Module 1 Practice Problems
Work through these 10 problems covering systems, augmented matrices, Gaussian elimination, RREF, and solution classification. Try each problem on paper before revealing the solution.
Problem 1
Write the augmented matrix for the system: 3x - y + 2z = 7, x + 4y - z = -2, 2x + y + 3z = 5.
Solution
[ 1 4 -1 |-2 ]
[ 2 1 3 | 5 ]
Each row corresponds to one equation; each column to one variable (x, y, z), with constants after the bar.
Problem 2
Perform R_2 - 3R_1 → R_2 on the matrix: [ 1 2 | 5 ] / [ 3 1 | 8 ].
Solution
New R_2: [3-3(1), 1-3(2) | 8-3(5)] = [0, -5 | -7].
[ 0 -5 |-7 ]
Problem 3
Use Gaussian elimination then back-substitution to solve: x + y = 3 and 2x - y = 3.
Solution
[ 2 -1 | 3 ]
R_2 - 2R_1 → R_2: [0, -3 | -3]. So -3y = -3, y = 1. Back-sub: x + 1 = 3, x = 2.
Solution: (2, 1).
Problem 4
Reduce to RREF: [ 2 4 | 10 ] / [ 1 3 | 7 ].
Solution
Swap rows to get a 1 in top-left: [ 1 3 | 7 ] / [ 2 4 | 10 ].
R_2 - 2R_1 → R_2: [0, -2 | -4]. Scale: (-1/2)R_2: [0, 1 | 2]. Eliminate above: R_1 - 3R_2: [1, 0 | 1].
[ 0 1 | 2 ]
Solution: x = 1, y = 2.
Problem 5
Solve the 3x3 system using Gauss-Jordan: x + y + z = 6, 2x + 3y + z = 14, x - y + 2z = 2.
Solution
Augmented: [ 1 1 1 | 6 ] / [ 2 3 1 | 14 ] / [ 1 -1 2 | 2 ].
R_2 - 2R_1, R_3 - R_1: [ 1 1 1 | 6 ] / [ 0 1 -1 | 2 ] / [ 0 -2 1 | -4 ].
R_3 + 2R_2: [ 0 0 -1 | 0 ]. Scale R_3 by -1: [0 0 1 | 0]. z = 0.
Eliminate upward: R_2 + R_3: [0 1 0 | 2]. R_1 - R_2 - R_3: [1 0 0 | 4].
Solution: (4, 2, 0).
Problem 6
Determine whether the system is consistent or inconsistent: x + 2y = 3, 2x + 4y = 7.
Solution
[ 1 2 | 3 ] / [ 2 4 | 7 ]. R_2 - 2R_1: [0 0 | 1]. This says 0 = 1, a contradiction.
Inconsistent -- no solution.
Problem 7
Solve: x + 2y + 3z = 1, 2x + 4y + 6z = 2. Write the solution in parametric vector form.
Solution
[ 1 2 3 | 1 ] / [ 2 4 6 | 2 ]. R_2 - 2R_1: [0 0 0 | 0]. RREF: [ 1 2 3 | 1 ].
Free variables: x_2 = s, x_3 = t. x_1 = 1 - 2s - 3t.
x = (1, 0, 0) + s(-2, 1, 0) + t(-3, 0, 1).
Problem 8
Solve the homogeneous system: x + 2y - z = 0, 3x + 6y - 3z = 0.
Solution
[ 1 2 -1 | 0 ] / [ 3 6 -3 | 0 ]. R_2 - 3R_1: [0 0 0 | 0]. RREF: [ 1 2 -1 | 0 ].
Free: y = s, z = t. x = -2s + t.
x = s(-2, 1, 0) + t(1, 0, 1). The solution set is a plane through the origin.
Problem 9
How many pivots can a 3x5 augmented matrix (representing a system with 4 unknowns) have at most? What does this imply about free variables?
Solution
At most 3 pivots (limited by the number of rows). With 4 unknowns and at most 3 pivots, there is at least 1 free variable. If the system is consistent, it has infinitely many solutions.
Problem 10
Given the RREF [ 1 0 -2 0 | 3 ] / [ 0 1 1 0 | -1 ] / [ 0 0 0 1 | 5 ], write the complete solution in parametric vector form.
Solution
Pivot columns: 1, 2, 4. Free variable: x_3 = t.
x_1 = 3 + 2t, x_2 = -1 - t, x_3 = t, x_4 = 5.
x = (3, -1, 0, 5) + t(2, -1, 1, 0).