Comprehension

निर्देश : निम्नलिखित प्रश्नों के लिए निम्नलिखित को ध्यान में रखें :  

माना \(\rm A=\begin{bmatrix}3&-3&4\\\ 2&-3&4\\\ 0&-1&1\end{bmatrix}\)

A-1 किसके बराबर है?

This question was previously asked in
NDA-II 2024 (Maths) Official Paper (Held On: 01 Sept, 2024)
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  1. \(\rm \begin{bmatrix}1&-1&0\\\ -2&3&-4\\\ -2&3&-3\end{bmatrix}\)
  2. \(\rm \begin{bmatrix}1/2&-1/2&0\\\ -1&3/2&-2\\\ -1&3/2&-3/2\end{bmatrix}\)
  3. \(\rm \begin{bmatrix}2&-2&0\\\ -4&6&-8\\\ -4&6&-6\end{bmatrix}\)
  4. \(\rm \begin{bmatrix}1/5&-1/5&0\\\ -2/5&3/5&-4/5\\\ -2/5&3/5&-3/5\end{bmatrix}\)

Answer (Detailed Solution Below)

Option 1 : \(\rm \begin{bmatrix}1&-1&0\\\ -2&3&-4\\\ -2&3&-3\end{bmatrix}\)
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NDA 01/2025: English Subject Test
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Detailed Solution

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व्याख्या:

दिया गया है

\(adj(A) = \rm \begin{bmatrix}1&-1&0\\\ -2&3&-4\\\ -2&3&-3\end{bmatrix}\)

अब, A-1 = \(\frac{1}{|A|} (Adj(A))\)

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

∴ विकल्प (a) सही है।

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