About Gaussian Elimination II:

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
Gaussian Elimination II Practice Test:

#1:

Instructions: Solve each linear system using matrix methods.

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


#2:

Instructions: Solve each linear system using matrix methods.

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


#3:

Instructions: Solve each linear system using matrix methods.

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


#4:

Instructions: Solve each linear system using matrix methods.

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


#5:

Instructions: Solve each linear system using matrix methods.

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


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


#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


#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


#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: ∅


#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