Determinants, and Interpretation Using Areas

Suppose that we have the following $2\times 2$ matrix: $$A=\left[ \begin{array}{ccc} a & b \\ c & d \end{array} \right]$$ The determinant of $A$, denoted $\det(A)$, is calculated as follows: $$\det(A)=\left| \begin{array}{ccc} a & b \\ c & d \end{array} \right|=ad-bc$$ The determinant is an extremely important property of a matrix and has many interesting applications, which will not be discussed here but rather on pages where they will hold more relevance.

The product of two square matrices has a determinant equal to the product of the determinants of the two matrices. Thus, while matrix multiplication is not commutative, determinant multiplication certainly is: $$\det(AB)=\det(A)\det(B)$$ We will only prove this for $2\times 2$ matrices here. To prove this, denote $A$ and $B$ as follows: $$A=\left[ \begin{array}{ccc} a_1 & a_2 \\ a_3 & a_4 \end{array} \right]$$ $$B=\left[ \begin{array}{ccc} b_1 & b_2 \\ b_3 & b_4 \end{array} \right]$$ Now, calculate the products $AB$ and $BA$: $$AB=\left[ \begin{array}{ccc} a_1b_1+a_2b_3 & a_1b_2+a_2b_4 \\ a_3b_1+a_4b_3 & a_3b_2+a_4b_4 \end{array} \right]$$ $$BA=\left[ \begin{array}{ccc} a_1b_1+a_3b_2 & a_2b_1+a_4b_2 \\ a_1b_3+a_3b_4 & a_2b_3+a_4b_4 \end{array} \right]$$ Now, calculate all of the desired determinants: $$\det(A)=a_1a_4-a_2a_3$$ $$\det(B)=b_1b_4-b_2b_3$$ $$\det(AB)=(a_1b_1+a_2b_3)(a_3b_2+a_4b_4)-(a_1b_2+a_2b_4)(a_3b_1+a_4b_3)$$ $$\det(BA)=(a_1b_1+a_3b_2)(a_2b_3+a_4b_4)-(a_2b_1+a_4b_2)(a_1b_3+a_3b_4)$$ Calculating the product of the determinants of $A$ and $B$, $$\det(A)\det(B)=(a_1a_4-a_2a_3)(b_1b_4-b_2b_3)$$ Some straightforward (though somewhat tedious) algebra gives us our desired result.

Using this fact, we can derive a very important corollary, often recognized on its own: $$AA^{-1}=A^{-1}A=I$$ $$\det(I)=\det(A)\det(A^{-1})$$ $$1=\det(A)\det(A^{-1})$$ $$\det(A^{-1})=\frac{1}{\det(A)}$$ This theorem, along with the corollary we just showed above, will serve a great purpose in many of the applications of the determinant.

Here is a worksheet to practice finding $2x2$ determinants: http://cdn.kutasoftware.com/Worksheets/PreAlg/Translations of Shapes.pdf

Since there is no explanation posted yet, I will do question 1 here as an example. Call the element in the top left corner ($0$) $a$, the element in the top right corner ($-4$) $b$, the element in the bottom left corner ($-6$) $c$, and the element in the bottom right corner ($-2$) $d$. So the determinant is $ad - bc = 0(-2) - (-4)(-6) = {-24}$.