Chapter 11.2, Problem 65AYU

To determine

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

Answer to Problem 65AYU

Solution:

Consistent system, infinitely many solutions.

Explanation of Solution

Given:

$x+2y+z=1$

$2x-y+2z=2$

$3x+y+3z=3$

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 $z$.

$x+2y+z=1$

$2x-y+2z=2$

$3x+y+3z=3$

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

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

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

$\approx \left[\begin{array}{ccc}1& 2& 1\\ 0& 1& 0\\ 0& 1& 0\end{array}|\begin{array}{c}1\\ 0\\ 0\end{array}\right]R_2^{\text{'}}=-\frac{1}{5}{R}_2;R_3^{\text{'}}=-\frac{1}{5}{R}_3$

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

We see that the rank of the coefficient matrix is 2 and rank of augmented matrix is also 2; whereas the number of variables and the number of equations are both 3.

Since the rank of coefficient matrix and the rank of augmented matrix are both equal to 2, which is less than 3; the given system of equations is consistent and has infinitely many solutions.