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

If A is a square matrix such that A2=A,{{A}^{2}}=A, then write the value of 7A−(I+A)3,7A-{{\left( I+A \right)}^{3}}, where I is an identity matrix.

Explanation

Solution

We have that A is a square matrix with the property A2=A.{{A}^{2}}=A. We have to find 7A−(I+A)3.7A-{{\left( I+A \right)}^{3}}. We will first expand (I+A)3{{\left( I+A \right)}^{3}} using the formula (a+b)3=a3+b3+3a2b+3b2a.{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3{{b}^{2}}a. Then we will use A2=A{{A}^{2}}=A to simplify. Once, we have the value of (I+A)3{{\left( I+A \right)}^{3}} we will put this in 7A−(I+A)37A-{{\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 A2=A.{{A}^{2}}=A. Now, we have to find the value of 7A−(I+A)3.7A-{{\left( I+A \right)}^{3}}. First, we will solve for (I+A)3.{{\left( I+A \right)}^{3}}. We know that,
(a+b)3=a3+b3+3a2b+3b2a{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3{{b}^{2}}a
So using this for (I+A)3,{{\left( I+A \right)}^{3}}, we will get,
(I+A)3=I3+A3+3I2A+3A2I{{\left( I+A \right)}^{3}}={{I}^{3}}+{{A}^{3}}+3{{I}^{2}}A+3{{A}^{2}}I
We know that In=I,{{I}^{n}}=I, So,
I3=I{{I}^{3}}=I
I2=I{{I}^{2}}=I
And,
A3=A2.A{{A}^{3}}={{A}^{2}}.A
So, we get,
⇒(I+A)3=I+A2.A+3IA+3A2I\Rightarrow {{\left( I+A \right)}^{3}}=I+{{A}^{2}}.A+3IA+3{{A}^{2}}I
We are given that, A2=A.{{A}^{2}}=A. So, we get,
⇒(I+A)3=I+A.A+3I.A+3A.I\Rightarrow {{\left( I+A \right)}^{3}}=I+A.A+3I.A+3A.I
Now, multiplying I with any matrix in that matrix itself
⇒I.A=A\Rightarrow I.A=A
So, we get,
⇒(I+A)3=I+A2+3A+3A\Rightarrow {{\left( I+A \right)}^{3}}=I+{{A}^{2}}+3A+3A
Again, as A2=A,{{A}^{2}}=A, so we get,
⇒(I+A)3=I+A+3A+3A\Rightarrow {{\left( I+A \right)}^{3}}=I+A+3A+3A
Simplifying, we get,
⇒(I+A)3=I+7A.....(i)\Rightarrow {{\left( I+A \right)}^{3}}=I+7A.....\left( i \right)
Now, we will find the value of 7A−(I+A)37A-{{\left( I+A \right)}^{3}} as (I+A)3=I+7A.{{\left( I+A \right)}^{3}}=I+7A. So, using this, we will get,
7A−(I+A)3=7A−(I+7A)7A-{{\left( I+A \right)}^{3}}=7A-\left( I+7A \right)
Opening the brackets, we get,
⇒7A−(I+A)3=7A−I−7A\Rightarrow 7A-{{\left( I+A \right)}^{3}}=7A-I-7A
Canceling the like terms, we get,
⇒7A−(I+A)3=−I\Rightarrow 7A-{{\left( I+A \right)}^{3}}=-I

Hence, we get the value of 7A−(I+A)37A-{{\left( I+A \right)}^{3}} as – I.

Note: Students need to remember that (I+A)3{{\left( I+A \right)}^{3}} is not equal to I3+A3.{{I}^{3}}+{{A}^{3}}. We will use (I+A)3=I3+A3+3I2A+3A2I{{\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 A3{{A}^{3}} as A.A2A.{{A}^{2}} and then we will use A2=A{{A}^{2}}=A to simplify further. Similarly, we have that In=I{{I}^{n}}=I so it gives us, I2=I,I3=I.{{I}^{2}}=I,{{I}^{3}}=I.