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MHTCET 18 Apr 2026 Shift 1 Paper

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Question : 115 of 150

Marks: +1, -0

Let

A=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]

B=\operatorname{adj} A

, then the matrix

B^{-1}

is equal to...

$I$
$A^{-1}$
$-A$
$A$

Solution:

Concept:

For any square matrix

A

, the relation between adjoint and inverse is

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

.

Explanation:

Given matrix

A

is a rotation matrix, so its determinant is:

|A| = (\cos^2 \alpha + \sin^2 \alpha)(1) = 1

.

Using

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

, we get:

B = \operatorname{adj} A = 1 \cdot A^{-1} = A^{-1}

.

Now take inverse on both sides:

B^{-1} = (A^{-1})^{-1} = A

.

Thus,

B^{-1}

is the matrix

A

itself.

Answer:

B^{-1} = A

Hence, the correct option is D. $A$ .

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