Today we covered two 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 & 21\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 & 155\\0 & 1 & 0 & 88\\0 & 0 & 1 & 21\end{bmatrix} This gives the solution x_1=155, x_2=88, x_3=21.
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.