Q.Evaluate $\begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}$.

Concept: Orthogonal Matrix Verification – The given matrix is orthogonal (its rows form an orthonormal set), so its determinant must be $\pm 1$.

Step 1: Check orthogonality. Compute the dot product of row 1 and row 2:

$(\cos\alpha\cos\beta)(-\sin\beta) + (\cos\alpha\sin\beta)(\cos\beta) + (-\sin\alpha)(0)$

$= \cos\alpha(-\sin\beta\cos\beta + \sin\beta\cos\beta) = 0$.

Step 2: Dot product of row 1 and row 3:

$(\cos\alpha\cos\beta)(\sin\alpha\cos\beta) + (\cos\alpha\sin\beta)(\sin\alpha\sin\beta) + (-\sin\alpha)(\cos\alpha)$

$= \sin\alpha\cos\alpha(\cos^2\beta + \sin^2\beta) - \sin\alpha\cos\alpha = 0$.

Step 3: Dot product of row 2 and row 3:

$(-\sin\beta)(\sin\alpha\cos\beta) + (\cos\beta)(\sin\alpha\sin\beta) + (0)(\cos\alpha)$

$= \sin\alpha(-\sin\beta\cos\beta + \cos\beta\sin\beta) = 0$.

Each row has unit length (check: row 1 squared = $\cos^2\alpha + \sin^2\alpha = 1$, etc.). Hence the matrix is orthogonal, so its determinant is $\pm 1$. The determinant of an orthogonal matrix with a $3\times3$ rotation-like structure is $+1$ (orientation-preserving).

Why this approach works

When you see a $3 \times 3$ matrix whose entries are all sines and cosines of $\alpha$ and $\beta$, your first instinct might be to expand the determinant directly. That works, but it’s messy. A much cleaner path: recognise the matrix as orthogonal.

An orthogonal matrix $Q$ satisfies $Q^T Q = I$, which implies $\det(Q) = \pm 1$. If we can show the given matrix is orthogonal, we only need to decide the sign — and that’s often easier than computing the full determinant.

Let’s name the matrix:

We’ll verify orthogonality by checking that the rows are unit vectors and mutually perpendicular.

Step-by-step verification

1. Check the first row with itself (norm squared)

Row 1: $(\cos\alpha\cos\beta,\; \cos\alpha\sin\beta,\; -\sin\alpha)$

So row 1 is a unit vector.

2. Check the second row with itself

Row 2: $(-\sin\beta,\; \cos\beta,\; 0)$

Row 2 is also a unit vector.

3. Check the third row with itself

Row 3: $(\sin\alpha\cos\beta,\; \sin\alpha\sin\beta,\; \cos\alpha)$

All three rows are unit vectors.

4. Check dot product of row 1 and row 2

The two terms cancel exactly. So rows 1 and 2 are orthogonal.

5. Check dot product of row 1 and row 3

Rows 1 and 3 are orthogonal.

6. Check dot product of row 2 and row 3

All three rows are pairwise orthogonal. Since each row is a unit vector, the rows form an orthonormal set. Therefore $M$ is an orthogonal matrix.

Determining the sign

We know $\det(M) = \pm 1$. To decide which, evaluate the determinant at a convenient pair of angles. Choose $\alpha = 0$, $\beta = 0$:

The determinant of $I$ is $1$. Since the determinant is a continuous function of $\alpha$ and $\beta$, and it never jumps between $+1$ and $-1$ (the matrix is always orthogonal, so the determinant is always $\pm 1$), it must be $+1$ for all $\alpha, \beta$.

Method: Evaluating a Determinant by Recognising the Matrix Is Orthogonal

Some $3\times3$ trigonometric determinants can be evaluated far faster by recognising the matrix as a rotation (orthogonal) matrix rather than expanding directly.

Steps

Step 1: Check whether each row (or column) is a unit vector

Sum the squares of each row's entries; if the sum simplifies to $1$ using $\sin^2+\cos^2=1$ for every row, that row is a unit vector.

Step 2: Check that every pair of rows is mutually perpendicular

Compute the dot product of each pair of rows; if every pair simplifies to $0$, the rows are pairwise orthogonal.

Step 3: Conclude the matrix is orthogonal, so $\det = \pm 1$

If both Step 1 and Step 2 hold for all rows, the matrix satisfies $A^TA = I$, which forces $\det(A)^2 = 1$, i.e. $\det(A) = \pm 1$ — you no longer need to expand the full $3\times3$ determinant.

Step 4: Determine the sign by testing a convenient special case

Since the determinant is a continuous function of the parameters and can only ever be $+1$ or $-1$ for an orthogonal matrix of this family, substitute simple values (e.g. all angles $=0$) into the original matrix. If that substitution gives the identity matrix (determinant $+1$), the determinant is $+1$ for all values of the parameters.

Common Mistakes

Mistake 1: Assuming an orthogonal matrix's determinant is always $+1$

Why it's wrong: $\det = \pm 1$ for ANY orthogonal matrix — a reflection-type orthogonal matrix has determinant $-1$, so the sign must always be checked, never assumed. Correct approach: after confirming orthogonality, test a simple special case of the parameters to pin down the actual sign.

Mistake 2: Expanding the full $3\times3$ determinant directly instead of spotting the structure

Why it's wrong: a brute-force expansion of a matrix full of products of sines and cosines is long and error-prone, when checking orthogonality gets to the same answer with far less arithmetic. Correct approach: before expanding, check whether the rows look like they could be unit vectors / mutually perpendicular — this is a strong hint the orthogonal-matrix shortcut applies.

Mistake 3: Making an arithmetic slip while verifying a row's dot product with itself

Why it's wrong: forgetting to distribute a square correctly (e.g. missing a cross term) can make a genuinely unit-length row look like it isn't, derailing the whole shortcut. Correct approach: expand $(\text{entry}_1)^2+(\text{entry}_2)^2+(\text{entry}_3)^2$ term by term and group the trig terms before applying the Pythagorean identity.