Systems of linear equations: System of linear equations and matrices
Systems with a parameter
So far we have only looked at systems of linear equations in which the coefficients are real numbers. But in practice we regularly run against systems in which one or more parameters appear in the coefficient matrix and/or at the right hand side of an equation.
If we need to solve such a system, we in fact need to solve an infinite number of systems simultaneously, namely a system for each value of the parameter(s). For many parameters row operations in Gaussian elimination result in similar systems, but for special parameter values we may need to take care so as not to divide by zero. Solution by means of case distinction may be unavoidable.
We give three examples.
Under the condition \(c=2a+b\)
The augmented matrix can also be written down in the presence of parameters \[\left(\begin{array}{rrr|r}1 & 2 & -3 & a \\ 2 & -1 & 4 & b\\ 4 & 3 & -2 & c\end{array}\right)\] Moreover, we can apply elementary row operations. \[\begin{array}{rcll} \left(\begin{array}{rrr|c}1 & 2 & -3 & a \\ 2 & -1 & 4 & b\\ 4 & 3 & -2 & c\end{array}\right) & \sim & \left(\begin{array}{rrr|c} 1 & 2 & -3 & a\\ 0 & -5 & 10 & b-2a \\ 0 & -5 & 10 & c-4a \end{array}\right) &\blue{\begin{array}{r} \phantom{x}\\ R_2-2R_1\\ R_3-4R_1\end{array}}
\\ \\ &\sim & \left(\begin{array}{rrr|c} 1 & 2 & -3 & a\\ 0 & -5 & 10 & b-2a \\ 0 & 0 & 0 & c-b-2a \end{array}\right) &\blue{\begin{array}{r} \phantom{x}\\ \phantom{x}\\ R_3- R_2\end{array}}\end{array}\] Now that the augmented matrix for the system has been reduced to echelon form, we see that there is no solution if \(c-b-2a\neq 0\). The system only has a solution provided that \(c-b-2a=0\), in other words: \(c=2a+b\). In this case, the system will have an infinite number of solutions.
Under the condition \(c=2a+b\), Gaussian elimination can be continued. We find a general solution: \[\begin{array}{rcll} \left(\begin{array}{rrr|c} 1 & 2 & -3 & a\\ 0 & -5 & 10 & b-2a \\ 0 & 0 & 0 & 0 \end{array}\right) & \sim & \left(\begin{array}{rrr|c} 1 & 2 & -3 & a\\ 0 & 1 & -2 & \frac{2}{5}a-\frac{1}{5}b \\ 0 & 0 & 0 & 0 \end{array}\right) & \blue{\begin{array}{r} \phantom{x}\\ -\frac{1}{5}R_2\\ \phantom{x}\end{array}}\\ \\ &\sim & \left(\begin{array}{rrr|c} 1 & 0 & 1 & \frac{1}{5}a+\frac{2}{5}b\\ 0 & 1 & -2 & \frac{2}{5}a-\frac{1}{5}b \\ 0 & 0 & 0 & 0 \end{array}\right) &\blue{\begin{array}{r} R_1- 2R_2\\ \phantom{x}\\ \phantom{x} \end{array}}\end{array}\] Thus, provided \(c=2a+b\), the general solution is \[\cv{x\\y\\z} = \cv{\frac{1}{5}a+\frac{2}{5}b \\ \frac{2}{5}a-\frac{1}{5}b\\0} + r\cv{-1\\2\\1}\] with free parameter \(r\).
The augmented matrix can also be written down in the presence of parameters \[\left(\begin{array}{rrr|r}1 & 2 & -3 & a \\ 2 & -1 & 4 & b\\ 4 & 3 & -2 & c\end{array}\right)\] Moreover, we can apply elementary row operations. \[\begin{array}{rcll} \left(\begin{array}{rrr|c}1 & 2 & -3 & a \\ 2 & -1 & 4 & b\\ 4 & 3 & -2 & c\end{array}\right) & \sim & \left(\begin{array}{rrr|c} 1 & 2 & -3 & a\\ 0 & -5 & 10 & b-2a \\ 0 & -5 & 10 & c-4a \end{array}\right) &\blue{\begin{array}{r} \phantom{x}\\ R_2-2R_1\\ R_3-4R_1\end{array}}
\\ \\ &\sim & \left(\begin{array}{rrr|c} 1 & 2 & -3 & a\\ 0 & -5 & 10 & b-2a \\ 0 & 0 & 0 & c-b-2a \end{array}\right) &\blue{\begin{array}{r} \phantom{x}\\ \phantom{x}\\ R_3- R_2\end{array}}\end{array}\] Now that the augmented matrix for the system has been reduced to echelon form, we see that there is no solution if \(c-b-2a\neq 0\). The system only has a solution provided that \(c-b-2a=0\), in other words: \(c=2a+b\). In this case, the system will have an infinite number of solutions.
Under the condition \(c=2a+b\), Gaussian elimination can be continued. We find a general solution: \[\begin{array}{rcll} \left(\begin{array}{rrr|c} 1 & 2 & -3 & a\\ 0 & -5 & 10 & b-2a \\ 0 & 0 & 0 & 0 \end{array}\right) & \sim & \left(\begin{array}{rrr|c} 1 & 2 & -3 & a\\ 0 & 1 & -2 & \frac{2}{5}a-\frac{1}{5}b \\ 0 & 0 & 0 & 0 \end{array}\right) & \blue{\begin{array}{r} \phantom{x}\\ -\frac{1}{5}R_2\\ \phantom{x}\end{array}}\\ \\ &\sim & \left(\begin{array}{rrr|c} 1 & 0 & 1 & \frac{1}{5}a+\frac{2}{5}b\\ 0 & 1 & -2 & \frac{2}{5}a-\frac{1}{5}b \\ 0 & 0 & 0 & 0 \end{array}\right) &\blue{\begin{array}{r} R_1- 2R_2\\ \phantom{x}\\ \phantom{x} \end{array}}\end{array}\] Thus, provided \(c=2a+b\), the general solution is \[\cv{x\\y\\z} = \cv{\frac{1}{5}a+\frac{2}{5}b \\ \frac{2}{5}a-\frac{1}{5}b\\0} + r\cv{-1\\2\\1}\] with free parameter \(r\).
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