Exercise 2 · System of three planes — discussion, geometric interpretation and solution

Discussion of a parametric linear system, geometric interpretation of the three planes and solving the consistent indeterminate case.

Maximum score · 2.5 points

Consider the following system of linear equations, formed by three planes in space and depending on the real parameter $m$: $$\left\{\begin{aligned} x+my+z&=4 \\ x+3y+z&=5 \\ mx+y+z&=4 \end{aligned}\right.$$

Discuss the system for the different values of the parameter $m$. 1 p
Interpret this system geometrically for all values of the parameter $m$ and solve it, if possible, for the case $m=1$. 1 p
For $m=1$, is it possible to add a fourth equation so that the resulting system is consistent and determinate and has $(x,y,z)=\left(3,\tfrac{1}{2},\tfrac{1}{2}\right)$ as its solution? Reason your answer. 0.5 p

Science Bachillerato · Block C — Geometry and linear algebra Discussion of systems Planes in space Solution and reasoning

Step-by-step solution

Key idea: the Rouché–Frobenius theorem relates the ranks of the coefficient matrix $A$ and the augmented matrix $A^{*}$ to the type of solution. The determinant of $A$ marks where the rank may drop; after that, each critical value must be examined separately.

a) Discussion according to $m$

The coefficient matrix is $A=\begin{pmatrix}1&m&1\\1&3&1\\m&1&1\end{pmatrix}$. We compute its determinant:

\det A=1\,(3-1)-m\,(1-m)+1\,(1-3m)=2-m+m^2+1-3m=m^2-4m+3=(m-1)(m-3).

Hence $\det A=0$ only if $m=1$ or $m=3$.

If $m\neq 1$ and $m\neq 3$: $\det A\neq 0$, so $\operatorname{rank}(A)=\operatorname{rank}(A^{*})=3=$ number of unknowns. The system is consistent and determinate (unique solution).

If $m=1$: the system is $\{x+y+z=4,\ x+3y+z=5,\ x+y+z=4\}$. The 1st and 3rd equations are identical, so only two of them are independent: $\operatorname{rank}(A)=\operatorname{rank}(A^{*})=2<3$. The system is consistent and indeterminate (infinitely many solutions, one degree of freedom).

If $m=3$: the system is $\{x+3y+z=4,\ x+3y+z=5,\ 3x+y+z=4\}$. The 1st and 2nd have the same left-hand side but different constant terms ($4\neq 5$): they are mutually inconsistent, so $\operatorname{rank}(A)=2$ but $\operatorname{rank}(A^{*})=3$. The system is inconsistent (no solution).

$m\neq 1,3$: consistent determinate (unique solution) · $m=1$: consistent indeterminate (infinitely many solutions) · $m=3$: inconsistent (no solution).

b) Geometric interpretation and solution for $m=1$

Each equation is a plane in space. The discussion above translates as follows:

$m\neq 1,3$: the three planes meet at a single point (the unique solution).
$m=1$: the 1st and 3rd planes are coincident ($x+y+z=4$) and the 2nd cuts them; the common intersection is a line.
$m=3$: the 1st and 2nd planes are parallel and distinct ($x+3y+z=4$ and $=5$); there is no point common to all three, which is why the system is inconsistent.

We solve the case $m=1$. Two independent equations remain:

\left\{\begin{aligned} x+y+z&=4 \\ x+3y+z&=5 \end{aligned}\right.

Subtracting the first from the second: $2y=1\Rightarrow y=\tfrac{1}{2}$. We take $z=\lambda$ as a parameter; from the first, $x=4-y-z=\tfrac{7}{2}-\lambda$.

(x,y,z)=\left(\tfrac{7}{2}-\lambda,\ \tfrac{1}{2},\ \lambda\right),\qquad \lambda\in\mathbb{R}.

Geometry according to $m$ as indicated. For $m=1$ the solution is the line $\left(\tfrac{7}{2}-\lambda,\ \tfrac{1}{2},\ \lambda\right)$, with direction vector $(-1,0,1)$.

c) Fourth equation with solution $\left(3,\tfrac12,\tfrac12\right)$

For $m=1$ the solutions form the line $\left(\tfrac{7}{2}-\lambda,\ \tfrac{1}{2},\ \lambda\right)$. We first check whether the proposed point lies on it: with $\lambda=\tfrac{1}{2}$ we get $x=\tfrac{7}{2}-\tfrac{1}{2}=3$, $y=\tfrac{1}{2}$, $z=\tfrac{1}{2}$. Yes: the point $\left(3,\tfrac12,\tfrac12\right)$ lies on the line of solutions.

Adding a fourth equation is equivalent to adding a fourth plane. If we choose a plane that cuts the line exactly at this point (that is, one passing through $\left(3,\tfrac12,\tfrac12\right)$ and not containing the line, for example not parallel to the direction $(-1,0,1)$), the system now has a unique solution and it is precisely this point.

\text{For example } \;z=\tfrac{1}{2}\;\text{ fixes } \lambda=\tfrac{1}{2}\;\Longrightarrow\; (x,y,z)=\left(3,\tfrac12,\tfrac12\right).

Yes, it is possible. It suffices to add a plane through $\left(3,\tfrac12,\tfrac12\right)$ that does not contain the line of solutions (e.g. $z=\tfrac12$): the system becomes consistent and determinate with this solution.