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ISC 2027
All chaptersMaths · Unit 2

Determinants

5 articles20 formulas26 ways the board asks it
MATEvaluating Determinants

Evaluating Determinants & Using Properties to Prove Results

This subtopic covers expanding determinants and, more importantly, using row/column properties to simplify a determinant before expansion and to prove standard identities. The key idea is that operations like Ri→Ri+kRjR_i \to R_i + kR_j leave a determinant unchanged, letting you create zeros or extract common factors so that a 3×33\times3 proof collapses neatly.

ISC examines this with proof-type questions where elegant use of properties is rewarded over brute-force expansion.

Expansion along a row
Δ=∣a1b1c1a2b2c2a3b3c3∣=a1C11+b1C12+c1C13\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = a_1 C_{11} + b_1 C_{12} + c_1 C_{13}
Cij=(−1)i+jMijC_{ij} = (-1)^{i+j} M_{ij} is the cofactor and MijM_{ij} the minor of the element in row ii, column jj.
Interchange and equal rows
Ri↔Rj⇒Δ→−Δ;Ri=Rj⇒Δ=0R_i \leftrightarrow R_j \Rightarrow \Delta \to -\Delta;\qquad R_i = R_j \Rightarrow \Delta = 0
Swapping two rows (or columns) changes the sign; two identical or proportional rows make Δ=0\Delta = 0.
Scalar factor of a line
∣ka1kb1kc1a2b2c2a3b3c3∣=k∣a1b1c1a2b2c2a3b3c3∣\begin{vmatrix} ka_1 & kb_1 & kc_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = k\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}
A common factor kk from any one row or column can be taken outside the determinant.
Invariance property (key tool)
Ri→Ri+k Rj(i≠j)⇒Δ unchangedR_i \to R_i + k\,R_j \quad(i \ne j) \Rightarrow \Delta \text{ unchanged}
Adding a multiple of one row to another (similarly for columns) does not alter Δ\Delta; used to create zeros or common factors.
Vandermonde-type result
∣1aa21bb21cc2∣=(a−b)(b−c)(c−a)\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = (a-b)(b-c)(c-a)
A standard ISC identity proved by R1→R1−R2, R2→R2−R3R_1 \to R_1 - R_2,\ R_2 \to R_2 - R_3 and factorising.
  • Expand along the row or column containing the most zeros to minimise arithmetic.
  • ∣A∣=∣AT∣|A| = |A^T|: a determinant is unchanged by transposing, so every row property has a matching column property.
  • To prove an identity, use Ri→Ri+kRjR_i \to R_i + kR_j to create a row/column of zeros or a common factor, take that factor out, then expand the simplified determinant.
  • Adding all rows (or columns), e.g. C1→C1+C2+C3C_1 \to C_1 + C_2 + C_3, often produces a common factor like (a+b+c)(a+b+c) in symmetric problems.
  • If after operations two rows become identical or proportional, Δ=0\Delta = 0 immediately, which proves many 'show =0=0' results without expansion.
  • Difference operations R1→R1−R2R_1 \to R_1 - R_2 create factors like (a−b)(a-b), the basis of the Vandermonde proof.
  • Apply operations to rows OR columns within a single step, and write each step explicitly for full marks.
Where the marks go
  • Applying Ri→kRiR_i \to kR_i (scaling a row in place) thinking Δ\Delta is unchanged; this actually multiplies Δ\Delta by kk.
  • Forgetting the alternating sign (−1)i+j(-1)^{i+j} in cofactors, which flips the sign of the whole expansion.
  • Doing two row operations that reference each other simultaneously instead of one at a time, leading to wrong values.
  • Stopping at a factored form that does not match the required target (e.g. leaving −(a−b)-(a-b) when (a−b)(a-b) is needed) by mishandling the sign on a row interchange.
How the board asks it
  • Derive / proveadding all columns to extract a common factor
    Using properties of determinants, prove that ∣a+b+2cabcb+c+2abcac+a+2b∣=2(a+b+c)3\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} = 2(a+b+c)^3.
  • Derive / provedifference operations and the vandermonde-type result
    Using properties of determinants, prove that ∣111abca2b2c2∣=(a−b)(b−c)(c−a)\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (a-b)(b-c)(c-a).
  • Derive / provetwo identical or proportional rows giving zero
    Without expanding, show that ∣1ab+c1bc+a1ca+b∣=0\begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{vmatrix} = 0.
  • NumericalC1→C1+C2+C3C_1 \to C_1 + C_2 + C_3 producing a common factor
    Using properties of determinants, evaluate ∣x+42x2x2xx+42x2x2xx+4∣\begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} and express it as a product of linear factors.
  • Numericalreducing a symmetric determinant to a factored equation in xx
    Using properties of determinants, find the value(s) of xx for which ∣x222x222x∣=0\begin{vmatrix} x & 2 & 2 \\ 2 & x & 2 \\ 2 & 2 & x \end{vmatrix} = 0.

Practise this topic

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.