Q.Prove that bc−a2ca−b2ab−c2ca−b2ab−c2bc−a2ab−c2bc−a2ca−b2 is divisible by a+b+c and find the quotient.
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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 — recognise the circulant. With p=bc−a2, q=ca−b2, r=ab−c2 the determinant is the cyclic array
Δ=pqrqrprpq=3pqr−p3−q3−r3=−(p+q+r)(p2+q2+r2−pq−qr−rp).
Sum: p+q+r=(ab+bc+ca)−(a2+b2+c2)=−(a2+b2+c2−ab−bc−ca).
Differences: p−q=(b−a)(a+b+c), and cyclically, so
p2+q2+r2−pq−qr−rp=21[(p−q)2+(q−r)2+(r−p)2]=(a+b+c)2(a2+b2+c2−ab−bc−ca).
Multiplying, the two minus signs cancel: …
The determinant is a circulant in p=bc−a2,q=ca−b2,r=ab−c2, equal to (a+b+c)2(a2+b2+c2−ab−bc−ca)2; dividing by a+b+c leaves (a+b+c)(a2+b2+c2−ab−bc−ca)2.
The idea
Each row is a cyclic shift of p,q,r. Such a circulant has the standard value 3pqr−p3−q3−r3, which factors as −(p+q+r)(p2+q2+r2−pq−qr−rp). We then substitute back in a,b,c.
Step 1 — Name the entries
p=bc−a2,q=ca−b2,r=ab−c2,Δ=pqrqrprpq.
Step 2 — Circulant value
Δ=3pqr−p3−q3−r3=−(p+q+r)(p2+q2+r2−pq−qr−rp).
Step 3 — The sum p+q+r
p+q+r=(bc+ca+ab)−(a2+b2+c2)=−(a2+b2+c2−ab−bc−ca).
Write S=a2+b2+c2−ab−bc−ca, so p+q+r=−S.
Step 4 — The second factor
Compute one difference:
p−q=(bc−a2)−(ca−b2)=c(b−a)+(b−a)(b+a)=(b−a)(a+b+c).
Cyclically, q−r=(c−b)(a+b+c) and r−p=(a−c)(a+b+c). Then …
Method: Evaluating a Circulant Determinant to Test Divisibility
Use this method whenever a determinant's three rows are cyclic shifts of the same three expressions (a circulant) and you're asked to show it's divisible by some factor and find the quotient.
Steps
Step 1: Name the repeating entries
Give short names (like p, q, r) to the three distinct expressions that appear, shifted cyclically, in each row. This turns a messy-looking determinant into the standard circulant pattern pqrqrprpq.
Step 2: Apply the standard circulant identity
A 3×3 circulant has the known factored value
pqrqrprpq=3pqr−p3−q3−r3=−(p+q+r)(p2+q2+r2−pq−qr−rp),
which avoids expanding the determinant term by term.
Step 3: Compute the sum p+q+r and the pairwise differences
Substitute the original expressions back in and simplify p+q+r — it usually collapses to a recognisable symmetric expression. Then compute differences like p−q; these often factor neatly, revealing a common factor (such as a+b+c) shared by every pairwise difference.
Step 4: Rewrite the second bracket using the differences …
Common Mistakes
Mistake 1: Not recognizing the circulant structure and expanding directly
Why it's wrong: substituting p=bc−a2, q=ca−b2, r=ab−c2 back into a raw 3×3 expansion in a,b,c produces a huge, error-prone polynomial. The efficient route is to keep working in p,q,r using the standard identity 3pqr−p3−q3−r3=−(p+q+r)(p2+q2+r2−pq−qr−rp) and substitute back only at the end.
Mistake 2: Losing the sign when computing p+q+r
Why it's wrong: p+q+r=(ab+bc+ca)−(a2+b2+c2), which is −S where S=a2+b2+c2−ab−bc−ca — a student who writes p+q+r=S (dropping the sign flip) carries an incorrect sign through the rest of the derivation, potentially flipping the final answer's sign or making the "divisible by a+b+c" claim look false. …
Showing the 12 most recent of 18 on this concept.
- 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 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-2026-04224 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 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 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 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 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-2024-06064 marksMCQQ.Let A=2−11−10−212−1 and let B=∣A∣1A. Then the value of ∣B∣ is equal to (A) 91 (B) 111 (C) 811 (D) 1211 (E) 1
›Reveal solutionSolution
For a 3×3 matrix, scaling by 1/∣A∣ scales the determinant by (1/∣A∣)3. …
- 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 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 …
- 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: …
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