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Solve each system by Gaussian elimination. $$ \begin{array}{r} \frac{4}{5} x-\frac{7}{8} y+\frac{1}{2} z=1 \\ -\frac{4}{5} x-\frac{3}{4} y+\frac{1}{3} z=-8 \\ -\frac{2}{5} x-\frac{7}{8} y+\frac{1}{2} z=-5 \end{array} $$

Short Answer

Expert verified
The solution is approximate: \(x \approx 11.95\), \(y = 10\), \(z = \frac{83}{161}\).

Step by step solution

01

Write the Augmented Matrix

Identify the coefficients from the system of equations to form the augmented matrix.\[\begin{bmatrix}\frac{4}{5} & -\frac{7}{8} & \frac{1}{2} & \big| & 1 \-\frac{4}{5} & -\frac{3}{4} & \frac{1}{3} & \big| & -8 \-\frac{2}{5} & -\frac{7}{8} & \frac{1}{2} & \big| & -5\end{bmatrix}\]
02

Eliminate x from Row 2 and Row 3

Add Row 1 to Row 2 to eliminate \(x\) in Row 2 and add half of Row 1 to Row 3 to eliminate \(x\) in Row 3.\( R_2 = R_2 + R_1 \)\( R_3 = R_3 + \frac{1}{2}R_1 \)Resulting matrix:\[\begin{bmatrix}\frac{4}{5} & -\frac{7}{8} & \frac{1}{2} & \big| & 1 \0 & -\frac{29}{40} & \frac{11}{6} & \big| & -7 \0 & -\frac{15}{16} & \frac{3}{4} & \big| & -\frac{9}{2}\end{bmatrix}\]
03

Eliminate y from Row 3

Eliminate \(y\) from Row 3 by adjusting:\( R_3 = R_3 - \frac{10}{29}R_2 \)Resulting matrix:\[\begin{bmatrix}\frac{4}{5} & -\frac{7}{8} & \frac{1}{2} & \big| & 1 \0 & -\frac{29}{40} & \frac{11}{6} & \big| & -7 \0 & 0 & \frac{161}{87} & \big| & \frac{83}{29}\end{bmatrix}\]
04

Solve for z

With the matrix in upper triangular form, start solving from the last equation. \( \frac{161}{87}z = \frac{83}{29} \)Solve for \(z\):\[ z = \frac{83}{29} \times \frac{87}{161} = \frac{249}{161} = \frac{83}{161} \]
05

Back Substitute to find y

Substitute \(z = \frac{83}{161}\) back into the second equation:\[ -\frac{29}{40}y + \frac{11}{6}\left(\frac{83}{161}\right) = -7 \]Solve for \(y\):Re-arranging gives:\[ y = \frac{-7 \times 40 + \frac{11}{6} \times \frac{83}{161} \times 40}{-29} \approx \frac{-290}{29} = 10 \]
06

Back Substitute to find x

Finally, use \(y\) and \(z\) in the first equation to solve for \(x\):\[ \frac{4}{5}x - \frac{7}{8}(10) + \frac{1}{2}\left(\frac{83}{161}\right) = 1 \]Re-arranging gives:\[ \frac{4}{5}x = 1 + \frac{70}{8} - \frac{83}{322} \]Solve for \(x\):\[ x = \frac{5}{4} \left(1 + \frac{70}{8} - \frac{83}{322}\right) \approx 11.95 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Augmented Matrices
An augmented matrix is a central concept when dealing with systems of linear equations, especially when using Gaussian elimination. It represents the coefficients and constants of a system of linear equations in a matrix format. This allows us to apply systematic operations to solve the equations more easily.
The matrix usually appears like this: \\[\begin{bmatrix}a_{11} & a_{12} & ... & a_{1n} & \big| & b_1\a_{21} & a_{22} & ... & a_{2n} & \big| & b_2\... & ... & ... & ... & \big| & ...\a_{m1} & a_{m2} & ... & a_{mn} & \big| & b_m\end{bmatrix}\] Here, the part before the vertical bar represents the coefficients of the variables, and on the right side of the bar, the constants are placed. This visualization makes it easier to perform various matrix operations necessary to solve the system by methods such as Gaussian elimination.
  • Arrange the coefficients of variables in each row.
  • Separate constants by a vertical line.
  • This structure helps in applying row operations efficiently.
By using augmented matrices, you simplify the process of transitioning between equations and matrix operations, ensuring that each step in solving the equations is logically connected and clear.
Solving Linear Equations
Linear equations are equations of the first degree, meaning they involve terms where variables are raised only to the power of one. They are fundamental in algebra and have the general form \(ax + by + cz + ... = d\). Solving them involves finding the values of variables that satisfy all equations in a given system simultaneously.

These equations can often be tricky when there are several variables because there might be many ways to combine them. However, handling them becomes much simpler using methods like Gaussian elimination. This method involves transforming the system into an upper triangular form, making it easy to solve step by step from the last variable upwards.
  • Determine the coefficients and constants clearly.
  • Re-arrange equations if needed to facilitate easier operations.
  • Use systematic approaches like row operations to isolate variables.
By understanding the basic principles of solving linear equations, you can not only solve individual equations but also extend your methods to systems with multiple equations and variables.
Performing Matrix Operations
Matrix operations are crucial in speeding up the computations when solving systems of linear equations. These operations include tasks like adding two rows, multiplying a row by a constant, or swapping rows. Each of these operations modifies the matrix to bring it closer to a more convenient form for solving the system, like reduced row-echelon form.
Here is an outline of typical matrix operations used in Gaussian elimination:
  • Row Addition: Add a multiple of one row to another row to eliminate variables. For example, adding Row 1 to Row 2 to remove variable terms.
  • Scalar Multiplication: Multiply a row by a constant to simplify expressions.
  • Row Swapping: Swap two rows if needed to move a row with easier coefficients to a more advantageous position.
Through these matrix operations, the augmented matrix is transformed step-by-step into an upper triangular form or even a diagonal form, from which solutions for the variables can be directly calculated. Understanding and mastering these operations is a stepping stone to efficiently solving complex systems of equations.

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Most popular questions from this chapter

For the following exercises, solve the system by Gaussian elimination. $$ \begin{array}{c} x+y=2 \\ x+z=1 \\ -y-z=-3 \end{array} $$

For the following exercises, use this scenario: A health-conscious company decides to make a trail mix out of almonds, dried cranberries, and chocolate- covered cashews. The nutritional information for these items is shown in Table 1 . $$ \begin{array}{|c|c|c|c|} \hline & \text { Fat (g) } & \text { Protein (g) } & \text { Carbohydrates (g) } \\ \hline \text { Almonds (10) } & 6 & 2 & 3 \\ \hline \text { Cranberries (10) } & 0.02 & 0 & 8 \\ \hline \text { Cashews (10) } & 7 & 3.5 & 5.5 \\ \hline \end{array} $$ For the special "low-carb" trail mix, there are 1,000 pieces of mix. The total number of carbohydrates is \(425 \mathrm{~g},\) and the total amount of fat is \(570.2 \mathrm{~g}\). If there are 200 more pieces of cashews than cranberries, how many of each item is in the trail mix?

For the following exercises, create a system of linear equations to describe the behavior. Then, solve the system for all solutions using Cramer’s Rule. You invest \(\$ 80,000\) into two accounts, \(\$ 22,000\) in one account, and \(\$ 58,000\) in the other account. At the end of one year, assuming simple interest, you have earned \(\$ 2,470\) in interest. The second account receives half a percent less than twice the interest on the first account. What are the interest rates for your accounts?

For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. At an ice cream shop, three flavors are increasing in demand. Last year, banana, pumpkin, and rocky road ice cream made up \(12 \%\) of total ice cream sales. Th s year, the same three ice creams made up \(16.9 \%\) of ice cream sales. The rocky road sales doubled, the banana sales increased by \(50 \%,\) and the pumpkin sales increased by \(20 \%\). If the rocky road ice cream had one less percent of sales than the banana ice cream, fi \(\mathrm{d}\) out the percentage of ice cream sales each individual ice cream made last year.

For the following exercises, solve the system of linear equations using Cramer's Rule. $$ \begin{aligned} -5 x+2 y-4 z &=-47 \\ 4 x-3 y-z &=-94 \\ 3 x-3 y+2 z &=94 \end{aligned} $$

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