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Matrices & Determinants: Let Pmatrix Pmatrix Pmatrix Pmatrix Pmatrix Pmatrix Number

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

Question
Let A=(221252122)A = \begin{pmatrix} 2 & 2 & 1 \\ 2 & 5 & 2 \\ 1 & 2 & 2 \end{pmatrix} and B=(−x−yz0y2zx−yz)B = \begin{pmatrix} -x & -y & z \\ 0 & y & 2z \\ x & -y & z \end{pmatrix} where x,y,z∈Rx, y, z \in \mathbb{R}. If BTAB=(80002700042)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)(x, y, z) is:
A22
B66
C88correct
D99
Solution
Step 1: Compute ABAB
AB=(−x−y7z0y14zx−y7z)AB = \begin{pmatrix} -x & -y & 7z \\ 0 & y & 14z \\ x & -y & 7z \end{pmatrix}
 Step 2: Compute BT(AB)B^T(AB)
BTAB=(2x20003y200042z2)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
2x2=8,3y2=27,42z2=422x^2 = 8, \quad 3y^2 = 27, \quad 42z^2 = 42
x2=4, y2=9, z2=1  ⟹  x=±2, y=±3, z=±1x^2 = 4,\ y^2 = 9,\ z^2 = 1 \implies x = \pm 2,\ y = \pm 3,\ z = \pm 1
 Step 4: Count the ordered triplets
2×2×2=82 \times 2 \times 2 = 8
 Answer: (3)
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