## Precalculus (6th Edition) Blitzer

$\left[ \begin{matrix} 1 & -1 & 1 & 8 \\ 0 & 5 & -3 & -18 \\ 0 & 1 & -12 & -15 \\ \end{matrix} \right],\left[ \begin{matrix} 1 & -1 & 1 & 8 \\ 0 & 1 & \frac{-3}{5} & \frac{-18}{5} \\ 0 & 1 & -12 & -15 \\ \end{matrix} \right]$
A matrix in which all the diagonal elements are $1$ and all the elements lower than the principal diagonal are $0$ is known as a matrix in row echelon form. The given matrix can be converted into the row echelon form by applying some matrix operations. The operations to be performed are given below: Perform the operation $-2{{R}_{1}}+{{R}_{2}}$ to get, \begin{align} & \left[ \begin{matrix} 1 & -1 & 1 & 8 \\ 2 & 3 & -1 & -2 \\ 3 & -2 & -9 & 9 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 & 1 & 8 \\ -2+2 & 2+3 & -2-1 & -16-2 \\ 3 & -2 & -9 & 9 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & -1 & 1 & 8 \\ 0 & 5 & -3 & -18 \\ 3 & -2 & -9 & 9 \\ \end{matrix} \right] \end{align} Now perform the operation $-3{{R}_{1}}+{{R}_{3}}$ to get, $\left[ \begin{matrix} 1 & -1 & 1 & 8 \\ 0 & 5 & -3 & -18 \\ -3+3 & 3-2 & -3-9 & -24+9 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 & 1 & 8 \\ 0 & 5 & -3 & -18 \\ 0 & 1 & -12 & -15 \\ \end{matrix} \right]$ The matrix obtained above is similar to the matrix given with missing elements. Thus, the first solution matrix is, $\left[ \begin{matrix} 1 & -1 & 1 & 8 \\ 0 & 5 & -3 & -18 \\ 0 & 1 & -12 & -15 \\ \end{matrix} \right]$ Now, perform the operation, ${{R}_{2}}\to \frac{1}{5}{{R}_{2}}$ to get, $\left[ \begin{matrix} 1 & -1 & 1 & 8 \\ 0 & \frac{1}{5}\times 5 & \frac{1}{5}\times -3 & \frac{1}{5}\times -18 \\ 0 & 1 & -12 & -15 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 & 1 & 8 \\ 0 & 1 & \frac{-3}{5} & \frac{-18}{5} \\ 0 & 1 & -12 & -15 \\ \end{matrix} \right]$ Thus, the second solution matrix is, $\left[ \begin{matrix} 1 & -1 & 1 & 8 \\ 0 & 1 & \frac{-3}{5} & \frac{-18}{5} \\ 0 & 1 & -12 & -15 \\ \end{matrix} \right]$