Chapter 11.4, Problem 78AYU

To determine

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

Answer to Problem 78AYU

Solution:

  $x=-\frac{3}{2},y=\frac{5}{3},z=4$

Explanation of Solution

Given:

  $2x+3y-z=-2$

  $4x+3z=6$

  $6y-2z=2$

Calculation:

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

  $AX=B$

Where,

  $A=\left[\begin{array}{ccc}2& 3& -1\\ 4& 0& 3\\ 0& 6& -2\end{array}\right],X=\left[\begin{array}{l}x\hfill \\ y\hfill \\ z\hfill \end{array}\right]$ and $B=\left[\begin{array}{c}-2\\ 6\\ 2\end{array}\right]$

To solve the for x, y and $z$ 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 defmition 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 $A$ .

  $A^{-1}=\left[\begin{array}{rrr}\hfill \frac{1}{2}& \hfill 0& \hfill -\frac{1}{4}\\ \hfill -\frac{2}{9}& \hfill \frac{1}{9}& \hfill \frac{5}{18}\\ \hfill -\frac{2}{3}& \hfill \frac{1}{3}& \hfill \frac{1}{3}\end{array}\right]$

Solving x, y and $z$

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

  $=\left[\begin{array}{ccc}\frac{1}{2}& 0& -\frac{1}{4}\\ -\frac{2}{9}& \frac{1}{9}& \frac{5}{18}\\ -\frac{2}{3}& \frac{1}{3}& \frac{1}{3}\end{array}\right]\left[\begin{array}{c}-2\\ 6\\ 2\end{array}\right]$

  $=\left[\begin{array}{c}-1-\frac{1}{2}\\ \frac{4}{9}+\frac{6}{9}+\frac{5}{9}\\ \frac{4}{3}+\frac{6}{3}+\frac{2}{3}\end{array}\right]$

  $=\left[\begin{array}{c}-\frac{3}{2}\\ \frac{5}{3}\\ 4\end{array}\right]$

The solution for the given system of equations is $x=-\frac{3}{2},y=\frac{5}{3},z=4$ .