Finding det(A) with standard basis vectors For example, let $e_1  = [1 \quad 0]$ and $e_2 = [0 \quad 1] $be standard basis vectors. A is a $2 \times 2$ matrix. $Ae_1 = [-3\quad 7]$ and $Ae2 = [3 \quad 5] $ 
How do I find the $\det(A)$?
 A: Let $a,b,c,d \in \mathbb{R}$, then
\begin{align}
&A = \begin{bmatrix}
a&b \\ c&d
\end{bmatrix}\begin{bmatrix}
1 \\ 0
\end{bmatrix} = \begin{bmatrix}
-3 \\ 7
\end{bmatrix} \implies a = -3, \quad \text{and} \quad c = 7 \\
&A = \begin{bmatrix}
a&b \\ c&d
\end{bmatrix}\begin{bmatrix}
0 \\ 1
\end{bmatrix} = \begin{bmatrix}
3 \\ 5
\end{bmatrix} \implies b = 3, \quad \text{and} \quad d = 5.\\
& \qquad\text{therefore, }\det \begin{vmatrix}
A 
\end{vmatrix} = \det\begin{vmatrix}
-3 & 3 \\ 7 & 5
\end{vmatrix}  = -15-21 = -36.
\end{align}
A: Since $A$ is 2 by 2 matrix, its first column is $Ae_1=[-3, 7]$ and second column is $Ae_2=[3,5]$$Det(A) =\left|\matrix{-3 & 3\\5& 7}\right|$
A: Since A is a 2x2 matrix, simply use the formula for finding determinants for 2x2 matrices. 
\begin{bmatrix}a&b\\c&d\end{bmatrix}
The formula is ad-bc. In this case, A$e_1$ = \begin{bmatrix}-3\\7\end{bmatrix}
And A$e_2$ = \begin{bmatrix}3\\5\end{bmatrix}
Combine these vectors to get: \begin{bmatrix}-3&3\\7&5\end{bmatrix}
Using the formula, det(A) = (-3 $\cdot$ 5) - (3 $\cdot$ 7) = (-15) - (21) = -36.
