5.1 Two-by-Two Matrices: Introduction

5.1.1 Linear Equations of Two Unknowns

Consider the system of linear equations for the two unknowns x and y,

\begin{eqnarray} ax + by& =& e %& \\ cx + dy& =& f,%&(5.1) \\ \end{eqnarray}

where a,b,c,d,e,f are constant numbers. This system can be easily solved: solve the first equation for y,

\begin{eqnarray} y = {e − ax\over b} & & %&(5.2) \\ \end{eqnarray}

and insert it into the second equation,

\begin{eqnarray} cx + dy& =& cx + {de − adx\over b} = f ⇝ (cb − ad)x = fb − de %& \\ x& =& {de − fb\over ad − cb} %& \\ y& =& {e − ax\over b} = {e(ad − cb) − a(de − fb)\over b(ad − cb)} = {af − ec\over ad − cb}.%&(5.3) \\ \end{eqnarray}

For this general solution for x and y to be valid, the denominator ad − cb apparently has to be different from zero.

5.1.2 Two–by–Two Matrices: Definition

We write the two unknowns x and y as the components of a two–dimensional vector x,

\begin{eqnarray} x := \left (\array{ x\cr y } \right ).& & %&(5.4)\\ \end{eqnarray}

Then, we write the two constants e and f as the components of a two–dimensional vector v

\begin{eqnarray} v := \left (\array{ e\cr f } \right ).& & %&(5.5)\\ \end{eqnarray}

The two–by–two system of linear equations, Eq. (5.1), maps the vector x onto the vector v. We write this in the following abstract form:

\begin{eqnarray} Ax& =& v ⇔\left (\array{ a&b\cr c &d } \right )\left (\array{ x\cr y } \right ) = \left (\array{ e\cr f } \right ),%&(5.6)\\ \end{eqnarray}

where we defined the two–by–two matrix

\begin{eqnarray} A := \left (\array{ a&b\cr c &d } \right ).& & %&(5.7)\\ \end{eqnarray}

A two–by–two matrix is a quadratic scheme which, upon operating on a vector x on its right, transforms this vector into another vector v according to the rule

\begin{eqnarray} Ax = \left (\array{ a&b\cr c &d } \right )\left (\array{ x\cr y } \right ) = \left (\array{ ax + by\cr cx + dy } \right ) = v.& & %&(5.8)\\ \end{eqnarray}

By comparison we recognise that this matrix equation, Ax = v, is equivalent to the system Eq.(5.1).

5.1.3 Linear Mappings and Matrix Operatings

A linear mapping A from {ℝ}^{2} → {ℝ}^{2} maps a vector x onto the vector Ax. The mapping is represented by a two-by-two matrix A. The mapping must fulfill {x}_{1} +{ x}_{2} → A({x}_{1} +{ x}_{2}) = A{x}_{1} + A{x}_{2},\quad λx → A(λx) = λAx,\quad λ ∈ ℝ.

The above can be generalised (trivially) to complex matrices; the mapping is then from {ℂ}^{2} → {ℂ}^{2}, and λ can also be complex

Examples

\begin{eqnarray} A& =& \left (\array{ 1& 3\cr 2 & −1 } \right ),\quad {x}_{1} = \left (\array{ 1\cr −2 } \right ),\quad {x}_{2} = \left (\array{ 1\cr 1 } \right ),\quad {x}_{1} +{ x}_{2} = \left (\array{ 2\cr −1 } \right )%& \\ ⇝ A{x}_{1}& =& \left (\array{ 1& 3\cr 2 & −1 } \right )\left (\array{ 1\cr −2 } \right ) = \left (\array{ 1 ⋅ 1 + 3 ⋅ (−2)\cr 2 ⋅ 1 + (−1) ⋅ (−2) } \right ) = \left (\array{ − 5\cr 4 } \right ) %& \\ A{x}_{2}& =& \left (\array{ 1& 3\cr 2 & −1 } \right )\left (\array{ 1\cr 1 } \right ) = \left (\array{ 4\cr 1 } \right ). %&(5.9)\\ \end{eqnarray}

We compare this to

\begin{eqnarray} A& =& \left (\array{ 1& 3\cr 2 & −1 } \right ),\quad {x}_{1} = \left (\array{ 1\cr −2 } \right ),\quad {x}_{2} = \left (\array{ 1\cr 1 } \right ),\quad {x}_{1} +{ x}_{2} = \left (\array{ 2\cr 1 } \right ) %& \\ ⇝ A({x}_{1} +{ x}_{2})& =& \left (\array{ 1& 3\cr 2 & −1 } \right )\left (\array{ 2\cr −1 } \right ) = \left (\array{ − 1\cr 5 } \right ) = \left (\array{ − 5\cr 4 } \right ) + \left (\array{ 4\cr 1 } \right ) = A{x}_{1} + A{x}_{2} ⇝\text{OK.}%&\\ \end{eqnarray}