Binet's Theorem

Statement of Binet's Theorem

The determinant of the product of two square matrices, det(AB), is equal to the product of their determinants: $$ \det(A \cdot B) = \det(A) \cdot \det(B) $$

Let’s take two square matrices of the same order as an example:

$$ A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} $$

$$ B = \begin{pmatrix} 2 & 4 \\ 1 & 6 \end{pmatrix} $$

Their product is:

$$ AB = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \cdot \begin{pmatrix} 2 & 4 \\ 1 & 6 \end{pmatrix} $$

Expanding the multiplication:

$$ AB = \begin{pmatrix} (1 \cdot 2 + 2 \cdot 1) & (1 \cdot 4 + 2 \cdot 6) \\ (3 \cdot 2 + 4 \cdot 1) & (3 \cdot 4 + 4 \cdot 6) \end{pmatrix} $$

$$ AB = \begin{pmatrix} 2 + 2 & 4 + 12 \\ 6 + 4 & 12 + 24 \end{pmatrix} $$

$$ AB = \begin{pmatrix} 4 & 16 \\ 10 & 36 \end{pmatrix} $$

Now, calculating the determinant of the product:

$$ \det(AB) = \begin{vmatrix} 4 & 16 \\ 10 & 36 \end{vmatrix} = 4 \cdot 36 - 16 \cdot 10 = 144 - 160 = -16 $$

Next, let's compute the determinants of A and B separately:

$$ \det(A) = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 1 \cdot 4 - 2 \cdot 3 = 4 - 6 = -2 $$

$$ \det(B) = \begin{vmatrix} 2 & 4 \\ 1 & 6 \end{vmatrix} = 2 \cdot 6 - 4 \cdot 1 = 12 - 4 = 8 $$

Multiplying the two determinants:

$$ \det(A) \cdot \det(B) = -2 \cdot 8 = -16 $$

Since both calculations yield the same result, we confirm that:

$$ \det(AB) = \det(A) \cdot \det(B) = -2 \cdot 8 = -16 $$

This verifies Binet's theorem.