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GUJCET 2025 Math Paper

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Question : 18 of 40

Marks: +1, -0

If

A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}

(A+I)^3 + (A-I)^3 = \dots

$8A$
$8I$
$6A$
$6I$

Solution:

Concept:

For a matrix

A

with

A^2=I

, all odd powers equal

A

and even powers equal

I

.

The identity matrix

I

commutes with

A

, so binomial expansion is valid.

Explanation:

Given

A = \begin{bmatrix}0&1\\1&0\end{bmatrix}

.

First find

A^2

:

A^2 = \begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}=I

.

Hence

A^3 = A^2A = IA = A

.

Now expand

(A+I)^3

:

(A+I)^3 = A^3 + 3A^2I + 3AI^2 + I^3 = A + 3I + 3A + I = 4A+4I

.

Similarly, expand

(A-I)^3

:

(A-I)^3 = A^3 - 3A^2I + 3AI^2 - I^3 = A - 3I + 3A - I = 4A-4I

.

Add the two expressions:

(4A+4I)+(4A-4I)=8A

.

Answer:

Therefore,

(A+I)^3+(A-I)^3=8A

, which matches option A.

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