College Algebra, Chapter 7, 7.4, Section 7.4, Problem 32

Consider the system

$\left\{
\begin{equation}
\begin{aligned}

x + 2y + 6z =& 5
\\
-3x - 6y + 5z =& 8
\\
2x + 6y + 9z =& 7

\end{aligned}
\end{equation}
\right.$

a.) Check that $x = -1, y = 0, z = 1$ is a solution of the system


$
\left\{
\begin{equation}
\begin{aligned}

-1 + 2 (0) + 6(1) =& 5
\\
-3(-1) - 6(0) + 5(1) =& 8
\\
2 (-1) + 6(0) + 9(1) =& 7

\end{aligned}
\end{equation}
\right.
$



$
\left\{
\begin{equation}
\begin{aligned}

5 =& 5
\\
8 =& 8
\\
7 =& 7

\end{aligned}
\end{equation}
\right.
$


It shows that the given value of $x, y$ and $z$ is a solution of the system

b.) Determine the determinant of the coefficient matrix

For this system we have


$
\begin{equation}
\begin{aligned}

|D| =& \left| \begin{array}{ccc}
1 & 2 & 6 \\
-3 & -6 & 5 \\
2 & 6 & 9
\end{array} \right|
\\
\\
|D| =& 1 \left| \begin{array}{cc}
-6 & 5 \\
6 & 9
\end{array} \right| -2 \left| \begin{array}{cc}
-3 & 5 \\
2 & 9
\end{array} \right| + 6 \left| \begin{array}{cc}
-3 & -6 \\
2 & 6
\end{array} \right|
\\
\\
|D| =& 1 (-6 \cdot 9 - 5 \cdot 6) - 2 (-3 \cdot 9 - 5 \cdot 2) + 6 (-3 \cdot 6 - (-6) \cdot 2)
\\
\\
|D| =& -46

\end{aligned}
\end{equation}
$


c.) Without solving the system, determine whether there are any other solutions.

There aren't any other solutions in the given system, since that the lines are unique. Thus, there is only one solution in the system and that is in part (a).

d.) Can Cramer's Rule be used to solve this system? Why or why not?

Cramer's Rule can be used to solve the system because the given system has a determinant of the coefficient matrix.