Question

In Problems 20–25, solve each system of equations using matrices. If the system has no solution, say that it is inconsistent.

$\left\{\begin{array}{l}5x-6y-3z=6\\ 4x-7y-2z=-3\\ 3x+y-7z=1\end{array}\right.$

Short answer

The solutions to the given equation are$x=\frac{33}{4},y=\frac{25}{6},z=\frac{41}{12}$

Step-by-step solution

Step 1. Given information

The given system of equations is$\left\{\begin{array}{l}5x-6y-3z=6\\ 4x-7y-2z=-3\\ 3x+y-7z=1\end{array}\right.$

Step 2. Solve system of equations using matrix

$A=\left[\begin{array}{ccc}5& -6& -3\\ 4& -7& -2\\ 3& 1& -7\end{array}\right],X=\left[\begin{array}{l}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}6\\ -3\\ 5\end{array}\right]$

$$\begin{gathered} AX=B \\ X=A^{-1}B \end{gathered}$$

The inverse of the matrix $A$can now be determined by using the graphical utility.

So,

$A^{-1}=\left[\begin{array}{ccc}\frac{17}{16}& -\frac{15}{16}& -\frac{3}{16}\\ \frac{11}{24}& -\frac{13}{24}& -\frac{1}{24}\\ \frac{25}{48}& -\frac{23}{48}& -\frac{11}{48}\end{array}\right]$

$$\begin{gathered} X=A^{-1}B \\ \begin{array}{rl}X& =\left[\begin{array}{c}6\\ -3\\ 5\end{array}\right]\left[\begin{array}{ccc}\frac{17}{16}& -\frac{15}{16}& -\frac{3}{16}\\ \frac{11}{24}& -\frac{13}{24}& -\frac{1}{24}\\ \frac{25}{48}& -\frac{23}{48}& -\frac{11}{48}\end{array}\right]\\ & =\left[\begin{array}{c}6\times \frac{17}{16}+(-3)\times (-\frac{15}{16})+5\times (-\frac{3}{16})\\ 6\times \frac{11}{24}+(-3)\times (-\frac{13}{24})+5\times (-\frac{1}{24})\\ 6\times \frac{25}{48}+(-3)\times (-\frac{23}{48})+5\times (-\frac{11}{48})\end{array}\right]\\ & =\left[\begin{array}{c}\frac{102}{16}+\frac{45}{16}-\frac{15}{16}\\ \frac{66}{24}+\frac{39}{24}-\frac{5}{24}\\ \frac{150}{48}+\frac{69}{48}-\frac{55}{48}\end{array}\right]\\ & =\left[\begin{array}{c}\frac{33}{4}\\ \frac{25}{6}\\ \frac{41}{12}\end{array}\right]\end{array} \end{gathered}$$

The solutions to the given equation are$x=\frac{33}{4},y=\frac{25}{6},z=\frac{41}{12}$