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+### 9.24
+###### Original problem
+
+Let us present a rigorous proof. Let $\sigma(Z) = (\lambda_1,..,\lambda_n)$ denote the eigenvalues of Z.
+$${\rm Tr}(I + \varepsilon Z) = \prod_{i=1}^n (1 + \varepsilon \lambda_i)\approx1+\varepsilon\sum_{i=1}^n\lambda_i$$
+
+$$\Longrightarrow V(I+\varepsilon Z) = -\log\left(\prod_{i=1}^n (1 + \varepsilon \lambda_i)\right) \approx -\sum_{i=1}^n\log(1 + \varepsilon \lambda_i) \approx -\varepsilon\sum_{i=1}^n\lambda_i$$
+
+$$\Longrightarrow V(X + \varepsilon Y) = V(X) + V(I + \varepsilon X^{-1}Y) \approx V(X) - \varepsilon{\rm Tr}(X^{-1}Y)$$
+
+Thus, we have completed the proof.
+
+
+###### Cramer's rule
+Consider a linear system
+$$Ax = b,A\in\mathbb{R^{n\times n}},b\in\mathbb{R^n},\det(A)\neq 0$$
+Let $x$ denote its solution, and let $A_i$ represent the matrix obtained by replacing the $i$-th column of $A$ with $b$. We can derive:
+$$\det(A+\varepsilon b e_i^T) = \det(A)\det(I+\varepsilon A^{-1}b e_i^T)$$
+
+$$\approx\det(A)(1+\varepsilon{\rm Tr}(A^{-1}b e_i^T)) = \det(A)(1+x_i)$$
+
+$$\Longrightarrow x_i \approx \frac{\det(A+\varepsilon b e_i^T)-\det(A)}{\varepsilon\det(A)}\to\frac{\det(A_i)}{\det(A)}$$
+
+This completes our derivation of Cramer's rule.
+
+
+### 9.25
+This problem requires further clarification as its meaning is not immediately clear.
+
+### 9.27
+###### 1. 
+Let us define $x_0$ as an integer solution and 
+$$\ker_N = \ker(A)\cap\mathbb{N}^n$$
+
+We can then establish that $x_0+\ker_N$ forms a lattice.
+
+###### 2.
+Through Gaussian elimination, we can efficiently determine an integer solution $x_0$ and a rational basis $K_Q$ of $\ker(A)$ in polynomial time. By multiplying $K_Q$ by the least common multiple (LCM) of the denominators of all its elements (with all rational elements reduced to their simplest form), we obtain an integer basis $K_Z$. Consequently, we can construct the lattice $x_0+{\rm span}_N(K_Z)$ in polynomial time.
+
+###### 3.
+If
+$$\exists x\in K_N ~~s.t.~~c^Tx\neq 0$$
+
+then we can conclude that the optimal value of the LP must be $+\infty$.
+
+Conversely, if no such vector exists, then every vector in $x_0+{\rm span}(K_Z)$ constitutes an optimal solution of the LP.
+
+Therefore, we can conclude that the IP can be solved in polynomial time.