Let \(A\) be an \(n \times n\) matrix with entries from a field. Recall that the determinant of \(A\), denoted by \(\det(A)\), is defined to be \[\sum_{\sigma \in S_n} (-1)^{\text{#inv}(\sigma)} A_{1,\sigma(1)}\cdots A_{n,\sigma(n)}.\]
Here are some properties of the determinant.
\(\det(A^\mathsf{T}) = \det(A).\)
If \(A\) is invertible, then \(\displaystyle\det(A^{-1}) = \frac{1}{\det(A)}.\)
Where \(\alpha\) is a scalar, \(\det(\alpha A) = \alpha^n\det(A).\)
\(\det(A) \neq 0\) if and only if \(A\) is nonsingular.
\( \det(A^\mathsf{T}) = \det(A) = 3.\)
Since \(\det(A)\neq 0\), \(A\) is nonsingular and therefore invertible. Hence, \( \det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{3}.\)
\( \det(2A) = 2^2\det(A) = 4\cdot 3 = 12\) since \(A\) is \(2\times 2\).
\( \det(-A) = \det((-1)A) = (-1)^2\det(A) = \det(A) = 3\). Notice that in this case, \(\det(-A) = \det(A)\)!
We now give a proof of this important result.
Let \(R\) be the RREF of \(A\). Then there exist elementary matrices \(M_1,\ldots,M_k\) such that \(M_{k} M_{k-1} \cdots M_1 A = R\). Using the fact that the determinant of a product of square matrices is the same as the product of the determinants of the matrices, we get \[\det(M_k) \det(M_{k-1})\cdots \det(M_1) \det(A) = \det(R).\]
As \(R\) is a square matrix, \(A\) is nonsingular iff \(R\) does not contain a row of 0's. If \(R\) does contain a row of 0's, then \(\det(R) = 0\). Otherwise, \(R\) must be the identity matrix with determinant 1.
Hence, to complete the proof, it suffices to show that \(\det(M_i) \neq 0\) for all \(i = 1,\ldots,k\). Note that every elementary matrix is either a triangular matrix with nonzeros on the diagonal, or a permutation matrix which has nonzero determinant. Hence, \(\det(M_i) \neq 0\) for all \(i = 1,\ldots, k\) as desired.
With this result, we can state that if \(A \in \mathbb{F}^{n\times n}\) for some field \(\mathbb{F}\), then the following statements are equivalent:
\(A\) is invertible.
\(A\) is nonsingular.
\(\det(A) \neq 0\).
Compute the determinants of each of the following matrices:
\(2\begin{bmatrix} 2 & 3 \\ 0 & 2\end{bmatrix}\)
\(\begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{bmatrix}^\mathsf{T}\)
Prove that if \(A\) is invertible, then \(\displaystyle\det(A^{-1}) = \frac{1}{\det(A)}.\)