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MHTCET 25 Apr 2025 Shift 1 Paper

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Question : 120 of 150

Marks: +1, -0

If

A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]_{3 \times 3}

A^{-1}=

A
$A^{2}$
$A^{3}$
$A^{4}$

Solution:

Concept:

For a non-singular matrix,

A^{-1} = \frac{\operatorname{adj} A}{|A|}

, and the correct power of

A

is identified by comparing matrices.

Explanation:

First, compute the determinant:

|A|=\left|\begin{array}{ccc}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right|=1 \neq 0

So

A^{-1}

exists.

The cofactor matrix of

A

is:

\left[\begin{array}{ccc}1 & -2 & -2 \\ -1 & 3 & 3 \\ 0 & -4 & -3\end{array}\right]

Taking its transpose gives:

\operatorname{adj} A = \left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]

Since

|A|=1

,

A^{-1}=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]

Now compute powers of

A

:

A^2 = A \cdot A = \left[\begin{array}{ccc}3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3\end{array}\right]

A^3 = A^2 \cdot A = \left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]

This matches

A^{-1}

exactly.

Answer:

A^{-1} = A^3

, so the correct option is C.

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