If \( A \) is square matrix such that \( \mathrm{A}^2=\mathrm{A} \), then \( (2\mathrm{I}+\mathrm{A})^4-65\mathrm{A} \) is equal to (Where I is identity matrix)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 16I
Since \( \mathrm{A}^2=\mathrm{A} \), every higher power is also A: \( \mathrm{A}^3=\mathrm{A}^2\mathrm{A}=\mathrm{A}^2=\mathrm{A} \), and \( \mathrm{A}^4=\mathrm{A} \).
I and A commute, so the binomial theorem applies:
\( \displaystyle (2\mathrm{I}+\mathrm{A})^4=16\mathrm{I}+4(8)\mathrm{A}+6(4)\mathrm{A}^2+4(2)\mathrm{A}^3+\mathrm{A}^4 \)
\( \displaystyle =16\mathrm{I}+32\mathrm{A}+24\mathrm{A}+8\mathrm{A}+\mathrm{A} \)
\( \displaystyle =16\mathrm{I}+65\mathrm{A} \)
\( \displaystyle (2\mathrm{I}+\mathrm{A})^4-65\mathrm{A}=16\mathrm{I} \)
Shortcut: The result must hold for every idempotent matrix, so try A = O: \( (2\mathrm{I})^4-\mathrm{O}=16\mathrm{I} \). Only Option (2) gives 16I.
Tip: For an idempotent matrix (\( \mathrm{A}^2=\mathrm{A} \)), \( (x\mathrm{I}+\mathrm{A})^n=x^n\mathrm{I}+\left[(x+1)^n-x^n\right]\mathrm{A} \); here \( 3^4-2^4=65 \).