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(B) For a matrix $A$ to be orthogonal,the rows must be mutually orthogonal unit vectors. Let the rows be $\vec{r}_1, \vec{r}_2, \vec{r}_3$. $1$. For $

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$a= \pm \frac{1}{\sqrt{2}}, b= \pm \frac{1}{\sqrt{6}}, c= \pm \frac{1}{\sqrt{3}}$

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(B) For a matrix $A$ to be orthogonal,the rows must be mutually orthogonal unit vectors. Let the rows be $\vec{r}_1, \vec{r}_2, \vec{r}_3$. $1$. For $\vec{r}_1 = (0, a, a)$,we have $|\vec{r}_1|^2 = 0^2 + a^2 + a^2 = 2a^2 = 1 \Rightarrow a = \pm \frac{1}{\sqrt{2}}$. $2$. For $\vec{r}_2 = (2b, b, -b)$,we have $|\vec{r}_2|^2 = (2b)^2 + b^2 + (-b)^2 = 4b^2 + b^2 + b^2 = 6b^2 = 1 \Rightarrow b = \pm \frac{1}{\sqrt{6}}$. $3$. For $\vec{r}_3 = (c, -c, c)$,we have $|\vec{r}_3|^2 = c^2 + (-c)^2 + c^2 = 3c^2 = 1 \Rightarrow c = \pm \frac{1}{\sqrt{3}}$. Checking orthogonality: $\vec{r}_1 \cdot \vec{r}_2 = 0(2b) + a(b) + a(-b) = ab - ab = 0$. $\vec{r}_1 \cdot \vec{r}_3 = 0(c) + a(-c) + a(c) = -ac + ac = 0$. $\vec{r}_2 \cdot \vec{r}_3 = 2b(c) + b(-c) + (-b)(c) = 2bc - bc - bc = 0$. Thus,the values are $a = \pm \frac{1}{\sqrt{2}}, b = \pm \frac{1}{\sqrt{6}}, c = \pm \frac{1}{\sqrt{3}}$.

If the matrix $A = \begin{bmatrix} 0 & a & a \\ 2b & b & -b \\ c & -c & c \end{bmatrix}$ is orthogonal,then the values of $a, b, c$ are

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