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For fixed coefficient dimension and $Y=X{\beta }_0+\epsilon$, the ordinary least squares error satisfiesConsequently convergence in probability of $X^{\mathsf{T}}X/n$ to a positive-definite matrix $Q$, together with $X^{\mathsf{T}}\epsilon /n\to 0$ in probability, proves statistical consistency by continuity of matrix inversion.

For independent and identically distributed random-design pairs, finite second moments and $\mathbb{E}[x_i{\epsilon }_i]=0$ imply these limits by the law of large numbers, provided $Q=\mathbb{E}[x_i{x}_i^{\mathsf{T}}]$ is nonsingular. Conditional mean zero, $\mathbb{E}[{\epsilon }_i\mid x_i]=0$, is a familiar sufficient exogeneity condition. For a deterministic design matrix, the vanishing normalized design-error product must instead be justified from its errors and weights.

Finite moments and a nonsingular design alone are insufficient. In an intercept-only model with errors identically one, the estimated intercept is always ${\beta }_0+1$. The missing zero-expected value condition changes the limiting target.

Matrix inversion computes the matrix inverse $A^{-1}$ of a square matrix satisfying $A{A}^{-1}=A^{-1}A=I$. It is possible exactly when $A$ has full rank. One method applies Gauss-Jordan elimination to $[A\mid I]$; another factors $A$ and solves a linear system for each column of $I$.

For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with $ad-bc\ne 0$, the inverse is $(ad-bc{)}^{-1}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$. To solve a single numerical system $Ax=b$, a factorization followed by triangular solves is usually preferable to explicitly forming the inverse. A large condition number can amplify input errors regardless of the inversion method.

Even over the real numbers, matrix inversion does not preserve addition. For $A=B=I$ of any positive size,All three matrices $A$, $B$ and $A+B$ are invertible, so their invertibility does not rescue the proposed rule. The familiar scalar identity $1/(a+b)\ne 1/a+1/b$ already supplies this counterexample through scalar matrices.

The centered regressors make the first normal equation read $n{\hat{\beta }}_0={\sum }_i{Y}_i$. DefineThe remaining equations are $A{\hat{\beta }}_1+B{\hat{\beta }}_2=u$ and $B{\hat{\beta }}_1+C{\hat{\beta }}_2=v$. Provided $D=AC-B^2>0$, matrix inversion givesBy the Cauchy-Schwarz inequality, $D\ge 0$. If $D=0$, the two regressor columns are linearly dependent, and their separate slopes are not uniquely identifiable. One still has well-defined linear least squares fitted values, but cannot use the displayed divisions by $D$.