5.4. Systems with parameters

In this section we handle systems of linear equations with parameters. The difference between a parameter, a variable, and an unknown, is more of a philosophical or psychological difference, than a mathematical one. Mathematically, they are all simply letters that represent numbers. In these notes we will usually call the lower-case letters $x,y,z,w$ “variables”, and most other lower-case letters, like $a,b,c,\lambda ,\mu$ will be “parameters”. We will talk about linear systems of equations in the variables $x,y,$ and $z$, and as coefficients in these equations the parameters $a,b,$ or $c$ might appear. See Example 5.1, where $a$ and $k$ occur as parameters.

The crucial fact to bear in mind is that the solutions of a system of equations with parameters depends on the values that the parameters may take. Hence, we need to discuss each possible case. The following examples provide basic guidelines for doing this.

Example 5.4.1.

• Solve the system of linear equations in $x,y,$ and $z$:

From the start, the question has identified $x,y,$ and $z$ as the “variables”. So when we write the augmented matrix form we treat $k$ and $c$ as coefficients. The variables are the ones which disappear when you convert the system into augmented matrix form. And as usual, we begin the row operations:

Here $A_e$ is an echelon form of the matrix $A$, but it is not in reduced echelon form. Now, we must be very careful. Indeed, our first reflex may be to carry on as before and hence scale each row of $A_e$ in order to get leading coefficients equal to $1$. But this requires that the leading coefficients involving parameters are non-zero. This condition determines three cases, as follows:

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  $k+2\ne 0$,

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  $k+2=0$ and $c+4=0$, and

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  $k+2=0$ and $c+4\ne 0$.

At this stage, we proceed case by case, each leading to a different outcome.

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  $k+2\ne 0$ Then, we may divide $R_3$ by $(k+2)$ and carry on until we obtain the reduced echelon form of $A_e$ (hence of $A$). We thus divide $R_2$ by $2$, then do $R_2=r_2-\frac{1}{2}{r}_3$ and finally $R_1=r_1-5r_2$ and $R_1=r_1-2r_3$. The resulting augmented matrix has the identity matrix on the left and the right-hand side contains the solution, namely,

  $$x=\frac{-3k+c-2}{2(k+2)},y=\frac{k-c-2}{2(k+2)}\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}z=\frac{c+4}{k+2}.$$

  All of these are valid expressions, since we have assumed $k+2\ne 0$. Therefore, in this case, we have a unique solution.

  This can be interpreted geometrically in ${ℝ}^3$, as three 2-dimensional planes (corresponding to the three equations) having only a single point in common.

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  $k+2=0$ and $c+4=0$ Then, the last row is filled with zeros, and we get that the reduced echelon form of $A$ is

  Thus, we may freely choose the value for one variable (e.g. $z$), which will then determine the other two. Therefore there are infinitely many solutions.

  Another way to write this solution is to give one of the variables another name, and think of it as a parameter. For example, say that you give a value $\lambda \in ℝ$ to $z$. Then we can write $x$ and $y$ in terms of $\lambda$ as well. So the complete solution set is:

  $$\{(x,y,z)=(-\frac{3}{2}+\frac{1}{2}\lambda \mathit{\hspace{1em}}\text{,}\mathit{\hspace{1em}}\frac{1}{2}-\frac{1}{2}\lambda \mathit{\hspace{1em}}\text{,}\mathit{\hspace{1em}}\lambda ):\lambda \in ℝ\}.$$

  This is called a parametric equation of a line.

  This answer can be interpreted geometrically by saying that this line lies on all three planes.

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  $k+2=0$ and $c+4\ne 0$ In this case, the system is inconsistent, that is, there are no solutions. Indeed, the last row in the augmented matrix above is

  saying that $0x+0y+0z=c+4\ne 0$, which is a contradiction.

  This could be interpreted geometrically by saying that there is no point in common among the three planes.