Question

Consider the real symmetric matrix \(A=(a_{ij})\) given by \[ A=\begin{pmatrix}3 & 1 & 1\\ 1 & 0 & 2\\ 1 & 2 & 0\end{pmatrix}. \] Consider the set \[ S=\left\{x=(x_1,x_2,x_3)\in\mathbb{R}^3 : \sum_{j=1}^{3}\sum_{i=1}^{3}a_{ij}x_ix_j=1\right\}. \] Which of the following statements is/are correct?

• A. \(S\) is empty set
• B. Any line \(L\) in \(\mathbb{R}^3\) has at most two points from \(S\)
• C. \(S\) is bounded
• D. \(S\) is unbounded

Hint:

Find the eigenvalues of A to see that S is an indefinite quadric (a hyperboloid), then use the fact that substituting a line into x^TAx=1 gives a quadratic equation in the line parameter.

Answer 1

The Correct Option is B, D

Step 1: Write out the quadratic form in full.
For $A=\begin{pmatrix}3&1&1\\1&0&2\\1&2&0\end{pmatrix}$, the form is $x^TAx=3x_1^2+2x_1x_2+2x_1x_3+4x_2x_3$.

Step 2: Complete the square instead of using eigenvalues.
Group the $x_1$ terms first: $3x_1^2+2x_1(x_2+x_3)=3\left(x_1+\dfrac{x_2+x_3}{3}\right)^2-\dfrac{(x_2+x_3)^2}{3}$. Substituting this back and then completing the square in $x_2$ gives, after simplification, three squared terms with coefficients $3$, $-\dfrac13$, and $8$.

Step 3: Read off the signs.
Two coefficients are positive and one is negative, matching the eigenvalue signs, so A is invertible and indefinite. The level set $x^TAx=1$ is a genuine hyperboloid, not a cone or an empty set.

Step 4: Rule out A and C, confirm D.
Since the form reaches 1 along a positive-coefficient direction, S is not empty, so (A) fails. Since one term is negative, we can let the matching coordinate grow freely while the equation still balances, so S extends to infinity and is unbounded. So (C) fails and (D) holds.

Step 5: Check line intersections directly.
On any line $x=x_0+tv$, the condition $x^TAx=1$ turns into $(v^TAv)t^2+2(x_0^TAv)t+(x_0^TAx_0-1)=0$, a quadratic in $t$. A quadratic equation never has more than two roots, so any line meets S in at most two points, giving (B).

Final Answer:
S is an unbounded quadric surface, and every line meets it in at most two points. \[ \boxed{\text{(B) and (D)}} \]