Q.If cos2θ=0, then 0cosθsinθcosθsinθ0sinθ0cosθ2= ________ .
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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: …
Let a=cosθ, b=sinθ. Expanding along the first row,
0abab0b0a=−a(a2)+b(−b2)=−(a3+b3)=−(cos3θ+sin3θ).
The condition cos2θ=0 gives cos2θ=sin2θ=21. Taking the standard value θ=4π (so cosθ=sinθ=21): …
The determinant equals −(cos3θ+sin3θ); with cos2θ=0 its square is 21.
Set up
Write a=cosθ and b=sinθ, so the determinant is
Δ=0abab0b0a.
Expand along the first row
Δ=0⋅(…)−aab0a+babb0=−a(a2)+b(−b2)=−(a3+b3).
So Δ=−(cos3θ+sin3θ) and Δ2=(cos3θ+sin3θ)2.
Use the condition
cos2θ=0⇒cos2θ=sin2θ=21, hence cosθsinθ=±21. Expanding the square,
Δ2=cos6θ+sin6θ+2cos3θsin3θ=(1−3cos2θsin2θ)+2cos3θsin3θ. …
Method: Evaluating a Trigonometric Determinant Under a Given Angle Condition
When a determinant's entries are trig functions of a single angle and you're given a condition on that angle (like cos2θ=0), expand the determinant symbolically first, simplify using trig identities, and only substitute the angle condition at the very end — and always check whether the condition admits more than one essentially different case.
Steps
Step 1: Expand the determinant symbolically, keeping cosθ and sinθ as separate variables
Use cofactor expansion along the row or column with the most zeros. Track the sign of each cofactor carefully — with several zero entries in a 3×3 trig determinant, it's easy to drop a minus sign on one of the surviving terms.
Step 2: Simplify the resulting expression using standard identities
The expansion typically reduces to a sum/difference of sin3θ and cos3θ (or similar). Keep the expression in terms of sinθ,cosθ rather than immediately substituting numbers — this makes it easier to apply the given condition cleanly in the next step.
Step 3: Translate the given trig condition into a usable algebraic fact …
Common Mistakes
Mistake 1: Dropping a sign while cofactor-expanding a determinant with several zero entries
Why it's wrong: with three zeros scattered through a 3×3 trig determinant, it's tempting to write down only the surviving (non-zero) terms without also tracking each one's cofactor sign correctly — a single dropped minus sign flips the whole final expression's sign. Correct approach: write the full cofactor expansion with explicit (−1)i+j signs before cancelling any zero terms, not after.
Mistake 2: Assuming cos2θ=0 forces one specific value of cosθ and sinθ …
- AHSEC Higher Secondary (HS) Final Examination 2025Set ANNUAL1 markQ.Find the determinant of the matrix A=[−3]1×1.
›Reveal solutionSolution
The determinant of a 1×1 matrix is simply its single entry: −3.
For a 1×1 matrix A=[a], the determinant is defined as det(A)=a itself (there is no expansion needed since there's only o …
- AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL1 markQ.If A=[1422], then find ∣2nA∣, for some n∈N.
›Reveal solutionSolution
First find ∣A∣=−6, then use ∣kA∣=k2∣A∣ for a 2×2 matrix.
A=[1422], so ∣A∣=1(2)−2(4)=2−8=−6.
…
- AHSEC Higher Secondary (HS) Final Examination 2019Set ANNUAL1 markQ.Let A be a skew-symmetric matrix of odd order. Write the value of ∣A∣.
›Reveal solutionSolution
Any skew-symmetric matrix of odd order has determinant 0.
Let A be skew-symmetric of odd order n, so AT=−A.
Taking determinants: det(AT)=det(−A).
Now det(AT)=det(A) (transpose doesn't change the determinant), and det(−A)=(−1)ndet(A) (each of the n rows is multiplied by −1).
…
- AHSEC Higher Secondary (HS) Final Examination 2019Set ANNUAL1 markQ.Let A be a matrix of order 3, such that ∣A∣=−9. Find the value of ∣−3A−1∣.
›Reveal solutionSolution
Using ∣A−1∣=1/∣A∣ and ∣kA∣=kn∣A∣ for an n×n matrix, ∣−3A−1∣=(−3)3⋅−91=3.
A has order 3 and ∣A∣=−9.
Step 1. ∣A−1∣=∣A∣1=−91=−91.
…
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