Chapter 11.2, Problem 66AYU

To determine

To find: The solution to the given system of equations using matrices.

Answer to Problem 66AYU

Solution:

Inconsistent system, No solutions.

Explanation of Solution

Given:

$x+2y-z=3$

$2x-y+2z=6$

$x-\text{ 3}y+3z=4$

Calculation:

To solve a system of three equations in $x,y$ and $z$ using matrices:

Step 1: Write the corresponding matrix associated with the system of equations.

Step 2: Use elementary row operations to get equivalent matrix of the form:

$\left[\begin{array}{ccc}1& a& b\\ 0& 1& c\\ 0& 0& 1\end{array}|\begin{array}{c}d\\ e\\ f\end{array}\right]$; where $a,b,c,d,e,f$ are constants.

Step 3: Solve for $x,y$ and $z$.

$x+2y-z=3$

$2x-y+2z=6$

$x-\text{ 3}y+3z=4$

The corresponding matrix associated with the above system of equations is:

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

$\approx \left[\begin{array}{ccc}1& 2& -1\\ 0& -5& 4\\ 0& -5& 4\end{array}|\begin{array}{c}3\\ 0\\ 1\end{array}\right];R_2=-2r_1+r_2;R_3=r_3-r_1$

$\approx \left[\begin{array}{ccc}1& 2& -1\\ 0& -5& 4\\ 0& 0& 0\end{array}|\begin{array}{c}3\\ 0\\ 1\end{array}\right];R_3^{\text{'}}=R_3-R_2$

$\begin{array}{l}\approx \left[\begin{array}{ccc}1& 2& -1\\ 0& 1& -\frac{4}{5}\\ 0& 0& 0\end{array}|\begin{array}{c}3\\ 0\\ 1\end{array}\right];R_2^{\text{'}}=-\frac{1}{5}{R}_2\\ \end{array}$

We see that the rank of the coefficient matrix is 2 and rank of augmented matrix is 3.

Since the rank of coefficient matrix and the rank of augmented matrix are not equal; therefore the given system of equations has no solutions.

Hence the given system of equations is inconsistent.