**Proof plan.** We use the adjugate (classical adjoint) matrix. The key identity is $\operatorname{adj}(A - tI) \cdot (A - tI) = \det(A - tI) \cdot I = \chi_A(t) \cdot I$. Since $\operatorname{adj}(A - tI)$ is a matrix whose entries are polynomials in $t$ of degree at most $n - 1$, it can be written as a polynomial in $t$ with matrix coefficients. Substituting $t = A$ into the identity will yield the result. **Step 1: Expand the adjugate as a matrix polynomial in $t$.** Each entry of $\operatorname{adj}(A - tI)$ is a cofactor of $A - tI$, hence a polynomial in $t$ of degree at most $n-1$. Collecting powers of $t$, we can write: \begin{align*} \operatorname{adj}(A - tI) = B_0 + B_1 t + B_2 t^2 + \cdots + B_{n-1} t^{n-1}, \end{align*} where $B_0, B_1, \ldots, B_{n-1}$ are $n \times n$ matrices with entries depending only on $A$. **Step 2: Write the characteristic polynomial.** Let $\chi_A(t) = c_0 + c_1 t + c_2 t^2 + \cdots + c_n t^n$, where $c_n = (-1)^n$ and $c_0 = \det(A)$. **Step 3: Use the adjugate identity.** The fundamental identity $\operatorname{adj}(M) \cdot M = \det(M) \cdot I$ applied to $M = A - tI$ gives: \begin{align*} (B_0 + B_1 t + \cdots + B_{n-1} t^{n-1})(A - tI) = (c_0 + c_1 t + \cdots + c_n t^n) I. \end{align*} **Step 4: Equate coefficients of powers of $t$.** Expanding the left side and matching coefficients of $t^0, t^1, \ldots, t^n$: \begin{align*} t^0: \quad & B_0 A = c_0 I, \\ t^1: \quad & B_1 A - B_0 = c_1 I, \\ t^2: \quad & B_2 A - B_1 = c_2 I, \\ & \;\;\vdots \\ t^{n-1}: \quad & B_{n-1} A - B_{n-2} = c_{n-1} I, \\ t^n: \quad & -B_{n-1} = c_n I. \end{align*} **Step 5: Multiply and sum.** Multiply the $t^0$ equation by $I$ on the right, the $t^1$ equation by $A$ on the right, the $t^2$ equation by $A^2$ on the right, and so on — the $t^k$ equation by $A^k$: \begin{align*} B_0 A &= c_0 I, \\ B_1 A^2 - B_0 A &= c_1 A, \\ B_2 A^3 - B_1 A^2 &= c_2 A^2, \\ &\;\;\vdots \\ B_{n-1} A^n - B_{n-2} A^{n-1} &= c_{n-1} A^{n-1}, \\ -B_{n-1} A^n &= c_n A^n. \end{align*} Summing all equations, the left side telescopes: every $B_k A^{k+1}$ term appears once with a positive sign and once with a negative sign, leaving $0$. The right side gives: \begin{align*} 0 = c_0 I + c_1 A + c_2 A^2 + \cdots + c_n A^n = \chi_A(A). \qquad \square \end{align*}