Concept 04 / 09Matrices and Determinants
Properties of Determinants and Techniques for Evaluating Them
The effect of row and column operations, splitting a determinant along a row, factor-theorem evaluation of symmetric determinants, the product of determinants in row-by-row form, and determinants whose entries are functions (differentiation and summation).
Evaluating determinants by brute-force expansion is slow and error-prone. JEE rewards using properties to simplify first. Every property below is stated for rows, and each holds equally for columns, because .
1. Interchanging two rows changes the sign of the determinant. Consequently, a determinant with two identical rows is (swapping them changes the sign but not the determinant).
2. Multiplying one row by multiplies the determinant by . So a common factor of a row can be taken outside. And a determinant with two proportional rows is .
3. Adding a multiple of one row to another row, , does not change the determinant. This is the workhorse operation: use it to create zeros, then expand along the row or column with the zeros.
4. Linearity in each row: if every entry of one row is a sum of two terms, the determinant splits into the sum of two determinants, each containing one of the terms in that row and the other rows unchanged. Only one row is split at a time. Splitting two rows gives four determinants, not two.
5. If every entry of a row is , the determinant is .
A caution about operations applied simultaneously: , , all at once is not allowed; it replaces the matrix by one whose rows sum to zero and the determinant becomes incorrectly. Apply operations one after another, each using the current rows.
Standard strategies.
Row or column sums equal: if all rows (or columns) have the same sum, apply to make the first column constant, take the common factor out, then subtract rows to create zeros. For example, .
Factor theorem: treat the determinant as a polynomial in one variable. If setting makes two rows equal, then is a factor. Find all such factors, compare degrees, and determine the remaining constant by a single convenient substitution.
Recognise a product: a determinant may be the product of two simpler determinants. The product rule can be applied in 'row-by-row' form: since , the entry of the product may be formed as row of dotted with row of . For instance, a determinant with entries can be written as a product of two determinants built from and .
Determinants of functions. If the entries are functions of , the derivative of the determinant is the sum of determinants, each obtained by differentiating one row and leaving the others alone. This follows from the product rule applied to each term of the expansion. Similarly, if only one row depends on a summation index , then is the determinant with that row replaced by the sums of its entries, and the other rows unchanged (linearity in a single row).
Key Takeaways:
- Swap two rows: sign changes. Two equal or proportional rows: determinant .
- Factor out a common multiple from a single row.
- leaves the determinant unchanged; use it to create zeros.
- A determinant splits along one row at a time.
- Equal row sums: add all columns into one and take out the common sum.
- Factor theorem: equal rows when means is a factor.
- = sum of determinants with one row differentiated at a time.
- Pitfall: applying cyclic row operations simultaneously, which makes every determinant appear to be .
Brain-Teaser 1: Evaluate .
Solution: Subtract row 1 from rows 2 and 3: gives and gives . Now expand along the third column: the entries are . The determinant is , with signs at position and at . The first minor is , and the second is . Total: . In symmetric form, when . Creating zeros before expanding turned a messy expansion into two small determinants.
Brain-Teaser 2: If where is constant, find at .
Solution: Only the first row depends on , so differentiating three times differentiates only that row: . At the first row is . Expanding along it, the only non-zero term is at position with sign : . So . Recognising that a constant row stays fixed means you never need to expand the original determinant at all.
Brain-Teaser 3: Show that for all non-negative real .
Solution: Apply : the first column becomes in every row. Taking it out, the determinant is . Subtract row 1 from rows 2 and 3, then expand along the first column: . Simplifying the bracket gives . So the determinant is , which is when . Equality holds when or . The sign becomes obvious only after the determinant is factored into a sum and a sum of squares.
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