Given a transformation [math]T:\mathbb{R}^n\to \mathbb{R}^n[/math] such that [math]T(v)=v[/math] for any [math]v[/math] in [math]\mathbb{R}^n[/math] i.e. [math]T[/math] does "nothing". Then by definition, it is obviously a linear transformation and we call it an [b]identity transformation[/b], which is usually denoted by [math]Id[/math]. And the matrix for an identity transformation is called an [b]identity matrix[/b], denoted by [math]I_n[/math] (or just [math]I[/math] if there is no ambiguity). Explicitly, we have[br][br][math]I_n=\begin{pmatrix} 1 & 0 & \cdots & \cdots & 0 \\ 0 & 1 & \cdots & \cdots & 0 \\ \vdots & \vdots & \ddots & & \vdots \\ \vdots & \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & \cdots & 1\end{pmatrix}[/math][br][br]([u]Note[/u]: It is an n x n matrix if the domain and range of [math]Id[/math] is [math]\mathbb{R}^n[/math].)[br][br]In general, a matrix is called a [b]square matrix[/b] if its number of rows equals its number of columns. A square matrix is called a [b]diagonal matrix[/b] if its entries outside the main diagonal are zero. Therefore, an identity matrix is always a diagonal matrix.[br][br]Given any linear transformation [math]S:\mathbb{R}^n\to \mathbb{R}^n[/math]. We would like to know if there is a transformation that "reverses" the action of [math]S[/math]. If it exists, we call it the [b]inverse transformation[/b] of S, denoted by [math]S^{-1}[/math]. It can be shown that [math]S^{-1}[/math] is again a linear transformation (Why?) and [math]S^{-1}\circ S=S\circ S^{-1}=Id[/math].[br][br]Suppose [math]A[/math] is the n x n matrix for the linear transformation [math]S[/math]. Then we denote the matrix for its inverse transformation [math]S^{-1}[/math] by [math]A^{-1}[/math]. Since the composition of linear transformations corresponds to the matrix multiplication, we have[br][br][math]A^{-1}A=AA^{-1}=I[/math][br][br]From the viewpoint of matrix algebra, the inverse of a matrix plays a similar role as "reciprocal" for real numbers. You can see the analogy between "[math]A^{-1}A=I[/math]" and "[math]\frac 1a \cdot a = 1[/math]". When you solve a simple equation like "[math]2x=3[/math]", you can immediately write down the solution [math]\frac 32[/math]. But the underlying procedure to obtain such a solution goes like this:[br][br][math] \frac 12\cdot 2x = \frac 12 \cdot 3 \Rightarrow 1\cdot x = \frac 32 \Rightarrow x = \frac 32[/math][br][br]Recall that a system of n linear equation with n variables can be expressed as a matrix equation [math]Ax=b[/math], where [math]A[/math] is the n x n matrix containing all the coefficients in the system. We can use the same idea to solve for the solution if [math]A^{-1}[/math] exists:[br][br][math]A^{-1}Ax=A^{-1}b \Rightarrow Ix=A^{-1}b\Rightarrow x=A^{-1}b[/math][br][br]Therefore, solving such a system of linear equations boils down to finding the inverse of A. However, this works only when [math]A^{-1}[/math] exists. Usually, we say [math]A[/math] is [b]invertible[/b] (or [b]nonsingular[/b]) if [math]A^{-1}[/math] exists. Then [math]A[/math] is [b]not invertible[/b] (or [b]singular[/b]) if [math]A^{-1}[/math] does not exist. Even if [math]A[/math] is invertible, finding its inverse is usually not an easy task, especially when the size of the matrix is large. We will talk about the methods for computing inverses later in the course.[br][br] [br]

Although it is not easy to compute the inverse of a general matrix, we do have a simple formula for the inverse of an invertible 2 x 2 matrix. The following is the main result:[br][br][b]Theorem[/b]: Let [math]A=\begin{pmatrix} a & b\\c & d\end{pmatrix}[/math]. If [math]ad-bc\ne 0[/math], then [math]A[/math] is invertible and [math]A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d & -b\\-c & a\end{pmatrix}[/math][br]If [math]ad-bc=0[/math], then [math]A[/math] is not invertible.[br][br][b]Proof[/b]: We can do the direct verification to show the formula works. As for the case when [math]ad-bc=0[/math], what can you say about the column vectors [math]\begin{pmatrix}a\\c\end{pmatrix}[/math] and [math]\begin{pmatrix}b\\d\end{pmatrix}[/math]?[br][br]In the above theorem, we have not only a formula for the inverse of a 2 x 2 matrix, but also a condition for checking either a 2 x 2 matrix is invertible or not by checking if [math]ad-bc[/math] is non-zero. This important quantity [math]ad-bc[/math] is called the [b]determinant[/b] of [math]A[/math], denoted by [math]\det A[/math]. Determinants are so important that we will devote a whole chapter to them later.