Show that the determinant of any $2\times 2$ Matrix $A=\pmatrix {a& b\\c&d}$ is $ad-bc$ using the basic definitions.
Proof:
Perform the reduce row reduction algorithm on $A$ using the row vectors:
$\langle a, b \rangle = \dfrac{1}{a} \langle a, b \rangle$ Call this new matrix $A_1$
$\langle c, d\rangle = -c \langle a, b \rangle + \langle c, d \rangle$ Call this new matrix $A_2$
$\langle c, d\rangle = \dfrac{1}{d} \langle c, d \rangle$ Call this new matrix $A_3$
$\langle a, b \rangle = -b \langle c, d \rangle + \langle a, b \rangle$ Call this new matrix $A_4$
After performing all these operations we now have $I_{2 \times 2}$ and we can backtrack through the steps to find the determinant.
$\det A_3 = \det I_{2 \times 2}$ The determinant does not change when we add a linear combination of the row vectors.
$\dfrac{1}{d}\det A_2 = \det A_3$ Since we are scaling the row vector
$\det A_1 = \det A_2$ The determinant does not change when we add a linear combination of the row vectors.
$\dfrac{1}{a}\det A = \det A_1$ Since we are scaling the row vector
Solving we get $\det A = ad$
- Where did I go wrong?
- Why am I missing the $-bc$
I have a question in the definition of the determinant: Let $F$ be an arbitrary field. A determinant is a function which assigns to each n-tuple $\{a_1 \ldots a_n\}$ of vectors in $F_n$ an element of $F, D = D(a_1 \ldots a_n)$ such that the following hold...
What does each n-tuple mean? I think this is where I may have made a mistake.