Proof:

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. To determine if $A$ is invertible, augment $A$ with the identity matrix $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and row reduce: $$\left[ \begin{array}{cc|cc} a & b & 1 & 0 \\ c & d & 0 & 1 \end{array} \right].$$

First, divide row 1 by $a$ (assuming $a \neq 0$): $$\left[ \begin{array}{cc|cc} 1 & \frac{b}{a} & \frac{1}{a} & 0 \\ c & d & 0 & 1 \end{array} \right].$$ Next, subtract $c \times \text{row 1}$ from row 2: $$\left[ \begin{array}{cc|cc} 1 & \frac{b}{a} & \frac{1}{a} & 0 \\ 0 & d - \frac{bc}{a} & -\frac{c}{a} & 1 \end{array} \right].$$

Now, divide row 2 by $d - \frac{bc}{a}$: $$\left[ \begin{array}{cc|cc} 1 & \frac{b}{a} & \frac{1}{a} & 0 \\ 0 & 1 & \frac{-c}{ad - bc} & \frac{a}{ad - bc} \end{array} \right].$$ Finally, subtract $\frac{b}{a} \times \text{row 2}$ from row 1: $$\left[ \begin{array}{cc|cc} 1 & 0 & \frac{d}{ad - bc} & \frac{-b}{ad - bc} \\ 0 & 1 & \frac{-c}{ad - bc} & \frac{a}{ad - bc} \end{array} \right].$$

The left side is now the identity matrix, and since $A$ can be row reduced to the identity matrix, we know that A is invertible.

$$A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.$$ We also know that if $ad - bc = 0$, the denominator is undefined, and $A$ is not invertible. Thus, $A$ is invertible if and only if $ad - bc \neq 0$.

Proof:

Let $A$ be an $n \times n$ matrix. Suppose $A$ is invertible. Since $A$ is invertible then there exists a sequence of invertible elementary matrices $E_p, \dots, E_1$ such that: $$E_p \cdots E_1 A = I_n.$$ Each $E_i$ corresponds to an elementary row operation, which is invertible. Applying this sequence of operations reduces $A$ to $I_n$, so $A$ is row-equivalent to $I_n$.

Conversely, suppose $A$ is row-equivalent to $I_n$. Then there exists a sequence of invertible elementary row operation matrices $E_p, \dots, E_1$ such that: $$E_p \cdots E_1 A = I_n.$$ Since the product of invertible matrices is invertible, $A$ must also be invertible.

To find $A^{-1}$, augment $A$ with $I_n$ to form: $$\left[ \begin{array}{c|c} A & I_n \end{array} \right].$$ Apply the same sequence of elementary row operations $E_p, \dots, E_1$ that reduces $A$ to $I_n$. These operations simultaneously transform $I_n$ into $A^{-1}$, because the row operations correspond to multiplying $A$ and $I_n$ by the same sequence of invertible matrices. At the end of the process, the augmented matrix becomes: $$\left[ \begin{array}{c|c} I_n & A^{-1} \end{array} \right].$$ Thus, the sequence of row operations $E_p, \dots, E_1$ that reduces $A$ to $I_n$ simultaneously transforms $I_n$ into $A^{-1}$.

Therefore, $A$ is invertible if and only if $A$ is row-equivalent to $I_n$, and the sequence of elementary row operations $E_p, \dots, E_1$ transforms $I_n$ into $A^{-1}$.

Proof:

(a): Suppose $A$ is an invertible matrix. Then there exists a matrix $A^{-1}$ such that: $$A A^{-1} = I \quad \text{and} \quad A^{-1} A = I.$$ Let $B = (A^{-1})^{-1}$. Then $A^{-1} B = I$ and $B A^{-1} = I$. So then $B$ must equal $A$. Therefore, $(A^{-1})^{-1} = A$.

(b): Let $A$ and $B$ be $n \times n$ invertible matrices. Suppose $C = B^{-1} A^{-1}$. Then we have that: $$(AB)(B^{-1} A^{-1}) = A(BB^{-1})A^{-1}\\ = A I A^{-1} = A A^{-1} = I,$$ and: $$(B^{-1} A^{-1})(AB) = B^{-1} (A^{-1} A) B\\ = B^{-1} I B = B^{-1} B = I.$$ Since both conditions hold, $C = B^{-1} A^{-1}$. Thus, $(AB)^{-1} = B^{-1} A^{-1}$.

(c): Suppose $A$ is an invertible matrix. Then $A^{-1}$ exists such that: $$A A^{-1} = I \quad \text{and} \quad A^{-1} A = I.$$ Taking the transpose of both sides: $$(A A^{-1})^T = I^T \quad \text{and} \quad (A^{-1} A)^T = I^T.$$ By the properties of transposes: $$(A^{-1})^T A^T = I \quad \text{and} \quad A^T (A^{-1})^T = I.$$ Thus, $(A^T)^{-1} = (A^{-1})^T$.

Proof:

If $A$ is an invertible $n \times n$ matrix, then by definition, there exists $A^{-1}$ such that: $$A A^{-1} = I \quad \text{and} \quad A^{-1} A = I.$$ Suppose we are solving the equation $A \vec{x} = \vec{b}$ for $\vec{x}$. Since $A$ is invertible, we can multiply both sides by $A^{-1}$ on the left: $$A^{-1}(A \vec{x}) = A^{-1} \vec{b}.$$ By associativity of matrix multiplication, the left-hand side simplifies to: $$(A^{-1} A) \vec{x} = A^{-1} \vec{b}.$$ Since $A^{-1} A = I$, this becomes: $$I \vec{x} = A^{-1} \vec{b}.$$ Thus, $\vec{x} = A^{-1} \vec{b}$.

To show uniqueness, suppose there exist two solutions, $\vec{x}_1$ and $\vec{x}_2$, such that: $$A \vec{x}_1 = \vec{b} \quad \text{and} \quad A \vec{x}_2 = \vec{b}.$$ Subtracting these equations gives: $$A (\vec{x}_1 - \vec{x}_2) = \vec{0}.$$ Since $A$ is invertible, the only solution to $A \vec{z} = \vec{0}$ is $\vec{z} = \vec{0}$. Therefore: $$\vec{x}_1 - \vec{x}_2 = \vec{0},$$ which implies $\vec{x}_1 = \vec{x}_2$.

Thus, $A \vec{x} = \vec{b}$ has the unique solution $\vec{x} = A^{-1} \vec{b}$.