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Matrices & Determinants: Positive Integers Vmatrix Vmatrix Sum Possible Values

JEE Maths question with a full step-by-step solution.

Question

x, y, z

are positive integers and

\begin{vmatrix} x^3+1 & x^2y & x^2z \\ xy^2 & y^3+1 & y^2z \\ xz^2 & yz^2 & z^3+1 \end{vmatrix} = 30

then the sum of all possible values of

x

is:

5

correct

3

6

DNone of these

Solution

Step 1: Simplify the determinant Write the matrix as

I + \vec{v}\vec{u}^T

where

\vec{v} = (x^2, y^2, z^2)^T

and

\vec{u} = (x, y, z)^T

. By the matrix determinant lemma:

\det(I + \vec{v}\vec{u}^T) = 1 + \vec{u}^T\vec{v} = 1 + x^3 + y^3 + z^3

Step 2: Solve for integer solutions

1 + x^3 + y^3 + z^3 = 30 \implies x^3 + y^3 + z^3 = 29

Since

29 = 3^3 + 1^3 + 1^3

, the ordered positive-integer solutions are

(3,1,1)

(1,3,1)

(1,1,3)

. Step 3: Sum all possible values of

x

The values of

x

across these solutions are

3, 1, 1

, giving a sum of

\mathbf{3+1+1=5}

. Answer: (1)

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