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,

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

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

$AX=B$

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

$A^{-1}\left(AX\right)=A^{-1}B$

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

$(A^{-1}A)X=A^{-1}B$

By the definition of inverse matrix $(A^{-1}A)=I_n$.

$I_{n}X=A^{-1}B$

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

Hence,

$X=A^{-1}B$

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

Determining $A^{-1}$.

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

$\left[A|I_n\right]=[\begin{array}{ccc}2& 3& -1\\ 4& 0& 3\\ 0& 6& -2\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

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

$=[\begin{array}{ccc}2& 3& -1\\ 4& 0& 3\\ 0& 6& -2\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

$\underset{R_1=\left(\frac{1}{2}\right)r_1}{\to }[\begin{array}{ccc}1& \frac{3}{2}& -\frac{1}{2}\\ 4& 0& 3\\ 0& 6& -2\end{array}|\begin{array}{ccc}\frac{1}{2}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

$\underset{R_2=r_2-4r_1}{\to }[\begin{array}{ccc}1& \frac{3}{2}& -\frac{1}{2}\\ 0& -6& 5\\ 0& 6& -2\end{array}|\begin{array}{ccc}\frac{1}{2}& 0& 0\\ -2& 1& 0\\ 0& 0& 1\end{array}]$

$\underset{R_2=\left(-\frac{1}{6}\right)r_2}{\to }[\begin{array}{ccc}1& \frac{3}{2}& -\frac{1}{2}\\ 0& 1& -\frac{5}{6}\\ 0& 6& -2\end{array}|\begin{array}{ccc}\frac{1}{2}& 0& 0\\ \frac{1}{3}& -\frac{1}{6}& 0\\ 0& 0& 1\end{array}]$

$\underset{\begin{array}{c}{R}_1=r_1-\frac{3}{2}{r}_2\\ R_3=r_3-6r_2\end{array}}{\to }[\begin{array}{ccc}1& 0& \frac{3}{4}\\ 0& 1& -\frac{5}{6}\\ 0& 0& 3\end{array}|\begin{array}{ccc}0& \frac{1}{4}& 0\\ \frac{1}{3}& -\frac{1}{6}& 0\\ -2& 1& 1\end{array}]$

$\underset{R_3=\left( \frac{1}{3}\right)r_3}{\to }[\begin{array}{ccc}1& 0& \frac{3}{4}\\ 0& 1& -\frac{5}{6}\\ 0& 0& 1\end{array}|\begin{array}{ccc}0& \frac{1}{4}& 0\\ \frac{1}{3}& -\frac{1}{6}& 0\\ -\frac{2}{3}& \frac{1}{3}& \frac{1}{3}\end{array}]$

$\underset{\begin{array}{c}{R}_1=r_1-\left( \frac{3}{4}\right)r_3\\ R_2=R_2+\frac{5}{6}{R}_3\end{array}}{\to }[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}\frac{1}{2}& 0& -\frac{1}{4}\\ -\frac{2}{9}& \frac{1}{9}& \frac{5}{18}\\ -\frac{2}{3}& \frac{1}{3}& \frac{1}{3}\end{array}]$

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

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

Solving $x,y\text{ and}z$.

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

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

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

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

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