Sure, I'd be happy to help you with that math problem!

To find the $(3,2)$ entry of $A^{-1}$, we need to first find the inverse of matrix $A$. The formula for the inverse of a $3 \times 3$ matrix is:

A−1=1det⁡(A)[C11C21C31C12C22C32C13C23C33]TA^{-1} = \frac{1}{\det(A)} \begin{bmatrix}
C\_{11} & C\_{21} & C\_{31} \\
C\_{12} & C\_{22} & C\_{32} \\
C\_{13} & C\_{23} & C\_{33}
\end{bmatrix}^TA−1=det(A)1​⎣⎡​C11​C12​C13​​C21​C22​C23​​C31​C32​C33​​⎦⎤​Twhere $\det(A)$ is the determinant of matrix $A$, $C\_{ij}$ is the cofactor of the $(i,j)$ entry of $A$, and $^T$ denotes the transpose of the matrix.

So, the first step is to calculate the determinant of matrix $A$. We can use the formula for the determinant of a $3 \times 3$ matrix:

det⁡(A)=a∣a+1a+211∣−0+0=a(a+1)−a=a2.\det(A) = a\begin{vmatrix}
a+1 & a+2 \\
1 & 1
\end{vmatrix} - 0 + 0 = a(a+1) - a = a^2.det(A)=a∣∣​a+11​a+21​∣∣​−0+0=a(a+1)−a=a2.Next, we need to find the cofactors of each entry of $A$:

C11=∣a+1a+211∣=a+1−(a+2)=−1,C\_{11} = \begin{vmatrix}
a+1 & a+2 \\
1 & 1
\end{vmatrix} = a+1 - (a+2) = -1,C11​=∣∣​a+11​a+21​∣∣​=a+1−(a+2)=−1,C12=−∣a011∣=−a,C\_{12} = -\begin{vmatrix}
a & 0 \\
1 & 1
\end{vmatrix} = -a,C12​=−∣∣​a1​01​∣∣​=−a,C13=∣a0a+1a+2∣=−a(a+2),C\_{13} = \begin{vmatrix}
a & 0 \\
a+1 & a+2
\end{vmatrix} = -a(a+2),C13​=∣∣​aa+1​0a+2​∣∣​=−a(a+2),C21=−∣1100∣=0,C\_{21} = -\begin{vmatrix}
1 & 1 \\
0 & 0
\end{vmatrix} = 0,C21​=−∣∣​10​10​∣∣​=0,C22=∣a000∣=0,C\_{22} = \begin{vmatrix}
a & 0 \\
0 & 0
\end{vmatrix} = 0,C22​=∣∣​a0​00​∣∣​=0,C23=−∣a011∣=−a,C\_{23} = -\begin{vmatrix}
a & 0 \\
1 & 1
\end{vmatrix} = -a,C23​=−∣∣​a1​01​∣∣​=−a,C31=∣110a∣=a,C\_{31} = \begin{vmatrix}
1 & 1 \\
0 & a
\end{vmatrix} = a,C31​=∣∣​10​1a​∣∣​=a,C32=−∣a00a∣=−a2,C\_{32} = -\begin{vmatrix}
a & 0 \\
0 & a
\end{vmatrix} = -a^2,C32​=−∣∣​a0​0a​∣∣​=−a2,C33=∣a0a+1a+1∣=−a.C\_{33} = \begin{vmatrix}
a & 0 \\
a+1 & a+1
\end{vmatrix} = -a.C33​=∣∣​aa+1​0a+1​∣∣​=−a.Now, we can substitute these values into the formula for $A^{-1}$ and simplify:

A^{-1} = \frac{1}{a^2} \begin{bmatrix}
-1 & 0 & a \\
-a & 0 & -a^2 \\
-a(a+2) & -a & -a
\end{bmatrix}^T
= \begin{bmatrix}
-\frac{1}{a^2} & \frac{a}{a^2} & -\frac{a(a+2)}{a^2} \\
0 & 0 & -\frac{1}{a} \\
\frac{a
물론 기꺼이 수학 문제를 도와드리겠습니다!A^{-1}$의 $(3,2)$ 항목을 찾으려면 먼저 행렬 $A$의 역을 찾아야 합니다. 3 \times 3$ 행렬의 역의 공식은 다음과 같습니다:A-1=1det(A)[C11C21C31C12C22C32C13C23C33]TA^{-1} = \frac{1}{\det(A)} \begin{bmatrix}c\_{11} & c\_{21} & c\_{31} \\c\_{12} & c\_{22} & c\_{32} \\c\_{13} & c\_{23} & c\_{33}\end{bmatrix}^TA-1=det(A)1⎣⎡C11C12C13C21C22C23C31C32C33⎦⎤T 여기서 $\det(A)$는 행렬 $A$의 행렬식, $C\_{ij}$는 $A$의 $(i,j)$ 항목의 계승자, $^T$는 행렬의 치환을 나타냅니다.따라서 첫 번째 단계는 행렬 $A$의 행렬식을 계산하는 것입니다. 3 \times 3$ 행렬의 행렬식에 대한 공식을 사용할 수 있습니다:det⁡(A)=a∣a+1a+211∣−0+0=a(a+1)−a=a2.\det(A) = a\begin{vmatrix}a+1 & a+2 \\1 & 1\end{vmatrix} - 0 + 0 = a(a+1) - a = a^2.det(A)=a∣∣a+11a+21∣∣-0+0=a(a+1)-a=a2.다음으로, $A$의 각 항목의 계승을 구해야 합니다:C11=∣a+1a+211∣=a+1-(a+2)=-1,C\_{11} = \begin{vmatrix}a+1 & a+2 \\1 & 1\end{vmatrix} = a+1 - (a+2) = -1,C11=∣∣∣a+11a+21∣∣=a+1-(a+2)=-1,C12=-∣a011∣=-a,C\_{12} = -\begin{vmatrix}a & 0 \\1 & 1\end{vmatrix} = -a,C12=-∣∣∣a101∣∣=-a,C13=∣a0a+1a+2∣=-a(a+2),C\_{13} = \begin{vmatrix}a & 0 \\a+1 & a+2\end{vmatrix} = -a(a+2),C13=∣∣aa+10a+2∣∣=-a(a+2),C21=-∣1100∣=0,C\_{21} = -\begin{vmatrix}1 & 1 \\0 & 0\end{vmatrix} = 0,C21=-∣∣1010∣∣=0,C22=∣a000∣=0,C\_{22} = \begin{vmatrix}a & 0 \\0 & 0\end{vmatrix} = 0,C22=∣∣a000∣∣∣=0,C23=-∣a011∣=-a,C\_{23} = -\begin{vmatrix}a & 0 \\1 & 1\end{vmatrix} = -a,C23=-∣∣∣a101∣∣=-a,C31=∣110a∣=a,C\_{31} = \begin{vmatrix}1 & 1 \\0 & a\end{vmatrix} = a,C31=∣∣101a∣∣=a,C32=-∣a00a∣=-a2,C\_{32} = -\begin{vmatrix}a & 0 \\0 & a\end{vmatrix} = -a^2,C32=-∣∣a00a∣∣=-a2,C33=∣a0a+1a+1∣=-a.C\_{33} = \begin{vmatrix}a & 0 \\a+1 & a+1\end{vmatrix} = -a.C33=∣∣aa+10a+1∣∣=-a.이제 이 값을 $A^{-1}$의 공식에 대입하여 단순화할 수 있습니다:A^{-1} = \frac{1}{a^2} \begin{bmatrix}-1 & 0 & a \\-a & 0 & -a^2 \\-a(a+2) & -a & -a\end{bmatrix}^T= \begin{bmatrix}-\frac{1}{a^2} & \frac{a}{a^2} & -\frac{a(a+2)}{a^2} \\0 & 0 & -\frac{1}{a} \\\frac{a