Matrix Equations

The system of equations

a11x1 + a12x2 + … + a1nxn = b1  
a21x1 + a22x2 + … + a2nxn = b2  
    $\displaystyle \vdots$  
am1x1 + am2x2 + … + amnxn = bm  

is the same as the following matrix equation:

$\displaystyle \left[\vphantom{ \begin{array}{cccc} a_{11} & a_{12} & \ldots ... ... \ldots & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn} \end{array} }\right.$$\displaystyle \begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{2... ...& \vdots & \ldots & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn} \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{cccc} a_{11} & a_{12} & \ldots ... ... \ldots & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn} \end{array} }\right]$$\displaystyle \left[\vphantom{ \begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{array} }\right.$$\displaystyle \begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{array} }\right]$ = $\displaystyle \left[\vphantom{ \begin{array}{c} b_{1} \\ b_{2} \\ \vdots \\ b_{m} \end{array} }\right.$$\displaystyle \begin{array}{c} b_{1} \\ b_{2} \\ \vdots \\ b_{m} \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{c} b_{1} \\ b_{2} \\ \vdots \\ b_{m} \end{array} }\right]$

You can solve these systems using Exact on the Solve submenu. There are advantages to solving systems of equations in this way, and often you can best deal with systems of linear equations by solving the matrix version of the system. Two of the preceding examples correspond to the first two of the following matrix equations. Compare these results with the solutions obtained previously.


\begin{example} Multiply the coefficient matrix $\left[ \begin{array}{rrr} 1... ...y}{r} 1 \\ 0 \\ -1 \end{array} \right] \end{displaymath} \end{example}

$\blacktriangleright$ Solve + Exact

$\left[\vphantom{ \begin{array}{rrr} 1 & 1 & -2 \\ 2 & -4 & 1 \\ 0 & 2 & -3 \end{array} }\right.$$\begin{array}{rrr} 1 & 1 & -2 \\ 2 & -4 & 1 \\ 0 & 2 & -3 \end{array}$$\left.\vphantom{ \begin{array}{rrr} 1 & 1 & -2 \\ 2 & -4 & 1 \\ 0 & 2 & -3 \end{array} }\right]$$\left[\vphantom{ \begin{array}{c} x \\ y \\ z \end{array} }\right.$$\begin{array}{c} x \\ y \\ z \end{array}$$\left.\vphantom{ \begin{array}{c} x \\ y \\ z \end{array} }\right]$ = $\left[\vphantom{ \begin{array}{c} 1 \\ 0 \\ -1 \end{array} }\right.$$\begin{array}{c} 1 \\ 0 \\ -1 \end{array}$$\left.\vphantom{ \begin{array}{c} 1 \\ 0 \\ -1 \end{array} }\right]$, Solution is : $\left[\vphantom{ \begin{array}{c} \frac{17}{8}\vspace{6pt} \\ \vspace{6pt}\frac{11}{8} \\ \frac{5}{4} \end{array} }\right.$$\begin{array}{c} \frac{17}{8}\vspace{6pt} \\ \vspace{6pt}\frac{11}{8} \\ \frac{5}{4} \end{array}$$\left.\vphantom{ \begin{array}{c} \frac{17}{8}\vspace{6pt} \\ \vspace{6pt}\frac{11}{8} \\ \frac{5}{4} \end{array} }\right]$

 

$\left[\vphantom{ \begin{array}{rrr} 2 & -1 & 0 \\ 1 & 0 & 3 \end{array} }\right.$$\begin{array}{rrr} 2 & -1 & 0 \\ 1 & 0 & 3 \end{array}$$\left.\vphantom{ \begin{array}{rrr} 2 & -1 & 0 \\ 1 & 0 & 3 \end{array} }\right]$$\left[\vphantom{ \begin{array}{c} x \\ y \\ z \end{array} }\right.$$\begin{array}{c} x \\ y \\ z \end{array}$$\left.\vphantom{ \begin{array}{c} x \\ y \\ z \end{array} }\right]$ = $\left[\vphantom{ \begin{array}{c} 1 \\ 4 \end{array} }\right.$$\begin{array}{c} 1 \\ 4 \end{array}$$\left.\vphantom{ \begin{array}{c} 1 \\ 4 \end{array} }\right]$, Solution is : $\left[\vphantom{ \begin{array}{c} 4-3t_{1} \\ 7-6t_{1} \\ t_{1} \end{array} }\right.$$\begin{array}{c} 4-3t_{1} \\ 7-6t_{1} \\ t_{1} \end{array}$$\left.\vphantom{ \begin{array}{c} 4-3t_{1} \\ 7-6t_{1} \\ t_{1} \end{array} }\right]$

 

$\left[\vphantom{ \begin{array}{rrrc} 2 & -1 & 0 & 1 \\ 1 & 0 & 3 & 1 \end{array} }\right.$$\begin{array}{rrrc} 2 & -1 & 0 & 1 \\ 1 & 0 & 3 & 1 \end{array}$$\left.\vphantom{ \begin{array}{rrrc} 2 & -1 & 0 & 1 \\ 1 & 0 & 3 & 1 \end{array} }\right]$$\left[\vphantom{ \begin{array}{c} x \\ y \\ z \\ w \end{array} }\right.$$\begin{array}{c} x \\ y \\ z \\ w \end{array}$$\left.\vphantom{ \begin{array}{c} x \\ y \\ z \\ w \end{array} }\right]$ = $\left[\vphantom{ \begin{array}{c} 1 \\ 4 \end{array} }\right.$$\begin{array}{c} 1 \\ 4 \end{array}$$\left.\vphantom{ \begin{array}{c} 1 \\ 4 \end{array} }\right]$, Solution is : $\left[\vphantom{ \begin{array}{c} 4-3t_{1}-t_{2} \\ 7-6t_{1}-t_{2} \\ t_{1} \\ t_{2} \end{array} }\right.$$\begin{array}{c} 4-3t_{1}-t_{2} \\ 7-6t_{1}-t_{2} \\ t_{1} \\ t_{2} \end{array}$$\left.\vphantom{ \begin{array}{c} 4-3t_{1}-t_{2} \\ 7-6t_{1}-t_{2} \\ t_{1} \\ t_{2} \end{array} }\right]$

In the first case, you can also solve the equation by multiplying both the left and right sides of the equation by the inverse of the coefficient matrix, and evaluating the product.

$\blacktriangleright$ Evaluate

$\left[\vphantom{ \begin{array}{c} x \\ y \\ z \end{array} }\right.$$\begin{array}{c} x \\ y \\ z \end{array}$$\left.\vphantom{ \begin{array}{c} x \\ y \\ z \end{array} }\right]$ = $\left[\vphantom{ \begin{array}{rrr} 1 & 1 & -2 \\ 2 & -4 & 1 \\ 0 & 2 & -3 \end{array} }\right.$$\begin{array}{rrr} 1 & 1 & -2 \\ 2 & -4 & 1 \\ 0 & 2 & -3 \end{array}$$\left.\vphantom{ \begin{array}{rrr} 1 & 1 & -2 \\ 2 & -4 & 1 \\ 0 & 2 & -3 \end{array} }\right]^{{-1}}_{}$$\left[\vphantom{ \begin{array}{r} 1 \\ 0 \\ -1 \end{array} }\right.$$\begin{array}{r} 1 \\ 0 \\ -1 \end{array}$$\left.\vphantom{ \begin{array}{r} 1 \\ 0 \\ -1 \end{array} }\right]$ =  $\left[\vphantom{ \begin{array}{c} \vspace{6pt}\frac{17}{8} \\ \vspace{6pt}\frac{11}{8} \\ \frac{5}{4}\vspace*{6pt} \end{array} }\right.$$\begin{array}{c} \vspace{6pt}\frac{17}{8} \\ \vspace{6pt}\frac{11}{8} \\ \frac{5}{4}\vspace*{6pt} \end{array}$$\left.\vphantom{ \begin{array}{c} \vspace{6pt}\frac{17}{8} \\ \vspace{6pt}\frac{11}{8} \\ \frac{5}{4}\vspace*{6pt} \end{array} }\right]$