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If A is a square matrix such that A2=AA^2=A and II is identity matrix of same order, then (2I+A)3−19A(2I+A)^3-19A is equal to:

Solution

✅ Correct Option: 1

Since A2=AA^2=A, we also have A3=AA^3=A.

Expanding, (2I+A)3=8I+12A+6A2+A3=8I+12A+6A+A=8I+19A(2I+A)^3=8I+12A+6A^2+A^3=8I+12A+6A+A=8I+19A.

Subtracting 19A19A leaves 8I8I.

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