The determinant od the $2 \times 2$ matrix \(A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix}\) is $det(A) = a_{11} a_{22} - a_{12} a_{21}$
(Note: You already knew this, and you already know how to calculate determinants for larger matrices too, so I’m skipping this)
a) If $A$ has a zero row, then $det(A) = 0$
b) If $B$ is obtained by interchanging two rows (or columns) of A, then $det(B) = -det(A)$
c) If $A$ has two identical rows (or columns), then $det(A) = 0$
d) If $B$ is obtained by multiplying a row (or column) of $A$ by $k$ then $det(B) = kdet(A)$
e) If $C$ is identical to two matrices $A$ and $B$ in all but one row (or column), and that column is the sum of the corresponding columns in $A$ and $B$, then $det(C) = det(A) + det(B)$
f) If $B$ is obtained by adding a multiple of one row (or column) of A to another row (or column), then $det(B) = det(A)$
AN elementary matrix is one that differs from the identity matrix by a single elementary row transformation, (row exchange, multiplication of row by a constant, or addition of a multiple of another row to the current row).
As a result, the determinant of the corresponding elementary matrix must either be $-1, k$, or $1$ respectively for these three types of elementary matrices.
Premultiplying any other matrix by an elementary matrix corresponds to applying the corresponding row transformation on a matrix itself.
Theorem: Let $B$ be an $n \times n$ matrix and let $E$ be an $n \times n$ elementary matrix, then $det(EB) = det(E)\ det(B)$
Proof:
Since we know that premultiplying by an elementary matrix corresponds to applying the corresponding row transformation on a matrix itself, and that the elementary matrix can have determinant equal to either $-1, k$ or $1$, we can prove this be case analysis.
Case 1: $E$ is an exchange matrix with $det(E) = -1$
In this case, $EB$ would correspond to the matrix $B$ with a single row exchange performed, and thus $det(EB) = -det(B)$.
Since $det(E) = -1$, we can see that $det(E)\ det(B) = - det(B)$.
Case 2: $E$ be a row multiplication matrix with $det(E) = k$
In this case, $EB$ correspons to the matrix $B$ with a single row multiplied by the constant $k$,
and thus $det(B) = k\ det(B)$
Since $det(E) = k$, then here too we can see that $det(E)\ det(B) = k\ det(B)$
Case 3: $E$ be a row addition matrix with $det(E) = 1$
In this case, $EB$ would correspond to the matrix $B$ with a single row addition, and thus $det(EB) = det(B)$.
Since $det(B) = 1$, then here again we can see that $det(E)\ det(B) = det(B)$
Theorem: A square matrix is invertible iff $det(A) \neq 0$
Proof:
Let $A$ be a $n \times n$ matrix and $R$ be the RREF of $A$. Then by the theorem of invertible matrices, $R = I_n$.
We also know that we can perform the conversion to RREF through a set of elementary row operations,
which in turn can be represented as a set of premultiplications with the corresponding elementary matrices, i.e.
And we know all of these $det(E_i)$s to be non-zero, thus
\[c\ det(A) = det(R)\]where $c \neq 0$, and thus,
\[det(A) \neq 0 \iff det(R) \neq 0\]Conversely, suppose that we know that $det(A) \neq 0$, then this implies that it’s RREF $R$ must be $I_n$. Again we can say
Thus $A^{-1} = (E_r E_{r-1} … E_2 E_1)$, the inverse exists and is given.