The Properties NCERT Dropped — and JEE Kept

Older editions of the textbook carried a full section on determinant properties. The current edition dropped it, but JEE Main did not — property-based questions appear nearly every session, and the properties also make many evaluations near-instant. This corner restores the toolkit.

Throughout, "row" statements hold equally for columns, because of the first property.

P1 — Transpose invariance. $\vert A' \vert = \vert A \vert$. Whatever is true for rows is true for columns.

P2 — Swap rule. Interchanging any two rows (or columns) multiplies the determinant by $-1$.

P3 — Repetition collapse. If two rows are identical or proportional, the determinant is $0$. (Swap the two identical rows: the value must both stay the same and change sign — so it is $0$.)

P4 — Single-row scaling. Multiplying one row by $k$ multiplies the determinant by $k$. Applied to all $n$ rows, this recovers $\vert kA \vert = k^n\vert A \vert$.

P5 — Split rule. If each entry of one row is a sum of two terms, the determinant splits into a sum of two determinants (that row separated, all other rows repeated).

P6 — The workhorse. The operation $R_i \to R_i + kR_j$ (add a multiple of another row) does not change the determinant. This is the property behind almost every elegant evaluation: create zeros, extract common factors, collapse to triangular form.

The strategy

A property-based evaluation almost always follows the same script:

1. Look for a constant sum — if every row (or column) sums to the same expression, do $C_1 \to C_1 + C_2 + C_3$ and pull the common factor out (P4).
2. Create zeros with $R_i \to R_i - R_j$ (P6).
3. Watch for proportional rows at every stage — the moment two appear, the answer is $0$ (P3).

A determinant that looks like a ten-minute expansion usually collapses in three lines under this script.

The Product and Adjoint Family

The second half of the toolkit is a cluster of one-line formulas that JEE Main tests directly. For $A$, $B$ square of order $n$, with $A$ nonsingular where inverses appear:

For $n = 3$ the last three read: $\vert \operatorname{adj} A \vert = \vert A \vert^2$, $\;\operatorname{adj}(\operatorname{adj} A) = \vert A \vert A$, $\;\vert \operatorname{adj}(\operatorname{adj} A) \vert = \vert A \vert^4$.

Two structural zero/one facts

Skew-symmetric, odd order. If $A' = -A$ and $n$ is odd, then $$\vert A \vert = \vert A' \vert = \vert -A \vert = (-1)^n \vert A \vert = -\vert A \vert \;\Rightarrow\; \boxed{\vert A \vert = 0}$$ Every $3 \times 3$ skew-symmetric determinant is $0$ — no expansion needed. (Even order escapes: $\begin{vmatrix} 0 & 1 \\ -1 & 0 \end{vmatrix} = 1$.)

Orthogonal matrices. If $AA' = I$, taking determinants gives $\vert A \vert^2 = 1$, so $$\boxed{\vert A \vert = \pm 1}$$ Rotation matrices have $\vert A \vert = 1$; reflections have $\vert A \vert = -1$.

These two facts, plus the table above, settle a large share of JEE determinant one-liners before any arithmetic begins.

Worked Examples

Example 1 — Constant-sum collapse

Evaluate $\begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{vmatrix}$.

Step 1 — spot the constant sum: in every row, the last two entries add to $a + b + c$. Do $C_3 \to C_3 + C_2$: $$\begin{vmatrix} 1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c \end{vmatrix}$$

Step 2 — extract and collapse: pull out $(a+b+c)$ from $C_3$ (P4); columns $C_1$ and $C_3$ are now identical, so the determinant is $0$ (P3).

Answer: $0$ — this is also why the points $(a, b+c)$, $(b, c+a)$, $(c, a+b)$ from Section 2 are collinear.

Example 2 — Rows that sum to zero

Evaluate $\begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix}$.

Step 1 — add all columns: each row's three entries sum to $(a-b) + (b-c) + (c-a) = 0$, so $C_1 \to C_1 + C_2 + C_3$ creates a column of zeros.

Step 2 — expand along the zero column: every term carries a factor $0$.

Answer: $0$ — one glance at the telescoping sum finishes a determinant that brute expansion would take half a page over.

Example 3 — Difference rows create zeros

Evaluate $\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix}$.

Step 1 — subtract the first row (P6): $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: $$\begin{vmatrix} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & y \end{vmatrix}$$

Step 2 — triangular: the determinant is the diagonal product.

Answer: $xy$.

Example 4 — The Vandermonde pattern

Prove that $\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (a-b)(b-c)(c-a)$.

Step 1 — create differences: $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$: $$\begin{vmatrix} 0 & 0 & 1 \\ a-b & b-c & c \\ a^2-b^2 & b^2-c^2 & c^2 \end{vmatrix}$$

Step 2 — extract factors (P4): pull $(a-b)$ from $C_1$ and $(b-c)$ from $C_2$: $$(a-b)(b-c)\begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & c \\ a+b & b+c & c^2 \end{vmatrix}$$

Step 3 — expand along $R_1$: the single term is $1 \cdot \left[(b+c) - (a+b)\right] = c - a$.

Answer: $(a-b)(b-c)(c-a)$ — the memorable "product of differences", and the parent of a whole family of JEE items.

Example 5 — A symmetric classic

Prove that $\begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix} = 4abc$.

Step 1 — one subtraction up front (P6): $R_1 \to R_1 - R_2 - R_3$: $$\begin{vmatrix} 0 & -2c & -2b \\ b & c+a & b \\ c & c & a+b \end{vmatrix}$$

Step 2 — expand along $R_1$: $$2c\begin{vmatrix} b & b \\ c & a+b \end{vmatrix} - 2b\begin{vmatrix} b & c+a \\ c & c \end{vmatrix} = 2c\left[ab + b^2 - bc\right] - 2b\left[bc - c^2 - ac\right]$$

Step 3 — simplify: $2abc + 2b^2c - 2bc^2 - 2b^2c + 2bc^2 + 2abc = 4abc$. ∎

Answer: $4abc$.

Example 6 — Quick-fire with the formula table

$A$ is a nonsingular matrix of order $3$ with $\vert A \vert = -2$. Find $\vert 2A \vert$, $\vert A^2 \vert$, $\vert \operatorname{adj} A \vert$, $\det(A^{-1})$ and $\vert \operatorname{adj}(\operatorname{adj} A) \vert$.

Step 1 — apply each formula: $$\vert 2A \vert = 2^3(-2) = -16, \qquad \vert A^2 \vert = (-2)^2 = 4, \qquad \vert \operatorname{adj} A \vert = (-2)^2 = 4$$ $$\det(A^{-1}) = -\tfrac{1}{2}, \qquad \vert \operatorname{adj}(\operatorname{adj} A) \vert = (-2)^{(3-1)^2} = (-2)^4 = 16$$

Answer: $-16$, $4$, $4$, $-\frac{1}{2}$, $16$ — five JEE one-liners from one number.

Example 7 — Skew-symmetric of odd order

Show without expanding that $\begin{vmatrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{vmatrix} = 0$.

Step 1 — identify the structure: the matrix satisfies $A' = -A$ — skew-symmetric — and its order $3$ is odd.

Step 2 — the parity argument: $\vert A \vert = \vert A' \vert = \vert -A \vert = (-1)^3\vert A \vert = -\vert A \vert$, forcing $\vert A \vert = 0$.

Answer: $0$, with no expansion — the argument works for every odd order.

Example 8 — Orthogonal determinants

If $A$ is a square matrix with $AA' = I$, prove that $\det A = \pm 1$.

Step 1 — take determinants of both sides: $\vert AA' \vert = \vert I \vert = 1$.

Step 2 — apply the product rule and P1: $\vert A \vert \vert A' \vert = \vert A \vert^2 = 1$.

Answer: $\det A = \pm 1$. ∎ (The orthogonal specimen $\frac{1}{3}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{pmatrix}$ from the Matrices JEE Corner has determinant exactly $1$ — worth checking once by hand.)