If A is a square matrix such that \[{{A}^{2}}=A,\] then write the value of \[7A-{{\left( I+A \right)}^{3}},\] where I is an identity matrix.

Hint: We have that A is a square matrix with the property \[{{A}^{2}}=A.\] We have to find \[7A-{{\left( I+A \right)}^{3}}.\] We will first expand \[{{\left( I+A \right)}^{3}}\] using the formula \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3{{b}^{2}}a.\] Then we will use \[{{A}^{2}}=A\] to simplify. Once, we have the value of \[{{\left( I+A \right)}^{3}}\] we will put this in \[7A-{{\left( I+A \right)}^{3}}\] to get the required solution.

Complete step by step answer:
We are given that A is a square matrix such that \[{{A}^{2}}=A.\] Now, we have to find the value of \[7A-{{\left( I+A \right)}^{3}}.\] First, we will solve for \[{{\left( I+A \right)}^{3}}.\] We know that,
\[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3{{b}^{2}}a\]
So using this for \[{{\left( I+A \right)}^{3}},\] we will get,
\[{{\left( I+A \right)}^{3}}={{I}^{3}}+{{A}^{3}}+3{{I}^{2}}A+3{{A}^{2}}I\]
We know that \[{{I}^{n}}=I,\] So,
\[{{I}^{3}}=I\]
\[{{I}^{2}}=I\]
And,
\[{{A}^{3}}={{A}^{2}}.A\]
So, we get,
\[\Rightarrow {{\left( I+A \right)}^{3}}=I+{{A}^{2}}.A+3IA+3{{A}^{2}}I\]
We are given that, \[{{A}^{2}}=A.\] So, we get,
\[\Rightarrow {{\left( I+A \right)}^{3}}=I+A.A+3I.A+3A.I\]
Now, multiplying I with any matrix in that matrix itself
\[\Rightarrow I.A=A\]
So, we get,
\[\Rightarrow {{\left( I+A \right)}^{3}}=I+{{A}^{2}}+3A+3A\]
Again, as \[{{A}^{2}}=A,\] so we get,
\[\Rightarrow {{\left( I+A \right)}^{3}}=I+A+3A+3A\]
Simplifying, we get,
\[\Rightarrow {{\left( I+A \right)}^{3}}=I+7A.....\left( i \right)\]
Now, we will find the value of \[7A-{{\left( I+A \right)}^{3}}\] as \[{{\left( I+A \right)}^{3}}=I+7A.\] So, using this, we will get,
\[7A-{{\left( I+A \right)}^{3}}=7A-\left( I+7A \right)\]
Opening the brackets, we get,
\[\Rightarrow 7A-{{\left( I+A \right)}^{3}}=7A-I-7A\]
Canceling the like terms, we get,
\[\Rightarrow 7A-{{\left( I+A \right)}^{3}}=-I\]

Hence, we get the value of \[7A-{{\left( I+A \right)}^{3}}\] as – I.

Note: Students need to remember that \[{{\left( I+A \right)}^{3}}\] is not equal to \[{{I}^{3}}+{{A}^{3}}.\] We will use \[{{\left( I+A \right)}^{3}}={{I}^{3}}+{{A}^{3}}+3{{I}^{2}}A+3{{A}^{2}}I\] to simplify and solve. Also, remember that we can express \[{{A}^{3}}\] as \[A.{{A}^{2}}\] and then we will use \[{{A}^{2}}=A\] to simplify further. Similarly, we have that \[{{I}^{n}}=I\] so it gives us, \[{{I}^{2}}=I,{{I}^{3}}=I.\]