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 set up 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
