§ 19. Determinants of Order 2¶

Definition and Properties of Determinants of Order 2¶

Definition 19.1 : Determinant of a \(2 \times 2\) Matrix

If

\[ A=\left(\begin{array}{ll} a & b \\ c & d \end{array}\right) \]

is a \(2 \times 2\) matrix with entries from a field \(F\), then we define the determinant of \(A\), denoted \(\operatorname{det}(A)\) or \(|A|\), to be the scalar \(a d-b c\).

Theorem 19.2 : Linearity of determinant in each row

The function \(\operatorname{det}: \mathrm{M}_{2 \times 2}(F) \rightarrow F\) is a linear function of each row of a \(2 \times 2\) matrix when the other row is held fixed. That is, if \(u\), \(v\), and \(w\) are in \(F^{2}\) and \(k\) is a scalar, then

\[ \operatorname{det}\binom{u+k v}{w}=\operatorname{det}\binom{u}{w}+k \operatorname{det}\binom{v}{w} \]

and

\[ \operatorname{det}\binom{w}{u+k v}=\operatorname{det}\binom{w}{u}+k \operatorname{det}\binom{w}{v} . \]

Proof

Let \(u=\left(a_{1}, a_{2}\right), v=\left(b_{1}, b_{2}\right)\), and \(w=\left(c_{1}, c_{2}\right)\) be in \(F^{2}\) and \(k\) be a scalar. Then

\[ \operatorname{det}\binom{u}{w}+k \operatorname{det}\binom{v}{w}=\operatorname{det}\left(\begin{array}{ll} a_{1} & a_{2} \\ c_{1} & c_{2} \end{array}\right)+k \operatorname{det}\left(\begin{array}{ll} b_{1} & b_{2} \\ c_{1} & c_{2} \end{array}\right) \]

\[ \begin{aligned} & =\left(a_{1} c_{2}-a_{2} c_{1}\right)+k\left(b_{1} c_{2}-b_{2} c_{1}\right) \\ & =\left(a_{1}+k b_{1}\right) c_{2}-\left(a_{2}+k b_{2}\right) c_{1} \\ & =\operatorname{det}\left(\begin{array}{cc} a_{1}+k b_{1} & a_{2}+k b_{2} \\ c_{1} & c_{2} \end{array}\right) \\ & =\operatorname{det}\binom{u+k v}{w} . \end{aligned} \]

A similar calculation shows that

\[ \operatorname{det}\binom{w}{u}+k \operatorname{det}\binom{w}{v}=\operatorname{det}\binom{w}{u+k v} . \]

Theorem 19.3 : Invertibility and the determinant

Let \(A \in \mathrm{M}_{2 \times 2}(F)\). Then the determinant of \(A\) is nonzero if and only if \(A\) is invertible. Moreover, if \(A\) is invertible, then

\[ A^{-1}=\frac{1}{\operatorname{det}(A)}\left(\begin{array}{rr} A_{22} & -A_{12} \\ -A_{21} & A_{11} \end{array}\right) . \]

Proof

If \(\operatorname{det}(A) \neq 0\), then we can define a matrix

\[ M=\frac{1}{\operatorname{det}(A)}\left(\begin{array}{rr} A_{22} & -A_{12} \\ -A_{21} & A_{11} \end{array}\right) . \]

A straightforward calculation shows that \(A M=M A=I\), and so \(A\) is invertible and \(M=A^{-1}\).

Conversely, suppose that \(A\) is invertible. Theorem 16.2 shows that the rank of

\[ A=\left(\begin{array}{ll} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right) \]

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

\[ \left(\begin{array}{cc} A_{11} & A_{12} \\ 0 & A_{22}-\frac{A_{12} A_{21}}{A_{11}} \end{array}\right) \]

Because elementary row operations are rank-preserving by Corollary 16.5, it follows that

\[ A_{22}-\frac{A_{12} A_{21}}{A_{11}} \neq 0 \]

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

Theorem 19.4 : Characterization of the \(2 \times 2\) determinant

Let \(\delta: \mathrm{M}_{2 \times 2}(F) \rightarrow F\) be a function with the following three properties.

(a) \(\delta\) is a linear function of each row of the matrix when the other row is held fixed.
(b) If the two rows of \(A \in \mathrm{M}_{2 \times 2}(F)\) are identical, then \(\delta(A)=0\).
(c) If \(I\) is the \(2 \times 2\) identity matrix, then \(\delta(I)=1\).

Prove that \(\delta(A)=\operatorname{det}(A)\) for all \(A \in \mathrm{M}_{2 \times 2}(F)\).

Proof

Let

\[ A=\left(\begin{array}{cc} a & b \\ c & d \end{array}\right) . \]

Write the first row as \(a(1,0)+b(0,1)\) and the second row as \(c(1,0)+d(0,1)\). Using (a) twice, \(\delta\) is linear in each row, so

\[ \delta(A)=ac\,\delta\!\left(\begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array}\right) +ad\,\delta\!\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) +bc\,\delta\!\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) +bd\,\delta\!\left(\begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array}\right) . \]

By (b), the first and last terms are zero (the two rows are identical), so

\[ \delta(A)=ad\,\delta\!\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) +bc\,\delta\!\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) . \]

By (c),

\[ \delta\!\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)=\delta(I)=1 . \]

To compute \(\delta\!\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)\), apply (b) to the matrix whose two rows are both \((1,1)\):

\[ 0=\delta\!\left(\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}\right) . \]

Using (a) to expand in the first row and then the second row gives

\[ \delta\!\left(\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}\right) =\delta\!\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) +\delta\!\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) \]

because the two remaining terms have identical rows and vanish by (b). Hence

\[ 0=1+\delta\!\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right), \]

so

\[ \delta\!\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)=-1 . \]

Substituting back, we obtain

\[ \delta(A)=ad(1)+bc(-1)=ad-bc=\operatorname{det}(A) . \]

Theorem 19.2, Theorem 19.3, and Theorem 19.4 will be generalized to determinants of order \(n\) in section 22.

The Area of a Parallelogram¶

Definition 19.5 : Angle Between Two Vectors

By the angle between two vectors in \(\mathbb{R}^{2}\), we mean the angle with measure \(\theta(0 \leq \theta<\pi)\) that is formed by the vectors having the same magnitude and direction as the given vectors but emanating from the origin. (See Figure 19.1.)

Figure 19.1: Angle between two vectors in \(\mathbb{R}^{2}\).

Definition 19.6 : Orientation of an Ordered Basis

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

\[ \mathrm{O}\binom{u}{v}=\frac{\operatorname{det}\binom{u}{v}}{\left|\operatorname{det}\binom{u}{v}\right|} \]

(The denominator of this fraction is nonzero by Theorem 19.3.) Clearly

\[ \mathrm{O}\binom{u}{v}= \pm 1 \]

For convenience, we also define

\[ \mathrm{O}\binom{u}{v}=1 \]

if \(\{u, v\}\) is linearly dependent.

Definition 19.7 : Right-Handed and Left-Handed Coordinate Systems

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.

Figure 19.2: A right-handed coordinate system.

Figure 19.3: A left-handed coordinate system.

Theorem 19.8 : Orientation and right-handed coordinate systems

In general,

\[ \mathrm{O}\binom{u}{v}=1 \]

if and only if the ordered basis \(\{u, v\}\) forms a right-handed coordinate system.

Definition 19.9 : Parallelogram Determined by Two Vectors

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

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.

We denote

\[ \mathrm{A}\binom{u}{v}, \]

the area of the parallelogram determined by \(u\) and \(v\).

Theorem 19.10 : Area of a parallelogram

\[ \mathrm{A}\binom{u}{v}=\mathrm{O}\binom{u}{v} \cdot \operatorname{det}\binom{u}{v}, \]

from which it follows that

\[ \mathrm{A}\binom{u}{v}=\left|\operatorname{det}\binom{u}{v}\right| . \]

Proof

Our argument that

\[ \mathrm{A}\binom{u}{v}=\mathrm{O}\binom{u}{v} \cdot \operatorname{det}\binom{u}{v} \]

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

\[ \mathrm{O}\binom{u}{v}= \pm 1 \]

we may multiply both sides of the desired equation by

\[ \mathrm{O}\binom{u}{v} \]

to obtain the equivalent form

\[ \mathrm{O}\binom{u}{v} \cdot \mathrm{~A}\binom{u}{v}=\operatorname{det}\binom{u}{v} . \]

We establish this equation by verifying that the three conditions of Theorem 19.4 are satisfied by the function

\[ \delta\binom{u}{v}=\mathrm{O}\binom{u}{v} \cdot \mathrm{~A}\binom{u}{v} \]

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

\[ \delta\binom{u}{c v}=c \cdot \delta\binom{u}{v} . \]

Observe that this equation is valid if \(c=0\) because

\[ \delta\binom{u}{c v}=\mathrm{O}\binom{u}{0} \cdot \mathrm{~A}\binom{u}{0}=1 \cdot 0=0 \]

So assume that \(c \neq 0\). Regarding \(c v\) as the base of the parallelogram determined by \(u\) and \(c v\), we see that

\[ \mathrm{A}\binom{u}{c v}=\text { base } \times \text { altitude }=|c|(\text { length of } v)(\text { altitude })=|c| \cdot \mathrm{A}\binom{u}{v} \]

since the altitude \(h\) of the parallelogram determined by \(u\) and \(c v\) is the same as that in the parallelogram determined by \(u\) and \(v\). (see Figure 19.5)

Figure 19.5.

Hence

\[ \begin{aligned} \delta\binom{u}{c v} & =\mathrm{O}\binom{u}{c v} \cdot \mathrm{~A}\binom{u}{c v}=\left[\frac{c}{|c|} \cdot \mathrm{O}\binom{u}{v}\right]\left[|c| \cdot \mathrm{A}\binom{u}{v}\right] \\ & =c \cdot \mathrm{O}\binom{u}{v} \cdot \mathrm{~A}\binom{u}{v}=c \cdot \delta\binom{u}{v} \end{aligned} \]

A similar argument shows that

\[ \delta\binom{c u}{v}=c \cdot \delta\binom{u}{v} . \]
(b) We next prove that

\[ \delta\binom{u}{a u+b w}=b \cdot \delta\binom{u}{w} \]

for any \(u, w \in \mathbb{R}^{2}\) 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 19.6),

Figure 19.6.

it follows that

\[ \mathrm{A}\binom{u}{w}=\mathrm{A}\binom{u}{u+w} . \]

If \(a=0\), then

\[ \delta\binom{u}{a u+b w}=\delta\binom{u}{b w}=b \cdot \delta\binom{u}{w} \]

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

\[ \delta\binom{u}{a u+b w}=a \cdot \delta\binom{u}{u+\frac{b}{a} w}=a \cdot \delta\binom{u}{\frac{b}{a} w}=b \cdot \delta\binom{u}{w} . \]

So the desired conclusion is obtained in either case. We are now able to show that

\[ \delta\binom{u}{v_{1}+v_{2}}=\delta\binom{u}{v_{1}}+\delta\binom{u}{v_{2}} \]

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

\[ \delta\binom{u}{v_{1}+v_{2}}=\delta\binom{u}{\left(a_{1}+a_{2}\right) u+\left(b_{1}+b_{2}\right) w}=\left(b_{1}+b_{2}\right) \delta\binom{u}{w} \]

\[ =\delta\binom{u}{a_{1} u+b_{1} w}+\delta\binom{u}{a_{2} u+b_{2} w}=\delta\binom{u}{v_{1}}+\delta\binom{u}{v_{2}} . \]

A similar argument shows that

\[ \delta\binom{u_{1}+u_{2}}{v}=\delta\binom{u_{1}}{v}+\delta\binom{u_{2}}{v} \]

for all \(u_{1}, u_{2}, v \in \mathbb{R}^{2}\).
(c) Since

\[ \mathrm{A}\binom{u}{u}=0, \quad \text { it follows that } \quad \delta\binom{u}{u}=\mathrm{O}\binom{u}{u} \cdot \mathrm{~A}\binom{u}{u}=0 \]

for any \(u \in \mathbb{R}^{2}\). Because the parallelogram determined by \(e_{1}\) and \(e_{2}\) is the unit square,

\[ \delta\binom{e_{1}}{e_{2}}=\mathrm{O}\binom{e_{1}}{e_{2}} \cdot \mathrm{~A}\binom{e_{1}}{e_{2}}=1 \cdot 1=1 \]

Therefore \(\delta\) satisfies the three conditions of Theorem 19.4, and hence \(\delta=\operatorname{det}\). So the area of the parallelogram determined by \(u\) and \(v\) equals

\[ \mathrm{O}\binom{u}{v} \cdot \operatorname{det}\binom{u}{v} \]

Example 19.11 : Area of a parallelogram

The area of the parallelogram determined by \(u=(-1,5)\) and \(v=(4,-2)\) is

\[ \left|\operatorname{det}\binom{u}{v}\right|=\left|\operatorname{det}\left(\begin{array}{rr} -1 & 5 \\ 4 & -2 \end{array}\right)\right|=18 \]

Exercise¶

Exercise 19.10

The classical adjoint of a \(2 \times 2\) matrix \(A \in \mathrm{M}_{2 \times 2}(F)\) is the matrix

\[ C=\left(\begin{array}{rr} A_{22} & -A_{12} \\ -A_{21} & A_{11} \end{array}\right) \]

Prove that

(a) \(C A=A C=[\operatorname{det}(A)] I\).
(b) \(\operatorname{det}(C)=\operatorname{det}(A)\).
(c) The classical adjoint of \(A^{t}\) is \(C^{t}\).
(d) If \(A\) is invertible, then \(A^{-1}=[\operatorname{det}(A)]^{-1} C\).