Let A, B=A−I, ω=(√3i−1)/2: Count n With Aⁿ+(ωB)ⁿ=A+B

JEE Main 2022 — 25 July, Evening Shift. Previous Year Question.

Problem

Let $A = \begin{pmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{pmatrix}$ and $B = A – I$. If $\omega = \dfrac{\sqrt{3}i-1}{2}$, then the number of elements in the set

$$\left\{\, n \in \{1,2,\dots,100\} : A^n + (\omega B)^n = A+B \,\right\}$$

is equal to _.

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Approach

Everything turns on the structure of $B=A-I$. Once $B$’s own algebra is pinned down — what $B^2$ equals in terms of $B$ itself — both $A^n$ and $B^n$ collapse into simple closed forms, and the matrix equation reduces to a plain scalar condition on $n$. Since $\omega$ is a complex cube root of unity, its powers cycle with period 3, which is what turns that scalar condition into a residue condition on $n$.

Solution

Step 1 — Find $B$ and spot its structure

$$B = A – I = \begin{pmatrix} 1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{pmatrix}$$

Every row of $B$ is identical, so $B$ has rank 1. That’s the structural clue: a rank-1 matrix built from a single row vector always satisfies a clean self-multiplication rule.

Step 2 — Compute $B^2$

Multiplying $B$ by itself (each entry of $B^2$ is a row of $B$ dotted with a column of $B$, and every row is $(1,-1,-1)$):

$$(1)(1)+(-1)(1)+(-1)(1) = -1$$

so every entry of $B^2$ is $-1$ times the corresponding entry of $B$:

$$B^2 = -B$$

Step 3 — $A$ is idempotent

Since $B = A – I$, substitute into $B^2=-B$:

$$(A-I)^2 = -(A-I) \implies A^2 – 2A + I = -A + I \implies A^2 = A$$

So $A$ is idempotent, which means $A^n = A$ for every $n \geq 1$ — raising it to any power just gives $A$ back.

Step 4 — Powers of $B$

From $B^2=-B$: $B^3 = B^2\cdot B = -B\cdot B = -B^2 = B$, and the pattern repeats with period 2:

$$B^n = (-1)^{n-1}B \quad \text{for } n \geq 1$$

Step 5 — Reduce the original equation

Substituting $A^n=A$ into $A^n+(\omega B)^n = A+B$ cancels $A$ from both sides:

$$\omega^n B^n = B \implies \omega^n(-1)^{n-1}B = B$$

Since $B$ is a genuine nonzero matrix, this forces the scalar coefficient to equal 1:

$$\omega^n(-1)^{n-1} = 1$$

Step 6 — Identify $\omega$ and use its periodicity

$\omega = \dfrac{\sqrt{3}i-1}{2} = \cos 120^\circ + i\sin 120^\circ$ is a primitive complex cube root of unity: $\omega^3=1$, and $\omega,\omega^2$ are not real. The right-hand side of $\omega^n(-1)^{n-1}=1$ is real, so $\omega^n$ must be real too — which only happens when $3\mid n$, giving $\omega^n=1$. The condition then becomes:

$$(-1)^{n-1}=1 \implies n \text{ is odd}$$

So $n$ must be an odd multiple of 3.

Step 7 — Count valid $n$ from 1 to 100

Odd multiples of 3 up to 100: $3, 9, 15, \dots, 99$ — an arithmetic sequence with common difference 6.

$$\frac{99-3}{6}+1 = 17$$

(Checked directly too: substituting $n=3$ and $n=9$ back into $\omega^n(-1)^{n-1}=1$ both give exactly $1$; $n=1,2,6$ all fail, matching $n\equiv 0\pmod 3$ and $n$ odd.)

Answer

Common mistakes

• Assuming $A$ and $B$ commute without checking, when combining $A^n$ and $(\omega B)^n$ — matrix multiplication isn’t commutative in general, and this problem only works cleanly because $A=I+B$ and $B$ commute with themselves, not because of a general assumption.
• Forgetting that $\omega^n$ must be real before comparing it to the real right-hand side. It’s tempting to solve $\omega^n(-1)^{n-1}=1$ by taking logs or guessing residues, but the fast route is noticing $\omega^n$ can only be real when $n$ is a multiple of 3 — that alone kills two-thirds of the cases immediately.

Practise next

• Let $M$ be a matrix whose every row equals the same row vector $r$. Show that $M^2 = (r\cdot\mathbf{1})\,M$, where $\mathbf{1}$ is the all-ones vector, and say what that makes $M$ when $r\cdot\mathbf{1}$ is $1$, and when it is $0$.