Q.Using the properties of determinants, evaluate: 0x2yx2zxy20zy2xz2yz20
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 entry carries factors of x,y,z in a pattern; pull them out of the columns and then the rows, leaving a plain numerical determinant.
Step 1 — factor x from C1, y from C2, z from C3:
Δ=xyz0xyxzxy0yzxzyz0.
Step 2 — factor x from R1, y from R2, z from R3:
Δ=x2y2z20xxy0yzz0.
Step 3 — expand along the first row:
0xxy0yzz0=−y(0−zx)+z(xy−0)=xyz+xyz=2xyz.
So Δ=x2y2z2⋅2xyz=2x3y3z3.
2x3y3z3
Pulling x,y,z from the three columns and then from the three rows leaves a small determinant equal to 2xyz, so the value is 2x3y3z3.
Intuition
Each entry is a single monomial in x,y,z, so instead of a brute expansion we strip the common factors out of every column and every row. Each strip multiplies out front, and the leftover determinant is tiny.
Setting up
Δ=0x2yx2zxy20zy2xz2yz20.
Working the steps
1. Factor the columns. Column 1 has common factor x, column 2 has y, column 3 has z:
Δ=xyz0xyxzxy0yzxzyz0.
2. Factor the rows. Now row 1 has common factor x, row 2 has y, row 3 has z:
Δ=xyz⋅xyz0xxy0yzz0=x2y2z20xxy0yzz0.
3. Expand the small determinant along the first row:
0xxy0yzz0=0−y(0⋅0−z⋅x)+z(x⋅y−0⋅x)=−y(−zx)+z(xy)=2xyz.
4. Multiply back:
Δ=x2y2z2⋅2xyz=2x3y3z3.
Check with x=1, y=2, z=3: the original determinant is 02340129180=432, and 2⋅13⋅23⋅33=432. ✓
2x3y3z3
Method: Factoring Common Variables Out of Rows and Columns Before Expanding
This method applies to determinants whose entries are monomials sharing common variable factors across rows and/or columns — instead of expanding a messy 3×3 directly, you strip out every shared factor first, leaving a tiny, easy determinant.
Steps
Step 1: Factor a common term out of each column
Scan each column for a variable common to every entry in it (treating a 0 entry as compatible with any factor) and pull it out in front of the determinant, dividing every entry in that column by the factor as you do:
Δ=(column factors)×⋯.
Step 2: Repeat for rows if a further common factor remains
After factoring columns, check whether each row of what's left also shares a common variable. If so, factor that out too — it's legitimate to factor rows and columns in sequence, as long as each factor is multiplied back in outside the determinant.
Step 3: Expand the small remaining determinant
What's left after two rounds of factoring is usually a determinant with simple 0s and single variables — expand this by cofactor expansion along whichever row/column has the most zeros.
Step 4: Multiply every factored term back together
Combine all the factors pulled out in Steps 1–2 with the value of the small determinant from Step 3 to get the final answer, and sanity-check by plugging in small numeric values for the variables into both the original and final expressions.
Common Mistakes
Mistake 1: Attempting a direct cofactor expansion instead of factoring first
Why it's wrong: expanding this 3×3 determinant directly (without first pulling x,y,z out of the columns and rows) means juggling six degree-5 monomial terms at once, which is slow and highly error-prone. Correct approach: always scan for a common monomial factor in each column (and then each row) before expanding — here x,y,z factor cleanly from the three columns, then again from the three rows.
Mistake 2: Mixing up which factor belongs to which row or column
Why it's wrong: factoring happens in two separate passes (columns, then rows), and assigning the wrong variable to the wrong row/column in the second pass gives a wrong overall power of x, y, or z in the final answer. Correct approach: track each factoring step explicitly — column factors give xyz, and the row factors on the new matrix independently give another xyz, for a combined x2y2z2.
Mistake 3: Sign error in the small 3×3 expansion
Why it's wrong: expanding 0xxy0yzz0 along the first row involves a double-negative in the middle cofactor (−y(0⋅0−z⋅x)=−y(−zx)=+xyz), and dropping one of the two negative signs gives −2xyz or 0 instead of 2xyz. Correct approach: write out 0⋅0−z⋅x explicitly before applying the cofactor's minus sign, rather than combining the signs mentally.
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 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 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 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 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 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 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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