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

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(C) Let $A = \begin{bmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{bmatrix}$.Given conditions imply that the rows (or columns) of

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$1$

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(C) Let $A = \begin{bmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{bmatrix}$.Given conditions imply that the rows (or columns) of the matrix $A$ are orthonormal vectors.Specifically,the condition $a_i^2 + b_i^2 + c_i^2 = 1$ means the sum of squares of elements in each column is $1$.The condition $a_i a_j + b_i b_j + c_i c_j = 0$ for $i 
e j$ means the dot product of any two distinct columns is $0$.Thus,$A^T A = I$,where $I$ is the identity matrix of order $3$.Taking the determinant on both sides,$|A^T A| = |I| = 1$.Since $|A^T| = |A|$,we have $|A|^2 = 1$.Therefore,the value of the square of the determinant is $1$.

If $a_i^2 + b_i^2 + c_i^2 = 1$ for $(i = 1, 2, 3)$ and $a_i a_j + b_i b_j + c_i c_j = 0$ for $(i \ne j, i, j = 1, 2, 3)$,then the value of $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|^2$ is

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