Q11Written answer
If \[A\] is a square matrix such that \[{{A}^{2}}=A\] , then \[{{\left( I+A \right)}^{3}}-7A\] is equal to
\[A\]
\[I-A\]
\[I\]
\[3A\]
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Worked solution
Ans: \[{{\left( I+A \right)}^{3}}-7A={{I}^{3}}+{{A}^{3}}+3A+3{{A}^{2}}-7A\]
\[\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+{{A}^{3}}+3A+3{{A}^{2}}-7A\]
Given that \[{{A}^{2}}=A\] ,
\[\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+{{A}^{2}}\cdot A+3A+3{{A}^{2}}-7A\]
\[\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+A\cdot A-A\]
\[\Rightarrow {{\left( I+A \right)}^{3}}-7A=I\]
Thus, option (C) is correct.