Question

For \(i=1,2,3\) and \(j=1,2,3\), if
\[ a_i^2+b_i^2+c_i^2=1 \] and
\[ a_ia_j+b_ib_j+c_ic_j=0 \quad \forall \; i\neq j \] and
\[ A= \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} \] then \(\det(AA^T)=\)

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(3\)

Hint:

If the columns (or rows) of a matrix are orthonormal, then the matrix is orthogonal and satisfies \(A^TA=I\). For an orthogonal matrix, determinant is always \(+1\) or \(-1\).

Answer 1

The Correct Option is B

Step 1: Interpret the given conditions.
The columns of matrix \(A\) are
\[ \begin{bmatrix} a_1\\ b_1\\ c_1 \end{bmatrix}, \quad \begin{bmatrix} a_2\\ b_2\\ c_2 \end{bmatrix}, \quad \begin{bmatrix} a_3\\ b_3\\ c_3 \end{bmatrix} \]
The condition
\[ a_i^2+b_i^2+c_i^2=1 \] means that each column vector has unit length.
The condition
\[ a_ia_j+b_ib_j+c_ic_j=0 \quad (i\neq j) \] means that distinct column vectors are mutually orthogonal.
Thus, the columns of \(A\) form an orthonormal set.

Step 2: Compute \(A^TA\).
Since the columns are orthonormal, we have
\[ A^TA=I \] where \(I\) is the \(3\times 3\) identity matrix.
Therefore, \(A\) is an orthogonal matrix.

Step 3: Use determinant properties.
We know that
\[ \det(AA^T)=\det(A)\det(A^T) \]
Also,
\[ \det(A^T)=\det(A) \]
Hence,
\[ \det(AA^T)=(\det A)^2 \]
Since \(A\) is orthogonal,
\[ A^TA=I \] Taking determinants on both sides,
\[ \det(A^TA)=\det(I) \]
\[ (\det A)^2=1 \]
Therefore,
\[ \det(AA^T)=1 \]

Step 4: Final conclusion.
Hence,
\[ \boxed{1} \]