If \( A=\begin{bmatrix} 0 & x & x\\ 2y & y & -y\\ z & -z & z\end{bmatrix} \) such that \( AA^{T}=2I \) (where I is an identity matrix of order 3), then which of the following are True?
A. \( x^{2}=1,3y^{2}=2 \)
B. \( 6y^{2}=4,2z^{2}=3 \)
C. \( 9z^{2}=6,x^{2}=1 \)
D. \( x^{2}:y^{2}:z^{2}=3:1:2 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → C and D only
The (i, j) entry of \( AA^{T} \) is (row i of A) · (row j of A). With rows \( R_{1}=(0,x,x) \), \( R_{2}=(2y,y,-y) \), \( R_{3}=(z,-z,z) \), the condition \( AA^{T}=2I \) makes each diagonal entry 2:
\( R_{1}\cdot R_{1}=x^{2}+x^{2}=2x^{2}=2 \)
\( x^{2}=1 \)
\( R_{2}\cdot R_{2}=4y^{2}+y^{2}+y^{2}=6y^{2}=2 \)
\( \displaystyle y^{2}=\frac{1}{3} \)
\( R_{3}\cdot R_{3}=z^{2}+z^{2}+z^{2}=3z^{2}=2 \)
\( \displaystyle z^{2}=\frac{2}{3} \)
The off-diagonal entries are 0 for all x, y, z: \( R_{1}\cdot R_{2}=xy-xy=0 \), \( R_{1}\cdot R_{3}=-xz+xz=0 \), \( R_{2}\cdot R_{3}=2yz-yz-yz=0 \).
So \( x^{2}=1 \), \( y^{2}=\frac{1}{3} \) and \( z^{2}=\frac{2}{3} \).
Checking the statements
(A) \( x^{2}=1,3y^{2}=2 \) — Incorrect. \( 3y^{2}=3\cdot\frac{1}{3}=1 \), not 2.
(B) \( 6y^{2}=4,2z^{2}=3 \) — Incorrect. \( 6y^{2}=2 \) and \( 2z^{2}=\frac{4}{3} \).
(C) \( 9z^{2}=6,x^{2}=1 \) — Correct. \( 9z^{2}=9\cdot\frac{2}{3}=6 \) and \( x^{2}=1 \).
(D) \( x^{2}:y^{2}:z^{2}=3:1:2 \) — Correct. \( 1:\frac{1}{3}:\frac{2}{3}=3:1:2 \) on multiplying by 3.