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5 votes

Let $n \geq 1$ and let $A$ be an $n \times n$ matrix with real entries such that $A^{k}=0$, for some $k \geq 1$. Let $I$ be the identity $n \times n$ matrix. Then.

- $I+A$ need not be invertible.
- Det $(I+A)$ can be any non-zero real number.
- Det $(I+A) = 1$
- $A^{n}$ is a non-zero matrix.

4 votes

**By cayley hamilton theorem**

A^n = 0 so

$\lambda ^n = 0$

solving for $\lambda$ we get 0 for every eigen value

now by properties of Eigen value

Eigen values of $(A+I)$ = Individual Eigen value of A+1

so the eigen values of (A+I) all are 1

so Det(I+A) =1

* Option C* is the Answer

4 votes

$A^k = 0$ means $A$ is a Nilpotent Matrix.

And all eigen values of a Nilpotent Matrix are $0$. So All, Eigen values of $(I+A)$ are $1$. So $\tt det(I+A)$ = Product of eigen values $= 1$

This can be seen from an example : Let $A = \begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$

$I+ A = \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$

So, $det(I+A) = 1$