Chapter 9.9, Problem 24WE

To determine

To find:the general expression for the solution other than $0$ if its exist.

Answer to Problem 24WE

There is no others solution.

Explanation of Solution

Given:

  $\begin{array}{l}x+2y-2z=0.\\ 2x+5y+2z=0.\\ 3x+4y-2z=0.\end{array}$ .

Concept used:

Composite equations can be converted into matrices in the form of:

  $AX=B$ .

  $A$ and $B$ are matrices.

  $X$ is variable.

Matrices multiplication:

  $\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ b_{21}& b_{22}& b_{23}\\ c_{31}& c_{32}& c_{33}\end{array}\right]\left[\begin{array}{c}{d}_1\\ d_2\\ d_3\end{array}\right]=\left[\begin{array}{c}{a}_{11}.d_1+a_{12}{d}_2+a_{13}{d}_3\\ b_{21}.d_1+b_{22}.d_2+b_{23}.d_3\\ c_{31}.d_1+c_{32}.d_2+c_{33}.d_3\end{array}\right]$ .

Use row and column operation to solve the matrices.

Calculation:

According to the given:

  $\begin{array}{l}x+2y-2z=0.\\ 2x+5y+2z=0.\\ 3x+4y-2z=0.\end{array}$

It can be written in the form of matrices as:

  $\left[\begin{array}{ccc}1& 2& -2\\ 2& 5& 2\\ 3& 4& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right].$

It can solve further by row column operation lets take raw operation only:

  $R_2\to R_2-2R_1$ and

  $\left[\begin{array}{ccc}1& 2& -2\\ 0& 1& 6\\ 3& 4& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right].$

Again, using raw operation as $R_3\to R_3-3R_1$ :

  $\left[\begin{array}{ccc}1& 2& -2\\ 0& 1& 6\\ 0& -2& 4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right].$

Raw operation as $R_3\to R_3+2R_2$ :

  $\left[\begin{array}{ccc}1& 2& -2\\ 0& 1& 6\\ 0& 0& 16\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right].$

The multiplying the matrices:

  $\begin{array}{l}x+2y-2z=0.\\ 0+y+6z=0.\\ 0+0-16z=0.\end{array}$

From the third equations:

  $\begin{array}{l}-16z=0.\\ z=0.\end{array}$

Third equation implies that $z=0$ and that there is no other solution except beside $\left(0,0,0\right)$ .

Hence, there is no other solutions.