Let A=(\beginarrayrrr2 & -1 & -1 1 & 0 & -1 1 & -1 & 0\endarray) and B=A-I . If \omega=\frac\sqrt3 i-12 , then the number of elements in the \operatornameset\n \in\1,2, \ldots, 100\: A^n+(\omega B)^n=A+B\ is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2022 · 25 Jul · Shift 1 · Q36 with Solution

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Correct answer: 17

Given matrices

$A=\begin{pmatrix}2&-1&-1\\1&0&-1\\1&-1&0\end{pmatrix},\qquad B=A-I$

So $B=\begin{pmatrix}1&-1&-1\\1&-1&-1\\1&-1&-1\end{pmatrix}.$

We need the number of integers $n\in\{1,2,\dots,100\}$ such that $A^n+(\omega B)^n=A+B,$ where $\omega=\frac{\sqrt3 i-1}{2}=-\frac12+\frac{\sqrt3}{2}i.$

Notice that $\omega=e^{2\pi i/3}$ which is a primitive cube root of unity, so $\omega^3=1,\qquad 1+\omega+\omega^2=0.$

Find simple algebraic relations for $A$ and $B$

Let us compute $A^2$ .

Using matrix multiplication,

\begin{pmatrix} 2&-1&-1\\ 1&0&-1\\ 1&-1&0 \end{pmatrix} \begin{pmatrix} 2&-1&-1\\ 1&0&-1\\ 1&-1&0 \end{pmatrix} = \begin{pmatrix} 2&-1&-1\\ 1&0&-1\\ 1&-1&0 \end{pmatrix}=A.$$ Hence $$A^2=A.$$ So $A$ is idempotent, and therefore for every $n\ge 1$, $$A^n=A.$$ Now compute $B^2$. Since all rows of $B$ are equal to $(1,-1,-1)$, $$B=\begin{pmatrix}1\\1\\1\end{pmatrix}(1,-1,-1).$$ Thus if $u=\begin{pmatrix}1\\1\\1\end{pmatrix}$ and $v=(1,-1,-1)$, then $B=uv$ and $$B^2=(uv)(uv)=u(vu)v.$$ Now $$vu=1-1-1=-1.$$ Therefore $$B^2=-uv=-B.$$ So $$B^2=-B.$$ This gives a power pattern: $$B^1=B, \quad B^2=-B, \quad B^3=B, \quad B^4=-B,\dots$$ Hence for all $n\ge1$, $$B^n=(-1)^{n-1}B.$$ Therefore $$(\omega B)^n=\omega^n B^n=\omega^n(-1)^{n-1}B.$$ --- 3. **Substitute into the required equation** We want $$A^n+(\omega B)^n=A+B.$$ Using $A^n=A$, $$A+\omega^n(-1)^{n-1}B=A+B.$$ So $$\omega^n(-1)^{n-1}B=B.$$ Since $B\neq 0$, this requires $$\omega^n(-1)^{n-1}=1.$$ Thus the problem reduces to solving $$(-1)^{n-1}\omega^n=1.$$ --- 4. **Solve the scalar equation** Since $-1=e^{i\pi}$ and $\omega=e^{2\pi i/3}$, we can inspect by parity. ### Case 1: $n$ odd Then $(-1)^{n-1}=1$, so we need $$\omega^n=1.$$ Because $\omega$ has order $3$, $$\omega^n=1 \iff 3\mid n.$$ Thus for odd $n$, solutions are $$n\equiv 3 \pmod{6}.$$ ### Case 2: $n$ even Then $(-1)^{n-1}=-1$, so we need $$-\omega^n=1\iff \omega^n=-1.$$ But powers of $\omega$ are only $$1,\omega,\omega^2,$$ and none of these equals $-1$. So there are **no even solutions**. Hence the only solutions are $$n\equiv 3\pmod{6}.$$ --- 5. **Count such integers from 1 to 100** The numbers are $$3,9,15,\dots,99.$$ This is an arithmetic progression with first term $3$, last term $99$, common difference $6$. Number of terms $$=\frac{99-3}{6}+1=\frac{96}{6}+1=16+1=17.$$ --- 6. **Final answer** The required number of elements is $$\boxed{17}.$$ --- 7. **Comparison with stored answer** Stored correct answer = $17$. Our derived answer also equals $17$, so it agrees.

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View written solutionFree

Correct answer: 17

Given matrices

$A=\begin{pmatrix}2&-1&-1\\1&0&-1\\1&-1&0\end{pmatrix},\qquad B=A-I$

So $B=\begin{pmatrix}1&-1&-1\\1&-1&-1\\1&-1&-1\end{pmatrix}.$

We need the number of integers $n\in\{1,2,\dots,100\}$ such that $A^n+(\omega B)^n=A+B,$ where $\omega=\frac{\sqrt3 i-1}{2}=-\frac12+\frac{\sqrt3}{2}i.$

Notice that $\omega=e^{2\pi i/3}$ which is a primitive cube root of unity, so $\omega^3=1,\qquad 1+\omega+\omega^2=0.$

Find simple algebraic relations for $A$ and $B$

Let us compute $A^2$ .

Using matrix multiplication,

\begin{pmatrix} 2&-1&-1\\ 1&0&-1\\ 1&-1&0 \end{pmatrix} \begin{pmatrix} 2&-1&-1\\ 1&0&-1\\ 1&-1&0 \end{pmatrix} = \begin{pmatrix} 2&-1&-1\\ 1&0&-1\\ 1&-1&0 \end{pmatrix}=A.$$ Hence $$A^2=A.$$ So $A$ is idempotent, and therefore for every $n\ge 1$, $$A^n=A.$$ Now compute $B^2$. Since all rows of $B$ are equal to $(1,-1,-1)$, $$B=\begin{pmatrix}1\\1\\1\end{pmatrix}(1,-1,-1).$$ Thus if $u=\begin{pmatrix}1\\1\\1\end{pmatrix}$ and $v=(1,-1,-1)$, then $B=uv$ and $$B^2=(uv)(uv)=u(vu)v.$$ Now $$vu=1-1-1=-1.$$ Therefore $$B^2=-uv=-B.$$ So $$B^2=-B.$$ This gives a power pattern: $$B^1=B, \quad B^2=-B, \quad B^3=B, \quad B^4=-B,\dots$$ Hence for all $n\ge1$, $$B^n=(-1)^{n-1}B.$$ Therefore $$(\omega B)^n=\omega^n B^n=\omega^n(-1)^{n-1}B.$$ --- 3. **Substitute into the required equation** We want $$A^n+(\omega B)^n=A+B.$$ Using $A^n=A$, $$A+\omega^n(-1)^{n-1}B=A+B.$$ So $$\omega^n(-1)^{n-1}B=B.$$ Since $B\neq 0$, this requires $$\omega^n(-1)^{n-1}=1.$$ Thus the problem reduces to solving $$(-1)^{n-1}\omega^n=1.$$ --- 4. **Solve the scalar equation** Since $-1=e^{i\pi}$ and $\omega=e^{2\pi i/3}$, we can inspect by parity. ### Case 1: $n$ odd Then $(-1)^{n-1}=1$, so we need $$\omega^n=1.$$ Because $\omega$ has order $3$, $$\omega^n=1 \iff 3\mid n.$$ Thus for odd $n$, solutions are $$n\equiv 3 \pmod{6}.$$ ### Case 2: $n$ even Then $(-1)^{n-1}=-1$, so we need $$-\omega^n=1\iff \omega^n=-1.$$ But powers of $\omega$ are only $$1,\omega,\omega^2,$$ and none of these equals $-1$. So there are **no even solutions**. Hence the only solutions are $$n\equiv 3\pmod{6}.$$ --- 5. **Count such integers from 1 to 100** The numbers are $$3,9,15,\dots,99.$$ This is an arithmetic progression with first term $3$, last term $99$, common difference $6$. Number of terms $$=\frac{99-3}{6}+1=\frac{96}{6}+1=16+1=17.$$ --- 6. **Final answer** The required number of elements is $$\boxed{17}.$$ --- 7. **Comparison with stored answer** Stored correct answer = $17$. Our derived answer also equals $17$, so it agrees.

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