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Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$, then show that the equation

enter image description here

represents a straight line.

Expansion of determinant is tedious. I can see that matrix is symmetric but not able to use this fact. Could someone help me with this?

Mathematics
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2 Answers2

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Notice that the inner matrix can be written as

$$T = \mathrm{a}\mathrm{x}^T + \mathrm{x}\mathrm{a}^T - \mathrm{a}^T\mathrm{x},$$

where $\mathrm{a} = (a, b, c)$ and $\mathrm{x} = (x, y, 1)$ are column vectors. This gives

$$T\mathrm{a} = \mathrm{x}, \qquad T\mathrm{x} = \|\mathrm{x}\|^2 \mathrm{a}, \qquad T\mathrm{v} = -\langle \mathrm{a}, \mathrm{x} \rangle \mathrm{v} \tag{*}$$

whenever $\mathrm{v}$ is perpendicular to both $\mathrm{a}$ and $\mathrm{x}$. So if $\mathrm{a}$ and $\mathrm{x}$ are linearly independent, $\text{(*)}$ completely determines the representation of $T$ w.r.t. the basis $\{\mathrm{a}, \mathrm{x}, \mathrm{a}\times\mathrm{x}\}$, from which we compute as

$$\det T = \|\mathrm{x}\|^2 \langle a, x\rangle.$$

By appealing to the continuity or by a similar idea as before, we can check that this relation continues to hold when they are dependent. Now the conclusion can be easily drawn from this formula.

Sangchul Lee
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    Another way to use (*) is to note that we can form two eigenvectors from the first two equations, and we automatically have $v$ as an eigenvector from the third. From the corresponding eigenvalues we can compute the desired determinant. – Semiclassical May 04 '17 at 14:27
  • More precisely, the first two eigenvectors are $x\pm\Vert x\Vert a$ and the eigenvalues are $\pm \Vert x \Vert$. From this and the obvious third eigenvalue the conclusion follows. – Semiclassical May 04 '17 at 14:35
  • @Semiclassical, Nice observation, those combinations are nice and clear enough to simply the argument! – Sangchul Lee May 04 '17 at 14:37
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\begin{eqnarray*} Det\left[ \begin{array}{ccc} ax-by-c&bx+ay&cx+a \\ bx+ay&-ax+by-c&cy+b\\ cx+a&cy+b&-ax-by+c \\ \end{array} \right]=(x^2+y^2+1)(a^2+b^2+c^2)(ax+by+c). \end{eqnarray*} The first two factors are positive so $\color{red}{ax+by+c=0}$.

Donald Splutterwit
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