Determinants, Invertibility and Cramer's Rule

Definition A real $n \times n$ matrix $A$ is invertible if and only if $\det A \ne 0$. In that case

If $A$ is invertible, then $AA^{-1} = I$, and the product rule gives $\det A \, \det(A^{-1}) = \det I = 1$, so neither factor is $0$. If $A$ is not invertible, its columns are linearly dependent, and a matrix with dependent columns has determinant $0$. So a nonzero determinant forces $A$ to be invertible. Elimination tells the same story. Reducing $A$ with row swaps and replacements to an echelon form $U$, which is upper triangular, gives $\det A = \pm u_{11} u_{22} \cdots u_{nn}$, which is nonzero exactly when every column has a pivot. For an $n \times n$ matrix $A$ the following statements are therefore equivalent.

• The determinant $\det A$ is not zero.
• The matrix $A$ is invertible.
• The matrix $A$ has $n$ pivots, so $\operatorname{rank} A = n$.
• The null space $\operatorname{Null} A$ is $\{\mathbf{0}\}$.
• The columns of $A$ form a basis of $\mathbb{R}^n$.
• The system $A\mathbf{x} = \mathbf{b}$ has exactly one solution for every $\mathbf{b}$ in $\mathbb{R}^n$.

Only square matrices have determinants, so a rectangular matrix or an augmented matrix $[A \mid \mathbf{b}]$ is studied with rank instead.

Definition A square system $A\mathbf{x} = \mathbf{b}$, with $A$ of size $n \times n$, is singular when $A$ is singular, that is, when $\det A = 0$. Such a system never has exactly one solution. Either it has no solution, or it has a solution $\mathbf{x}_p$, and then its solution set is

which is infinite, because $\operatorname{Null} A$ contains a nonzero vector $\mathbf{w}$ and all of its multiples $t\mathbf{w}$.

Definition A right-hand side $\mathbf{b}$ is reachable for a matrix $A$ if $A\mathbf{x} = \mathbf{b}$ for some $\mathbf{x}$, that is, if $\mathbf{b}$ lies in the image $\operatorname{Col} A$. Otherwise it is unreachable, and the system has no solution. For an $n \times n$ matrix with $\det A \ne 0$, every $\mathbf{b}$ in $\mathbb{R}^n$ is reachable. When $\det A = 0$, the image $\operatorname{Col} A$ is a subspace of dimension $\operatorname{rank} A < n$, so some right-hand sides are reachable and others are not. The number $\det A$ depends on $A$ alone, so it cannot decide which. Row reduction of the augmented matrix $[A \mid \mathbf{b}]$ does decide, since the system is inconsistent exactly when an echelon form has a row $[\,0 \ \cdots \ 0 \mid c\,]$ with $c \ne 0$.

Definition Let $A$ be a real $n \times n$ matrix with $\det A \ne 0$, let $\mathbf{b}$ be in $\mathbb{R}^n$, and for each $i$ let $A_i$ be the matrix obtained from $A$ by replacing column $i$ with $\mathbf{b}$, keeping every other column in place. Then the unique solution $\mathbf{x} = (x_1, \dots, x_n)$ of $A\mathbf{x} = \mathbf{b}$ is given by

To see why, write the columns of $A$ as $\mathbf{a}_1, \dots, \mathbf{a}_n$, so that $\mathbf{b} = x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n$. By linearity in column $i$, the number $\det A_i$ is the sum over $j$ of $x_j$ times the determinant of $A$ with column $i$ replaced by $\mathbf{a}_j$. For $j \ne i$ that matrix has two equal columns, so its determinant is $0$. For $j = i$ it is $A$ itself. Hence $\det A_i = x_i \det A$, and dividing by $\det A \ne 0$ gives the formula.

Definition For an $n \times n$ system with $\det A \ne 0$, Cramer’s rule and elimination give the same unique solution, but for larger $n$ at very different cost. Cramer’s rule needs the $n + 1$ determinants $\det A, \det A_1, \dots, \det A_n$. Computed by elimination, each of them costs about as much as solving the whole system once by elimination, so the rule does roughly $n + 1$ times the work. Computed by full cofactor expansion, each is a sum of $n!$ products. Elimination also handles $\det A = 0$, where it decides consistency and describes every solution. A factorization $A = LU$, or $PA = LU$ when rows must be swapped, is computed once and then solves each new right-hand side with two triangular solves. Cramer’s rule suits small systems, especially when only one unknown is needed or the entries are symbolic, and it shows how each coordinate of the solution depends on $\mathbf{b}$.