Answer
(a) $(\frac{5}{2},\frac{5}{2},0)$
(b) no solution.
Work Step by Step
(a) Write the augmented matrix of the system of equation and use Gaussian elimination:
$\begin{bmatrix} 1&-1&2&0\\2&-4&5&-5\\0&2&-3&5\end{bmatrix}
\begin{array} \\2R_1-R_2\to R_2 \\ \end{array}
\begin{bmatrix} 1&-1&2&0\\0&2&-1&5\\0&2&-3&5\end{bmatrix}
\begin{array} \\ \\ \\R_2-R_3\to R_3 \end{array}
\begin{bmatrix} 1&-1&2&0\\0&2&-1&5\\0&0&2&0\end{bmatrix}$
Row 3 gives $z=0$, use it in row 2 to get $2y-0=5$ or $y=\frac{5}{2}$, thus $x=y-2z=\frac{5}{2}$ and the solution is $(\frac{5}{2},\frac{5}{2},0)$
(b) Write the augmented matrix of the system of equation and use Gaussian elimination:
$\begin{bmatrix} 2&-3&1&3\\1&2&2&-1\\4&1&5&4\end{bmatrix}
\begin{array} \\ \\ \\R_3-R_1\to R_3 \end{array}
\begin{bmatrix} 2&-3&1&3\\1&2&2&-1\\2&4&4&1\end{bmatrix}
\begin{array} \\ \\ \\R_3-2R_2\to R_3 \end{array}
\begin{bmatrix} 2&-3&1&3\\1&2&2&-1\\0&0&0&3\end{bmatrix}$
As the last row gives $0=3$, there is no solution to the system.
