# Recitation 1B

Today we covered three examples.

Example 1: Solve the given system of equations. \begin{aligned}x_1 & - & 2x_2 & + & x_3 & = & 0\\& & 2x_2 & - & 8x_3 & = & 0\\-4x_1 & + & 5x_2 & + & 9x_3 & = & -9.\end{aligned}

Solution 1: We turn the system into an augmented matrix. \begin{bmatrix}1 & -2 & 1 & 0\\0 & 2 & -8 & 8\\-4 & 5 & 9 & -9\end{bmatrix} Add 4 times the first row to the third row. \begin{bmatrix}1 & -2 & 1 & 0\\0 & 2 & -8 & 8\\0 & -3 & 13 & -9\end{bmatrix} Add the second row to the first row.\begin{bmatrix}1 & 0 & -7 & 8\\0 & 2 & -8 & 8\\0 & -3 & 13 & -9\end{bmatrix} Divide the second row by 2. \begin{bmatrix}1 & 0 & -7 & 8\\0 & 1 & -4 & 4\\0 & -3 & 13 & -9\end{bmatrix} Add 3 times the second row to the third row. \begin{bmatrix}1 & 0 & -7 & 8\\0 & 1 & -4 & 4\\0 & 0 & 1 & 3\end{bmatrix} Add 7 times the third row to the first row and add 4 times the third row to the second row. \begin{bmatrix}1 & 0 & 0 & 29\\0 & 1 & 0 & 16\\0 & 0 & 1 & 3\end{bmatrix} This gives the solution x_1=29, x_2=16, x_3=3.

Example 2: Determine if the following system is consistent. \begin{aligned}& & x_2 & - & 4x_3 & = & 8\\2x_1& - & 3x_2 &+ & 2x_3& = & 1\\5x_1 & - & 8x_2 & + & 7x_3 & = & 1.\end{aligned}

Solution 2: First we write down the augmented matrix. \begin{bmatrix}0 & 1 & -4 & 8\\2 & -3 & 2 & 1\\5 & -8 & 7 & 1\end{bmatrix} Interchange the first row and the second row because we want to obtain x_1 in the first equation. \begin{bmatrix}2 & -3 & 2 & 1\\0 & 1 & -4 & 8\\5 & -8 & 7 & 1\end{bmatrix} Add -5/2 times row 1 to row 3. \begin{bmatrix}2 & -3 & 2 & 1\\0 & 1 & -4 & 8\\0 & -1/2 & 2 & -3/2\end{bmatrix} Add 1/2 times row 2 to row 3. \begin{bmatrix}2 & -3 & 2 & 1\\0 & 1 & -4 & 8\\0 & 0 & 0 & 5/2\end{bmatrix} The last row in the augmented matrix says 0=5/2 which is impossible. Therefore the original system is inconsistent, i.e., has no solution.

Example 3: Row reduce the matrix A below to echelon form, and locate the pivot columns of A. A=\begin{bmatrix}0 & -3 & -6 & 4 & 9\\-1 & -2 & -1 & 3 & 1\\-2 & -3 & 0 & 3 & -1 \\ 1 & 4 & 5 & -9 & -7\end{bmatrix}

Solution 3: Interchange rows 1 and 2. \begin{bmatrix} [-1] & -2 & -1 & 3 & 1\\0 & -3 & -6 & 4 & 9\\-2 & -3 & 0 & 3 & -1 \\1 & 4 & 5 & -9 & -7\end{bmatrix} Create zeros below the pivot [-1], by adding multiple of the first row to the rows below, and obtain the next matrix. \begin{bmatrix} [-1] & -2 & -1 & 3 & 1\\0 & -3 & -6 & 4 & 9\\0 & 1 & 2 & -3 & -3 \\0 & 2 & 4 & -6 & -6\end{bmatrix} Interchange rows 2 and 3. \begin{bmatrix} [-1] & -2 & -1 & 3 & 1\\0 & [1] & 2 & -3 & -3 \\0 & -3 & -6 & 4 & 9\\0 & 2 & 4 & -6 & -6\end{bmatrix} Add multiple of the second row to the rows below and obtain.\begin{bmatrix} [-1] & -2 & -1 & 3 & 1\\0 & [1] & 2 & -3 & -3 \\0 & 0 & 0 & [-5] & 0\\0 & 0 & 0 & 0 & 0\end{bmatrix} This is an echelon form with the pivots in the brackets.