যদি A= \(\rm \begin{bmatrix} 2 &1 \\ 1& 2 \end{bmatrix}\) , f(x) = x 2 - 2x - I হয়, তাহলে f(A) এর মান কী হবে?

  1. \(\rm \begin{bmatrix} 0 &2 \\ 2& 0 \end{bmatrix}\)
  2. \(\rm \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}\)
  3. \(\rm \begin{bmatrix} 1 &2 \\ 2& 1 \end{bmatrix}\)
  4. \(\rm \begin{bmatrix} 1 &1 \\ 1& 1 \end{bmatrix}\)

Answer (Detailed Solution Below)

Option 1 : \(\rm \begin{bmatrix} 0 &2 \\ 2& 0 \end{bmatrix}\)
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অনুসৃত ধারণা:

ধরা যাক, f(x) = a0 xn + axn-1 + a2 xn-2 + ... + an-1 x + an একটি বহুপদী এবং ধরি ম্যাট্রিক্স A n ক্রমের একটি বর্গ ম্যাট্রিক্স, তাহলে,

f(A) = a0 An + a1 An-1 + a2 An-2 + ... + an-1 A + an In কে ম্যাট্রিক্স বহুপদী বলা হয়।

গণনা:

প্রদত্ত , A = \(\rm \begin{bmatrix} 2 &1 \\ 1& 2 \end{bmatrix}\) এবং f(x) = x 2 - 2x - I

ম্যাট্রিক্স A হল ক্রম 2 এর বর্গ ম্যাট্রিক্স, তাহলে এটি প্রদত্ত বহুপদকে সন্তুষ্ট করছে 

⇒ f(A) = A2 - 2A - I

⇒ f(A) = AA - 2A - I

⇒ f(A) = \(\rm \begin{bmatrix} 2 &1 \\ 1& 2 \end{bmatrix}\) \(\rm \begin{bmatrix} 2 &1 \\ 1& 2 \end{bmatrix}\) - 2 \(\rm \begin{bmatrix} 2 &1 \\ 1& 2 \end{bmatrix}\) - \(\rm \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}\)

⇒ f(A) = \(\rm \begin{bmatrix} 5 &4 \\ 4& 5 \end{bmatrix}\) - \(\rm \begin{bmatrix} 4 &2 \\ 2& 4 \end{bmatrix}\) - \(\rm \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}\)

⇒ f(A) = \(\rm \begin{bmatrix} 1 &2 \\ 2& 1 \end{bmatrix}\) - \(\rm \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}\)

⇒ f(A) = \(\rm \begin{bmatrix} 0 &2 \\ 2& 0 \end{bmatrix}\)

সঠিক বিকল্প হল 1

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