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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.

1. $I+A$ need not be invertible.
2. Det $(I+A)$ can be any non-zero real number.
3. Det $(I+A) = 1$
4. $A^{n}$ is a non-zero matrix.
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+1
I think, ans should be option A.

??
0
(C) is correct.

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

by (213 points)
+1
0

A^k is 0 not A^n. Will A^n be 0 if for some k<n, A^k is zero??

$A^k = 0$ 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, $det(I+A) = 1$

by Boss (28.8k points)