The Special Families

Why this section exists. NCERT stops at symmetric, skew-symmetric and invertible. JEE Main goes further: idempotent, involutory, nilpotent and orthogonal matrices, the trace, and $A^n$ pattern problems appear nearly every session. This section catalogues the toolkit.

The four families

Standard specimens (all verified by direct multiplication): $$A = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix} \text{ is idempotent}, \qquad B = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix} \text{ satisfies } B^3 = O$$ $$C = \frac{1}{3}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{pmatrix} \text{ is orthogonal}, \qquad \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \text{ is orthogonal for every } \theta$$

The trace

Definition. For a square matrix, $\operatorname{tr}(A) = \sum a_{ii}$, the sum of the diagonal entries.

$$\operatorname{tr}(A + B) = \operatorname{tr}A + \operatorname{tr}B, \qquad \operatorname{tr}(kA) = k\operatorname{tr}A, \qquad \boxed{\operatorname{tr}(AB) = \operatorname{tr}(BA)}$$

The boxed identity holds even when $AB \neq BA$ — and it powers a famous JEE-style impossibility: no matrices satisfy $AB - BA = I$, because the left side has trace $0$ while $\operatorname{tr}(I_n) = n \neq 0$.

Common mistakes to avoid

Mistake 1 — confusing the families: $A^2 = A$ (idempotent) vs $A^2 = I$ (involutory) — one letter apart, wholly different behaviour.

Mistake 2 — inverting a nilpotent matrix: if $A^k = O$ and $A^{-1}$ existed, multiplying $A^k = O$ by $(A^{-1})^k$ would give $I = O$ — contradiction. Nilpotent matrices are never invertible.

Mistake 3 — checking orthogonality with $A^2 = I$: the test is $AA' = I$; only symmetric orthogonal matrices also satisfy $A^2 = I$.

Powers and Matrix Equations

$A^n$ by pattern and induction

Pattern 1 — the shear/triangular family: $$\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}^n = \begin{pmatrix} a^n & n\,a^{n-1}b \\ 0 & a^n \end{pmatrix}$$ (check: $\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}^3 = \begin{pmatrix} 8 & 12 \\ 0 & 8 \end{pmatrix}$, and $3 \cdot 2^2 \cdot 1 = 12$ ✓).

Pattern 2 — the all-ones family: $J = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ satisfies $J^2 = 2J$, hence by induction $$J^n = 2^{n-1}J$$ More generally any matrix with $A^2 = cA$ has $A^n = c^{n-1}A$ — one relation collapses all powers.

Pattern 3 — rotations: $F(\theta)^n = F(n\theta)$ (angles add), so powers of rotation matrices cycle; e.g. $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ has period 4.

The method: compute $A^2$ (maybe $A^3$), conjecture the pattern, prove it by induction with one multiplication. JEE often just wants the conjecture evaluated at a specific $n$.

Matrix equations and inverses without determinants

If a square matrix satisfies a polynomial equation with nonzero constant term, its inverse falls out by algebra alone.

Template. Suppose $A^2 - 4A + I = O$. Then $$I = 4A - A^2 = A(4I - A) \implies A^{-1} = 4I - A$$

Worked instance. $A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$: direct computation gives $A^2 = \begin{pmatrix} 7 & 12 \\ 4 & 7 \end{pmatrix} = 4A - I$, so $$A^{-1} = 4I - A = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}$$ (matching the inverse verified by multiplication in the Invertible Matrices section). Higher powers also collapse: $A^3 = A \cdot A^2 = A(4A - I) = 4A^2 - A = 4(4A - I) - A = 15A - 4I$ — every power of $A$ is a linear combination of $A$ and $I$.

Common mistakes to avoid

Mistake 1 — conjecturing from one power. Compute at least $A^2$ and $A^3$ before trusting a pattern.

Mistake 2 — the equation trick without a constant term: $A^2 - 4A = O$ gives $A(A - 4I) = O$, which does NOT yield an inverse (indeed such an $A$ with $A \neq 4I$ is a zero divisor, hence singular).

Mistake 3 — treating $(I + A)^n$ with the binomial theorem when $A$ does not commute with the other term. With $I$ it is legal ($I$ commutes with everything); with two general matrices it is not.

JEE-Pattern Worked Examples

Example 1 — Certifying an idempotent

Show that $A = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}$ is idempotent.

Step 1 — compute $A^2$ row by row: e.g. the $(1,1)$ entry is $4 + 2 - 4 = 2$, the $(1,2)$ entry $-4 - 6 + 8 = -2$; continuing, every entry of $A^2$ reproduces $A$.

Answer: $A^2 = A$ — idempotent, and hence $A^{2026} = A$ as well: idempotency freezes all powers.

Example 2 — Certifying a nilpotent

Show that $B = \begin{pmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{pmatrix}$ is nilpotent of index 3.

Step 1 — square: $$B^2 = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\ -1 & -1 & -3 \end{pmatrix} \neq O$$

Step 2 — cube: $B^3 = B^2 \cdot B = O$ (each row of $B^2$ is a multiple of $(1, 1, 3)$, which $B$ annihilates on the left).

Answer: $B^3 = O$ with $B^2 \neq O$ — nilpotent of index exactly 3, and therefore not invertible.

Example 3 — Involutory reflections

Show that $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ is involutory, and evaluate $A^{57}$.

Step 1 — square: $A^2 = I$ (swapping twice restores order).

Step 2 — reduce the exponent mod 2: $A^{57} = (A^2)^{28} \cdot A = A$.

Answer: $A^{57} = A$ — involutory powers alternate $I, A, I, A, \ldots$; only the exponent's parity matters.

Example 4 — Certifying an orthogonal matrix

Show that $C = \dfrac{1}{3}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{pmatrix}$ is orthogonal.

Step 1 — compute $CC'$: each diagonal entry is $\frac{1}{9}(1 + 4 + 4) = 1$; each off-diagonal entry is $\frac{1}{9}(2 + 2 - 4) = 0$ (and its mates likewise).

Step 2 — conclude: $CC' = I$, and $C'C = I$ by the same computation.

Answer: orthogonal — the rows are unit vectors, mutually perpendicular; and for free, $C^{-1} = C'$: no inversion work needed.

Example 5 — Trace kills $AB - BA = I$

Prove that no square matrices $A, B$ of order $n$ satisfy $AB - BA = I$.

Step 1 — trace both sides, using $\operatorname{tr}(AB) = \operatorname{tr}(BA)$: $$\operatorname{tr}(AB - BA) = \operatorname{tr}(AB) - \operatorname{tr}(BA) = 0$$

Step 2 — but $\operatorname{tr}(I_n) = n \neq 0$ — contradiction.

Answer: impossible — a one-line trace argument disposing of what looks like a hard existence question; the single most elegant trace application in JEE.

Example 6 — Inverse from a matrix equation

For $A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$, show $A^2 - 4A + I = O$ and hence find $A^{-1}$.

Step 1 — verify the equation: $A^2 = \begin{pmatrix} 7 & 12 \\ 4 & 7 \end{pmatrix}$ and $4A - I = \begin{pmatrix} 7 & 12 \\ 4 & 7 \end{pmatrix}$ — equal ✓.

Step 2 — factor the identity out: $$I = 4A - A^2 = A(4I - A) \implies A^{-1} = 4I - A = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}$$

Answer: $A^{-1} = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}$ — inverses without determinants, provided the constant term is nonzero.

Example 7 — Collapsing powers with $J^2 = 2J$

For $J = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, prove $J^n = 2^{n-1}J$ and evaluate $J^{10}$.

Step 1 — base: $J^1 = 2^0 J$ ✓.

Step 2 — induction: $J^{n+1} = J^n J = 2^{n-1}J \cdot J = 2^{n-1}(2J) = 2^n J$.

Step 3 — evaluate: $J^{10} = 2^9 J = \begin{pmatrix} 512 & 512 \\ 512 & 512 \end{pmatrix}$.

Answer: as above — one relation $J^2 = 2J$ turns exponentiation into scalar bookkeeping.

Example 8 — The triangular power pattern

Conjecture and prove a formula for $\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}^n$.

Step 1 — experiment: the square is $\begin{pmatrix} a^2 & 2ab \\ 0 & a^2 \end{pmatrix}$, the cube $\begin{pmatrix} a^3 & 3a^2 b \\ 0 & a^3 \end{pmatrix}$ — conjecture $\begin{pmatrix} a^n & n\,a^{n-1}b \\ 0 & a^n \end{pmatrix}$.

Step 2 — induction: $$\begin{pmatrix} a^n & n a^{n-1} b \\ 0 & a^n \end{pmatrix}\begin{pmatrix} a & b \\ 0 & a \end{pmatrix} = \begin{pmatrix} a^{n+1} & a^n b + n a^n b \\ 0 & a^{n+1} \end{pmatrix} = \begin{pmatrix} a^{n+1} & (n+1)a^n b \\ 0 & a^{n+1} \end{pmatrix}$$

Answer: proved — the derivative-like factor $n a^{n-1}$ in the corner is the signature of this family (check: $\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}^3 = \begin{pmatrix} 8 & 12 \\ 0 & 8 \end{pmatrix}$ ✓).