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 $\det(AA^T)=$

Question

If \[ A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \] and $\theta = \frac{2\pi}{7}$, then $A^{100} = A \times A \times \ldots$ (100 times) is equal to:

Answer 1

Step 1: Read what the conditions say.
Each column of $A$ is a vector $(a_i,b_i,c_i)$. The condition $a_i^2+b_i^2+c_i^2=1$ says every column has length $1$, and $a_ia_j+b_ib_j+c_ic_j=0$ for $i\ne j$ says different columns are perpendicular.
Step 2: Recognize an orthonormal set.
Unit length plus mutual perpendicularity means the three columns form an orthonormal set in space.
Step 3: Compute $A^TA$.
The $(i,j)$ entry of $A^TA$ is exactly the dot product of column $i$ with column $j$. By the conditions, these dot products are $1$ on the diagonal and $0$ off it, so $A^TA=I$.
Step 4: A matrix with $A^TA=I$ is orthogonal.
For such a matrix it is also true that $AA^T=I$.
Step 5: Take the determinant.
From $AA^T=I$, $\det(AA^T)=\det(I)=1$.
Step 6: Confirm with the product rule.
Also $\det(AA^T)=\det(A)\det(A^T)=(\det A)^2$, and since $(\det A)^2=1$ the value is $1$, agreeing with the direct computation.
\[ \boxed{1} \]

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 \(\det(AA^T)=\)

Show Hint

The Correct Option is B

Solution and Explanation

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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 \(\det(AA^T)=\)