MATRIX

In the derivation of the vector space the components of the versors, (c_j's) were determined as the solutions to the sets of linear systems of equations. Linear systems of equations appear frequently and their solutions depend on the number of the unknowns and the number of the equations. In general, there may be a unique set of solutions, infinitely many or no solutions at all.

The simplest linear equation consists of one unknown:

$a\; x\; =\; b,$

where a and b are the known quantities and x is the unknown. If a ≠ 0, then the unique solution is

$x\; =\; b\; /\; a={\mathrm{a}}^{\mathrm{-1}} b$, where ${\mathrm{a}}^{\mathrm{-1}}$ is the multiplicative inverse of $a$ (its reciprocal).

If $a\; =\; 0$, then there are two possibilities: if $b\; =\; 0$, then there are infinitely many solutions: x can be any number. If, on the other hand, $b\; \ne \; 0$, then there is no solution at all. So, the possibilities are: no solution, one solution or infinitely many.

Next, consider a slightly more complicated case of a coupled linear system of two equations and two unknowns:

$a\; x\; +\; b\; y\; =\; c$
$d\; x\; +\; e\; y\; =\; f,$

where the coefficients $a,\; b,\; c,\; d,\; e,\; f$ are the known quantities, with $x$ and $y$ as the unknowns. It will be instructive to express the two coupled equations compactly in a form that resembles the previous case involving one unknown. To that end, the four known quantities on the left hand side can be collected and arranged into a square array of two rows and two columns:

$A\; =$ $\left(\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{d}& \mathrm{e}\end{array}\right)$.

Similarly, the two unknowns on the left, and the two known quantities on the right, each, forms a column of length two:

$X\; =$ $\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$

$B\; =$ $\left(\begin{array}{c}\mathrm{c}\\ \mathrm{f}\end{array}\right)$.

Putting them all together, the linear system of two equations and two unknowns can now be expressed in the following compact form:

$AX\; =\; B$.

These rectangular arrays of numbers are called matrices. A square matrix has equal number of rows and columns. In special cases where the matrix consists of only one row or one column they are called row/column matrices or equivalently row/column vectors. The original linear system of equations is thus transformed into an equivalent compact matrix equation.

The matrix equation also reveals an important property of matrices: they can be multiplied just like numbers as indicated on the left hand side. The product of the two matrices A and X is defined in the only natural way possible: it is defined to reproduce the left hand side of the original system of equations it was obtained from:

$\left(\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{d}& \mathrm{e}\end{array}\right)$ $\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)$ := $\left(\begin{array}{c}\mathrm{a\; x\; +\; b\; y}\\ \mathrm{d\; x\; +\; e\; y}\end{array}\right)$.

The multiplication can be readily extended to arbitrary matrices of compatible dimensions: the number of the columns of the first matrix must always be equal to the number of the rows of the second matrix.

The following matrix

$I\; =$ $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

serves as the neutral element for the multiplication of 2×2 matrices: $IM\; =\; MI\; =\; M$, for arbitrary 2×2 matrix M, the same way 1 is the neutral element for multiplication among numbers: $1a\; =\; a1\; =\; a$, for arbitrary a. Note that the product of matrices, in general, is not commutative: $MN\; \ne \; NM$. Matrices can also be added component wise:

$\left(\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right)$ + $\left(\begin{array}{cc}\mathrm{e}& \mathrm{f}\\ \mathrm{g}& \mathrm{h}\end{array}\right)$ = $\left(\begin{array}{cc}\mathrm{a+e}& \mathrm{b+f}\\ \mathrm{c+g}& \mathrm{d+h}\end{array}\right)$.

Notice that for this to be possible, the two matrices must be of the same dimensions.

Solving the matrix equation $AX\; =\; B$ follows a similar procedure to that of $ax\; =\; b$. When $a\; \ne \; 0$ the unique solution was obtained by multiplying both sides with its multiplicative inverse (its reciprocal), $x\; =a^{\mathrm{-1}}b$. Note that if a number $a$ has inverse, ${\mathrm{a}}^{\mathrm{-1}}$ then $a{a}^{\mathrm{-1}}=a^{\mathrm{-1}}a\; =\; 1$. Analogously, if the coefficient matrix $A$ has multiplicative inverse, $A^{-1}$, then $A{A}^{-1}=A^{-1}A\; =\; I$, where $I$ is the identity matrix, the multiplicative neutral element. It behaves similar to number 1. The solution to the matrix equation is now obtained as follows:

$AX\; =\; B$
${\mathrm{A}}^{-1}AX\; =A^{-1}B$,   (multiplying from the left on both sides of the eq. by ${\mathrm{A}}^{-1}$)
$IX\; =A^{-1}B$
$X\; =A^{-1}B$.

When the coefficient matrix $A$ has no inverse, (a singular matrix), it is similar to the case $a\; =\; 0$; there may be infinitely many or no solution at all.