Finding inverse using inverse formula Given matrix A
$$ A=
\begin{bmatrix}
1&2&3\\
0&2&3\\
0&0&3\\
\end{bmatrix}
$$
I calculated $\det(A)= 6$ and below:


*

*$(A^{-1})_{11} = 6$

*$(A^{-1})_{12} = 0$

*$(A^{-1})_{13} = 0$

*$(A^{-1})_{21} = -6$

*$(A^{-1})_{22} = 3$

*$(A^{-1})_{23} = 0$

*$(A^{-1})_{31} = 0$

*$(A^{-1})_{32} = -3$

*$(A^{-1})_{33} = 2$


Which lead me to:
$$ (A^{-1})= \frac 16\
\begin{bmatrix}
6&0&0\\
-6&3&0\\
0&-3&2\\
\end{bmatrix}
$$
but the answer is $$ (A^{-1})= \frac16\
\begin{bmatrix}
6&-6&0\\
0&3&-3\\
0&0&2\\
\end{bmatrix}
$$
I did not know what I did wrong. I'd appreciate any help from you guys!
 A: Your lecture notes seem to be incorrect. Cramer's rule states that
$$A^{-1}=\frac{1}{\det(A)}\cdot\mathrm{adj}(A)$$
where $\mathrm{adj}(A)=C^{\top}$ and $C_{ij}=(-1)^{i+j}M_{ij}$ as described here. You (or your professor) seem to have missed the transpose.
A: To address the question writer's question on why do we need to take inverse, I derive from Laplace expansion formula, which he/she knows well and he/she has used in the above calculations.  In this way, he/she will really understand how the transpose of the cofactor matrix interacts with the matrix itself through matrix multiplication to give a diagonal matrix.
The unordered list of numbers are call cofactors.  I'll prefer using $C_{ij}$ to denote "$(A^{-1})_{ij}$ in the question".  I hope following classical argument is accessible to any interested high school students.
By the well-known Laplace expansion formula for calculating determinants, for any $i \in \lbrace1,\dots,n\rbrace$
\begin{align}
\det(A) &= \sum_{j=1}^n (-1)^{i+j} a_{ij} \det(M_{ij}) \\
&= \sum_{j=1}^n a_{ij} C_{ij} \label{1}\tag{1}
\end{align}
But a matrix product of $AB$ has entries of the form
$$(AB)_{ij} = \sum_{k=1}^n a_{ik} b_{kj} \label{2}\tag{2}.$$
To make \eqref{1} resembles more \eqref{2}, we consider the transpose of the cofactor matrix, whose $(j,i)$-th entry is $C_{ij}$.  For any $i$ fixed,
$$
\det(A) = \sum_{j=1}^n a_{ij} (C^T)_{ji} \label{1'}\tag{1'}
$$
We change the $j$ in \eqref{1'} to $k$.  For each $i$ fixed,
$$
\det(A) = \sum_{k=1}^n a_{ik} (C^T)_{ki} \label{1''}\tag{1''}
$$
\eqref{1''} represents any diagonal entry of $AC^T$.  To completely address the question writer's addition question in the comments, one needs to justify that the non-diagonal entries of $AC^T$ equals zero.
For each $(i,j)$ fixed with $i\ne j$,
$$
(AC^T)_{ij} = \sum_{k=1}^n a_{ik} (C^T)_{kj} = \sum_{k=1}^n a_{ik} C_{jk} \label{3}\tag{3}
$$
This is in fact the Laplace expansion of the determinant
$$
\begin{vmatrix}
 a_{11} & a_{12} & \cdots & a_{1n} \\
 a_{21} & a_{22} & \cdots & a_{2n} \\
 \vdots& \vdots & \ddots & \vdots \\
 a_{i1} & a_{i2} & \cdots & a_{in} \\
 \vdots& \vdots & \ddots & \vdots \\
 \color{red}{a_{i1}} & \color{red}{a_{i2}} & \cdots & \color{red}{a_{in}} \\
 \vdots& \vdots & \ddots & \vdots \\
 a_{n1} & a_{n2} & \cdots & a_{nn}    
 \end{vmatrix}
$$
Note that the $\color{red}{j\text{-th row}}$ of the above matrix is replaced by the $i$-th row of $A$ as the Laplace expansion formula suggests.  Since any matrix with two identical rows has zero determinant, we conclude the following useful formula.
$$\bbox[2px, border: 1px solid red]{\det(A)I_n=AC^T=C^TA}$$
Uptil this step, this works for entries defined on any commutative ring (equipped with addition and abelian multiplication, I don't know whether this works for non-commutative rings.)
To get the well-known matrix formula, we divide both sides by $\det(A)$.  This last step requires the entries to be defined on a field.
