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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 $\det A = ad$

  1. Where did I go wrong?
  2. 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.

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  • $\begingroup$ Thanks for the catch, I fixed the matrix. $\endgroup$ Nov 9, 2012 at 0:39

1 Answer 1

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$$ \det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = a\det\begin{bmatrix} 1 & b/a \\ c & d \end{bmatrix} = a\det\begin{bmatrix} 1 & b/a \\ 0 & (-cb/a)+d \end{bmatrix} $$ In the second step I added $-c$ times the first row to the second; that does not change the determinant. Now we can add a constant times the second row to the first to get \begin{align} & = a\det\begin{bmatrix} 1 & 0 \\ 0 & (-cb/a)+d \end{bmatrix} \\[8pt] & = a\cdot\left(\frac{-cb}{a}+d\right)\det\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[8pt] & = (ad-bc)\det\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \end{align}

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  • $\begingroup$ You can also use the relation $^{t}Com(A)A=\det(A)I_{2}$ to catch the value of $\det(A)$. $\endgroup$ Aug 1, 2022 at 16:32

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