CUET UG 2025 — Mathematics Algebra

Given that $A$A is a square matrix, $I$I is the identity matrix of the same order, and $A^2 = I$A2=I.

To find: $(A - I)^3 + (A + I)^3 - 3A$(A−I)3+(A+I)3−3A

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Expanding $(A - I)^3$(A−I)3:

$$(A - I)^3 = A^3 - 3A^2I + 3AI^2 - I^3$$(A−I)3=A3−3A2I+3AI2−I3

Since $AI = IA = A$AI=IA=A, $I^2 = I$I2=I, and $I^3 = I$I3=I:

$$(A - I)^3 = A^3 - 3A^2 + 3A - I$$(A−I)3=A3−3A2+3A−I

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Expanding $(A + I)^3$(A+I)3:

$$(A + I)^3 = A^3 + 3A^2I + 3AI^2 + I^3$$(A+I)3=A3+3A2I+3AI2+I3

$$(A + I)^3 = A^3 + 3A^2 + 3A + I$$(A+I)3=A3+3A2+3A+I

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Adding both expansions:

$$(A - I)^3 + (A + I)^3 = (A^3 - 3A^2 + 3A - I) + (A^3 + 3A^2 + 3A + I)$$(A−I)3+(A+I)3=(A3−3A2+3A−I)+(A3+3A2+3A+I)

$$= 2A^3 + 6A$$=2A3+6A

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Substituting into the original expression:

$$(A - I)^3 + (A + I)^3 - 3A = 2A^3 + 6A - 3A$$(A−I)3+(A+I)3−3A=2A3+6A−3A

$$= 2A^3 + 3A$$=2A3+3A

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Using the condition $A^2 = I$A2=I:

$$A^3 = A \times A^2$$A3=A×A2

$$= A \times I$$=A×I

$$= A$$=A

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Therefore:

$$2A^3 + 3A = 2A + 3A$$2A3+3A=2A+3A

$$= 5A$$=5A

The answer is $5A$5A.