If

is a $2\times 2$ matrix with entries from a field F, then we define the determinant of A, denoted det(A) or $\left|A\right|,$ to be the scalar $ad-bc$.

Let $u=(a_1,a_2),v=(b_1,b_2),$ and $w=(c_1,c_2)$ be in ${\text{F}}^2$ and k be a scalar. Then

A similar calculation shows that

For the $2\times 2$ matrices A and B in Example 1, it is easily checked that A is invertible but B is not. Note that $\det (A)\ne 0$ but $\det (B)=0.$ We now show that this property is true in general.

If $\det (A)\ne 0,$, then we can define a matrix

A straightforward calculation shows that $AM=MA=I,$ and so A is invertible and $M=A^{-1}$.

Conversely, suppose that A is invertible. A remark on page 152 shows that the rank of

must be 2. Hence $A_{11}\ne 0$ or $A_{21}\ne 0.$ If $A_{11}\ne 0,$ add $-A_{21}/A_{11}$ times row 1 of A to row 2 to obtain the matrix

Because elementary row operations are rank-preserving by the corollary to Theorem 3.4 (p. 152), it follows that

Therefore $\det (A)=A_{11}{A}_{22}-A_{12}{A}_{21}\ne 0.$. On the other hand, if $A_{21}\ne 0,$ we see that $\det (A)\ne 0$ by adding $-A_{11}/A_{21}$ times row 2 of A to row 1 and applying a similar argument. Thus, in either case, $\det (A)\ne 0$.

In Sections 4.2 and 4.3, we extend the definition of the determinant to $n\times n$ matrices and show that Theorem 4.2 remains true in this more general context. In the remainder of this section, which can be omitted if desired, we explore the geometric significance of the determinant of a $2\times 2$ matrix. In particular, we show the importance of the sign of the determinant in the study of orientation.

By the angle between two vectors in ${\text{R}}^2,$ we mean the angle with measure $\theta (0\le \theta <\pi )$ that is formed by the vectors having the same magnitude and direction as the given vectors but emanating from the origin. (See Figure 4.1.)

If $\beta =\{u,v\}$ is an ordered basis for ${\text{R}}^2,$ we define the orientation of $\beta$ to be the real number

(The denominator of this fraction is nonzero by Theorem 4.2.) Clearly

Notice that

Recall that a coordinate system {u, v} is called right-handed if u can be rotated in a counterclockwise direction through an angle $\theta (0<\theta <\pi )$ to coincide with v. Otherwise {u, v} is called a left-handed system. (See Figure 4.2.) In general (see Exercise 12),

if and only if the ordered basis {u, v} forms a right-handed coordinate system. For convenience, we also define

if {u, v} is linearly dependent.

Any ordered set {u, v} in ${\text{R}}^2$ determines a parallelogram in the following manner. Regarding u and v as arrows emanating from the origin of ${\text{R}}^2$, we call the parallelogram having u and v as adjacent sides the parallelogram determined by u and v. (See Figure 4.3.) Observe that if the set {u, v}

is linearly dependent (i.e., if u and v are parallel), then the “parallelogram” determined by u and v is actually a line segment, which we consider to be a degenerate parallelogram having area zero.

There is an interesting relationship between

the area of the parallelogram determined by u and v, and

which we now investigate. Observe first, however, that since

may be negative, we cannot expect that

But we can prove that

from which it follows that

Our argument that

employs a technique that, although somewhat indirect, can be generalized to ${\text{R}}^n.$ First, since

we may multiply both sides of the desired equation by

to obtain the equivalent form

We establish this equation by verifying that the three conditions of Exercise 11 are satisfied by the function

(a) We begin by showing that for any real number c

Observe that this equation is valid if $c=0$ because

So assume that $c\ne 0.$ Regarding cv as the base of the parallelogram determined by u and cv, we see that

since the altitude h of the parallelogram determined by u and cv is the same as that in the parallelogram determined by u and v. (See Figure 4.4.) Hence

A similar argument shows that

We next prove that

for any u, $w\in {\text{R}}^2$ and any real numbers a and b. Because the parallelograms determined by u and w and by u and $u+w$ have a common base u and the same altitude (see Figure 4.5), it follows that

If $a=0,$ then

by the first paragraph of (a). Otherwise, if $a\ne 0,$ then

So the desired conclusion is obtained in either case.

We are now able to show that

for all u, $v_1,v_2\in {\text{R}}^2.$ Since the result is immediate if $u=0,$ we assume that $u\ne 0.$ Choose any vector $w\in {\text{R}}^2$ such that {u, w} is linearly independent. Then for any vectors $v_1,v_2\in {\text{R}}^2$ there exist scalars $a_i$ and $b_i$ such that $v_i=a_{i}u+b_{i}w(i=1,2).$ Thus

A similar argument shows that

for all $u_1,u_2,v\in {\text{R}}^2$.

(b) Since

for any $u\in {\text{R}}^2$.

(c) Because the parallelogram determined by $e_1$ and $e_2$ is the unit square,

Therefore $\delta$ satisfies the three conditions of Exercise 11, and hence $\delta =\det .$. So the area of the parallelogram determined by u and v equals

Thus we see, for example, that the area of the parallelogram determined by $u=(-1,5)$ and $v=(4,-2)$ is