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UPSC NDA II 2024 Mathematics Paper

Question Numbers: 49-50

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}

Question : 50 of 120

Marks: +1, -0

What is

A^{-1}

equal to?

$\begin{bmatrix}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{bmatrix}$
$\begin{bmatrix}\frac{1}{2} & -\frac{1}{2} & 0 \\ -1 & \frac{3}{2} & -2 \\ -1 & \frac{3}{2} & -\frac{3}{2}\end{bmatrix}$
$\begin{bmatrix}2 & -2 & 0 \\ -4 & 6 & -8 \\ -4 & 6 & -6\end{bmatrix}$
$\begin{bmatrix}\frac{1}{5} & -\frac{1}{5} & 0 \\ -\frac{2}{5} & \frac{3}{5} & -\frac{4}{5} \\ -\frac{2}{5} & \frac{3}{5} & -\frac{3}{5}\end{bmatrix}$

Solution:

Concept:

The inverse of a matrix

A

is given by

A^{-1} = \frac{1}{|A|} \text{adj}(A)

Explanation:

We are given

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

It is known (from the problem context) that

|A| = 1

Therefore,

A^{-1} = \frac{1}{1} \cdot \text{adj}(A) = \text{adj}(A)

A^{-1}

equals the given adjoint matrix.

Answer:

Option A:

\begin{bmatrix}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{bmatrix}

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