Show that the system of equations 2x+6y=-11, 6x+20y-6z=-3, 6y-18z=-1 is not consistent.
To show that the system is not consistent, we can use the method of elimination. We'll start by eliminating the variable z.
Multiplying the first equation by 3, we get 6x + 18y = -33.
Adding this equation to the second equation, we get:
(6x + 20y) + (6x + 18y) - 6z = -3 + (-33)
12x + 38y - 6z = -36.
But this equation is inconsistent with the third equation 6y - 18z = -1, as we cannot find values of x, y, and z that satisfy all three equations simultaneously. Therefore, the system of equations is not consistent.
To show that the system of equations is not consistent, we need to find a contradiction or inconsistency in the given equations.
First, let's write the given system of equations in matrix form:
$\begin{bmatrix}
2 & 6 & 0 & -11 \\
6 & 20 & -6 & -3 \\
0 & 6 & -18 & -1 \\
\end{bmatrix} \begin{bmatrix}
x \\
y \\
z \\
\end{bmatrix} = \begin{bmatrix}
0 \\
0 \\
0 \\
\end{bmatrix}$
Now, let's row-reduce the augmented matrix:
$\begin{bmatrix}
2 & 6 & 0 & -11 \\
6 & 20 & -6 & -3 \\
0 & 6 & -18 & -1 \\
\end{bmatrix} \rightarrow \begin{bmatrix}
1 & 3 & 0 & -\frac{11}{2} \\
0 & 8 & -6 & 8 \\
0 & 6 & -18 & -1 \\
\end{bmatrix} \rightarrow \begin{bmatrix}
1 & 3 & 0 & -\frac{11}{2} \\
0 & 1 & -\frac{3}{4} & 1 \\
0 & 0 & -\frac{9}{2} & -\frac{5}{2} \\
\end{bmatrix}$
From the row-reduced form, we can see that the last row of the matrix represents the equation $0x + 0y + (-\frac{9}{2})z = -\frac{5}{2}$. This equation simplifies to $-\frac{9}{2}z = -\frac{5}{2}$.
However, this equation has no solution since the left-hand side represents a multiple of $z$, while the right-hand side is a constant.
Hence, we have reached a contradiction and thus, the system of equations is inconsistent.