Q.Using the properties of determinants, evaluate: a+xxxya+yyzza+z
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.
Key idea: every row of the determinant adds to the same value a+x+y+z, so fold all columns into the first and pull that factor out.
Step 1 — C1→C1+C2+C3. Each new first-column entry is a+x+y+z:
a+x+y+za+x+y+za+x+y+zya+yyzza+z=(a+x+y+z)111ya+yyzza+z.
Step 2 — R2→R2−R1 and R3→R3−R1 give a triangular determinant:
(a+x+y+z)100ya0z0a=(a+x+y+z)a2.
a2(a+x+y+z)
Adding all columns into the first exposes the common factor a+x+y+z; reducing to triangular form leaves the diagonal 1,a,a, so the determinant is a2(a+x+y+z).
Intuition
Whenever every row of a determinant adds up to the same thing, that common sum is hiding as a factor. You reveal it by folding all the columns into one, then clear the rest to a triangle whose diagonal you read straight off.
Setting up
Δ=a+xxxya+yyzza+z.
Working the steps
1. Fold the columns in: apply C1→C1+C2+C3. Row by row the first entry becomes (a+x)+y+z, x+(a+y)+z, x+y+(a+z) — all equal to a+x+y+z:
Δ=a+x+y+za+x+y+za+x+y+zya+yyzza+z.
2. Pull the factor out of the first column:
Δ=(a+x+y+z)111ya+yyzza+z.
3. Make zeros: R2→R2−R1 and R3→R3−R1:
Δ=(a+x+y+z)100ya0z0a.
4. Triangular determinant = product of the diagonal =1⋅a⋅a=a2:
Δ=a2(a+x+y+z).
Check with a=1, x=1, y=2, z=3: the formula gives 1⋅7=7, which matches a direct expansion.
a+xxxya+yyzza+z=a2(a+x+y+z)
Method: Column-Sum Trick (C₁→C₁+C₂+C₃) for Determinants Where Every Row Sums Alike
This method is the standard shortcut whenever every row (or column) of a determinant adds up to the same expression — a strong visual signal that a common factor is hiding inside the matrix.
Steps
Step 1: Check whether every row's entries add to a common expression
Add across each row and see if the sums match. If they do (here, every row of a+xxxya+yyzza+z sums to a+x+y+z), that common sum is the factor this trick will expose.
Step 2: Fold all columns into one via C₁→C₁+C₂+C₃
This operation does not change the determinant's value, but it replaces every entry of the first column with the common row-sum, since C1+C2+C3 is exactly what you summed in Step 1.
Step 3: Factor the common expression out of the column
Since every entry in the new first column is identical, pull that common factor outside the determinant (the "scaling a row/column" property, used in reverse):
(a+x+y+z)111ya+yyzza+z.
Step 4: Zero out the first column and read off the triangular determinant
Apply R2→R2−R1 and R3→R3−R1 to turn the remaining first-column entries into zeros, reaching an upper-triangular matrix. The determinant of a triangular matrix is just the product of its diagonal entries, so multiply that product by the factor pulled out in Step 3 for the final answer.
Common Mistakes
Mistake 1: Adding the wrong rows/columns together
Why it's wrong: the trick only works because every ROW happens to sum to the same value a+x+y+z when the three COLUMNS are added into one — a student who adds rows instead of columns (or checks only one row's sum instead of all three) may pull out a factor that doesn't actually apply uniformly. Correct approach: verify each of the three rows gives the same sum a+x+y+z after C1→C1+C2+C3 before factoring it out.
Mistake 2: Sign slip while zeroing out rows 2 and 3
Why it's wrong: applying R2→R2−R1 and R3→R3−R1 to the reduced matrix [1,y,z;1,a+y,z;1,y,a+z] requires subtracting the same first row from each — mixing up which row is subtracted from which turns the diagonal a,a into the wrong values, giving a wrong power of a in the final answer. Correct approach: always subtract R1 from R2 and R3 (never the reverse), and check that the first column becomes all zeros below the top row.
Mistake 3: Forgetting to reattach the factored constant
Why it's wrong: after reducing to the triangular determinant 1⋅a⋅a=a2, a student can report just a2 and forget the (a+x+y+z) factor pulled out earlier. Correct approach: always carry the factored-out term through to the final line — the answer is a2(a+x+y+z), not a2 alone.
Showing the 12 most recent of 18 on this concept.
- 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 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-2025-04294 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 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 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 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 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-2026-04184 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 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:
∣A∣=a11C11+a12C12+a13C13=1(−40)+3(10)+(−2)(35)=−40+30−70=−80.
✓Final answerThe correct option is (B).
- 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 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 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,
each second-column entry is r times the first-column entry (a2=ra1, a4=ra3, a6=ra5). Columns 1 and 2 are therefore linearly dependent, so the determinant equals 0.
✓Final answerThe correct option is (E).
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