<p>If \(x = 31\), \(y = 32\) and \(z = 33\), then the value of \[\begin{vmatrix} (x^2+1)^2 & (xy+1)^2 & (xz+1)^2 \\ (xy+1)^2 & (y^2+1)^2 & (yz+1)^2 \\ (xz+1)^2 & (yz+1)^2 & (z^2+1)^2 \end{vmatrix}\] is ________.</p>
Step-by-Step Solution
Key Concept: Recognize that the determinant has a special structure forming a Gram matrix of the form ||u⃗||² where u⃗ = [x, y, z, 1]. The determinant equals zero when vectors are linearly dependent, which occurs here because the matrix represents dot products of vectors in a lower-dimensional space.
<p><strong>Step 1: Identify the Matrix Structure</strong></p><p>The matrix can be written as M = GG<sup>T</sup> where:<br>G = [x, y, z, 1]<br>This is because (a²+1)² = a⁴ + 2a² + 1 can be expressed as dot product of [a², a, 1] with itself, and (ab+1)² as cross terms.</p><p><strong>Step 2: Recognize as Gram Determinant</strong></p><p>The matrix entries represent: M<sub>ij</sub> = (u<sub>i</sub> · u<sub>j</sub>)² where u₁=[x,1], u₂=[y,1], u₃=[z,1] in ℝ². This forms a Gram matrix from 3 vectors in 2-dimensional space.</p><p><strong>Step 3: Apply Rank Deficiency</strong></p><p>Since we have 3 vectors in ℝ², they must be linearly dependent. A Gram determinant of linearly dependent vectors equals zero.</p><p><strong>Step 4: Verification Using Properties</strong></p><p>The determinant of a Gram matrix G = V<sup>T</sup>V where V has rank r in dimension n is zero when r < n. Here V is 3×2 (rank ≤ 2), so det(V<sup>T</sup>V) = 0.</p><p><strong>∴ Answer: 0</strong></p>
Correct Answer: 0