Question

If ${A^3} = 0$, then $I + A + {A^2}$ equals
A) $I - A$
B) $\left( {I + {A^{ - 1}}} \right)$
C) ${\left( {I - A} \right)^{ - 1}}$
D) None of these

Answer 1

Hint:
We can take $I - A$ as an invertible matrix. Now we can find its product with the given expression. Then we can simplify the equation using properties of matrix multiplication. Then we can multiply with the inverse of $I - A$ on both sides. On further simplification, we can obtain the required value of the expression.

Complete step by step solution:
We are given that ${A^3} = 0$ and we need to find the value of $I + A + {A^2}$.
As ${A^3} = 0$, we can say that $\left| A \right| = 0$. Therefore, A is not an invertible matrix.
So, we can take $I - A$ as an invertible matrix. Now we can find its product with the given expression.
Let $X = \left( {I - A} \right)\left( {I + A + {A^2}} \right)$,
We can open the brackets and multiply. So, we get,
 $ \Rightarrow X = I\left( {I + A + {A^2}} \right) - A\left( {I + A + {A^2}} \right)$
On further expansion, we get,
 $ \Rightarrow X = I + A + {A^2} - A - {A^2} - {A^3}$
On simplification, we get,
 $ \Rightarrow X = I - {A^3}$
We are given that ${A^3} = 0$. So, we get,
 $ \Rightarrow X = I$
 $ \Rightarrow \left( {I - A} \right)\left( {I + A + {A^2}} \right) = I$
As $\left( {I - A} \right)$ is an invertible matrix we can multiply with its inverse on both sides of the above equation.
 $ \Rightarrow {\left( {I - A} \right)^{ - 1}}\left( {I - A} \right)\left( {I + A + {A^2}} \right) = {\left( {I - A} \right)^{ - 1}}I$
We know that when a matrix is multiplied with its inverse, we get an identity matrix.
 $ \Rightarrow I\left( {I + A + {A^2}} \right) = {\left( {I - A} \right)^{ - 1}}I$
We also know that, when a matrix is multiplied with an identity matrix, we will get the same matrix. So, we can write,
 $ \Rightarrow \left( {I + A + {A^2}} \right) = {\left( {I - A} \right)^{ - 1}}$
Now we have the required value of the expression ${\left( {I - A} \right)^{ - 1}}$

Therefore, the correct answer is option C, ${\left( {I - A} \right)^{ - 1}}$

Note:
We cannot multiply the given expression with A as A is not an invertible matrix. A matrix is non invertible if its determinant is equal to zero. Then its inverse does not exist. We must note that matrix multiplication is not commutative. So, we must multiply each matrix on the same side of the terms on both sides of the equation. Even though it is not given, A is a square matrix as identity matrix determinant and powers of matrix is possible for only square matrices.