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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Matrices

NDA 2021 · Mathematics · Matrices

If A= [ array ll 1 & a 0 & 1 array ] where a ∈ ℕ , then what is A^...

If A=A=\left[1a01\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right]wherewherea \in \mathbb{N},thenwhatis, then what isA^{100}-A^{50}-2 A^{25}equalto?whereequal to? whereI is the identity matrix.

Options

  1. A−2I-2 I
  2. B−I-I
  3. C2I2 I
  4. DII

Correct answer

A. −2I-2 I

Step-by-step solution

A=[1a01]⇒A2=[12a01]⇒A3=[13a01]⇒An=[1na01]\begin{aligned} & A=\left[\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right] \Rightarrow A^2=\left[\begin{array}{cc}1 & 2 a \\ 0 & 1\end{array}\right] \Rightarrow A^3=\left[\begin{array}{cc}1 & 3 a \\ 0 & 1\end{array}\right] \\ & \Rightarrow A^n=\left[\begin{array}{cc}1 & n a \\ 0 & 1\end{array}\right]\end{aligned} Now, A100−A50−2A25=[1100a01]−[150a01]−2[125a01]=[−200−2]=−2I.\begin{aligned} & \text{ Now, } A^{100}-A^{50}-2 A^{25} \\ & =\left[\begin{array}{cc}1 & 100 a \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}1 & 50 a \\ 0 & 1\end{array}\right]-2\left[\begin{array}{cc}1 & 25 a \\ 0 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}-2 & 0 \\ 0 & -2\end{array}\right]=-2 I.\end{aligned}
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