What is A^-1 equal to?

Question

Question

Direction : Consider the following for the items that follow :

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

This question was previously asked in

NDA-II 2024 (Maths) Official Paper (Held On: 01 Sept, 2024)

\(\rm \begin{bmatrix}1&-1&0\\\ -2&3&-4\\\ -2&3&-3\end{bmatrix}\)
\(\rm \begin{bmatrix}1/2&-1/2&0\\\ -1&3/2&-2\\\ -1&3/2&-3/2\end{bmatrix}\)
\(\rm \begin{bmatrix}2&-2&0\\\ -4&6&-8\\\ -4&6&-6\end{bmatrix}\)
\(\rm \begin{bmatrix}1/5&-1/5&0\\\ -2/5&3/5&-4/5\\\ -2/5&3/5&-3/5\end{bmatrix}\)

Answer 1

Explanation:

Given

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

Now, A^-1 = \(\frac{1}{|A|} (Adj(A))\)

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

∴ Option (a) is correct.

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