### About Gaussian Elimination:

We previously learned how to solve a system of linear equations using graphing, substitution, and elimination. Now we will move on and learn how to perform this task using a matrix. A matrix is an ordered array of numbers. We can transform our matrix using row operations to gain a solution for our system.

Test Objectives

- Demonstrate the ability to setup an augmented matrix
- Demonstrate the ability to place a matrix in row echelon form
- Demonstrate the ability to solve a linear system using matrix methods

#1:

Instructions: Solve each linear system using matrix methods.

a) $$-48y = 20x - 8$$ $$-14 + 84y = -35x$$

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#2:

Instructions: Solve each linear system using matrix methods.

a) $$-6y + 2x = 14$$ $$9x - 12 = 10y$$

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#3:

Instructions: Solve each linear system using matrix methods.

a) $$4x = 26 - 10y$$ $$-6x = -12y - 12$$

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#4:

Instructions: Solve each linear system using matrix methods.

a) $$x - 1 = -\frac{8}{7}y$$ $$10x = -5y + 10$$

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#5:

Instructions: Solve each linear system using matrix methods.

a) $$105 - 9x = 6y$$ $$-5y = -41 - 8x$$

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Written Solutions:

#1:

Solutions:

a) $$\left[ \begin{array}{cc|c} 1&\frac{12}{5}&\frac{2}{5}\\ 0&0&0 \end{array} \right] $$

Infinite number of solutions

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#2:

Solutions:

a) $$\left[ \begin{array}{cc|c} 1&-3&7\\ 0&1&-3 \end{array} \right] $$

{(-2,-3)}

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#3:

Solutions:

a) $$\left[ \begin{array}{cc|c} 1&\frac{5}{2}&\frac{13}{2}\\ 0&1&1 \end{array} \right] $$

{(4,1)}

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#4:

Solutions:

a) $$\left[ \begin{array}{cc|c} 1&\frac{8}{7}&1\\ 0&1&0 \end{array} \right] $$

{(1,0)}

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#5:

Solutions:

a) $$\left[ \begin{array}{cc|c} 1&\frac{2}{3}&\frac{35}{3}\\ 0&1&13 \end{array} \right] $$

{(3,13)}