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Chapter 8: Problem 60

Use matrices to solve the system of equations, if possible. Use Gaussian elimination with back-substitution. $$\left\\{\begin{aligned} x+2 y &=0 \\ x+y &=6 \\ 3 x-2 y &=8 \end{aligned}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = -26, y = -40\).

Step by step solution

01

Write down the system of equations as a matrix

First, express the given system of equations as a matrix. The coefficients of the variables form the elements of the matrix, and the constants on the right-hand side of the equations form the resulting column vector:\[\begin{bmatrix}1 & 2 \1 & 1 \3 & -2 \\end{bmatrix}\begin{bmatrix}x \y \\end{bmatrix}=\begin{bmatrix}0 \6 \8 \\end{bmatrix}\]
02

Apply Gaussian elimination

Next, perform Gaussian elimination to convert the augmented matrix to an upper triangular matrix (Row echelon form). The augmented matrix (joining the coefficient matrix and the constant column vector) would look like: \[\left[\begin{array}{cc|c}1 & 2 & 0 \1 & 1 & 6 \3 & -2 & 8 \\end{array}\right]\]Subtract row1 from row2 to create 0 under the first pivot (1), and subtract 3 * row1 from row3 for the same purpose: \[\left[\begin{array}{cc|c}1 & 2 & 0 \0 & -1 & 6 \0 & -8 & 8 \\end{array}\right]\]Now, multiply row2 by -1, and add 8 * row2 to row3 to create 0 under the second pivot (-1), which results in:\[\left[\begin{array}{cc|c}1 & 2 & 0 \0 & 1 & -6 \0 & 0 & -40 \\end{array}\right]\]
03

Conduct back-substitution

Finally, perform back-substitution: Start from the last equation, and put its value into the previous equation and solve it. Repeat the process until reaching the first equation:From the third equation, \(y = -40\),Put \(y = -40\) into the second equation to find \(x = -26\). Therefore, the solution to the system is \(x = -26, y = -40\).

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Most popular questions from this chapter

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