Determinant and matrix multiplication¶

Let $A$ and $B$ be $2 \times 2$ matrices, and let $a > 0$ be an area of some 2D shape. If $B$ is applied to each point of the shape, the area multiplies by $\det(B)$ (TODO), so we get $a\det(B)$. If we also apply $A$, the area then becomes $a\det(B)\det(A)$. On the other hand, first applying $B$ and then $A$ is what the matrix $AB$ does, so $$ a\det(B)\det(A)=a\det(AB), $$ and dividing by $a$ gives $$ \det(AB)=\det(A)\det(B). $$ On the rest of this page, we show that the same works for square matrices of any size.

If $\det(B) \ne 0$¶

Let $f$ be a function that takes in a square matrix and outputs a number, defined by $$ f(A) = \frac{\det(AB)}{\det(B)}. $$ For this to work, we have to assume that $\det(B) \ne 0$. Our goal is to show that this function computes $\det(A)$. Let's check whether it satisfies the conditions in the definition of determinant:

• $f(I) = 1$, because it divides $\det(B)$ with itself.
• Let's consider how $AB$ changes if two rows of $A$ are swapped. For example, with $2 \times 2$ matrices, if $$ A = \begin{bmatrix} \blue1 & \blue2 \\ \green3 & \green4 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} \red5 & \magenta6 \\ \red7 & \magenta8 \end{bmatrix}, $$ then $$ AB = \begin{bmatrix} \blue1 & \blue2 \\ \green3 & \green4 \end{bmatrix} \begin{bmatrix} \red5 & \magenta6 \\ \red7 & \magenta8 \end{bmatrix} = \begin{bmatrix} \blue{(1,2)}\cdot\red{(5,7)} & \blue{(1,2)}\cdot\magenta{(6,8)} \\ \green{(3,4)}\cdot\red{(5,7)} & \green{(3,4)}\cdot\magenta{(6,8)} \end{bmatrix}, $$ and with the rows of $A$ swapped, we would get $$ \begin{bmatrix} \green3 & \green4 \\ \blue1 & \blue2 \end{bmatrix} \begin{bmatrix} \red5 & \magenta6 \\ \red7 & \magenta8 \end{bmatrix} = \begin{bmatrix} \green{(3,4)}\cdot\red{(5,7)} & \green{(3,4)}\cdot\magenta{(6,8)} \\ \blue{(1,2)}\cdot\red{(5,7)} & \blue{(1,2)}\cdot\magenta{(6,8)} \end{bmatrix}. $$ So swapping two rows in a matrix $A$ swaps the corresponding rows in $AB$: $$ (\text{$A$ with two rows swapped})B = \text{$AB$ with two rows swapped} $$ This works the same with square matrices of any size, and we get $$ \begin{align} f(\text{$A$ with two rows swapped}) &= \frac{\det(\text{$AB$ with two rows swapped})}{\det(B)} \\ &= \frac{-\det(AB)}{\det(B)} \\ &= -f(A). \end{align} $$
• Let $m$ be any number, and let $A$ be a square matrix. Then, for any row in $A$, there is a matrix $E$ that multiplies that row by $m$: $$ EA = \text{$A$ with one of the rows multiplied by $m$} $$ Because the determinant is linear as a function of each row, this multiplies the determinant by $m$, so $\det(EA)=m\det(A)$, and we get $$ f(EA)=\frac{\det(EAB)}{\det(B)} =\frac{m\det(AB)}{\det(B)} = mf(A). $$ This almost shows that the function $f$ is linear as a function of each row in $A$, but not quite; we still need to show that $$ f(\text{matrix with row $\vec v + \vec w$}) = f(\text{matrix with row $\vec v$}) + f(\text{matrix with row $\vec w$}). $$
• Let $A$ be a matrix where one of the rows is $\vec v + \vec w$, and there are also other unrelated rows, say $(1,2,3)$ and $(4,5,6)$. Let $\vec{c_1},\vec{c_2},\vec{c_3}$ denote the columns of $B$. $$ A = \begin{bmatrix} 1~~~~2~~~~3 \\ \vec v + \vec w \\ 4~~~~5~~~~6 \end{bmatrix}, \quad B = \begin{bmatrix} && \\ \vec{c_1} & \vec{c_2} & \vec{c_3} \\ && \end{bmatrix} $$ Now (TODO: explain why $(\vec v + \vec w) \cdot \vec c = \vec v \cdot \vec c + \vec w \cdot \vec c$) $$ \begin{align} AB &= \begin{bmatrix} (1,2,3)\cdot\vec{c_1}&(1,2,3)\cdot\vec{c_2}&(1,2,3)\cdot\vec{c_3} \\ (\vec v + \vec w) \cdot \vec{c_1} &(\vec v + \vec w) \cdot \vec{c_2} &(\vec v + \vec w) \cdot \vec{c_3} \\ (4,5,6)\cdot\vec{c_1}&(4,5,6)\cdot\vec{c_2}&(4,5,6)\cdot\vec{c_3} \end{bmatrix} \\ &= \begin{bmatrix} (1,2,3)\cdot\vec{c_1}&(1,2,3)\cdot\vec{c_2}&(1,2,3)\cdot\vec{c_3} \\ \vec v \cdot \vec{c_1} + \vec w \cdot \vec{c_1} & \vec v \cdot \vec{c_2} + \vec w \cdot \vec{c_2} &\vec v\cdot \vec{c_3} + \vec w\cdot \vec{c_3} \\ (4,5,6)\cdot\vec{c_1}&(4,5,6)\cdot\vec{c_2}&(4,5,6)\cdot\vec{c_3} \end{bmatrix}. \end{align} $$ By the linearity of determinant, we get $$ \begin{align} \det(AB) &= \det\begin{bmatrix} (1,2,3)\cdot\vec{c_1}&(1,2,3)\cdot\vec{c_2}&(1,2,3)\cdot\vec{c_3} \\ \vec v \cdot \vec{c_1} & \vec v \cdot \vec{c_2} &\vec v\cdot \vec{c_3} \\ (4,5,6)\cdot\vec{c_1}&(4,5,6)\cdot\vec{c_2}&(4,5,6)\cdot\vec{c_3} \end{bmatrix} \\ &\qquad+ \det\begin{bmatrix} (1,2,3)\cdot\vec{c_1}&(1,2,3)\cdot\vec{c_2}&(1,2,3)\cdot\vec{c_3} \\ \vec w \cdot \vec{c_1} & \vec w \cdot \vec{c_2} &\vec w\cdot \vec{c_3} \\ (4,5,6)\cdot\vec{c_1}&(4,5,6)\cdot\vec{c_2}&(4,5,6)\cdot\vec{c_3} \end{bmatrix} \\ &= \det(A_{\vec v} ~ B)+\det(A_{\vec w} ~ B), \end{align} $$ where $A_{\vec v}$ and $A_{\vec w}$ are just like $A$, but with $\vec v$ or $\vec w$ instead of the row $\vec v + \vec w$. Dividing both sides by $\det(B)$ shows that $$ f(A)=f(A_{\vec v})+f(A_{\vec w}). $$

Because the function $f$ satisfies the conditions in the definition of determinant, it must be the determinant, and therefore $$ \det(A) = f(A) = \frac{\det(AB)}{\det(B)} $$ for all square matrices $A$ of the same size as $B$.

If $\det(B) = 0$¶

Let's now see what happens when the above calculations don't work due to division by zero. Because $\det(B)=0$, the columns of $B$ are linearly dependent, so $B$ does not always produce different outputs for different inputs, i.e. there are different vectors $\vec v \ne \vec w$ so that $$ B\vec v = B\vec w. $$ Multiplying from left by $A$ gives $$ AB\vec v = AB\vec w, $$ so $AB$ is not invertible, and $$ \det(AB) = 0. $$ This means that $\det(AB)=\det(A)\det(B)$ works even if $\det(B)=0$; in that case, it says that $0=0$.