$A = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}$: $\det = 6 - 5 = 1 \neq 0$. Invertible. $A^{-1} = \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}$.

$B = \begin{pmatrix} 2 & 6 \\ 1 & 3 \end{pmatrix}$: $\det = 6 - 6 = 0$. Singular.

$A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$: $\det = 0$ (the third row is the sum of the other two rows minus the first, or compute directly: $1(45-48) - 2(36-42) + 3(32-35) = -3 + 12 - 9 = 0$). Singular.

$B = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{pmatrix}$: $\det = 1(50-48) - 2(40-42) + 3(32-35) = 2 + 4 - 9 = -3 \neq 0$. Invertible.

$\det(AB) = \det(A)\det(B)$. So:

• $AB$ is invertible iff both $A$ and $B$ are invertible.
• $\det(A^n) = (\det A)^n$.
• $\det(A^{-1}) = 1/\det(A)$.

$\det(A^T) = \det(A)$, so $A$ is invertible iff $A^T$ is invertible. This means: the rows of $A$ are linearly independent iff the columns are linearly independent (which also follows from row rank = column rank).

$\det\begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix} = \det(A) \cdot \det(B).$

So this block diagonal matrix is invertible iff both $A$ and $B$ are invertible.

For which values of $t$ is $A = \begin{pmatrix} 1 & t \\ t & 1 \end{pmatrix}$ invertible?

$\det(A) = 1 - t^2 = (1-t)(1+t)$. So $A$ is invertible for all $t \neq \pm 1$.

$\det(A) = 0$ means the columns of $A$ are linearly dependent. Geometrically, the column vectors lie in a lower-dimensional subspace, so the parallelepiped they span has zero volume.

$\det(A) = \prod_{i=1}^n \lambda_i$ where $\lambda_i$ are the eigenvalues (counted with algebraic multiplicity). So $\det(A) = 0$ iff some eigenvalue is $0$ iff $A$ has a nontrivial kernel.

Over $\mathbb{F}_2 = \{0, 1\}$: $\det\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = 0 - 1 = -1 = 1 \neq 0$ (in $\mathbb{F}_2$). The matrix is invertible over $\mathbb{F}_2$. Its inverse is $\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$.

If $A$ is orthogonal ($A^T A = I$), then $\det(A)^2 = \det(A^T)\det(A) = \det(I) = 1$, so $\det(A) = \pm 1$. The special orthogonal group $\text{SO}(n)$ consists of orthogonal matrices with $\det = +1$ (rotations).

The set $\{A \in M_{n \times n}(\mathbb{R}) \mid \det(A) = 0\}$ is a single polynomial equation in $n^2$ variables. It is a hypersurface of dimension $n^2 - 1$ in $M_{n \times n}(\mathbb{R}) \cong \mathbb{R}^{n^2}$. The invertible matrices form the complement -- a dense open set.

If $B = P^{-1}AP$, then $\det(B) = \det(P)^{-1}\det(A)\det(P) = \det(A)$. The determinant is a similarity invariant, hence an invariant of the underlying linear operator $T$ (independent of the choice of basis).