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Recitation 1B

Today we covered three examples.

Example 1: Solve the given system of equations. x1−2x2+x3=02x2−8x3=0−4x1+5x2+9x3=−9.\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. [1−21002−88−459−9]\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. [1−21002−880−313−9]\begin{bmatrix}1 & -2 & 1 & 0\\0 & 2 & -8 & 8\\0 & -3 & 13 & -9\end{bmatrix} Add the second row to the first row. [10−7802−880−313−9]\begin{bmatrix}1 & 0 & -7 & 8\\0 & 2 & -8 & 8\\0 & -3 & 13 & -9\end{bmatrix} Divide the second row by 2. [10−7801−440−313−9]\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. [10−7801−440013]\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. [10029010160013]\begin{bmatrix}1 & 0 & 0 & 29\\0 & 1 & 0 & 16\\0 & 0 & 1 & 3\end{bmatrix} This gives the solution x1=29,x2=16,x3=3.x_1=29, x_2=16, x_3=3.

Example 2: Determine if the following system is consistent. x2−4x3=82x1−3x2+2x3=15x1−8x2+7x3=1.\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. [01−482−3215−871]\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 x1x_1 in the first equation. [2−32101−485−871]\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. [2−32101−480−1/22−3/2]\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. [2−32101−480005/2]\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=5/20=5/2 which is impossible. Therefore the original system is inconsistent, i.e., has no solution.

Example 3: Row reduce the matrix AA below to echelon form, and locate the pivot columns of AA. A=[0−3−649−1−2−131−2−303−1145−9−7]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. [[−1]−2−1310−3−649−2−303−1145−9−7]\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][-1], by adding multiple of the first row to the rows below, and obtain the next matrix. [[−1]−2−1310−3−649012−3−3024−6−6]\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. [[−1]−2−1310[1]2−3−30−3−649024−6−6]\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. [[−1]−2−1310[1]2−3−3000[−5]000000]\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.