Let A be a 10×10 matrix such that  A⁵ is null matrix and let I be the 10 × 10 identity matrix. The determinant of A+ I is _____.

The correct answer is 1.

To understand why the determinant of the matrix is equal to 1, we can analyze the eigenvalues of the matrix.

First, we are given that $A$ is a $10\times 10$ matrix such that:

$A^5=O$

where $O$ represents the null (zero) matrix. A matrix with this property is called a nilpotent matrix.

Let $\lambda$ be an eigenvalue of $A$, and let $v$ be the corresponding non-zero eigenvector. By definition:

$Av=\lambda v$

Applying the matrix $A$ repeatedly to both sides of the equation yields:

$A^{2}v={\lambda }^{2}v$

Continuing this process up to the fifth power, we get:

$A^{5}v={\lambda }^{5}v$

Since $A^5=O$, we substitute it into the relation:

$Ov={\lambda }^{5}v$

$0={\lambda }^{5}v$

Since the eigenvector $v$ is non-zero, the scalar must be zero:

${\lambda }^5=0\Rightarrow \lambda =0$

Thus, all eigenvalues of the nilpotent matrix $A$ are equal to 0.

Next, we determine the eigenvalues of the matrix $A+I$. If $\lambda$ is an eigenvalue of $A$ with eigenvector $v$, we have:

$(A+I)v=Av+Iv=\lambda v+v=(\lambda +1)v$

This demonstrates that the eigenvalues of $A+I$ are of the form $\lambda +1$.

Since all eigenvalues of $A$ are 0, every eigenvalue of the $10\times 10$ matrix $A+I$ is:

$0+1=1$

Finally, we use the property that the determinant of a matrix is equal to the product of its eigenvalues. Since all 10 eigenvalues of $A+I$ are 1, we have:

$det(A+I)=1^{10}=1$