Matrices & DeterminantsmediumFree

Matrices & Determinants — JEE Maths practice question

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

Question

Let

A = \begin{pmatrix} 2 & 2 & 1 \\ 2 & 5 & 2 \\ 1 & 2 & 2 \end{pmatrix}

and

B = \begin{pmatrix} -x & -y & z \\ 0 & y & 2z \\ x & -y & z \end{pmatrix}

where

x, y, z \in \mathbb{R}

. If

B^T A B = \begin{pmatrix} 8 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 42 \end{pmatrix}

, then the number of ordered triplets

(x, y, z)

is:

2

6

8

correct

9

Solution

Step 1: Compute

AB

AB = \begin{pmatrix} -x & -y & 7z \\ 0 & y & 14z \\ x & -y & 7z \end{pmatrix}

Step 2: Compute

B^T(AB)

B^T A B = \begin{pmatrix} 2x^2 & 0 & 0 \\ 0 & 3y^2 & 0 \\ 0 & 0 & 42z^2 \end{pmatrix}

Step 3: Equate corresponding entries

2x^2 = 8, \quad 3y^2 = 27, \quad 42z^2 = 42

x^2 = 4,\ y^2 = 9,\ z^2 = 1 \implies x = \pm 2,\ y = \pm 3,\ z = \pm 1

Step 4: Count the ordered triplets

2 \times 2 \times 2 = 8

Answer: (3)

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