Q. 82 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 75– 82, algebraically solve each system of equations using any method you wish.

$\left\{\begin{array}{l}-4x+3y+2x=6\\ 3x+y-z=-2\\ x+9y+z=6\end{array}\right.$

Step-by-Step Solution

Answer

The given equation $\left\{\begin{array}{l}-4x+3y+2x=6\\ 3x+y-z=-2\\ x+9y+z=6\end{array}\right.$ has determinant is zero,matrix A is singular.Hence the system has no solution 

Step 1. Given information

The given information$\left\{\begin{array}{l}-4x+3y+2x=6\\ 3x+y-z=-2\\ x+9y+z=6\end{array}\right.$

Step 2. Finding the values of x , y   a n d   z

Consider the equations $-4x+3y+2z=6,3x+y-z=-2 and x+9y+z=6$.To solve these equations we first convert them into matrix equation of the form $AX=B$

$\left[\begin{array}{ccc}-4& 3& 2\\ 3& 1& -1\\ 1& 9& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}6\\ -2\\ 6\end{array}\right]$

Now solve this matrix equation we have to find the inverse matrix A.So,to check whether its inverse is possible or not,we find its determinant

$$\begin{gathered} \left|\begin{array}{ccc}-4& 3& 2\\ 3& 1& -1\\ 1& 9& 1\end{array}\right|=-4\left\{1\left(1\right)-9(-1)\right\}-3\left\{3\left(1\right)-1(-1)\right\}+2\left\{3\left(9\right)-1\left(1\right)\right\} \\ =-4(1+9)-3(3+1)+2(27-1) \\ =-40-12+52 \\ \left|\begin{array}{ccc}-4& 3& 2\\ 3& 1& -1\\ 1& 9& 1\end{array}\right|=0 \end{gathered}$$

As the determinant is zero,matrix A is singular.Hence the system has no solution