Answer
No solution
Work Step by Step
Given: $A=\begin{bmatrix}
1&-2 & -1 & 3\\
-2 & 4 & 5 & -5\\
3& -6&-6&8
\end{bmatrix}$
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
1&-2 & -1 & 3 | 0\\
-2 & 4 & 5 & -5 | 3\\
3& -6&-6&8 | 2
\end{bmatrix} \approx^1\begin{bmatrix}
1&-2 & -1 & 3 | 0\\
0 & 0 & 3 & 1 | 3\\
0& 0&-3&-1 | 2
\end{bmatrix} \approx^2 \begin{bmatrix}
1&-2 & -1 & 3 | 0\\
0 & 0 & 3 & 1 | 3\\
0& 0&0&0| 5
\end{bmatrix} \approx^3 \begin{bmatrix}
1&-2 & -1 & 3 | 0\\
0 & 0 & 1 &\frac{1}{3} | 1\\
0& 0&0&0| 1
\end{bmatrix}$
$1.A_{12}(2),A_{13}(-3)$
$2.A_{23}(1)$
$3. M_2(\frac{1}{3}),M_3(\frac{1}{5})$
The equivalent system for this matrix is now:
$0x_1+0x_2+0x_3=1$
Hence, the system has no solution.
