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,

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

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 $\left[A|I_n\right]$.

$\left[A|I_n\right]=[\begin{array}{ccc}-4& 3& 2\\ 3& 1& -1\\ 1& 9& 1\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

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

$=[\begin{array}{ccc}-4& 3& 2\\ 3& 1& -1\\ 1& 9& 1\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

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

$\underset{R_1\leftrightarrow R_3}{\to }[\begin{array}{ccc}1& 9& 1\\ 3& 1& -1\\ -4& 3& 2\end{array}|\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}]$

$\underset{\begin{array}{c}{R}_2=r_2-3r_1\\ R_3=r_3+4r_1\end{array}}{\to }[\begin{array}{ccc}1& 9& 1\\ 0& -26& -4\\ 0& 39& 6\end{array}|\begin{array}{ccc}0& 0& 1\\ 0& 1& -3\\ 1& 0& 4\end{array}]$

$\underset{R_2=\left(-\frac{1}{26}\right)r_2}{\to }[\begin{array}{ccc}1& 9& 1\\ 0& 1& \frac{2}{13}\\ 0& 39& 6\end{array}|\begin{array}{ccc}0& 0& 1\\ 0& -\frac{1}{26}& \frac{3}{26}\\ 1& 0& 4\end{array}]$

$\underset{\begin{array}{c}{R}_1=r_1-9r_2\\ R_3=r_3-39r_2\end{array}}{\to }[\begin{array}{ccc}1& 0& -\frac{5}{13}\\ 0& 1& \frac{2}{13}\\ 0& 0& 0\end{array}|\begin{array}{ccc}0& \frac{9}{26}& -\frac{1}{26}\\ 0& -\frac{1}{26}& \frac{3}{26}\\ 1& \frac{3}{2}& -\frac{1}{2}\end{array}]$

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 system of equations has infinite solutions.