Solve System of Equations: Unique Solution & Geometric Explanation

[Guest2]

I'm trying to answer the question below in the attachment. Could someone please check my answer to part (c), as I'm not sure. Is it correct? Is that that the right geometric explanation for the planes?

(a) The equation is $\begin{pmatrix}1 & -1& 2& 1 \\ 2 & 1 & 1 & -1 \\ 1 & -2 & \lambda & 3 \end{pmatrix} $

(b) $\begin{pmatrix}1 & -1& 2& 1 \\ 2 & 1 & 1 & -1 \\ 1 & -2 & \lambda & 3 \end{pmatrix} \to \begin{pmatrix}1 & -1& 2& 1 \\ 0 & 3 & -3 & -3\\ 0 & -1 & \lambda -2 & 2 \end{pmatrix} \to \begin{pmatrix}1 & -1& 2& 1 \\ 0 & 1 & -1 & -1\\ 0 & 0 & \lambda -3 & 1 \end{pmatrix}$.

Thus the condition for the system to have a unique solution is $\lambda - 3 \ne 0 \implies \lambda \ne 3$.

And if $\lambda \ne 3$ then $ z = \frac{1}{\lambda-3}$ then $y = \frac{1}{\lambda-3}-1$ and $x = -\frac{1}{\lambda-3}$ is the unique solution to the system.

(c) If $\lambda = 3$ then the system has no solution. For all other values of $\lambda$ there's exactly one unique solution (given above). Geometrically it means their intersection is a point.

 

[Evgeny.Makarov]

(a) The equation is $\begin{pmatrix}1 & -1& 2& 1 \\ 2 & 1 & 1 & -1 \\ 1 & -2 & \lambda & 3 \end{pmatrix} $

An equation must have the $=$ sign. In this case it is
\[
\begin{pmatrix}1 & -1& 2\\ 2 & 1 & 1 \\ 1 & -2 & \lambda \end{pmatrix}
\begin{pmatrix}x\\y\\z\end{pmatrix}=
\begin{pmatrix}1\\-1\\3\end{pmatrix}.
\]
The matrix you wrote is the augmented matrix of the system of equations.

(c) If $\lambda = 3$ then the system has no solution. For all other values of $\lambda$ there's exactly one unique solution (given above). Geometrically it means their intersection is a point.

Whose intersection? It is clear from the question that you mean the planes defined by the equations, but I would not use only a pronoun in the answer.

You should also probably say explicitly that having infinitely many solutions is not possible. (It would be if the right-hand side of the last equation were 0.) If $\lambda\ne3$, then the tree planes intersect in a single point, and if $\lambda=3$, then the third plane is parallel to the intersection line of the first two planes, so the intersection of all three planes is empty.