Let A = ( \beginmatrix 1 + i & 1 - i & 0 \endmatrix ) where i = \sqrt - 1 . Then, the number of elements in the set n \in 1, 2, ......, 100 : An = A is \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2022 · 28 Jun · Shift 2 · Q47 with Solution

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Correct answer: 25

We need to find the number of integers $n \in \{1,2,\dots,100\}$ such that $A^n=A,$ where $A=\begin{pmatrix}1+i&1\\-i&0\end{pmatrix}.$
Rewrite the condition: $A^n=A \iff A(A^{n-1}-I)=0.$ A direct power computation would be tedious, so we first study the matrix algebraically.
Find the characteristic polynomial of $A$ :

=\begin{vmatrix}1+i-\lambda&1\\-i&-\lambda\end{vmatrix}.$$ So, $$\chi_A(\lambda)=(1+i-\lambda)(-\lambda)-1(-i) =-\lambda(1+i-\lambda)+i.$$ Expanding, $$\chi_A(\lambda)=\lambda^2-(1+i)\lambda+i.$$ Factor it: $$\lambda^2-(1+i)\lambda+i=(\lambda-1)(\lambda-i).$$ Hence the eigenvalues are $1$ and $i$, which are distinct. Therefore $A$ is diagonalizable. 4. Since $A$ is diagonalizable with eigenvalues $1$ and $i$, it is similar to $$\begin{pmatrix}1&0\\0&i\end{pmatrix}.$$ Thus $$A^n \sim \begin{pmatrix}1^n&0\\0&i^n\end{pmatrix}=egin{pmatrix}1&0\\0&i^n\end{pmatrix}.$$ For $A^n=A$, the corresponding eigenvalues must match, so we need $$1^n=1 \quad \text{and} \quad i^n=i.$$ The first is always true. The second gives $$i^n=i \iff i^{n-1}=1.$$ Now powers of $i$ are periodic with period $4$, so $$i^{n-1}=1 \iff n-1 \equiv 0 \pmod{4} \iff n\equiv 1 \pmod{4}.$$ 5. Count such $n$ in $\{1,2,\dots,100\}$: $$n=1,5,9,\dots,97.$$ This is an arithmetic progression with first term $1$, last term $97$, common difference $4$. Number of terms: $$\frac{97-1}{4}+1=\frac{96}{4}+1=24+1=25.$$ 6. Therefore, the required number of integers is $$\boxed{25}.$$ 7. Comparison with stored answer: Stored correct answer = $25$, which matches our derived answer.

View written solutionFree

Correct answer: 25

We need to find the number of integers $n \in \{1,2,\dots,100\}$ such that $A^n=A,$ where $A=\begin{pmatrix}1+i&1\\-i&0\end{pmatrix}.$
Rewrite the condition: $A^n=A \iff A(A^{n-1}-I)=0.$ A direct power computation would be tedious, so we first study the matrix algebraically.
Find the characteristic polynomial of $A$ :

=\begin{vmatrix}1+i-\lambda&1\\-i&-\lambda\end{vmatrix}.$$ So, $$\chi_A(\lambda)=(1+i-\lambda)(-\lambda)-1(-i) =-\lambda(1+i-\lambda)+i.$$ Expanding, $$\chi_A(\lambda)=\lambda^2-(1+i)\lambda+i.$$ Factor it: $$\lambda^2-(1+i)\lambda+i=(\lambda-1)(\lambda-i).$$ Hence the eigenvalues are $1$ and $i$, which are distinct. Therefore $A$ is diagonalizable. 4. Since $A$ is diagonalizable with eigenvalues $1$ and $i$, it is similar to $$\begin{pmatrix}1&0\\0&i\end{pmatrix}.$$ Thus $$A^n \sim \begin{pmatrix}1^n&0\\0&i^n\end{pmatrix}=egin{pmatrix}1&0\\0&i^n\end{pmatrix}.$$ For $A^n=A$, the corresponding eigenvalues must match, so we need $$1^n=1 \quad \text{and} \quad i^n=i.$$ The first is always true. The second gives $$i^n=i \iff i^{n-1}=1.$$ Now powers of $i$ are periodic with period $4$, so $$i^{n-1}=1 \iff n-1 \equiv 0 \pmod{4} \iff n\equiv 1 \pmod{4}.$$ 5. Count such $n$ in $\{1,2,\dots,100\}$: $$n=1,5,9,\dots,97.$$ This is an arithmetic progression with first term $1$, last term $97$, common difference $4$. Number of terms: $$\frac{97-1}{4}+1=\frac{96}{4}+1=24+1=25.$$ 6. Therefore, the required number of integers is $$\boxed{25}.$$ 7. Comparison with stored answer: Stored correct answer = $25$, which matches our derived answer.