If a_i^2 + b_i^2 + c_i^2 = 1 for i = 1, 2, 3 | vedclass

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(D) Let $A = \begin{bmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{bmatrix}$.The given conditions $a_i^2 + b_i^2 + c_i^2 = 1$ and

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$1$ or $-1$

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(D) Let $A = \begin{bmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{bmatrix}$.The given conditions $a_i^2 + b_i^2 + c_i^2 = 1$ and $a_ia_j + b_ib_j + c_ic_j = 0$ for $i 
e j$ imply that the columns of matrix $A$ are orthonormal vectors.Specifically,if we consider the product $A^T A$,we get:$A^T A = \begin{bmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{bmatrix} \begin{bmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = I$.Taking the determinant of both sides,we have $\det(A^T A) = \det(I) = 1$.Since $\det(A^T) = \det(A)$,we have $(\det(A))^2 = 1$.Therefore,$\det(A) = \pm 1$.

If $a_i^2 + b_i^2 + c_i^2 = 1$ for $i = 1, 2, 3$ and $a_ia_j + b_ib_j + c_ic_j = 0$ for $i \ne j$ where $i, j = 1, 2, 3$,then the value of the determinant $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|$ is:

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