Finding det(A) with standard basis vectors

Question

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 $\det(A)$?

Answer 1

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}

Answer 2

Since $A$ is 2 by 2 matrix, its first column is $Ae_1=[-3, 7]$ and second column is $Ae_2=[3,5]$
$Det(A) =\left|\matrix{-3 & 3\\5& 7}\right|$

Answer 3

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.