Q.Find the value of the following: 1000cosαsinα0sinα−cosα
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Determinant Evaluation Using Identities
Expanding a 4×4 or 5×5 determinant term by term is painful and error-prone. The smarter route is to transform the determinant into an easy form using properties (the "identities") that change its value in a known, controlled way — then read the answer off a triangular matrix.
The geometric intuition
A determinant measures the signed "volume" of the box spanned by the rows in n-dimensional space. Sliding one row parallel to another doesn't change that volume; swapping two rows flips its sign; scaling a row scales the volume. The algebraic identities are just these facts translated into rules.
The three row (or column) operations
- Swap two rows: det→−det (sign flips).
- Scale a row by k: det→kdet (the factor comes out).
- Add a multiple of one row to a different row (Ri→Ri+λRj, i=j): det unchanged.
The identical rules hold for columns. There is also row-wise linearity: if a row is a sum Ri=Ri′+Ri′′, the determinant splits into the sum of two determinants with all other rows fixed.
Row-wise linearity is not det(A+B)=detA+detB — that is false. The splitting works one row at a time.
The strategy
- Use operation 3 to create zeros in a row or column (value unchanged).
- Factor out common factors with operation 2.
- Swap rows if needed to reach upper-triangular form (track the sign change).
- The determinant is then the product of the diagonal entries.
Worked example
det1472583610.
Apply R2→R2−4R1 and R3→R3−7R1 (no change), then R3→R3−2R2: …
"Find the value" here means evaluate the determinant 1000cosαsinα0sinα−cosα.
Expand along the first row (only the top-left entry is non-zero): …
Expanding the determinant and using cos2α+sin2α=1 gives −1.
The instruction "find the value" means evaluate this determinant. The first row is (1,0,0), so expanding along it collapses to a single 2×2 determinant.
Expand along row 1
1000cosαsinα0sinα−cosα=1⋅cosαsinαsinα−cosα−0+0.
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 3×3 expansion into a single 2×2 determinant, since every term multiplied by a 0 entry vanishes automatically.
Step 2: Expand along that row/column using the cofactor formula
detA=∑jaijCij,Cij=(−1)i+jMij
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 2×2 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 2×2 block
Why it's wrong: the 2×2 determinant acbd=ad−bc subtracts the cross product — here that gives −cos2α−sin2α; writing it as −cos2α+sin2α (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. …
- COMEDK 2025Set 2025-A1 markMCQQ.The cofactor of the element a21 in the expansion of Δ=1−32451492 is (A) 5 (B) −24 (C) −4 (D) −5
›Reveal solutionSolution
The cofactor of a21 is found by taking (−1)2+1 times the determinant of the submatrix obtained by deleting row 2 and column 1. The result is −4, so the correct option is (C).
The cofactor of an element in a matrix is not just the minor (the determinant of the submatrix left after removing that element’s row and column). It also includes a sign factor (−1)i+j, where i and j are the row and column indices. This sign alternates like a chessboard pattern. For a21 (row 2, column 1), the sign is negative because 2+1=3 is odd. So we compute the minor and then flip its sign.
- Identify the element and its position. The element a21 is in row 2, column 1. In the given matrix
Δ=1−32451492,
a21=−3. But the cofactor depends only on position, not on the value of the element itself.
- Delete row 2 and column 1. Removing row 2 and column 1 leaves the submatrix:
(4142).
- Compute the minor M21. The minor is the determinant of that 2×2 submatrix:
M21=4142=(4)(2)−(4)(1)=8−4=4.
- Apply the sign factor. …
- COMEDK 2024Set 2024-A1 markMCQQ.cos(α+β)sinα−cosα−sin(α+β)cosαsinαcos2βsinβcosβ is independent of (A) β (B) α and β (C) Neither α nor β (D) α
›Reveal solutionSolution
Expanding along the first row collapses the determinant to 1+cos2β, which contains no α — so it is independent of α: option (D).
Cofactor expansion along row 1
The three minors are
M11=cosαsinαsinβcosβ=cosαcosβ−sinαsinβ=cos(α+β),
M12=sinα−cosαsinβcosβ=sinαcosβ+cosαsinβ=sin(α+β),
M13=sinα−cosαcosαsinα=sin2α+cos2α=1.
With the cofactor sign pattern (+,−,+) and the row-1 entries cos(α+β), −sin(α+β), cos2β: …
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