Determinants of 2 × 2 Matrices

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There is a simple way for determining whether a $2\times 2$ matrix $A$ is invertible and there is a simple formula for finding $A\inv$ . First, we present the formula. Let $A=\mattwo {a}{b}{c}{d}.$ and suppose that $ad-bc\neq 0$ . Then

$\begin{equation} \label{e:formAinv} A\inv = \frac{1}{ad-bc} \mattwo{d}{-b}{-c}{a}. \end{equation}$ This is most easily verified by directly applying the formula for matrix multiplication. So $A$ is invertible when $ad-bc\neq 0$ . We shall prove below that $ad-bc$ must be nonzero when $A$ is invertible.

From this discussion it is clear that the number $ad-bc$ must be an important quantity for $2\times 2$ matrices. So we define:

The determinant of the $2\times 2$ matrix $A$ is $\begin{equation} \label{D:determinant} \det(A) = ad - bc. \end{equation}$

As a function on $2\times 2$ matrices, the determinant satisfies the following properties.

The determinant of an upper triangular matrix is the product of the diagonal elements.
The determinants of a matrix and its transpose are equal.
$\det (AB) = \det (A)\det (B)$ .

Proof: Both (a) and (b) are easily verified by direct calculation. Property (c) is also verified by direct calculation — but of a more extensive sort. Note that $\left (\begin {array}{cc} a & b\\ c & d \end {array}\right ) \left (\begin {array}{cc} \alpha & \beta \\ \gamma & \delta \end {array}\right ) = \left (\begin {array}{cc} a\alpha +b\gamma & a\beta +b\delta \\ c\alpha +d\gamma & c\beta +d\delta \end {array}\right ).$ Therefore,
$\begin{eqnarray*} \det(AB) & = & (a\alpha+b\gamma)(c\beta+d\delta) - (a\beta+b\delta)(c\alpha+d\gamma)\\ & = & (ac\alpha\beta+bc\beta\gamma+ad\alpha\delta+bd\gamma\delta) \\ & & -(ac\alpha\beta+bc\alpha\delta+ad\beta\gamma+bd\gamma\delta)\\ & = & bc(\beta\gamma-\alpha\delta) + ad(\alpha\delta-\beta\gamma) \\ & = & (ad-bc)(\alpha\delta-\beta\gamma) \\ & = & \det(A)\det(B), \end{eqnarray*}$
as asserted. $\blacksquare$

A $2\times 2$ matrix $A$ is invertible if and only if $\det (A)\neq 0$ .

Proof: If $A$ is invertible, then $AA\inv = I_2$ . Proposition ?? implies that $\det (A)\det (A\inv ) = \det (I_2) = 1.$ Therefore, $\det (A)\neq 0$ . Conversely, if $\det (A)\neq 0$ , then (??) implies that $A$ is invertible. $\blacksquare$

Determinants and Area

Suppose that $v$ and $w$ are two vectors in $\R ^2$ that point in different directions. Then, the set of points $z=\alpha v + \beta w \quad \mbox {where } 0\leq \alpha ,\beta \leq 1$ is a parallelogram, that we denote by $P$ . We denote the area of $P$ by $|P|$ . For example, the unit square $S$ , whose corners are $(0,0)$ , $(1,0)$ , $(0,1)$ , and $(1,1)$ , is the parallelogram generated by the unit vectors $e_1$ and $e_2$ .

Next let $A$ be a $2\times 2$ matrix and let $A(P) = \{Az:z\in P\}.$ It follows from linearity (since $Az=\alpha Av+\beta Aw$ ) that $A(P)$ is the parallelogram generated by $Av$ and $Aw$ .

Let $A$ be a $2\times 2$ matrix and let $S$ be the unit square. Then $\begin{equation} \label{e:det&area2} |A(S)| = |\det A|. \end{equation}$

Proof: Note that $A(S)$ is the parallelogram generated by $u_1=Ae_1$ and $u_2=Ae_2$ , and $u_1$ and $u_2$ are the columns of $A$ . It follows that $(\det A)^2=\det (A^t)\det (A)=\det (A^tA) = \det \mattwo {u_1^tu_1}{u_1^tu_2}{u_2^tu_1}{u_2^tu_2}.$ Hence $(\det A)^2= \det \mattwo {||u_1||^2}{u_1\cdot u_2}{u_1\cdot u_2}{||u_2||^2} = ||u_1||^2||u_2||^2-(u_1\cdot u_2)^2.$ Recall that (??) of Chapter ?? states that $|P|^2 = ||v||^2||w||^2 - (v\cdot w)^2.$ where $P$ is the parallelogram generated by $v$ and $w$ . Therefore, $(\det A)^2 = |A(S)|^2$ and (??) is verified. $\blacksquare$

Let $P$ be a parallelogram in $\R ^2$ and let $A$ be a $2\times 2$ matrix. Then $\begin{equation} \label{e:det&area} |A(P)| = |\det A||P|. \end{equation}$

Proof: First note that (??) a special case of (??), since $|S|=1$ . Next, let $P$ be the parallelogram generated by the (column) vectors $v$ and $w$ , and let $B=(v|w)$ . Then $P=B(S)$ . It follows from (??) that $|P|=|\det B|$ . Moreover,
$\begin{eqnarray*} |A(P)| & = & |(AB)(S)| \\ & = & |\det(AB)| \\ & = & |\det A||\det B| \\ & = & |\det A||P|, \end{eqnarray*}$
as desired. $\blacksquare$

Exercises

Find the inverse of the matrix $\mattwo {2}{1}{3}{2}.$

Find the inverse of the shear matrix $\mattwo {1}{K}{0}{1}$ .

Show that the $2\times 2$ matrix $A=\mattwo {a}{b}{c}{d}$ is row equivalent to $I_2$ if and only if $ad-bc\neq 0$ . Hint: Prove this result separately in the two cases $a\neq 0$ and $a=0$ .

Let $A$ be a $2\times 2$ matrix having integer entries. Find a condition on the entries of $A$ that guarantees that $A\inv$ has integer entries.

Let $A$ be a $2\times 2$ matrix and assume that $\det (A)\neq 0$ . Then use the explicit form for $A\inv$ given in (??) to verify that $\det (A\inv ) = \frac {1}{\det (A)}.$

Sketch the triangle whose vertices are $0$ , $p=(3,0)^t$ , and $q=(0,2)^t$ ; and find the area of this triangle. Let $M=\mattwo {-4}{-3}{5}{-2}.$ Sketch the triangle whose vertices are $0$ , $Mp$ , and $Mq$ ; and find the area of this triangle.

Cramer’s rule provides a method based on determinants for finding the unique solution to the linear equation $Ax=b$ when $A$ is an invertible matrix. More precisely, let $A$ be an invertible $2\times 2$ matrix and let $b\in \R ^2$ be a column vector. Let $B_j$ be the $2\times 2$ matrix obtained from $A$ by replacing the $j^{th}$ column of $A$ by the vector $b$ . Let $x=(x_1,x_2)^t$ be the unique solution to $Ax=b$ . Then Cramer’s rule states that $\begin{equation} \label{E:cramer} x_j = \frac{\det(B_j)}{\det(A)}. \end{equation}$ Prove Cramer’s rule. Hint: Write the general system of two equations in two unknowns as

$\begin{eqnarray*} a_{11}x_1+a_{12}x_2 & = & b_1\\ a_{21}x_1+a_{22}x_2 & = & b_2. \end{eqnarray*}$

Subtract $a_{11}$ times the second equation from $a_{21}$ times the first equation to eliminate $x_1$ ; then solve for $x_2$ , and verify (??). Use a similar calculation to solve for $x_1$ .

In Exercises ?? – ?? use Cramer’s rule (??) to solve the given system of linear equations.

Solve $\begin {array}{rcl} 2x+3y & = & 2 \\ 3x - 5y & = & 1 \end {array}$ for $x$ .

Solve $\begin {array}{rcl} 4x-3y & = & -1 \\ x + 2y & = & 7 \end {array}$ for $y$ .

Use MATLAB to choose five $2\times 2$ matrices at random and compute their inverses. Do you get the impression that ‘typically’ $2\times 2$ matrices are invertible? Try to find a reason for this fact using the determinant of $2\times 2$ matrices.

In Exercises ?? – ?? use the unit square icon in the program map to test Proposition ??, as follows. Enter the given matrix $A$ into map and map the unit square icon. Compute $\det (A)$ by estimating the area of $A(S)$ — given that $S$ has unit area. For each matrix, use this numerical experiment to decide whether or not the matrix is invertible.

$A=\mattwo {0}{-2}{2}{0}$ .

$A=\mattwo {-0.5}{-0.5}{0.7}{0.7}$ .

$A=\mattwo {-1}{-0.5}{-2}{-1}$ .

$A=\mattwo {0.7071}{0.7071}{-0.7071}{0.7071}$ .