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Complex Numbers question

2021 · 25 Jul · Shift 1 · Q45
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  5. /2021 · 25 Jul · Shift 1 · Q45

Complex Numbers question

2021 · 25 Jul · Shift 1 · Q45

JEE MainMathematicsComplex NumbersNumerical+4 / −1
Let S={n∈N∣(0i10)n(abcd)=(abcd)∀a,b,c,d∈R}S = \left\{ {n \in N\left| {{{\left( {\begin{matrix} 0 & i \\ 1 & 0 \\ \end{matrix} } \right)}^n}\left( {\begin{matrix} a & b \\ c & d \\ \end{matrix} } \right) = \left( {\begin{matrix} a & b \\ c & d \\ \end{matrix} } \right)\forall a,b,c,d \in R} \right.} \right\}S={n∈N​(01​i0​)n(ac​bd​)=(ac​bd​)∀a,b,c,d∈R}, where i = −1\sqrt { - 1}−1​. Then the number of 2-digit numbers in the set S is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 11

  1. Let A=(0i10),M=(abcd).A=\begin{pmatrix}0&i\\1&0\end{pmatrix},\qquad M=\begin{pmatrix}a&b\\c&d\end{pmatrix}.A=(01​i0​),M=(ac​bd​).

We need all natural numbers nnn such that AnM=M∀a,b,c,d∈R.A^n M = M \quad \forall a,b,c,d\in \mathbb{R}.AnM=M∀a,b,c,d∈R.

Since this must hold for every matrix MMM, it implies An=I,A^n = I,An=I, because only the identity matrix leaves every 2×22\times 22×2 matrix unchanged under left multiplication.

So the problem reduces to finding all n∈Nn\in\mathbb Nn∈N such that ((0i10))n=I.\left(\begin{pmatrix}0&i\\1&0\end{pmatrix}\right)^n=I.((01​i0​))n=I.

  1. Compute powers of AAA.

First,

=\begin{pmatrix}i&0\\0&i\end{pmatrix}=iI.$$ Hence, $$A^4=(A^2)^2=(iI)^2=i^2I=-I,$$ and $$A^8=(A^4)^2=(-I)^2=I.$$ Thus the powers repeat with period dividing $8$. --- 3. Find exactly when $A^n=I$. Since $$A^2=iI,$$ we get $$A^{2k}=(A^2)^k=(iI)^k=i^k I.$$ For this to equal $I$, we need $$i^k=1 \iff k\equiv 0\pmod 4.$$ So $$2k\equiv 0\pmod 8 \implies n\equiv 0\pmod 8.$$ Now check odd powers: if $n=2k+1$, then $$A^{2k+1}=A\cdot A^{2k}=A\cdot i^k I=i^k A,$$ which can never be $I$ because $A$ is not a scalar multiple giving identity. Therefore, $$A^n=I \iff 8\mid n.$$ So, $$S=\{n\in\mathbb N: n\text{ is divisible by }8\}.$$ --- 4. Count 2-digit numbers in $S$. Two-digit multiples of $8$ are: $$16,24,32,40,48,56,64,72,80,88,96.$$ Number of such terms: $$\frac{96-16}{8}+1=11.$$ --- 5. Final answer: The number of 2-digit numbers in $S$ is $$\boxed{11}.$$ The derived answer matches the stored correct answer.
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