Q.Evaluate 111xx+yxyyx+y.
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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: Determinant Evaluation Using Identities (Row Operations)
Step 1: Apply R2→R2−R1 and R3→R3−R1:
100xy0y0x …
Subtracting the first row from the other two clears the first column and leaves a triangular-style determinant equal to xy.
We evaluate
Δ=111xx+yxyyx+y.
1. Row operations R2→R2−R1 and R3→R3−R1:
Δ=100xy0y0x. …
Method: Row-Elimination to Reach a Near-Triangular Determinant
This method evaluates a determinant by using one row (often one with simple entries like 1's) to zero out the corresponding column in the other rows, avoiding a full cofactor expansion.
Steps
Step 1: Look for a row or column that repeats a simple value
Here, the first column is (1,1,1)T — three identical, simple entries. A repeated column entry is the cue to eliminate it using row operations rather than expanding directly.
Step 2: Subtract the "reference" row from the others
Using operation 3 (Ri→Ri+λRj, value unchanged), subtract row 1 from every other row that shares the same first-column value:
R2→R2−R1,R3→R3−R1
This leaves the first column as (1,0,0)T.
Step 3: Expand along the now-simple first column …
Common Mistakes
Mistake 1: Subtracting the wrong row (e.g. R1→R1−R2 instead of R2→R2−R1)
Why it's wrong: row operations must keep the row being modified on the left of the arrow — swapping which row is the "reference" changes every entry's sign and can silently flip the final answer's sign. Correct approach: always keep row 1 (with its 1's in the first column) fixed as the reference, and only modify rows 2 and 3.
Mistake 2: Expanding the full 3×3 before simplifying …
- COMEDK 2024Set 2024-A1 markMCQQ.cos(α+β)sinα−cosα−sin(α+β)cosαsinαcos2βsinβcosβ is independent of (A) β (B) α and β (C) Neither α nor β (D) α
›Reveal solutionSolution
Expanding along the first row collapses the determinant to 1+cos2β, which contains no α — so it is independent of α: option (D).
Cofactor expansion along row 1
The three minors are
M11=cosαsinαsinβcosβ=cosαcosβ−sinαsinβ=cos(α+β),
M12=sinα−cosαsinβcosβ=sinαcosβ+cosαsinβ=sin(α+β),
M13=sinα−cosαcosαsinα=sin2α+cos2α=1.
With the cofactor sign pattern (+,−,+) and the row-1 entries cos(α+β), −sin(α+β), cos2β: …
- COMEDK 2025Set 2025-A1 markMCQQ.The cofactor of the element a21 in the expansion of Δ=1−32451492 is (A) 5 (B) −24 (C) −4 (D) −5
›Reveal solutionSolution
The cofactor of a21 is found by taking (−1)2+1 times the determinant of the submatrix obtained by deleting row 2 and column 1. The result is −4, so the correct option is (C).
The cofactor of an element in a matrix is not just the minor (the determinant of the submatrix left after removing that element’s row and column). It also includes a sign factor (−1)i+j, where i and j are the row and column indices. This sign alternates like a chessboard pattern. For a21 (row 2, column 1), the sign is negative because 2+1=3 is odd. So we compute the minor and then flip its sign.
- Identify the element and its position. The element a21 is in row 2, column 1. In the given matrix
Δ=1−32451492,
a21=−3. But the cofactor depends only on position, not on the value of the element itself.
- Delete row 2 and column 1. Removing row 2 and column 1 leaves the submatrix:
(4142).
- Compute the minor M21. The minor is the determinant of that 2×2 submatrix:
M21=4142=(4)(2)−(4)(1)=8−4=4.
- Apply the sign factor. …
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