Let \( A=\begin{bmatrix}0 & 2y & z\\x & y & -z\\x & -y & z\end{bmatrix} \) and \( I \) be an identity matrix of order 3. If \( A^TA=I \) , then \( x^2+y^2+z^2 \) is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1
The (j, j) entry of \( A^TA \) is the sum of the squares of the entries in column j of A. Since \( A^TA=I \), every diagonal entry is 1:
\( \displaystyle A^TA=\begin{bmatrix}0 & x & x\\2y & y & -y\\z & -z & z\end{bmatrix}\begin{bmatrix}0 & 2y & z\\x & y & -z\\x & -y & z\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}2x^2 & 0 & 0\\0 & 6y^2 & 0\\0 & 0 & 3z^2\end{bmatrix} \)
\( \displaystyle 2x^2=1\ \Rightarrow\ x^2=\frac{1}{2} \)
\( \displaystyle 6y^2=1\ \Rightarrow\ y^2=\frac{1}{6} \)
\( \displaystyle 3z^2=1\ \Rightarrow\ z^2=\frac{1}{3} \)
\( \displaystyle x^2+y^2+z^2=\frac{1}{2}+\frac{1}{6}+\frac{1}{3} \)
\( \displaystyle =\frac{3+1+2}{6}=1 \)