Chapter 11.2, Problem 53AYU

To determine

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

Answer to Problem 53AYU

Solution:

Consistent system of equations, infinitely many solutions.

Explanation of Solution

Given:

$-x+y+z=-1$

$-x+2y-3z=-4$

$3x-2y-7z=0$

Formula used:

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 $z$.

Calculation:

$-x+y+z=-1$

$-x+2y-3z=-4$

$3x-2y-7z=0$

The corresponding augmented matrix is:

$\left[\begin{array}{ccc}-1& 1& 1\\ -1& 2& -3\\ 3& -2& -7\end{array}|\begin{array}{c}-1\\ -4\\ 0\end{array}\right]$

$\approx \left[\begin{array}{ccc}1& -1& -1\\ -1& 2& -3\\ 3& -2& -7\end{array}|\begin{array}{c}1\\ -4\\ 0\end{array}\right];R_1=-1\cdot r_1$

$\approx \left[\begin{array}{ccc}1& -1& -1\\ 0& 1& -4\\ 0& 1& -4\end{array}|\begin{array}{c}1\\ -3\\ -3\end{array}\right];R_2=R_1+r_2;R_3=-3R_1+r_3$

$\approx \left[\begin{array}{ccc}1& 0& -5\\ 0& 1& -4\\ 0& 1& -4\end{array}|\begin{array}{c}-2\\ -3\\ -3\end{array}\right];R_1\text{'}=R_1+R_2$

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

Now we see that the rank of the coefficient matrix is 2, and the rank of augmented matrix is also 2, which is less than 3.

Since rank of the coefficient matrix and the rank of augmented matrix are equal and is less than 3, therefore the given system of equations has infinitely many solutions.

Hence this is a consistent system.