Consider the system \(Ax = b\) where \(A\) is \(n\times n\) and invertible. Cramer's rule says that the unique solution is given by \[ x_i = \frac{\det(B_i)}{\det(A)},~ i =1,\ldots, n,\] where \(B_i\) is obtained from \(A\) by replacing column \(i\) with \(b\).
For example, suppose that \(A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\) and \(b = \begin{bmatrix} 4 \\ 5\end{bmatrix}\). Then the unique solution is given by \[x_1 = \frac{\left|\begin{matrix} 4 & 2 \\5 & 3\end{matrix}\right|} {\left|\begin{matrix} 1 & 2 \\2 & 3\end{matrix}\right|} = \frac{ 2}{-1} = -2,\] \[x_2 = \frac{\left|\begin{matrix} 1 & 4 \\2 & 5\end{matrix}\right|} {\left|\begin{matrix} 1 & 2 \\2 & 3\end{matrix}\right|} = \frac{ -3}{-1} = 3.\]
We now obtain a formula for the inverse of an invertible matrix \(A\) in terms of cofactors. Observe that \(AB = I_n\) can be written as the following \(n\) equations: \(AB_j = e_j\) where \(B_j\) denotes the \(j\)th column of \(B\) and \(e_j\) denotes the \(j\)th column of \(I_n\).
By Cramer's rule, we get that \[B_{i,j} = \frac{\det(M_i)}{\det(A)}\] where \(M_i\) denotes the matrix obtained from \(A\) by replacing column \(i\) with \(e_j\).
Since column \(i\) of \(M_i\) is \(e_j\), expanding \(\det(M_i)\) along column \(i\), we get \(\det(M_i) = (-1)^{j+i}A(j \mid i) = C_{j,i}\). (Beware of the indexing!)
Hence, \(B_{i,j}\) is given by \(\displaystyle\frac{C_{j,i}}{\det(A)}\). So a formula for \(A^{-1}\) is \(\displaystyle\frac{1}{\det(A)} \begin{bmatrix} C_{1,1} & C_{2,1} & \cdots & C_{n,1} \\ C_{1,2} & C_{2,2} & \cdots & C_{n,2} \\ \vdots & \vdots & \ddots & \vdots \\ C_{1,n} & C_{2,n} & \cdots & C_{n,n}\end{bmatrix}.\)
Let \(A = \begin{bmatrix} 1 & -1 \\ -2 & 1\end{bmatrix}\) Let \(b = \begin{bmatrix} -1 \\ -1\end{bmatrix}\) Solve the system \(Ax = b\) using Cramer's rule.
Let \(A\in \mathbb{Z}^{n\times n}\). Let \(b\in \mathbb{Z}^n\). Prove that if \(\lvert\det(A)\rvert = 1\), then the solution to \(Ax = b\) has only integer entries.