Chapter 11.4, Problem 57AYU

To determine

To solve: system of equation using matrix inverse.

Answer to Problem 57AYU

Solution:

  $x=-\frac{34}{7},y=\frac{85}{7}$ and $z=\frac{12}{7}$

Explanation of Solution

Given:

  $x+y+z=9$

  $3x+2y-z=8$

  $3x+y+2z=1$

Calculation:

The onginal system of equations can be written compactly as a matrix equation.

  $AX=B$

Where,

  $A=\left[\begin{array}{ccc}1& 1& 1\\ 3& 2& -1\\ 3& 1& 2\end{array}\right],X=\left[\begin{array}{l}x\hfill \\ y\hfill \\ z\hfill \end{array}\right]\mathrm{and}B=\left[\begin{array}{l}9\hfill \\ 8\hfill \\ 1\hfill \end{array}\right]$

To solve the for $x$ and $y$ , the solution of $X$ has to be found as given below:

  $AX=B$

Multiply both sides by $A^{-1}$

  $A^{-1}(AX)=A^{-1}B$

By association property of matrix multiplication, the above equation can be rewnitten as,

  $\left(A^{-1}A\right)X=A^{-1}B$

By the defmition of inverse $\mathrm{matrix}\left(A^{-1}A\right)=I_n$

  $I_{n}X=A^{-1}B$

By the property of Identity matrix $I_{n}X=X$ .

Hence,

  $X=A^{-1}B$

ie $\left[\begin{array}{l}x\hfill \\ y\hfill \\ z\hfill \end{array}\right]=A^{-1}B$

Determining $A^{-1}$

Step 1: Construction $\mathrm{ef}\left[A\mid I_n\right]$

  

Step 2: Sequentially transforming the matrix $\left[A\mid I_n\right]$ into reduced row echelon fom.

  

Step 3: Separate the matrix on the right of the vertical bar which is the inverse of $A$ .

  $A^{-1}=\left[\begin{array}{ccc}-\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]$

Solving $x_p,x$ and z.

  $\left[\begin{array}{l}x\hfill \\ y\hfill \\ z\hfill \end{array}\right]=A^{-1}B$

  $=\left[\begin{array}{ccc}-\frac{5}{7}& \frac{1}{7}& \frac{2}{7}\\ \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]\left[\begin{array}{l}9\hfill \\ 8\hfill \\ 1\hfill \end{array}\right]$

  $=\left[\begin{array}{c}-\frac{45}{7}+\frac{8}{7}+\frac{3}{7}\\ \frac{91}{7}+\frac{8}{7}-\frac{4}{7}\\ \frac{27}{7}-\frac{16}{7}+\frac{1}{7}\end{array}\right]$

  $=\left[\begin{array}{c}-\frac{34}{7}\\ \frac{85}{7}\\ \frac{12}{7}\end{array}\right]$

The solution for the given system of equations is $x=-\frac{24}{7},y=\frac{85}{7}$ and $z=\frac{12}{7}$ .