MCQ
If $A$ is a square matrix such that $A^2 = A$, then $(I + A)^3– 7A$ is equal to:
- A$A$
- B$I - A$
- ✓$I$
- D$3A$
✓
Answer
Correct option: C.
$I$
Given : $A^2 = A...(i)$
Multiplying both sides by $A, A^3 = A^2= A \ [$From eq. $(i)]...(ii)$
Also given $(I + A)^3 – 7A = I^3+ A^3 + 3I^2A + 3IA^2 – 7A$
Putting $A^2 = A \ [$from eq. $(i)]$ and $A^3 = A \ [$From eq. $(ii)],$
$= I + A + 3IA + 3IA – 7A = I + A + 3A + 3A – 7A \ [\because IA = A]$
$= I + 7A – 7A$
$= I$
Multiplying both sides by $A, A^3 = A^2= A \ [$From eq. $(i)]...(ii)$
Also given $(I + A)^3 – 7A = I^3+ A^3 + 3I^2A + 3IA^2 – 7A$
Putting $A^2 = A \ [$from eq. $(i)]$ and $A^3 = A \ [$From eq. $(ii)],$
$= I + A + 3IA + 3IA – 7A = I + A + 3A + 3A – 7A \ [\because IA = A]$
$= I + 7A – 7A$
$= I$
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