Q.Prove that the determinant x−sinθcosθsinθ−x1cosθ1x is independent of θ.
Concept understanding — Determinant Evaluation Using Identities
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:
det1002−303−61=1×(−3)×1=−3.
No cofactor was ever expanded — we just slid rows around.
Aim your zeros at a row or column that already contains a 1 to keep the arithmetic clean. And remember operation 3 needs a different row: adding a multiple of a row to itself rescales it and changes the value.
Evaluating determinants using row and column operations rather than direct expansion is a core skill in the CBSE Class 12 Determinants chapter, and "properties of determinants class 12 with examples" is one of the most searched topics for board exam revision. This technique of reducing a determinant to triangular form is also a favourite approach in JEE Main and JEE Advanced problems involving higher-order determinants.
Concept: Determinant Evaluation Using Identities — we expand and simplify using sin2θ+cos2θ=1 to show the θ terms cancel.
Step 1: Expand the determinant along the first row:
Δ=x−x11x−sinθ−sinθcosθ1x+cosθ−sinθcosθ−x1
Step 2: Compute each 2×2 determinant:
−x11x=(−x)(x)−(1)(1)=−x2−1
−sinθcosθ1x=(−sinθ)(x)−(1)(cosθ)=−xsinθ−cosθ
−sinθcosθ−x1=(−sinθ)(1)−(−x)(cosθ)=−sinθ+xcosθ
Step 3: Substitute back:
Δ=x(−x2−1)−sinθ(−xsinθ−cosθ)+cosθ(−sinθ+xcosθ)
=−x3−x+xsin2θ+sinθcosθ−cosθsinθ+xcos2θ
Step 4: The terms sinθcosθ cancel. Using sin2θ+cos2θ=1:
Δ=−x3−x+x(sin2θ+cos2θ)=−x3−x+x=−x3
The determinant equals −x3, which is independent of θ.
The determinant simplifies to a constant expression in x alone — all θ terms cancel out — proving it is independent of θ. The simplified value is −x3.
The key idea is to treat the determinant as an expression in θ and see if it actually depends on θ at all. Often, determinants with trigonometric entries simplify using identities like sin2θ+cos2θ=1, or by expanding and grouping terms. Here, a direct expansion will work cleanly — no row operations needed.
Let’s go step by step.
- Write the determinant We have
Δ=x−sinθcosθsinθ−x1cosθ1x.
- Expand along the first row (or any row — first row is fine because it has x, sinθ, cosθ). Using the standard formula for a 3×3 determinant:
Δ=x⋅−x11x−sinθ⋅−sinθcosθ1x+cosθ⋅−sinθcosθ−x1.
- Compute each 2×2 determinant
- First minor:
−x11x=(−x)(x)−(1)(1)=−x2−1.
- Second minor:
−sinθcosθ1x=(−sinθ)(x)−(1)(cosθ)=−xsinθ−cosθ.
- Third minor:
−sinθcosθ−x1=(−sinθ)(1)−(−x)(cosθ)=−sinθ+xcosθ.
- Substitute back into the expansion
Δ=x(−x2−1)−sinθ(−xsinθ−cosθ)+cosθ(−sinθ+xcosθ).
Simplify term by term:
- First term: x(−x2−1)=−x3−x.
- Second term: −sinθ(−xsinθ−cosθ)=sinθ⋅(xsinθ+cosθ)=xsin2θ+sinθcosθ.
- Third term: cosθ(−sinθ+xcosθ)=−sinθcosθ+xcos2θ.
- Combine everything
Δ=(−x3−x)+(xsin2θ+sinθcosθ)+(−sinθcosθ+xcos2θ).
Notice sinθcosθ and −sinθcosθ cancel each other exactly.
So we are left with:
Δ=−x3−x+xsin2θ+xcos2θ.
- Use the Pythagorean identity
sin2θ+cos2θ=1.
Hence,
xsin2θ+xcos2θ=x(sin2θ+cos2θ)=x.
Therefore,
Δ=−x3−x+x=−x3.
A common mistake is to forget the sign pattern when expanding: the second term has a minus sign in front of sinθ, and then the minor itself is multiplied. Always double-check the (−1)i+j factor.
If you ever see sinθ and cosθ paired with x in a determinant, suspect that sin2θ+cos2θ=1 will simplify things. Expanding directly is often faster than trying clever row operations.
The final expression contains no θ at all — it is simply −x3, a function of x alone. So the determinant is independent of θ.
The determinant equals −x3, which does not involve θ; hence it is independent of θ.
Method: Proving a Determinant Is Independent of a Parameter (e.g. θ)
When a question asks you to show a determinant does NOT depend on some angle or variable, the strategy is to expand it fully and show every occurrence of that variable cancels out algebraically.
Steps
Step 1: Expand the determinant along the row or column that looks simplest
Choose the row/column with the fewest or simplest trigonometric entries to minimise the number of terms you carry forward.
Step 2: Compute every 2×2 minor carefully, keeping the trig terms unexpanded
Write out each minor as a product/difference of sines and cosines without simplifying yet — premature simplification is where sign errors creep in.
Step 3: Substitute the minors back and collect like terms
Group terms that are pure functions of the "other" variable (here x) separately from terms that still carry the parameter (here θ).
Step 4: Use a trigonometric identity to eliminate the parameter
Look for a combination like sin2θ+cos2θ hiding in the collected terms — replacing it with 1 is usually what makes the parameter vanish and confirms independence. If a sinθcosθ term appears twice with opposite signs, note that it cancels directly without needing any identity.
Common Mistakes
Mistake 1: Expanding along a row that leaves the messiest arithmetic
Why it's wrong: some rows/columns lead to far more terms to track than others; picking a "hard" row makes it much easier to drop a sign or a term. Correct approach: scan all three rows/columns first and expand along the one with the fewest distinct trig products.
Mistake 2: Missing the cancelling sinθcosθ terms
Why it's wrong: in problems like this, two cross-terms with opposite signs cancel exactly — if you simplify too aggressively or too early, it's easy to lose track of one of them and end up with a leftover θ-term that shouldn't be there. Correct approach: keep all terms explicit until the very end, then cancel matching pairs deliberately, one at a time.
Mistake 3: Forgetting to apply sin2θ+cos2θ=1 to finish the proof
Why it's wrong: stopping right after collecting terms like xsin2θ+xcos2θ without simplifying them to x leaves the expression looking like it still depends on θ, even though it doesn't. Correct approach: always scan the final expression for a sin2+cos2 pattern and apply the identity before declaring the proof complete.
Showing the 12 most recent of 18 on this concept.
- KEAM 2022Set eng-2022-P2-B14 marksMCQQ.sinαsinβsinγcos(α+θ)cos(β+θ)cos(γ+θ)cosαcosβcosγ= (A) −1 (B) 1 (C) 2 (D) 4 (E) 0
›Reveal solutionSolution
The determinant equals 0.
Concept and Intuition
A determinant is zero when one column is a linear combination of the others.
Step-by-Step Solution
- Expand cos(ϕ+θ)=cosθcosϕ−sinθsinϕ for each row's middle entry.
- So column 2 =cosθ(column of cosϕ)−sinθ(column of sinϕ).
- Column of cosϕ is column 3, column of sinϕ is column 1.
- Thus C2=cosθC3−sinθC1: columns are linearly dependent.
- Determinant =0.
Common Mistakes
- Trying a brute-force cofactor expansion instead of spotting dependence.
- Sign error in the cosine addition formula.
✓Final answerThe correct option is (E) — 0.
ANSWER: E
- KEAM 2026Set eng-2026-04224 marksMCQQ.The value of sin30∘sin45∘sin60∘cos30∘cos45∘cos60∘sin(30∘+75∘)sin(45∘+75∘)sin(60∘+75∘) is equal to (A) −2 (B) −1 (C) 0 (D) 1 (E) 2
›Reveal solutionSolution
The third column is a fixed linear combination of the first two, forcing a zero determinant.
Each entry of column 3 is sin(θ+75∘)=sinθcos75∘+cosθsin75∘.
So C3=cos75∘C1+sin75∘C2, i.e. column 3 is linearly dependent on columns 1 and 2.
A determinant with linearly dependent columns is 0.
✓Final answerThe correct option is (C).
- KEAM 2021Set eng-2021-P2-B14 marksMCQQ.Let x−1212x−1x+212x−1=ax3+bx2+cx+d, where a,b,c and d are constants. Then the value of d is (A) −8 (B) 6 (C) 0 (D) −6 (E) 16
›Reveal solutionSolution
Setting x=0 gives the constant term d=16.
Concept and Intuition
Writing the determinant as ax3+bx2+cx+d, the constant d equals the value at x=0, since all x-bearing terms vanish there.
Step-by-Step Solution
- At x=0 the matrix is −1212−1212−1.
- Expand: −1((−1)(−1)−2⋅2)−2(2⋅(−1)−2⋅1)+1(2⋅2−(−1)⋅1).
- =−1(1−4)−2(−2−2)+1(4+1)=3+8+5=16.
- So d=16.
Common Mistakes
- Trying to expand the full cubic in x instead of just substituting x=0.
✓Final answerThe correct option is (E) — 16.
ANSWER: E
- KEAM 2025Set eng-2025-04284 marksMCQQ.Let Δ=xx+y1yy+1x1x+1y. If x+y=−1, then the value of Δ is equal to (A) 3 (B) 2 (C) 1 (D) 0 (E) -3
›Reveal solutionSolution
Expanding Δ and substituting x+y=−1, every term reduces to 0 because it carries a factor x+y+1=0.
Expanding along the first row:
Δ=x[(y+1)y−(x+1)x]−y[(x+y)y−(x+1)]+[(x+y)x−(y+1)].
With x+y=−1:
- First term: x[(y2−x2)+(y−x)]=x(y−x)(x+y+1)=x(y−x)(0)=0.
- Second term: −y[(−1)y−x−1]=−y[−(x+y)−1]=−y(1−1)=0.
- Third term: (−1)x−y−1=−(x+y)−1=1−1=0.
Hence Δ=0.
✓Final answerThe correct option is (D).
- KEAM 2026Set eng-2026-04184 marksMCQQ.Let f(x)=−101x1−132x1. Then the value of f(−1) is equal to (A) 6 (B) -4 (C) -2 (D) 2 (E) 0
›Reveal solutionSolution
At x=−1 the determinant evaluates to 0.
At x=−1 the matrix is −101−11−13−21. Expanding along the first column:
f(−1)=−1[(1)(1)−(−2)(−1)]−0+1[(−1)(−2)−(3)(1)].
=−1(1−2)+1(2−3)=−1(−1)+1(−1)=1−1=0.
✓Final answerThe correct option is (E).
- KEAM 2021Set eng-2021-P2-B14 marksMCQQ.The value of the determinant 432423222433323 is (A) 52 (B) −24 (C) 24 (D) 48 (E) −48
›Reveal solutionSolution
The determinant equals −48.
Concept and Intuition
Each row (k,k2,k3) has a common factor k, which can be pulled out of the determinant. The remaining determinant is a small 3×3 evaluation.
Step-by-Step Solution
- Factor 4,3,2 from rows 1,2,3: value =4⋅3⋅21114321694=24D.
- Expand D=1(3⋅4−9⋅2)−4(1⋅4−9⋅1)+16(1⋅2−3⋅1).
- D=(−6)−4(−5)+16(−1)=−6+20−16=−2.
- Value =24×(−2)=−48.
Common Mistakes
- Forgetting the row factors, or sign slips in the cofactor expansion.
✓Final answerThe correct option is (E) — −48.
ANSWER: E
- KEAM 2025Set eng-2025-04294 marksMCQQ.If α+β+γ=0, then eαeβeγe2αe2βe2γe3α−1e3β−1e3γ−1= (A) e−1 (B) e (C) e2 (D) e3 (E) 0
›Reveal solutionSolution
Let a=eα,b=eβ,c=eγ, so abc=eα+β+γ=1. Split the third column a3−1=a3+(−1); the two resulting Vandermonde determinants are equal and cancel, giving 0.
Write the rows as (a,a2,a3−1) etc. By column-linearity the determinant splits as D1−D2 where
D1=abca2b2c2a3b3c3=abc111abca2b2c2=abc,V,
and D2=abca2b2c2111. A cyclic (even) column permutation turns D2 into the same Vandermonde V. Since abc=1, D1=V and D2=V, so the value is V−V=0.
✓Final answerThe correct option is (E).
- KEAM 2023Set eng-2023-P2-B24 marksMCQQ.The determinant of the matrix 111491682764 is (A) 13 (B) 208 (C) 104 (D) 26 (E) 52
›Reveal solutionSolution
Expanding along the first row gives determinant 52.
Concept and Intuition
A direct cofactor expansion along the first row is quickest for a 3×3 determinant.
Step-by-Step Solution
- det = 1(9·64 − 27·16) − 4(1·64 − 27·1) + 8(1·16 − 9·1).
- = 1(576 − 432) − 4(64 − 27) + 8(16 − 9).
- = 144 − 4·37 + 8·7 = 144 − 148 + 56.
- = 52.
Common Mistakes
- Arithmetic slips in the 2×2 minors.
✓Final answerThe correct option is (E) — 52.
ANSWER: E
- KEAM 2026Set eng-2026-04204 marksMCQQ.The value of the determinant (105+10−5)2(1006+100−6)2(6100+6−100)2(105−10−5)2(1006−100−6)2(6100−6−100)2111 is equal to (A) 100 (B) 200 (C) 0 (D) 6000 (E) 60600
›Reveal solutionSolution
In every row the two entries are (P+Q)2 and (P−Q)2 with PQ=1, so C1−C2=4PQ=4 for all rows. That makes column C1−C2 a multiple of the all-ones column C3; two proportional columns force the determinant to be 0.
In each row the entries have the form (P+Q)2, (P−Q)2, 1, where:
- Row 1: P=105, Q=10−5, PQ=1.
- Row 2: P=1006, Q=100−6, PQ=1.
- Row 3: P=6100, Q=6−100, PQ=1.
Apply the column operation C1→C1−C2. For every row,
(P+Q)2−(P−Q)2=4PQ=4⋅1=4.
So the new first column is (4,4,4)T=4(1,1,1)T, which is exactly 4 times the third column C3=(1,1,1)T.
A determinant with two proportional columns is 0.
✓Final answerThe correct option is (C).
- KEAM 2026Set eng-2026-04194 marksMCQQ.111112111131 is equal to (A) 7100 (B) 6800 (C) 7300 (D) 6900 (E) 6700
›Reveal solutionSolution
Cofactor expansion along the first row gives 7100.
Expansion. For 111112111131:
=11(21⋅31−1⋅1)−1(1⋅31−1⋅1)+1(1⋅1−21⋅1)
=11(651−1)−1(31−1)+1(1−21)=11⋅650−30−20.
=7150−50=7100.
✓Final answerThe correct option is (A).
- KEAM 2023Set eng-2023-P2-B24 marksMCQQ.Let A be (2n+1)×(2n+1) matrix with integer entries and positive determinant, where n∈N. If AAT=I=ATA, then which of the following statements always holds? (A) det(A)=0 (B) det(A+I)=0 (C) det(A+I)=0 (D) det(A−I)=0 (E) det(A−I)=0
›Reveal solutionSolution
An odd-dimensional orthogonal matrix with det +1 always has eigenvalue 1, so det(A−I) = 0.
Concept and Intuition
AA^T = I means A is orthogonal, and positive integer determinant forces det(A) = +1. In odd dimension a real rotation must fix an axis (eigenvalue 1) because complex eigenvalues occur in conjugate pairs and the leftover real eigenvalue, together with det = +1, must be +1.
Step-by-Step Solution
- AA^T = I ⇒ A orthogonal ⇒ eigenvalues have modulus 1 and det = ±1.
- Integer positive det ⇒ det(A) = +1.
- Size 2n+1 is odd; non-real eigenvalues pair as conjugates, leaving at least one real eigenvalue ±1.
- Product of all eigenvalues = +1 forces a +1 eigenvalue to exist.
- Eigenvalue 1 ⇒ det(A − I) = 0.
Common Mistakes
- Assuming det could be 0 — an orthogonal matrix is invertible.
✓Final answerThe correct option is (D) — det(A − I) = 0.
ANSWER: D
- KEAM 2024Set eng-2024-06074 marksMCQQ.If A=(−733−1), then det(A5) is equal to (A) 81 (B) -81 (C) 243 (D) -243 (E) -32
›Reveal solutionSolution
det(A5)=(detA)5.
detA=(−7)(−1)−(3)(3)=7−9=−2. Then det(A5)=(detA)5=(−2)5=−32.
✓Final answerThe correct option is (E).
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