Determinant of a matrix equation Let $A=\begin{bmatrix}
1 &amp;2  &amp; 4\\ 
 0&amp; 0 &amp;3 \\ 
0 &amp;0  &amp;-1 
\end{bmatrix}$, then we are asked to find the determinant of $A^{17}+A^{10}-I$
My approach is that the characteristic equation of this matrix gives $x^{3}=x$ so $A^{17}=A$ and $A^{10}=A^{2}$ hence $A^{17}+A^{10}-I$=$A^{2}+A-I$
Now $A^{2}$=$\begin{bmatrix}
1 &amp;2  &amp; 6\\ 
 0&amp; 0 &amp;-3 \\ 
0 &amp;0  &amp;1 
\end{bmatrix}$ and $A^{2}+A$=$\begin{bmatrix}
2 &amp;4  &amp; 10\\ 
 0&amp; 0 &amp;0 \\ 
0 &amp;0  &amp;0 
\end{bmatrix}$ and $A^{2}+A-I$=$\begin{bmatrix}
1 &amp;4  &amp; 10\\ 
 0&amp; -1 &amp;0 \\ 
0 &amp;0  &amp;-1 
\end{bmatrix}$ which, because its lower triangular has a determinant of $1(-1)(-1)=1$;
But apparently the answer in my textbook for this is 0, could it be an error in the textbook or am i wrong somewhere in the reasoning, any help would be much appreciated
Another way to get the determinant is to see that
$$A^{17}=\begin{bmatrix}
1 &amp;*  &amp; *\\ 
 0&amp; 0 &amp; * \\ 
0 &amp;0  &amp;-1 
\end{bmatrix} \quad \quad \text{and} \quad \quad A^{10}=\begin{bmatrix}
1 &amp;*  &amp; *\\ 
 0&amp; 0 &amp; * \\ 
0 &amp;0  &amp;1
\end{bmatrix}$$
so $$A^{17} + A^{10} - I =\begin{bmatrix}
1 &amp;*  &amp; *\\ 
 0&amp; -1 &amp; * \\ 
0 &amp;0  &amp; -1
\end{bmatrix}$$
Therefore the determinant is, indeed, equal to $1$.
Another way to get the determinant is to see that
$$A^{17}=\begin{bmatrix}
1 &amp;*  &amp; *\\ 
 0&amp; 0 &amp; * \\ 
0 &amp;0  &amp;-1 
\end{bmatrix} \quad \quad \text{and} \quad \quad A^{10}=\begin{bmatrix}
1 &amp;*  &amp; *\\ 
 0&amp; 0 &amp; * \\ 
0 &amp;0  &amp;1
\end{bmatrix}$$
so $$A^{17} + A^{10} - I =\begin{bmatrix}
1 &amp;*  &amp; *\\ 
 0&amp; -1 &amp; * \\ 
0 &amp;0  &amp; -1
\end{bmatrix}$$
Therefore the determinant is, indeed, equal to $1$.