Q.Evaluate Δ=0−sinαcosαsinα0−sinβ−cosαsinβ0.
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🔒 Start your 14-day free trial to unlock the full solution →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: …
Concept: Determinant Evaluation Using Identities — here, the matrix is skew-symmetric of odd order, so its determinant is zero.
Step 1: Check the property. A square matrix A is skew-symmetric if AT=−A. For the given matrix, each entry satisfies aij=−aji, and the diagonal entries are all zero. …
This determinant is zero because the matrix is skew-symmetric of odd order (3×3). The value is 0.
Why This Approach Works
When you see a determinant with zeros on the main diagonal and opposite signs across the diagonal, you're looking at a skew-symmetric matrix. For any skew-symmetric matrix of odd order, the determinant is always zero. This isn't a coincidence — it follows directly from the property AT=−A and the fact that det(AT)=det(A).
Let's see why.
Step-by-Step Solution
1. Identify the matrix type
The given matrix is:
Δ=0−sinαcosαsinα0−sinβ−cosαsinβ0
Notice:
- Every diagonal entry is 0.
- For any i=j, the entry at (i,j) is the negative of the entry at (j,i).
This is exactly the definition of a skew-symmetric matrix: AT=−A.
2. Apply the determinant property
For any square matrix A, we know det(AT)=det(A).
If A is skew-symmetric, then AT=−A, so:
det(A)=det(AT)=det(−A)
3. Factor out the scalar
For a 3×3 matrix, det(−A)=(−1)3det(A)=−det(A).
Thus:
det(A)=−det(A)
4. Solve the equation
The only number equal to its own negative is zero:
det(A)=0 …
Method: Recognizing a Skew-Symmetric Determinant of Odd Order
This method evaluates a determinant instantly, without any expansion, by recognizing that it belongs to a special structural class: a skew-symmetric matrix of odd order.
Steps
Step 1: Check the diagonal
A skew-symmetric matrix always has every diagonal entry equal to zero (since aii=−aii forces aii=0).
Step 2: Check the off-diagonal pairing
Confirm that for every pair of positions, aij=−aji — i.e. the matrix equals the negative of its own transpose, AT=−A.
Step 3: Check the order (size) of the matrix
This shortcut only applies when the matrix is odd-order (3×3, 5×5, etc.) — the argument breaks down for even order.
Step 4: Apply the determinant property
Since det(AT)=det(A) always, and AT=−A here, substitute:
det(A)=det(AT)=det(−A)=(−1)ndet(A) …
Common Mistakes
Mistake 1: Trying to expand the determinant directly instead of recognizing the skew-symmetric pattern
Why it's wrong: direct expansion works but is slower and riskier — it's easy to make a sign error among the several sinα,cosα,sinβ terms, when the entire determinant is actually forced to be zero by structure alone. Correct approach: first check the diagonal (all zero) and the off-diagonal pairing (aij=−aji) — if both hold and the order is odd, the determinant is 0 with no further computation needed.
Mistake 2: Applying the "skew-symmetric ⇒ determinant zero" shortcut to an even-order matrix …
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. …
- 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∘. …
- 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, …
- 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. …
- 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). …
- 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: …
- 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, …
- 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). …
- KEAM 2024Set eng-2024-06094 marksMCQQ.Let A=(aij) be a square matrix of order 3 and let Mij be the minors of aij. If M11=−40,M12=−10,M13=35 and a11=1,a12=3,a13=−2 then the value of ∣A∣ is equal to (A) -100 (B) -80 (C) 0 (D) 60 (E) 80
›Reveal solutionSolution
Convert minors to cofactors with alternating signs, then expand: ∣A∣=−80.
Cofactors: C11=+M11=−40, C12=−M12=10, C13=+M13=35.
Expanding along the first row: …
- 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) …
- 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). …
- KEAM 2024Set eng-2024-06074 marksMCQQ.Let A=a1a2a3b1b2b3c1c2c3 and B=a12a24a32b14b28b34c18c216c3. If ∣B∣=16, then the value of ∣A∣ is equal to (A) 4 (B) 41 (C) 8 (D) 81 (E) 16
›Reveal solutionSolution
Pull common factors out of each row and each column of B.
Rows of B are (a1,2b1,4c1), 2(a2,2b2,4c2), 4(a3,2b3,4c3), giving a row factor 1⋅2⋅4=8. The remaining matrix has columns with factors 1,2,4, giving a column factor 1⋅2⋅4=8, and what remains i …
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