## Linear Algebra and Its Applications (5th Edition)

The procedure is shown with the matrix notation for simpler understanding. $x_{1} - 3x_{2} + 4x_{3} = -4$ $3x_{1} - 7x_{2} + 7x_{3} = -8$ $-4x_{1} + 6x_{2} - x_{3} = 7$ This can be depicted in the augmented matrix notation as follows : $\begin{bmatrix} 1 &amp; -3 &amp; 4 &amp; -4 \\ 3 &amp; -7 &amp; 7 &amp; -8 \\ -4 &amp; 6 &amp; -1 &amp; 7 \end{bmatrix}$ To eliminate the $3x_{1}$ term in the second equation, add $-3$ times row 1 to row 2: $\begin{bmatrix} 1 &amp;-3 &amp; 4 &amp; -4 \\ 0 &amp; 2 &amp; -5 &amp; 4 \\ -4 &amp; 6 &amp; -1 &amp;7 \end{bmatrix}$ Next we use the $x_{1}$ term in the first equation to eliminate the $-4x_{1}$ term from the third equation. Add $4$ times row 1 to row 3: $\begin{bmatrix} 1 &amp; -3 &amp; 4 &amp; -4 \\ 0 &amp; 2 &amp; -5 &amp; 4 \\ 0 &amp; -6 &amp; 15 &amp; -9 \end{bmatrix}$ Next we use the $2x_{2}$ term in the second equation to eliminate the $6x_{2}$ term from the third equation. Add $3$ times row 2 to row 3: $\begin{bmatrix} 1 &amp; -3 &amp; 4 &amp; -4 \\ 0 &amp; 2 &amp; -5 &amp; 4 \\ 0 &amp; 0 &amp; 0 &amp; 3 \end{bmatrix}$ Now the augmented matrix is in a triangular form. For interpreting it, we go back to the equation notation: $x_{1} + -3x_{2} + 4x_{3} = -4$ $2x_{2} - 5x_{3} = 4$ $0 = 3$ The equation $0 = 3$ is a short form of $0x_{1} + 0x_{2} + 0x_{3} = 3$. This is a contradiction, and no values of $x_{1}, x_{2}$, and $x_{3}$ can satisfy this equation. Since the final and original equations have the same solution set, we conclude that the given system of equations $x_{1} - 3x_{2} + 4x_{3} = -4$ $3x_{1} - 7x_{2} + 7x_{3} = -8$ $-4x_{1} + 6x_{2} - x_{3} = 7$ has no solutions. It is inconsistent.