\blue{\begin{array}{c}\phantom{x} R_3-2R_1\end{array}}}\\&\sim\matrix{1&0&3&-5\\0&1&1&1\\0&1&1&1\\}&{\blue{\begin{array}{c}R_1-R_2\phantom{x}\end{array}}}\\&\sim\matrix{1&0&3&-5\\0&1&1&1\\0&0&0&0\\}&{\blue{\begin{array}{c}\phantom{x} R_3-R_2\end{array}}}\end{array}\]
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 worked-out solution, we have reduced the matrix #A# to the reduced echelon form, while it is sufficient to reduce #A# to an echelon form.
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