Q.Find the value of the following:
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.
(i) This is the 3×3 identity matrix; a diagonal determinant is the product of the diagonal entries: 1⋅1⋅1=1.
(ii) Expand along the first row of 1300514−12 (its middle entry is 0):
151−12−0+43051=1(10+1)+4(3−0)=11+12=23.
- 1;
- 23.
The identity determinant is 1; the second determinant expands to 23.
A 3×3 determinant can be expanded along any row or column, using the sign checkerboard +−+−+−+−+. Choosing a row or column that contains zeros saves work.
(i)
The matrix is the identity: 1's on the diagonal and 0's everywhere else. A diagonal (in fact triangular) determinant is the product of the diagonal entries, so the value is 1⋅1⋅1=1.
(ii)
1300514−12
Expand along row 1; the 0 in the middle kills that term:
151−12−0⋅(…)+43051.
The minors are 51−12=10−(−1)=11 and 3051=3−0=3.
So the value is 1(11)+4(3)=11+12=23.
- 1;
- 23.
Method: Recognising Special Matrix Structure Before Expanding
A time-saving check to run before committing to a full cofactor expansion.
Steps
Step 1: Check for special structure first
- Identity or diagonal matrix: determinant is the product of the diagonal entries (instantly).
- Triangular matrix: same shortcut — product of the diagonal entries.
Step 2: If no shortcut applies, choose the row/column with the most zeros
Fewer nonzero entries means fewer cofactor terms to compute.
Step 3: Expand using cofactors, skipping zero entries entirely
Δ=∑jaijCij,
where any term with aij=0 contributes nothing and can be omitted from the sum without computing its minor.
Step 4: Compute the remaining 2×2 minors and assemble the answer
Add up the nonzero contributions, tracking the (−1)i+j sign for each.
Common Mistakes
Mistake 1: Expanding the identity matrix's determinant the "long way" via full cofactor expansion
Why it's wrong: this wastes time and adds unnecessary arithmetic when the identity (or any diagonal) matrix's determinant is immediately 1 by the product-of-diagonal shortcut. Correct approach: check for identity/diagonal/triangular structure first and read the determinant off instantly when it applies.
Mistake 2: Still computing the 2×2 minor for a term whose coefficient is 0
Why it's wrong: multiplying a computed minor by 0 always gives 0, so working out that minor is wasted effort that only increases the chance of an unrelated arithmetic slip elsewhere. Correct approach: when expanding along a row/column with a zero entry, skip that term's minor entirely and move straight to the nonzero terms.
- 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. The cofactor C21 is given by
C21=(−1)2+1⋅M21=(−1)3⋅4=−4.
TipA quick check: the sign pattern for a 3×3 matrix starts with + in the top-left, so row 2, column 1 is a “−” position. So the cofactor is simply the negative of the minor.
Watch outA common mistake is to forget the sign and just give the minor (4), which is not among the options, or to accidentally use the value of the element itself. The cofactor is purely a function of position and the other entries.
✓Final answerThe correct option is (C).
ANSWER: C
- 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:
Δ=cos(α+β)M11+(−sin(α+β))(−M12)+cos2β⋅M13.
Δ=cos2(α+β)+sin2(α+β)+cos2β=1+cos2β.
The result depends only on β; every α term has cancelled.
✓Final answerΔ=1+cos2β, which is independent of α. Correct option: (D).
ANSWER: D
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