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

The procedure is shown with the matrix notation for simpler understanding. $x_{2} + 4x_{3} = -5$ $x_{1} + 3x_{2} + 5x_{3} = -2$ $3x_{1} + 7x_{2} + 7x_{3} = 6$ This can be depicted in the augmented matrix notation as follows : $\begin{bmatrix} 0 &amp; 1 &amp; 4 &amp; -5 \\ 1 &amp; 3 &amp; 5 &amp; -2 \\ 3 &amp; 7 &amp; 7 &amp; 6 \end{bmatrix}$ To obtain an $x_{1}$ in the ﬁrst equation,interchange rows 1 and 2: $\begin{bmatrix} 1 &amp; 3 &amp; 5 &amp; -2 \\ 0 &amp; 1 &amp; 4 &amp; -5 \\ 3 &amp; 7 &amp; 7 &amp; 6 \end{bmatrix}$ To eliminate the $3x_{1}$ term in the third equation, add $-3$ times row 1 to row 3: $\begin{bmatrix} 1 &amp; 3 &amp; 5 &amp; -2 \\ 0 &amp; 1 &amp; 4 &amp; -5 \\ 0 &amp; -2 &amp; -8 &amp; 12 \end{bmatrix}$ Next we use the $x_{2}$ term in the second equation to eliminate the $-2x_{2}$ term from the third equation. Add $2$ times row 2 to row 3: $\begin{bmatrix} 1 &amp; 3 &amp; 5 &amp; -2 \\ 0 &amp; 1 &amp; 4 &amp; -5 \\ 0 &amp; 0 &amp; 0 &amp; 2 \end{bmatrix}$ Now the augmented matrix is in a triangular form. For interpreting it, we go back to the equation notation: $x_{1} + 3x_{2} + 5x_{3} = -2$ $x_{2} + 4x_{3} = -5$ $0 = 2$ The equation $0 = 2$ is a short form of $0x_{1} + 0x_{2} + 0x_{3} = 2$. 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 $x_{2} + 4x_{3} = -5$ $x_{1} + 3x_{2} + 5x_{3} = -2$ $3x_{1} + 7x_{2} + 7x_{3} = 6$ has no solutions. It is inconsistent.