Q.If x+y+z=0, prove that xayczbybzaxczcxbya=xyzacbbaccba.
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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 — expand, then use x+y+z=0. Expanding the left determinant along the first row,
xayczbybzaxczcxbya=xyz(a3+b3+c3)−abc(x3+y3+z3).
Since x+y+z=0, the identity x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx) gives x3+y3+z3=3xyz. Hence
Δ=xyz(a3+b3+c3)−abc(3xyz)=xyz(a3+b3+c3−3abc). …
The left determinant expands to xyz(a3+b3+c3)−abc(x3+y3+z3); with x+y+z=0 we have x3+y3+z3=3xyz, leaving xyz(a3+b3+c3−3abc), which is xyz times the target determinant.
The idea
Expand the messy determinant into symmetric pieces. Everything collapses once x+y+z=0 is used through the identity x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx).
Step 1 — Expand the left side
Let Δ=xayczbybzaxczcxbya. Expanding along the first row and grouping terms,
Δ=xyza3+xyzb3+xyzc3−abcx3−abcy3−abcz3=xyz(a3+b3+c3)−abc(x3+y3+z3).
Step 2 — Apply x+y+z=0
From x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx) and x+y+z=0,
x3+y3+z3=3xyz.
Therefore
Δ=xyz(a3+b3+c3)−abc(3xyz)=xyz(a3+b3+c3−3abc). …
Method: Expanding a Determinant Identity and Applying a Symmetric-Function Constraint
Use this method whenever you must prove one determinant equals a scalar multiple of another, given an algebraic constraint (like x+y+z=0) on some of the variables involved.
Steps
Step 1: Expand the more complicated determinant fully
Expand the left-hand determinant by cofactor expansion and collect the result into natural groups — typically one group multiplying an expression like a3+b3+c3 and another multiplying an expression like x3+y3+z3, since the entries mix two sets of variables.
Step 2: Bring in the given constraint via a symmetric-function identity
Recall the standard identity
x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx).
When the problem states x+y+z=0, the right side vanishes, so this identity immediately gives x3+y3+z3=3xyz — substitute this simplification into the expansion from Step 1.
Step 3: Simplify to a single product
After substituting, the expression should collapse into a single product of a simple factor (like xyz) and a symmetric expression in the remaining variables (like a3+b3+c3−3abc).
Step 4: Evaluate the target determinant on the right-hand side independently …
Common Mistakes
Mistake 1: Expanding the left determinant directly instead of grouping symmetric terms
Why it's wrong: brute-force cofactor expansion of a determinant whose nine entries are all products (like xa, yc, zb) generates many cross terms; without deliberately grouping into xyz(a3+b3+c3) and abc(x3+y3+z3), it's very easy to mis-collect terms and lose the clean structure the identity depends on.
Mistake 2: Using x3+y3+z3=3xyz as if it were always true
Why it's wrong: this simplification only holds because x+y+z=0 is given (from the identity x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)); quoting it as a general algebraic fact without tying it to the given condition is a reasoning gap, and forgetting the condition entirely would make the whole proof invalid. …
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 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-2026-04224 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 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 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-2026-04204 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 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-2026-04194 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 2025Set eng-2025-04254 marksMCQQ.The numbers a1,a2,a3,a4,a5 and a6 are in G.P. If a1=2 and the common ratio r=21, then the value of a1a3a5a2a4a6111 is equal to (A) 1 (B) 2 (C) 21 (D) 4 (E) 0
›Reveal solutionSolution
Two columns are proportional, forcing the determinant to zero.
With a1=2,r=21: the terms are 2,1,21,41,81,161.
In the determinant
a1a3a5a2a4a6111, …
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