Q.Find the value of the following: $\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha \end{bmatrix}$

"Find the value" here means evaluate the determinant $\begin{vmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{vmatrix}$.

Expand along the first row (only the top-left entry is non-zero): …

The instruction "find the value" means evaluate this determinant. The first row is $(1,0,0)$, so expanding along it collapses to a single $2\times2$ determinant.

Expand along row 1

Evaluate the 2x2 block …

Method: Evaluating a Determinant Using Zero Entries and Known Identities

This method evaluates a determinant efficiently by expanding along a row/column that is mostly zero, then simplifying the leftover expression with an algebraic or trigonometric identity — rather than blindly expanding every term.

Steps

Step 1: Scan the determinant for a row or column with the most zeros

Expanding along a row/column of the form $(1,0,0)$ collapses the whole $3\times3$ expansion into a single $2\times2$ determinant, since every term multiplied by a $0$ entry vanishes automatically.

Step 2: Expand along that row/column using the cofactor formula

Only the non-zero entries in that row contribute a term — write down just those.

Step 3: Simplify the resulting expression using the relevant identity …

Common Mistakes

Mistake 1: Expanding along the wrong row/column, ignoring the zeros

Why it's wrong: expanding along a row or column that isn't mostly zero (e.g. column 2 or 3 here, instead of column 1 / row 1) means doing a full three-term cofactor expansion instead of collapsing immediately to one $2\times2$ determinant — much more room for arithmetic error, for no benefit. Correct approach: always scan for the row/column with the most zeros first, and expand along that one.

Mistake 2: A sign error in the $2\times2$ block

Why it's wrong: the $2\times2$ determinant $\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$ subtracts the cross product — here that gives $-\cos^2\alpha-\sin^2\alpha$; writing it as $-\cos^2\alpha+\sin^2\alpha$ (dropping the minus sign on the second term) changes the final answer completely. Correct approach: write out $ad-bc$ explicitly with all signs before simplifying. …