Evaluating Determinants & Using Properties to Prove Results

MATEvaluating Determinants

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 $R_i \to R_i + kR_j$ leave a determinant unchanged, letting you create zeros or extract common factors so that a $3\times3$ 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

\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}

C_{ij} = (-1)^{i+j} M_{ij}

is the cofactor and

M_{ij}

the minor of the element in row

i

, column

j

Interchange and equal rows

R_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

\Delta = 0

Scalar factor of a line

\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

k

from any one row or column can be taken outside the determinant.

Invariance property (key tool)

R_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

\Delta

; used to create zeros or common factors.

Vandermonde-type result

\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

R_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.
$|A| = |A^T|$ : a determinant is unchanged by transposing, so every row property has a matching column property.
To prove an identity, use $R_i \to R_i + kR_j$ 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. $C_1 \to C_1 + C_2 + C_3$ , often produces a common factor like $(a+b+c)$ in symmetric problems.
If after operations two rows become identical or proportional, $\Delta = 0$ immediately, which proves many 'show $=0$ ' results without expansion.
Difference operations $R_1 \to R_1 - R_2$ create factors like $(a-b)$ , 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 $R_i \to kR_i$ (scaling a row in place) thinking $\Delta$ is unchanged; this actually multiplies $\Delta$ by $k$ .
Forgetting the alternating sign $(-1)^{i+j}$ 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 $-(a-b)$ when $(a-b)$ 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 $\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} = 2(a+b+c)^3$ .
Derive / provedifference operations and the vandermonde-type result
Using properties of determinants, prove that $\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (a-b)(b-c)(c-a)$ .
Derive / provetwo identical or proportional rows giving zero
Without expanding, show that $\begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{vmatrix} = 0$ .
Numerical $C_1 \to C_1 + C_2 + C_3$ producing a common factor
Using properties of determinants, evaluate $\begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix}$ and express it as a product of linear factors.
Numericalreducing a symmetric determinant to a factored equation in $x$
Using properties of determinants, find the value(s) of $x$ for which $\begin{vmatrix} x & 2 & 2 \\ 2 & x & 2 \\ 2 & 2 & x \end{vmatrix} = 0$ .

Practise this topic

NextMinors, Cofactors & Adjoint

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.