Q.The determinant b2−abab−a2bc−acb−ca−bc−abc−acb2−abab−a2 equals
(A) abc(b−c)(c−a)(a−b)
(B) (b−c)(c−a)(a−b)
(C) (a+b+c)(b−c)(c−a)(a−b)
(D) None of these
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🔒 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: …
Concept — spot the column dependence. Factor (b−a) from columns 1 and 3:
C1=b(b−a)a(b−a)c(b−a)=(b−a)bac,C3=(b−a)cba,
while C2=b−ca−bc−a.
Now
C1−C3=(b−a)b−ca−bc−a=(b−a)C2, …
Columns 1 and 3 each carry a factor (b−a), and C1−C3=(b−a)C2 — a linear dependence among the columns — so the determinant is identically 0: option (D).
The idea
Every entry factors neatly, exposing a linear relation between the three columns. When columns are linearly dependent, the determinant is zero — no expansion needed.
Step 1 — Factor the columns
Δ=b2−abab−a2bc−acb−ca−bc−abc−acb2−abab−a2.
Column 1 entries are b(b−a),a(b−a),c(b−a), so C1=(b−a)(b,a,c)T. Column 3 entries are c(b−a),b(b−a),a(b−a), so C3=(b−a)(c,b,a)T. Column 2 is C2=(b−c,a−b,c−a)T.
Step 2 — Find the relation
Subtract:
C1−C3=(b−a)[(b,a,c)T−(c,b,a)T]=(b−a)(b−c,a−b,c−a)T=(b−a)C2.
So C1−C3−(b−a)C2=0: the columns are linearly dependent.
Step 3 — Conclude …
Method: Detecting Linear Dependence Among Rows or Columns
A determinant is zero whenever its rows or columns satisfy a linear relationship — spotting this relationship evaluates the determinant instantly, without any term-by-term expansion.
Steps
Step 1: Factor each row/column to expose its structure
Look at each row or column and factor out any common term from its entries, rewriting each as a scalar multiple of a simpler vector.
Step 2: Test simple combinations for a relation
Check whether one column (or row) minus another equals a scalar multiple of a third — e.g. compute C1−C3 and see if it matches λ⋅C2 for some expression λ.
Step 3: Conclude the determinant is identically zero
If a genuine linear relation among the rows/columns is found, the determinant equals zero for every value of the variables involved — no arithmetic expansion is required. …
Common Mistakes
Mistake 1: Attempting full term-by-term expansion instead of checking for column dependence
Why it's wrong: Directly expanding this 3×3 determinant, with entries like b2−ab and bc−ac, produces many terms and is highly error-prone; the entries are specifically constructed so that C1−C3=(b−a)C2. Correct approach: before expanding, check whether one column can be written as a combination of the others — if so, the determinant is immediately zero with no arithmetic needed.
Mistake 2: Confusing "zero for a specific numeric check" with "identically zero for all a,b,c" …
Showing the 12 most recent of 14 on this concept.
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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.
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›Reveal solutionSolution
Split the last column into 1 and x³; the determinant factors as (Vandermonde)(1+xyz).
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›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 …
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›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: …
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›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: …
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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. …
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›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).
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- GUJCET 2020Set 071 markMCQQ.Let f(t)=cost2tanttantttt12tt. Then limt→0t2f(t) is equal to ________. (A) 3 (B) 1 (C) −1 (D) 0
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Concept. Evaluate each 2×2 determinant ad−bc.
Steps.
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…
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