Q.Evaluate the determinant Δ=1−14231400.
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.
Concept: Determinant Evaluation Using Identities – we can expand along a row/column with zeros to simplify.
Step 1: Notice the third column has two zeros (at positions a23 and a33). Expanding along column 3 is efficient.
Step 2: The determinant is
Δ=4⋅(−1)1+3−1431+0+0
Step 3: Compute the 2×2 determinant:
(−1)(1)−(3)(4)=−1−12=−13
Step 4: Multiply: 4×(−13)=−52
The value is −52.
The determinant is found by expanding along the third column, which has two zeros, making the calculation trivial. The value is Δ=4×(−13)=−52.
The key insight here is not to blindly apply the full 3×3 formula. Instead, look for rows or columns with zeros — they make expansion much faster. In this determinant, the third column has two zeros (in the second and third rows). That means only one term survives when we expand along that column.
Let’s walk through it.
-
Choose the best expansion path.
The third column is (4,0,0)T. Expanding along this column means we multiply each entry by its cofactor and sum. Since the second and third entries are zero, only the first entry (4) contributes.
-
Write the expansion.
Expanding along column 3:
Δ=4⋅C13+0⋅C23+0⋅C33
where C13 is the cofactor of the entry in row 1, column 3.
- Find the cofactor C13. The cofactor is (−1)1+3=(−1)4=1 times the minor M13. The minor is the determinant of the 2×2 matrix left after deleting row 1 and column 3:
M13=−1431
Compute this:
M13=(−1)(1)−(3)(4)=−1−12=−13
So C13=1×(−13)=−13.
- Finish the calculation.
Δ=4×(−13)=−52
A common mistake is to forget the sign factor (−1)i+j when computing the cofactor. Here, i+j=1+3=4, which is even, so the sign is positive — but always check.
Whenever a row or column has two or more zeros, expand along it. It reduces the work to a single 2×2 determinant (or even simpler). This is a standard trick in JEE and board exams.
The value of the determinant is −52.
Method: Expansion Along the Row or Column with the Most Zeros
This method evaluates a 3×3 (or larger) determinant efficiently by choosing to expand along whichever row or column already contains the most zero entries, so most of the cofactor terms vanish automatically.
Steps
Step 1: Scan every row and column for zeros
Before expanding along the default first row, check every row and column of the determinant — the one with the most zeros needs the least computation.
Step 2: Choose that row/column for the expansion
If a column (or row) has two zero entries, only one cofactor term survives — the other two vanish because they're multiplied by 0.
Step 3: Write the expansion for the single surviving term
Δ=aij⋅Cij,Cij=(−1)i+jMij
where Mij is the 2×2 minor left after deleting row i and column j.
Step 4: Get the sign right
Compute (−1)i+j carefully — even i+j gives +, odd gives −. This is the step students most often get wrong.
Step 5: Evaluate the surviving 2×2 minor and multiply
Apply ad−bc to the minor, then multiply by the nonzero entry and its sign.
Always scan for zeros before committing to an expansion row — it turns a 3×3 (or bigger) determinant into a single 2×2 calculation whenever the matrix has that structure.
Common Mistakes
Mistake 1: Getting the cofactor sign wrong when expanding along the third column
Why it's wrong: the sign attached to the surviving term is (−1)i+j for its position, not always +1 — picking the wrong sign flips the final answer's sign (here it happens to be + since 1+3=4 is even, but this must be checked, not assumed). Correct approach: explicitly compute (−1)i+j for the exact row/column of the nonzero entry being expanded, every time.
Mistake 2: Expanding along the first row out of habit instead of scanning for zeros first
Why it's wrong: expanding along row 1 here requires evaluating three separate 2×2 minors instead of just one, tripling the arithmetic and the chances of a slip. Correct approach: always scan every row and column for zeros before choosing where to expand — here column 3 (with two zeros) is far faster.
Showing the 12 most recent of 14 on this concept.
- 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 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 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 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 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 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 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.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 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 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.
🎓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.