Matrix Equations and Linear Equations

Introduction

Abstract – We explain how every system of linear equations can be turned into a single matrix equation and vice versa. In the case where the unaugmented coefficient matrix is invertible, we show how to solve the matrix equation and, hence, the underlying system of linear equations using the inverse matrix.

Outline – Our starting point here is that a system of linear equations is equivalent to a single matrix equation of the form$A\cdot X=C$. Here $C$ is the augmentation column of the system, $A$ is its unaugmented coefficient matrix, and $X$ is the column matrix formed from its variables. This re-interpretation of linear equations leads to the following additional insights:

Structure of solution set: The solutions of a system $(E)$ of linear equations are always of the form $\Vect{u}+\Vect{z}$, where $\Vect{u}$ is some arbitrily chosen solution of $(E)$, and $\Vect{z}$ can be any solution of the associated system of homogeneous linear equations.
In the special case, where the coefficient matrix of the system is invertible, we obtain something really beautiful and simple: Let's go back to the situation where we worked only with numbers. Then we solve an equation of the form $ax=c$ by dividing both sides by $a$. If $A$ is invertible, we can do the same with this matrix equation!
\[X = A^{-1}\cdot C\]
is the one and only solution of the system of linear equations. – Note that we multiplied both sides of the equation by $A^{-1}$ on the left.

The method in (2) above is particularly effective if one needs to solve many systems of linear equations with the same unaugmented coefficient matrix $A$.

PropositionSolutions of linear matrix equations

Let $\Mtrx{A}$ and $\Mtrx{C}$ be matrices of size $(m,n)$ and $(m,1)$, respectively. For an unknown $(n,1)$-matrix $\Mtrx{X}$, consider the matrix equation

\[(E)\qquad \Mtrx{A}\cdot \Mtrx{X} = \Mtrx{C}\]

together with the associated homogeneous equation

\[(E_0)\qquad \Mtrx{A}\cdot \Mtrx{X} = \ZMtrx{(m,1)}.\]

Let $Z$ be the set of solutions of $(E_0)$, and let $\Mtrx{U}$ be a particular solution of $(E)$. Then $\Mtrx{X}$ solves $(E)$ if and only if

\[\Mtrx{X} = \Mtrx{U} + \Mtrx{Y}\]

for some $\Mtrx{Y}$ in $Z$.

Let us translate this result into the language of linear equations:

CorollarySolutions of linear matrix equations

Consider the system $(E)$ of linear equations

\[ \begin{array}{rcccrcr} \colorbox{lightgreen}{$a_{11}$} {\color{red} x_1} & + & \cdots & + & \colorbox{lightgreen}{$a_{1n}$} {\color{red} x_n} & = & c_1 \\ \vdots\ \ \ & & & & \vdots\ \ \ & & \vdots\ \ \\ \colorbox{lightgreen}{$a_{m1}$} {\color{red} x_1} & + & \cdots & + & \colorbox{lightgreen}{$a_{mn}$} {\color{red} x_n} & = & c_m \end{array} \]

together with the associated system $(E_0)$ of homogeneous linear equations.

\[ \begin{array}{rcccrcr} \colorbox{lightgreen}{$a_{11}$} {\color{red} x_1} & + & \cdots & + & \colorbox{lightgreen}{$a_{1n}$} {\color{red} x_n} & = & 0 \\ \vdots\ \ \ & & & & \vdots\ \ \ & & \vdots\ \ \\ \colorbox{lightgreen}{$a_{m1}$} {\color{red} x_1} & + & \cdots & + & \colorbox{lightgreen}{$a_{mn}$} {\color{red} x_n} & = & 0 \end{array} \]

If $\Vect{u}$ is a particular solution of $(E)$, then $\Vect{x}=(x_1,\dots,x_n)$ solves $(E)$ if and only if $\Vect{x} = \Vect{u} + \Vect{y}$ for some solution $\Vect{y}$ of $(E_0)$.

Proposition

Suppose the coefficient matrix $\Mtrx{A}$ of the system of $n$ linear equations in $n$ variables

\[ \begin{array}{rcccrcr} \colorbox{lightgreen}{$a_{11}$} {\color{red} x_1} & + & \cdots & + & \colorbox{lightgreen}{$a_{1n}$} {\color{red} x_n} & = & c_1 \\ \vdots\ \ \ & & & & \vdots\ \ \ & & \vdots\ \ \\ \colorbox{lightgreen}{$a_{n1}$} {\color{red} x_1} & + & \cdots & + & \colorbox{lightgreen}{$a_{nn}$} {\color{red} x_n} & = & c_n \end{array} \]

is invertible. Then this system has the unique solution

\[ {\color{red}\begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}}\ =\ A^{-1} \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix} \]

Study Materials

Matrix Equations and Linear Equations: Exercises