If A is a square matrix such that \( \mathrm{A^2=I} \), then \( \mathrm{(A+I)^3+(A-I)^3-7A} \) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \mathrm{A} \)
Since A and I commute, the binomial expansions hold:
\( (A+I)^3=A^3+3A^2+3A+I \)
\( (A-I)^3=A^3-3A^2+3A-I \)
Adding them, the even powers cancel:
\( (A+I)^3+(A-I)^3=2A^3+6A \)
Since \( A^2=I \), \( A^3=A\cdot A^2=A \). So
\( (A+I)^3+(A-I)^3-7A=2A+6A-7A \)
\( =A \)
NCERT Exemplar:
Matrices
