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, $x$ and $y$ can be written as $x=\frac{2(1-z)}{5}\text{ \&}y=\frac{z-6}{5}$ 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,

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

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 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.

The matrix $\left[A\mid I_n\right]$ 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 $y=\frac{z-6}{5}$ and $x=\frac{2(1-z)}{5}$ where $z$ is any real number.