Call the matrix

We have ${ a }^{ 3 }=a-1$ , hence $ab={ a }^{ 3 }+{ a }^{ 2 }+2a=a-1+{ a }^{ 2 }+2a={ a }^{ 2 }+3a-1$ and so ${ a }^{ 2 }b={ a }^{ 3 }+3{ a }^{ 2 }-a=3{ a }^{ 2 }-1$ .

As $(2-b)+a+{ a }^{ 2 }=0 , -1+(3-b)a+{ a }^{ 2 }=0 ,-1+(3-b){ a }^{ 2 }=0\\$ consider the system $(2-b){ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }=0 , -{ x }_{ 1 }+(3-b){ x }_{ 2 }+{ x }_{ 3 }=0 , -{ x }_{ 1 }+(3-b){ x }_{ 3 }=0\\$

We have $\left( \begin{matrix} 1 \\ a \\ { a }^{ 2 } \end{matrix} \right)$ is a non-zero solution of this system, hence the determinant of the matrix of this system is zero, and so $\left| \begin{matrix} 2-b & 1 & 1 \\ -1 & 3-b & 1 \\ -1 & 0 & 3-b \end{matrix} \right| =0$ .

But $\left| \begin{matrix} 2-b & 1 & 1 \\ -1 & 3-b & 1 \\ -1 & 0 & 3-b \end{matrix} \right|$ is also equal to $23-23b+8{ b }^{ 2 }-{ b }^{ 3 }$ (you can change the third column ${ c }_{ 3 }$ to ${ c }_{ 3 }+(3-b){ c }_{ 1 }$ and expand through the third row). Then, $23-23b+8{ b }^{ 2 }-{ b }^{ 3 }=0$

Thus the minimial polynomial is $b^3 - 8b^2 + 23 b - 23 = 0$ .