Question

P is an orthogonal matrix of order 3 and $\alpha, \beta, \gamma$ are direction angles of a straight line.

Let $A = \begin{bmatrix} \sin^2\alpha & \sin\alpha\sin\beta & \sin\alpha\sin\gamma \\ \sin\alpha\sin\beta & \sin^2\beta & \sin\beta\sin\gamma \\ \sin\alpha\sin\gamma & \sin\beta\sin\gamma & \sin^2\gamma \end{bmatrix}$ and $Q = P^TAP$. If $PQ^6P^T = 2^kA$, then $k =$

• A. 5
• B. 7
• C. 6
• D. 0

Answer 1

Answer: (A)

5

Answer 2

The problem asks us to find the value of $k$ given an orthogonal matrix $P$, direction angles $\alpha, \beta, \gamma$, and matrices $A$ and $Q$. The relationship to be satisfied is $PQ^6P^T = 2^kA$.

1. Simplify $PQ^6P^T$ using the properties of orthogonal matrices:

Given $Q = P^TAP$. We need to calculate $Q^n$. Let's find the first few powers of $Q$:

$Q^2 = (P^TAP)(P^TAP) = P^TA(PP^T)AP$. Since $P$ is an orthogonal matrix, $PP^T = I$ (identity matrix). So, $Q^2 = P^TA(I)AP = P^TA^2P$.

Similarly, $Q^3 = Q \cdot Q^2 = (P^TAP)(P^TA^2P) = P^TA(PP^T)A^2P = P^TA(I)A^2P = P^TA^3P$.

By induction, we can generalize this to $Q^n = P^TA^nP$. For $n=6$, we have $Q^6 = P^TA^6P$.

Now, substitute $Q^6$ back into the expression $PQ^6P^T$:

$PQ^6P^T = P(P^TA^6P)P^T$. Rearranging the terms: $PQ^6P^T = (PP^T)A^6(PP^T)$. Again, using $PP^T = I$: $PQ^6P^T = I A^6 I = A^6$.

So, the given equation simplifies to $A^6 = 2^kA$.

2. Analyze matrix A and calculate its powers:

The matrix $A$ is given as: $A = \begin{bmatrix} \sin^2\alpha & \sin\alpha\sin\beta & \sin\alpha\sin\gamma \\ \sin\alpha\sin\beta & \sin^2\beta & \sin\beta\sin\gamma \\ \sin\alpha\sin\gamma & \sin\beta\sin\gamma & \sin^2\gamma \end{bmatrix}$

This matrix can be expressed as an outer product of a vector $X$ with itself. Let $X = \begin{bmatrix} \sin\alpha \\ \sin\beta \\ \sin\gamma \end{bmatrix}$. Then $A = X X^T$.

$X X^T = \begin{bmatrix} \sin\alpha \\ \sin\beta \\ \sin\gamma \end{bmatrix} \begin{bmatrix} \sin\alpha & \sin\beta & \sin\gamma \end{bmatrix} = \begin{bmatrix} \sin^2\alpha & \sin\alpha\sin\beta & \sin\alpha\sin\gamma \\ \sin\beta\sin\alpha & \sin^2\beta & \sin\beta\sin\gamma \\ \sin\gamma\sin\alpha & \sin\gamma\sin\beta & \sin^2\gamma \end{bmatrix}$, which matches $A$.

Now, let's calculate $A^2$: $A^2 = (X X^T)(X X^T) = X (X^T X) X^T$. We need to calculate the scalar product $X^T X$:

$X^T X = \begin{bmatrix} \sin\alpha & \sin\beta & \sin\gamma \end{bmatrix} \begin{bmatrix} \sin\alpha \\ \sin\beta \\ \sin\gamma \end{bmatrix} = \sin^2\alpha + \sin^2\beta + \sin^2\gamma$.

We are given that $\alpha, \beta, \gamma$ are direction angles of a straight line. This means their cosines are direction cosines, and they satisfy the relation: $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.

Now, substitute $\sin^2\theta = 1 - \cos^2\theta$ into the expression for $X^T X$:

$X^T X = (1 - \cos^2\alpha) + (1 - \cos^2\beta) + (1 - \cos^2\gamma)$

$X^T X = 3 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma)$

Using the direction angles property: $X^T X = 3 - 1 = 2$.

Now substitute this value back into the expression for $A^2$: $A^2 = X (2) X^T = 2 (X X^T) = 2A$.

3. Generalize $A^n$ and find $A^6$:

We found $A^2 = 2A$. Let's find higher powers:

$A^3 = A \cdot A^2 = A \cdot (2A) = 2A^2 = 2(2A) = 2^2A$.

$A^4 = A \cdot A^3 = A \cdot (2^2A) = 2^2A^2 = 2^2(2A) = 2^3A$.

Following this pattern, we can see that $A^n = 2^{n-1}A$.

For $n=6$: $A^6 = 2^{6-1}A = 2^5A$.

4. Solve for k:

We have $PQ^6P^T = A^6$ and $A^6 = 2^5A$. The problem states $PQ^6P^T = 2^kA$. Therefore, $2^5A = 2^kA$.

Since $\alpha, \beta, \gamma$ are direction angles, it's impossible for all $\sin\alpha, \sin\beta, \sin\gamma$ to be zero (as that would imply $\cos^2\alpha = 1, \cos^2\beta = 1, \cos^2\gamma = 1$, leading to $1+1+1=1$, which is false). Thus, $X$ is not a zero vector, and $A$ is not a zero matrix.

Since $A \ne 0$, we can compare the coefficients of $A$: $2^5 = 2^k$. This implies $k=5$.