Chapter 11.4, Problem 82AYU

To determine

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

Answer to Problem 82AYU

Solution:

The system of equations has infinite solutions.

Explanation of Solution

Given:

  $-4x+3y+2z=6$

  $3x+y-z=-2$

  $x+9y+z=6$

Calculation:

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

  $AX=B$

Where,

  $A=\left[\begin{array}{ccc}-4& 3& 2\\ 3& 1& -1\\ 1& 9& 1\end{array}\right],X=\left[\begin{array}{l}x\hfill \\ y\hfill \\ z\hfill \end{array}\right]$ and $B=\left[\begin{array}{c}6\\ -2\\ 6\end{array}\right]$

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

Detemining $A^{-1}$ .

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.

  $\text{Interchange rows }{R}_1\text{ and }{R}_3$

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 system of equations has infinite solutions.