Let
\[ A = \left[ \begin{array}{r r r r} 1 & 2 & 1 & 0 \\ - 1 & 0 & 3 & 5 \\ 1 & - 2 & 1 & 1 \end{array} \right]. \]
Find a row-reduced echelon matrix R which is row-equivalent to A and an invertible \(3 \times 3\) matrix P such that R = PA.
Solution
A row-reduced echelon form of \(A\) is \[ R= \begin{bmatrix} 1&0&0&-\frac78\\ 0&1&0&-\frac14\\ 0&0&1&\frac{11}{8} \end{bmatrix}. \] Take \[ P= \begin{bmatrix} \frac38&-\frac14&\frac38\\ \frac14&0&-\frac14\\ \frac18&\frac14&\frac18 \end{bmatrix}. \] Direct multiplication gives \[ PA= \begin{bmatrix} 1&0&0&-\frac78\\ 0&1&0&-\frac14\\ 0&0&1&\frac{11}{8} \end{bmatrix}=R. \] Moreover \[ \det P=\frac1{16}\ne0, \] so \(P\) is invertible. Hence \(R=PA\) with the required properties.