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 14 on this concept.
- GUJCET 2020Set 071 markMCQQ.If x,y∈R and (ax+a−x)2(bx+b−x)2(cx+c−x)2(ax−a−x)2(bx−b−x)2(cx−c−x)2111=2y+6 then y= ________. (A) 0 (B) 3 (C) −3 (D) 6
›Reveal solutionSolution
Column 1 − Column 2 =4×(Column 3), so the determinant is 0; 2y+6=0 gives y=−3.
Concept — spot the linear dependence. For any base a, let p=ax, q=a−x, so pq=axa−x=1. Then
(ax+a−x)2−(ax−a−x)2=4axa−x=4
This holds for every row (with bases a,b,c). So in the matrix,
C1−C2=4=4C3
i.e. C1−C2−4C3=0 — the columns are linearly dependent, hence
⋯=0
Given the determinant equals 2y+6:
2y+6=0⇒y=−3
✓Final answerOption (C) −3
ANSWER: (C)
- GUJCET 2022Set 081 markMCQQ.For real numbers x,y,z such that x=y=z, xyzx2y2z21+x31+y31+z3=0 and 111xyzx2y2z2=0 then xyz= ______. (A) 2 (B) −1 (C) 0 (D) 1
›Reveal solutionSolution
Split the last column into 1 and x³; the determinant factors as (Vandermonde)(1+xyz).
Concept. xyzx2y2z21+x31+y31+z3=xyzx2y2z2111+xyzx2y2z2x3y3z3.
Solution. The first determinant, after cycling column 3 to the front (even number of swaps), equals the Vandermonde V=111xyzx2y2z2. The second =xyzV. So the total is V(1+xyz)=0.
Since V=0 (given), 1+xyz=0⇒xyz=−1.
✓Final answer(B) −1
ANSWER: (B)
- GUJCET 2020Set 071 markMCQQ.For △ABC, the value of 0−sin(B+C)tan(A+C)sinA0−cosCtanBcosC0= ________. (A) −1 (B) 0 (C) 1 (D) sinAcosC
›Reveal solutionSolution
Using A+B+C=π the matrix becomes skew-symmetric; a 3×3 (odd-order) skew-symmetric determinant is always 0.
Concept — trig identities in a triangle. Since A+B+C=π: sin(B+C)=sin(π−A)=sinA and tan(A+C)=tan(π−B)=−tanB.
Substituting, the matrix is
0−sinA−tanBsinA0−cosCtanBcosC0
Every aij=−aji with zero diagonal, i.e. it is skew-symmetric. For any odd-order skew-symmetric matrix det=0.
✓Final answer(B) 0
ANSWER: (B)
- GUJCET 2024Set 131 markMCQQ.If 2017201920182020+2021202320222024=2k, then k3= __________. (A) −64 (B) −8 (C) 0 (D) 8
›Reveal solutionSolution
Both determinants evaluate to −2; their sum −4=2k gives k=−2 and k3=−8.
Concept. Evaluate each 2×2 determinant ad−bc.
Steps.
2017201920182020=2017⋅2020−2018⋅2019=−2,
2021202320222024=2021⋅2024−2022⋅2023=−2.
Sum =−4=2k⇒k=−2⇒k3=−8.
✓Final answer(B) −8
ANSWER: (B)
- GUJCET 2025Set 031 markMCQQ.cos2θsin2θ−sin2θcos2θ= _____. (A) 21−21cos22θ (B) 41(3+cos4θ) (C) 1+21sin22θ (D) 1+2sin2θ⋅cos2θ
›Reveal solutionSolution
Expand the 2×2 determinant, then use double/quadruple-angle identities.
cos2θsin2θ−sin2θcos2θ=cos4θ+sin4θ=1−2sin2θcos2θ=1−21sin22θ.
Using sin22θ=21−cos4θ:
1−21⋅21−cos4θ=44−1+cos4θ=43+cos4θ.
✓Final answer(B) 41(3+cos4θ)
ANSWER: (B)
- GUJCET 2019Set 171 markMCQQ.If 1!2!3!2!3!4!3!4!5!=2016K, then K=. (A) 84 (B) 241 (C) 24 (D) 841
›Reveal solutionSolution
Evaluating the determinant of factorials gives 24; with 24=2016K, K=841.
Concept: Write out the values: 1!=1,2!=2,3!=6,4!=24,5!=120.
1262624624120=1(720−576)−2(240−144)+6(48−36)=144−192+72=24
Then 24=2016K⇒K=201624=841.
✓Final answer(D) 841
ANSWER: (D)
- GUJCET 2019Set 171 markMCQQ.sin2θ−cos2θcos2θsin2θ=. (A) 21(1+cos22θ) (B) 21(1−sin22θ) (C) cos2θ (D) 21sin22θ
›Reveal solutionSolution
sin2θ−cos2θcos2θsin2θ=sin4θ+cos4θ, which equals 21(1+cos22θ).
Concept: Expand: sin2θ⋅sin2θ−cos2θ⋅(−cos2θ)=sin4θ+cos4θ.
Now sin4θ+cos4θ=1−2sin2θcos2θ=1−21sin22θ. Using sin22θ=1−cos22θ:
1−21(1−cos22θ)=21+21cos22θ=21(1+cos22θ)
(Check θ=0: determinant =1, and 21(1+1)=1.)
✓Final answer(A) 21(1+cos22θ)
ANSWER: (A)
- GUJCET 2021Set 151 markMCQQ.For 21−130−2574, the sum of minor and cofactor of 7=. (A) 0 (B) 2 (C) −2 (D) −1
›Reveal solutionSolution
Element 7 sits at position (2,3); cofactor =(−1)2+3× minor.
Concept: Deleting row 2 and column 3:
M23=2−13−2=2(−2)−3(−1)=−1.
Cofactor C23=(−1)2+3M23=−(−1)=1. Sum =−1+1=0.
✓Final answer(A) 0
ANSWER: (A)
- GUJCET 2020Set 071 markMCQQ.Let f(t)=cost2tanttantttt12tt. Then limt→0t2f(t) is equal to ________. (A) 3 (B) 1 (C) −1 (D) 0
›Reveal solutionSolution
The determinant equals t(−tcost+tant), so f(t)/t2=−cost+ttant→−1+1=0.
Concept — simplify the determinant first. Column 2 is t[1,1,1]T, so pull out t:
f(t)=tcost2tanttant11112tt=tg(t)
Expand g(t) along the first column's cofactors (about row 1):
g(t)=cost(t−2t)−1(2ttant−2ttant)+1(2tant−tant)
=−tcost+0+tant
Therefore
t2f(t)=t2tg(t)=tg(t)=−cost+ttant
Taking t→0 (using ttant→1):
limt→0t2f(t)=−1+1=0
✓Final answerOption (D) 0
ANSWER: (D)
- GUJCET 2023Set 091 markMCQQ.sin3611πsin92πcos3611πcos92π= ______. (A) cos12π (B) sin92π (C) cos125π (D) sin127π
›Reveal solutionSolution
A determinant of this sin/cos form collapses to a single sine of the angle difference.
Concept. sinPsinQcosPcosQ=sinPcosQ−cosPsinQ=sin(P−Q).
Solution. P=3611π, Q=92π=368π.
sin(3611π−368π)=sin363π=sin12π.
Since cos125π=cos75∘=sin15∘=sin12π, the value equals cos125π.
✓Final answer(C) cos125π
ANSWER: (C)
- GUJCET 2019Set 171 markMCQQ.Matrix Ar=[rr−1r−1r]; r=1,2,3,… If ∑r=1100∣Ar∣=(10)K, then K=; (∣Ar∣=det(Ar)). (A) 6 (B) 4 (C) 2 (D) 8
›Reveal solutionSolution
∣Ar∣=r2−(r−1)2=2r−1; the sum of the first 100 odd numbers is 1002=10000=(10)8, so K=8.
Concept: ∣Ar∣=rr−1−˚1r=r2−(r−1)2=2r−1.
∑r=1100(2r−1)=1002=10000=104=(10)8
So K=8.
✓Final answer(D) 8
ANSWER: (D)
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.If f(θ)=cosθsinθ−sinθ−cosθ, then f(6π)= ______.(a) −21(b) 21(c) 23(d) −23
›Reveal solutionSolution
Evaluate the 2×2 determinant, simplify with a double-angle identity, then substitute.
f(θ)=cosθ(−cosθ)−(−sinθ)(sinθ)=−cos2θ+sin2θ=−cos2θ.
f(6π)=−cos3π=−21.
✓Final answerThe correct option is (a) −21.
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