Q.Evaluate xx−1x+1x.
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 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: …
The key idea is to evaluate a 2×2 determinant directly using the formula acbd=ad−bc.
Step 1: Identify the entries: a=x, b=x+1, c=x−1, d=x.
Step 2: Apply the formula: …
This determinant simplifies to a constant 1 by direct expansion — the x2 terms cancel, leaving only x2−(x2−1)=1.
The determinant of a 2×2 matrix (acbd) is ad−bc. That’s the only formula you need here — no row operations or factorisations required. The trick is to expand carefully and watch the signs.
-
Write down the determinant formula
For xx−1x+1x, we have a=x, b=x+1, c=x−1, d=x.
So the value is ad−bc=x⋅x−(x+1)(x−1).
-
Simplify the first term
x⋅x=x2.
-
Simplify the second term
(x+1)(x−1) is a difference of squares: (x+1)(x−1)=x2−1.
-
Subtract
x2−(x2−1)=x2−x2+1=1. …
Method: Evaluating a 2×2 Determinant with Algebraic Entries
This method evaluates a 2×2 determinant whose entries are algebraic expressions rather than plain numbers, using the same ad−bc formula but with careful expansion.
Steps
Step 1: Write down the formula with the algebraic entries substituted
acbd=ad−bc
Substitute the given expressions for a,b,c,d exactly as printed, keeping each in its own bracket.
Step 2: Expand each product separately
Expand ad and bc as separate algebraic products before subtracting — don't try to combine the subtraction and the expansion in one step, since that's where sign errors creep in.
Step 3: Recognize special product patterns
Watch for a difference-of-squares pattern like (x+1)(x−1)=x2−1 — spotting it saves a full expansion.
Step 4: Subtract and simplify …
Common Mistakes
Mistake 1: Failing to bracket (x+1)(x−1) before subtracting, leading to a sign error
Why it's wrong: writing the determinant as x⋅x−x+1⋅x−1 without parentheses around the product (x+1)(x−1) makes it easy to subtract only the first term of the product, not the whole product — turning x2−(x2−1)=1 into an incorrect expression. Correct approach: always keep the product bc in its own bracket, i.e. write ad−(bc), before expanding anything.
Mistake 2: Not recognizing the difference-of-squares pattern in (x+1)(x−1) …
Showing the 12 most recent of 14 on this concept.
- 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. …
- 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: …
- 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 …
- 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: …
- 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, …
- 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 …
- 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 …
- 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. …
- 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. …
- 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. …
- 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. …
- 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.
…
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.