Answer
Consistent
Solution set: $\left\{\left(x,y,z,w\right)|x=-0.6z-0.4w+1.4,y=1.4z+2.6w-1.6,z,w\text{ are any real numbers}\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
4x+y+z-w=4\\
x-y+2z+3w=3
\end{cases}$
Write the augmented matrix:
$\begin{bmatrix}
4&1&1&-1&|&4\\1&-1&2&3&|&3\end{bmatrix}$
Perform row operations to bring the matrix to the reduced row echelon form:
$R_1=-3r_2+r_1$
$\begin{bmatrix}
1&4&-5&-10&|&-5\\1&-1&2&3&|&3\end{bmatrix}$
$R_2=-r_1+r_2$
$\begin{bmatrix}1&4&-5&-10&|&-5\\0&-5&7&13&|&8\end{bmatrix}$
$R_2=-\dfrac{1}{5}r_2$
$\begin{bmatrix}1&4&-5&-10&|&-5\\0&1&-1.4&-2.6&|&-1.6\end{bmatrix}$
$R_1=-4r_2+r_1$
$\begin{bmatrix}1&0&0.6&0.4&|&1.4\\0&1&-1.4&-2.6&|&-1.6\end{bmatrix}$
The system has two equations and 4 variables; therefore it has infinitely many solutions.
Write the corresponding system of equations:
$\begin{cases}
x+0.6z+0.4w=1.4\\
y-1.4z-2.6w=-1.6
\end{cases}$
Express $x,y$ in terms of $z,w$:
$y-1.4z-2.6w=-1.6\Rightarrow y=1.4z+2.6w-1.6$
$x+0.6z+0.4w=1.4\Rightarrow x=-0.6z-0.4w+1.4$
The system is consistent. The solution set is:
$\left\{\left(x,y,z,w\right)|x=-0.6z-0.4w+1.4,y=1.4z+2.6w-1.6,z,w\text{ are any real numbers}\right\}$
