Chapter 11.4, Problem 77AYU

To determine

To solve: The system of equations algebraically by any method.

Answer to Problem 77AYU

Solution:

  $x=-4,y=2,z=\frac{5}{2}$

Explanation of Solution

Given:

  $x-2y+4z=2$

  $-3x+5y-2z=17$

  $4x-3y=-22$

Calculation:

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

  $AX=B$

Where,

  $A=\left[\begin{array}{ccc}1& -2& 4\\ -3& 5& -2\\ 4& -3& 0\end{array}\right],X=\left[\begin{array}{l}x\hfill \\ y\hfill \\ z\hfill \end{array}\right]$ and $B=\left[\begin{array}{c}2\\ 17\\ -22\end{array}\right]$

To solve the for x, y and = 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 rewritten as::

  $\left(A^{-1}A\right)X=A^{-1}B$

By the definition of inverse 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,

  $X=A^{-1}B$

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

Determining $A^{-1}$

Step 1: Construction of $\left[A\mid I_n\right]$

Step 2: Transform the matrix $\left[A\mid I_n\right]$ into reduced row echelon form.

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

  $A^{-1}=\left[\begin{array}{ccc}\frac{3}{17}& \frac{6}{17}& \frac{8}{17}\\ \frac{4}{17}& \frac{8}{17}& \frac{5}{17}\\ \frac{11}{34}& \frac{5}{34}& \frac{1}{34}\end{array}\right]$

Solving x, y and $z$

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

  $=\left[\begin{array}{ccc}\frac{3}{17}& \frac{6}{17}& \frac{8}{17}\\ \frac{4}{17}& \frac{8}{17}& \frac{5}{17}\\ \frac{11}{34}& \frac{5}{34}& \frac{1}{34}\end{array}\right]\left[\begin{array}{c}2\\ 17\\ -22\end{array}\right]$

  $=[\begin{array}{l}\frac{6}{17}+6-\frac{176}{17}\hfill \\ \frac{8}{17}+8-\frac{110}{17}\hfill \\ \frac{22}{34}+\frac{85}{34}-\frac{22}{34}\hfill \end{array}$

  $=\left[\begin{array}{c}-4\\ 2\\ 5\\ \frac{2}{2}\end{array}\right]$

The solution for the given system of equations is $x=-4,y=2,z=\frac{5}{2}$ .