Question

If $A$ is a square matrix such that $A^2$ = $A$ , then $(I-A{)}^3+A$ is equal to

• A. $A$
• B. $I-A$
• C. $I$
• D. $3A$

Answer 1

Answer: (C)

Given that, $A^2=A$
Then, $(I-A{)}^3+A$
$={\left(I\right)}^3+{\left(-A\right)}^3+3\left(I\right){\left(-A\right)}^2+3\left(-A\right){\left(I\right)}^2+A$
$=I-{\left(A\right)}^3+3A^2+3\left(-A\right){\left(I\right)}^2+A$
$=I-{\left(A\right)}^3+3A^2-3A+A$
$\because I^3=I,IA=A$
$=I-A{\left(A\right)}^2+3{\left(A\right)}^2-3A+A$
$=I-A\left(A\right)+3A-3A+A$ $\left(\because A^2=A\right)$
$=I-A^2+A$
$=\left(I-A\right)+A\left(\because A^2=A\right)$
$=I$