Chapter 11.4, Problem 81AYU

To determine

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

Answer to Problem 81AYU

Solution:

Unique solution of the given system of equation does not exist.

But, $x\text{ and }y$ can be written as $x=\frac{2\left(1-z\right)}{5}\&y=\frac{z-6}{5}$ and where $z$ is any real number.

Explanation of Solution

Given:

$2x-3y+z=4$

$-3x+2y-z=-3$

$-5y+z=6$

Calculation:

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

$AX=B$

Where,

$A=\left[\begin{array}{ccc}2& -3& 1\\ -3& 2& -1\\ 0& -5& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\text{ and }B=\left[\begin{array}{c}4\\ -3\\ 6\end{array}\right]$

To solve the for $x\text{, }y\text{ and }z$ 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 $\left[A|I_n\right]$.

$\left[A|I_n\right]=[\begin{array}{ccc}2& -3& 1\\ -3& 2& -1\\ 0& -5& 1\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

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

$=[\begin{array}{ccc}2& -3& 1\\ -3& 2& -1\\ 0& -5& 1\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

$\underset{R_1=\left(\frac{1}{2}\right)r_1}{\to }[\begin{array}{ccc}1& -\frac{3}{2}& \frac{1}{2}\\ -3& 2& -1\\ 0& -5& 1\end{array}|\begin{array}{ccc}\frac{1}{2}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

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

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

$\underset{\begin{array}{c}{R}_1=r_1+\left(\frac{3}{2}\right)r_2\\ R_3=r_3+5r_2\end{array}}{\to }[\begin{array}{ccc}1& 0& \frac{1}{5}\\ 0& 1& -\frac{1}{5}\\ 0& 0& 0\end{array}|\begin{array}{ccc}-\frac{2}{5}& -\frac{3}{5}& 0\\ -\frac{3}{5}& -\frac{2}{5}& 0\\ -3& -2& 1\end{array}]$

The matrix $\left[A|I_n\right]$ is sufficiently reduced for it to be clear that the identity matrix cannot appear to the left of the vertical bar. So $A$ is singular and has no inverse.

The given system of equation has no unique solution. Further, $-5y+z=6$ implies $y=\frac{z-6}{5}\text{ and }x=\frac{2\left(1-z\right)}{5}$ where $z$ is any real number.