Invertibility and determinants

We can use the properties of the determinant in Theorem 5.5 (and its proof) to obtain a useful criterion to decide when a matrix is invertible. Determinants can also be used to compute the inverse of an invertible matrix, however this is only of theoretical significance due to the complexity of the ensuing (iterative) algorithm.

Definition 5.14 (Related exercises: Exercise 5.1)

Let \(A \in {\mathrm {Mat}}_{n \times n}\). For \(1 \le i, j \le n\), denote by \(A_{ij}\) that is obtained from \(A\) by deleting the \(i\)-th row and the \(j\)-th column. The number

\[ c_{ij} := (-1)^{i+j} \det A_{ij} \]

is called the \((i,j)\)-cofactor of \(A\).

The adjugate of \(A\) is the \(n \times n\)-matrix defined as

\[ \mathrm{adj}(A) = (c_{ij}(A))^T = (c_{ji}(A)). \]

Theorem 5.15 (Related exercises: Exercise 5.6, Exercise 5.1, Exercise 5.3, Exercise 5.8, Exercise 6.14)

An \(n \times n\)-matrix \(A\) is invertible if and only if

\[ \det A \ne 0. \]

If this is the case, then the inverse can be computed using the so-called adjugate formula:

\[ A^{-1} = \frac 1 {\det A} \mathrm{adj}(A). \]

(5.16)

Proof. We revisit the proof of Theorem 5.5: say \(A \leadsto A'\), a reduced row echelon matrix, by means of elementary operations (Gaussian elimination). In this process, one does not multiply any row by zero, so that \(\det A = 0\) if and only if \(\det A' = 0\). We also know that \(A' = UA\), where \(U\) is an invertible matrix (namely, a product of elementary matrices). Moreover, \(A'\) is invertible if and only if \(A\) is invertible (since \(U\) is invertible). We therefore have

\[ A \text{ is invertible} \Leftrightarrow A' \text{ is invertible} \stackrel ? \Leftrightarrow \det A' \ne 0 \Leftrightarrow \det A \ne 0 \]

and it remains to show the middle equivalence.

The matrix \(A'\) is in reduced row echelon form. Thus, either \(A' = {\mathrm {id}}\) or \(A'\) contains a zero row. In the first event, \(A'\) is invertible, and \(\det A' = 1 \ne 0\). In the second event, \(A'\) is not invertible (by Corollary 4.94) and \(\det A' = 0\) as was noted around .

We skip the proof of the adjugate formula, cf. (Nicholson 1995, Theorem 3.2.4). ◻

Example 5.17 (Related exercises: Exercise 5.8)

For a \(2 \times 2\)-matrix \(A = \left ( \begin{array}{cc} a & b \\ c & d \end{array} \right )\), the adjugate matrix is

\[ \left ( \begin{array}{cc} d & -b \\ -c & a \end{array} \right ). \]

Therefore, the inverse can be computed as

\[ A^{-1} = \frac 1 {ad-bc} \left ( \begin{array}{cc} d & -b \\ -c & a \end{array} \right ). \]

Nicholson, W. K. 1995. *Linear Algebra with Applications*. Mathematics Series. PWS Publishing Company. .