### About Row Echelon Form:

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) $$-3x + 2y + 3z = -11$$ $$x + 2y - z = 9$$ $$2x - 4y - 4z = 8$$

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#2:

Instructions: Solve each linear system using matrix methods.

a) $$-4x - 2y + 2z = 0$$ $$-2x + 3y - 3z = -8$$ $$x - 4y + 4z = 9$$

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#3:

Instructions: Solve each linear system using matrix methods.

a) $$-2x + y - 5z = 10$$ $$7x - 7y - 5z = -11$$ $$-3x - 6y - 6z = -33$$

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#4:

Instructions: Solve each linear system using matrix methods.

a) $$-3x + 3y - 6z = -24$$ $$-7x + 7y - 4z = -32$$ $$-5x + 5y - 3z = 10$$

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#5:

Instructions: Solve each linear system using matrix methods.

a) $$-5x + 5y + 10z = -10$$ $$3x - y - 2z = -2$$ $$3x - 6y - 6z = -24$$

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Written Solutions:

#1:

Solutions:

a) $$\left[ \begin{array}{ccc|c} 1&2&-1&9\\ 0&1&0&2\\ 0&0&1&-3 \end{array} \right] $$

{(2,2,-3)}: x = 2, y = 2, z = -3

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#2:

Solutions:

a) $$\left[ \begin{array}{ccc|c} 1&-4&4&9\\ 0&1&-1&-2\\ 0&0&0&0 \end{array} \right] $$

Infinite number of solutions

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#3:

Solutions:

a) $$\left[ \begin{array}{ccc|c} 1&2&2&11\\ 0&1&\frac{19}{21}&\frac{88}{21}\\ 0&0&1&-2 \end{array} \right] $$

{(3,6,-2)}: x = 3, y = 6, z = -2

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#4:

Solutions:

a) $$\left[ \begin{array}{ccc|c} 1&-1&2&8\\ 0&0&7&50\\ 0&0&10&24 \end{array} \right] $$

No solution: ∅

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#5:

Solutions:

a) $$\left[ \begin{array}{ccc|c} 1&-1&-2&2\\ 0&1&0&10\\ 0&0&1&-7 \end{array} \right] $$

{(-2,10,-7)}: x = -2, y = 10, z = -7