जर A= \(\rm \begin{bmatrix} 2 &1 \\ 1& 2 \end{bmatrix}\), f(x) = x2 - 2x - I असेल, तर f(A) चे मूल्य काय आहे?

  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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Detailed Solution

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संकल्पना:

f(x) = a0 xn + axn-1 + a2 xn-2 + ... + an-1 x + an हे बहुपदी आहे असे समजा आणि सारणी A ही काेटी n ची चौरस सारणी आहे असे समजा. तर,

f(A) = a0 An + a1 An-1 + a2 An-2 + ... + an-1 A + an In  याला सारणी बहुपदी म्हणतात.

गणना:

दिलेले आहे, A = \(\rm \begin{bmatrix} 2 &1 \\ 1& 2 \end{bmatrix}\) आणि  f(x) = x2 - 2x - I . 

सारणी A ही काेटी 2 असलेली चौरस सारणी आहे, तर ती दिलेल्या बहुपदीला समाधान देते

⇒ f(A) = A2 - 2A - I 

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

योग्य पर्याय 1 आहे.

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