Math 251 — 3 practice exams · 8 questions each · ~50 minutes per exam
Sections covered: Matrix Operations · Lines & Planes · RREF & Systems · Span & Independence
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Practice Exam 1
8 questions — work without notes first, then check hints/solutions
Instructions: Show all work. For RREF problems, write every row operation. For independence/span, show the full row-reduction. Suggested time: 50 minutes.
Q1Matrix Operations
Let A =
2
1
0
3
and B =
−1
4
2
1
.
Compute: (a) 2A − B (b) AB (c) BA
Hint: For scalar multiplication, multiply every entry of A by 2, then subtract B entry-by-entry. For matrix products, the (i,j) entry of AB = dot product of row i of A with column j of B.
(a) 2A − B
2A =
4
2
0
6
→ 2A − B =
4−(−1)
2−4
0−2
6−1
=
5
−2
−2
5
(b) AB — row of A · column of B
AB =
2(−1)+1(2)
2(4)+1(1)
0(−1)+3(2)
0(4)+3(1)
=
0
9
6
3
(c) BA
BA =
(−1)(2)+4(0)
(−1)(1)+4(3)
2(2)+1(0)
2(1)+1(3)
=
−2
11
4
5
AB ≠ BA — matrix multiplication is NOT commutative
Q2Matrix Operations
Let A be 3×4, B be 4×3, and C be 4×2.
For each product, state whether it is defined. If so, give its size. (a) AB (b) BA (c) AC (d) BC
Hint: Product PQ is defined only when columns of P = rows of Q. The result has size: rows of P × cols of Q.
(a) AB: (3×4)(4×3) — inner dims 4=4 ✓ → result is 3×3
(b) BA: (4×3)(3×4) — inner dims 3=3 ✓ → result is 4×4
(c) AC: (3×4)(4×2) — inner dims 4=4 ✓ → result is 3×2
(d) BC: (4×3)(4×2) — inner dims 3 ≠ 4 → NOT DEFINED
Q3Lines & Planes
Find the parametric equations of the line in ℝ³ that passes through
P₁ = (1, 2, 0) and P₂ = (3, −1, 4).
Hint: Direction vector d = P₂ − P₁. Then write r = P₁ + td and expand into component form.
Step 1 — direction vector
d = P₂ − P₁ = (3−1, −1−2, 4−0) = (2, −3, 4)
Step 2 — parametric equations (using P₁ as base point)
x = 1 + 2t
y = 2 − 3t
z = 4t
where t ∈ ℝ
Note: The answer is not unique — any point on the line can be the base point, and −d is equally valid as a direction vector.
Q4Lines & Planes
Find the general equation of the plane through
P₁ = (1, 0, 0), P₂ = (0, 1, 0), P₃ = (0, 0, 1).
Hint: Find d₁ = P₂ − P₁ and d₂ = P₃ − P₁. The normal vector is n = d₁ × d₂. Then use the point-normal form.
Solve the following system by Gauss-Jordan elimination. Show all row operations.
x + y + z = 6
x − y + z = 2
x + y − z = 4
Hint: Form the 3×4 augmented matrix. Eliminate the x-column first (rows 2 and 3), then the y-column, then the z-column. Expect 3 pivots → unique solution.
Start — augmented matrix [A|b]
1
1
1
6
1
−1
1
2
1
1
−1
4
R₂ → R₂ − R₁ R₃ → R₃ − R₁
1
1
1
6
0
−2
0
−4
0
0
−2
−2
R₂ → R₂ ÷ (−2) R₃ → R₃ ÷ (−2)
1
1
1
6
0
1
0
2
0
0
1
1
R₁ → R₁ − R₃ (clear z from row 1)
1
1
0
5
0
1
0
2
0
0
1
1
R₁ → R₁ − R₂ (clear y from row 1) → RREF
1
0
0
3
0
1
0
2
0
0
1
1
x = 3, y = 2, z = 1 (unique solution)
Q6RREF & Systems
Solve the following system by Gauss-Jordan elimination. Write the general solution.
x₁ + 2x₂ − x₃ = 5
2x₁ + 4x₂ + x₃ = 7
Hint: This is a 2-equation, 3-variable system. After row reduction, expect 2 pivots and 1 free variable. Express leading variables in terms of the free parameter t.
Augmented matrix
1
2
−1
5
2
4
1
7
R₂ → R₂ − 2R₁
1
2
−1
5
0
0
3
−3
R₂ → R₂ ÷ 3
1
2
−1
5
0
0
1
−1
R₁ → R₁ + R₂ → RREF
1
2
0
4
0
0
1
−1
Pivots in columns 1 and 3 → leading variables: x₁, x₃. Free variable: x₂ = t
(x₁, x₂, x₃) = (4 − 2t, t, −1), t ∈ ℝ (infinitely many solutions)
Q7Span & Independence
Determine whether the set is linearly independent or dependent.
v₁ =
1
2
1
,
v₂ =
0
1
3
,
v₃ =
1
3
4
Hint: Form the matrix [v₁ | v₂ | v₃] and row-reduce. If a free variable appears → dependent. Also, try checking if one vector equals a combination of the others.
Form [v₁ | v₂ | v₃] and row-reduce
1
0
1
2
1
3
1
3
4
R₂ → R₂ − 2R₁ R₃ → R₃ − R₁
1
0
1
0
1
1
0
3
3
R₃ → R₃ − 3R₂
1
0
1
0
1
1
0
0
0
← zero row → free variable in column 3
Column 3 has no pivot → x₃ is free → the homogeneous system has non-trivial solutions.
Hint: Form the augmented matrix [v₁ | v₂ | b] and row-reduce. If the system is consistent → yes, b is in the span. If there is a contradiction row [0 0 | k] with k ≠ 0 → no.
i: (−1)(1) − (3)(1) = −4
j: (3)(1) − (−1)(1) = 4
k: (−1)(1) − (−1)(1) = 0 n = (−4, 4, 0) → simplify by dividing by −4: n = (1, −1, 0)
Plane equation using P₁ = (2,1,0)
1(x−2) − 1(y−1) + 0(z−0) = 0
x − 2 − y + 1 = 0
x − y = 1
Check: P₁: 2−1=1 ✓ P₂: 1−0=1 ✓ P₃: 3−2=1 ✓
Q5RREF & Systems
Solve the following system by Gauss-Jordan, or show it is inconsistent.
x + y + z = 3
x + y + 2z = 5
2x + 2y + 3z = 7
Hint: Perform row operations to eliminate the x-column. Watch for a contradiction row of the form [0 0 0 | k] with k ≠ 0.
Augmented matrix
1
1
1
3
1
1
2
5
2
2
3
7
R₂ → R₂ − R₁ R₃ → R₃ − 2R₁
1
1
1
3
0
0
1
2
0
0
1
1
R₃ → R₃ − R₂
1
1
1
3
0
0
1
2
0
0
0
−1
← CONTRADICTION: 0 = −1
INCONSISTENT — no solution exists
Q6RREF & Systems
The augmented matrix of a linear system has been row-reduced to the following RREF. Write the general solution.
1
2
0
1
5
0
0
1
−1
3
0
0
0
0
0
Hint: Pivot columns = leading variables. Non-pivot columns (excluding augmented) = free variables. Assign parameters s, t to the free variables and express each leading variable in terms of them.
Pivots in columns 1 and 3 → leading variables: x₁, x₃
No pivots in columns 2 and 4 → free variables: x₂ = s, x₄ = t
Hint: Form [v₁ | v₂ | b] and row-reduce. Check whether the last row produces a contradiction.
Form [v₁ | v₂ | b]
1
2
5
0
1
2
2
3
7
R₃ → R₃ − 2R₁
1
2
5
0
1
2
0
−1
−3
R₃ → R₃ + R₂
1
2
5
0
1
2
0
0
−1
← CONTRADICTION: 0 = −1
NO — b is NOT in Span{v₁, v₂}
The system c₁v₁ + c₂v₂ = b is inconsistent — no linear combination of v₁ and v₂ can produce b.
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Practice Exam 3
8 questions — includes conceptual/true-false and geometric interpretation
Q1Matrix Operations
Let A =
1
2
3
4
5
6
(3×2) and B =
1
0
1
0
1
2
(2×3).
Compute both AB and BA. State the size of each result.
Hint: AB is (3×2)(2×3) = 3×3. BA is (2×3)(3×2) = 2×2. Both products are defined even though neither is square.
AB — size 3×3
AB =
1(1)+2(0)
1(0)+2(1)
1(1)+2(2)
3(1)+4(0)
3(0)+4(1)
3(1)+4(2)
5(1)+6(0)
5(0)+6(1)
5(1)+6(2)
=
1
2
5
3
4
11
5
6
17
BA — size 2×2
BA =
1(1)+0(3)+1(5)
1(2)+0(4)+1(6)
0(1)+1(3)+2(5)
0(2)+1(4)+2(6)
=
6
8
13
16
Note AB is 3×3 and BA is 2×2 — they are completely different objects.
Q2Matrix Operations
State whether each is True or False. Justify your answer.
(a) If A is m×n and B is n×p, then BA is always defined. (b) (AB)ᵀ = AᵀBᵀ for any matrices A, B where AB is defined. (c) The product of two 3×3 matrices is always a 3×3 matrix.
Hint: For (a), check whether the dimensions of B and A are compatible for BA. For (b), recall the transpose reversal rule. For (c), think about what size a product of two 3×3 matrices always is.
(a)FALSE
If A is m×n and B is n×p, then AB is m×p (defined). For BA, B is n×p and A is m×n — inner dims are p and m, so BA is defined only if p = m. Counter-example: A is 2×3, B is 3×4. AB is 2×4 ✓. BA would be (3×4)(2×3) — inner 4 ≠ 2 → undefined.
(b)FALSE
The correct rule is (AB)ᵀ = BᵀAᵀ (order reverses). AᵀBᵀ has the wrong order.
(c)TRUE
(3×3)(3×3) — inner dims always match, result always 3×3.
Q3Lines & Planes
(a) Write the parametric equations of the line through P = (2, 1, −1) with direction d = (3, −2, 1).
(b) Does the point Q = (8, −3, 1) lie on this line?
Hint: For (a), use r = P + td and write component form. For (b), solve for t from one component equation, then check the others.
Solve the following system by Gauss-Jordan elimination. Write the general solution.
x₁ + x₂ + x₃ + x₄ = 2
x₁ + x₂ + 2x₃ + x₄ = 3
Hint: 2 equations, 4 unknowns. After row reduction expect 2 pivots and 2 free variables.
Augmented matrix
1
1
1
1
2
1
1
2
1
3
R₂ → R₂ − R₁
1
1
1
1
2
0
0
1
0
1
R₁ → R₁ − R₂ → RREF
1
1
0
1
1
0
0
1
0
1
Pivots in columns 1 and 3 → leading variables: x₁, x₃
Free variables: x₂ = s, x₄ = t
x₃ = 1 (from row 2)
x₁ = 1 − s − t (from row 1)
(x₁,x₂,x₃,x₄) = (1−s−t, s, 1, t), s,t ∈ ℝ
Q6RREF & Systems
Solve the system by Gauss-Jordan. Show every row operation.
x + 2y − z = 2
2x + y + z = 7
x − y + 2z = 5
Hint: 3 equations, 3 unknowns — expect a unique solution. Eliminate column by column, then back-substitute to clear entries above each pivot.
Augmented matrix
1
2
−1
2
2
1
1
7
1
−1
2
5
R₂ → R₂ − 2R₁ R₃ → R₃ − R₁
1
2
−1
2
0
−3
3
3
0
−3
3
3
R₂ → R₂ ÷ (−3) R₃ → R₃ − R₂ (now identical rows)
1
2
−1
2
0
1
−1
−1
0
0
0
0
← zero row → free variable!
R₁ → R₁ − 2R₂ → RREF
1
0
1
4
0
1
−1
−1
0
0
0
0
Free variable: x₃ = t → x₂ = −1 + t, x₁ = 4 − t
(x,y,z) = (4−t, −1+t, t), t ∈ ℝ (infinitely many solutions)
Q7Span & Independence
Describe Span{v₁, v₂, v₃} geometrically (line, plane, or all of ℝ³?).
v₁ =
1
1
0
,
v₂ =
0
1
1
,
v₃ =
1
2
1
Hint: Row-reduce [v₁ | v₂ | v₃] and count the rank. rank = 1 → line, rank = 2 → plane, rank = 3 → all of ℝ³. If it's a plane, find its normal vector using v₁ × v₂.
Normal vector: n = v₁ × v₂
i: (1)(1) − (0)(1) = 1
j: (0)(0) − (1)(1) = −1
k: (1)(1) − (1)(0) = 1 n = (1, −1, 1)
Span{v₁,v₂,v₃} = the plane x − y + z = 0 in ℝ³
Q8Span & Independence
Does the set {v₁, v₂, v₃} span ℝ³?
v₁ =
1
0
0
,
v₂ =
1
1
0
,
v₃ =
1
1
1
Hint: Form [v₁ | v₂ | v₃] and row-reduce. If rank = 3, the set spans ℝ³. A set of n vectors in ℝⁿ spans ℝⁿ if and only if the matrix they form has rank n.
Form [v₁ | v₂ | v₃]
1
1
1
0
1
1
0
0
1
already upper triangular — 3 pivots on diagonal
Rank = 3 = dimension of ℝ³
Since rank equals the number of dimensions, every vector in ℝ³ can be written as a linear combination of {v₁, v₂, v₃}.
YES — {v₁, v₂, v₃} spans ℝ³
The set is also linearly independent (3 pivots, no free variables) — it forms a basis for ℝ³.
⚡
Trick Questions
Short answers — these are the traps that catch people who know the material but miss the fine print
Most of these require almost no computation. The answer comes from knowing a rule cold. If you find yourself doing a long row-reduction, stop — there's probably a shortcut.
T1RREF
Is the following matrix in RREF? If not, state which property is violated.
1
0
3
5
0
0
1
2
0
0
0
0
NOT in RREF
Column 3 has a pivot (the 1 in row 2). RREF requires all entries above AND below a pivot to be zero. But row 1 has a 3 in column 3 — that's above the pivot and it's non-zero.
To fix it: R₁ → R₁ − 3R₂ gives [1, 0, 0 | −1].
Common mistake: Students check that zeros are below each pivot but forget to clear above. RREF requires both directions.
T2RREF
The augmented matrix of a 3-variable system is in RREF:
1
3
0
4
0
0
1
2
0
0
0
0
Write the general solution. Which variable is free?
Pivots are in columns 1 and 3 → leading variables: x₁ and x₃
Column 2 has no pivot → x₂ is the free variable
Trap: Students assume the free variable is always the last one (x₃). The free variable is whichever column has no pivot — here it's the middle one, x₂.
T3RREF
These two matrices are the final RREF of two different systems. For each one, state: inconsistent, unique solution, or infinitely many solutions.
(a)
1
0
2
3
0
1
−1
4
0
0
0
0
(b)
1
0
2
3
0
1
−1
4
0
0
0
5
(a) Last row is [0 0 0 | 0] — no contradiction, and column 3 has no pivot → free variable x₃ Infinitely many solutions
(b) Last row reads 0·x₁ + 0·x₂ + 0·x₃ = 5, i.e., 0 = 5 — this is impossible Inconsistent — no solution
The only difference is one entry. Both matrices look almost identical — the trap is assuming a zero row always means infinite solutions. It does only when the augmented entry is also 0.
T4Span & Independence
Is the following set linearly independent? Answer immediately — no row reduction needed.
v₁ =
1
2
3
,
v₂ =
0
0
0
,
v₃ =
4
5
6
DEPENDENT — immediately, no work needed
Any set containing the zero vector is automatically linearly dependent. You can always write:
0·v₁ + 1·v₂ + 0·v₃ = 0
This is a non-trivial solution (c₂ = 1 ≠ 0), so the set is dependent by definition.
Rule: Zero vector in the set → always dependent. No calculation required.
T5Span & Independence
Can 4 vectors in ℝ³ ever be linearly independent?
NO — never
Any set of more than n vectors in ℝⁿ is always linearly dependent. With 4 vectors in ℝ³, when you form the 3×4 matrix and row-reduce, you get at most 3 pivots (one per row). That leaves at least one free variable → dependent.
Intuitively: ℝ³ only has "room" for 3 independent directions. A 4th vector must fall in the span of the first three.
Flip side: Can 2 vectors span ℝ³? Also no — you need at least 3 vectors to span ℝ³.
T6Span & Independence
Suppose {v₁, v₂} is linearly independent.
Is the set {v₁, v₂, v₁ + v₂} linearly independent?
NO — always dependent
We can write a non-trivial combination that equals zero:
Since the coefficients (1, 1, −1) are not all zero, the set is dependent — regardless of what v₁ and v₂ actually are.
Trap: "But v₁ and v₂ are independent, so adding more vectors should be fine." Wrong — the moment you add a vector that is already expressible as a combination of existing vectors, the set becomes dependent. v₁+v₂ is always in Span{v₁, v₂}.
T7Lines & Planes
Are these two parametric equations the same line, or different lines?
Line 1: x = 1 + 2t, y = 3 − t, z = 4t Line 2: x = 5 + 6s, y = 1 − 3s, z = 8 + 12s
Step 2 — check if base point of Line 2 lies on Line 1
Does (5, 1, 8) satisfy Line 1's equations?
x: 1 + 2t = 5 → t = 2
y: 3 − 2 = 1 ✓
z: 4(2) = 8 ✓
The point (5,1,8) is on Line 1 (at t = 2), and both lines have parallel directions → same line.
Rule: Parametric forms of lines are NOT unique. Same line can be written with any point on it as base and any scalar multiple of d as direction.
T8Systems / RREF
True or false: "A linear system with fewer equations than unknowns always has infinitely many solutions."
FALSE
Fewer equations than unknowns means you will always have at least one free variable if the system is consistent. But the system might be inconsistent, giving no solution at all.
The correct statement is: if a system with fewer equations than unknowns is consistent, then it has infinitely many solutions. Consistency is not guaranteed.
T9Matrix Operations
Let A be a 3×2 matrix and B be a 2×4 matrix.
(a) Is BA defined? If so, what size? (b) True or false: (AB)ᵀ = AᵀBᵀ
(a) BA would require B (2×4) times A (3×2). Inner dimensions: 4 ≠ 3.
BA is NOT defined
(b)FALSE
The correct rule is: (AB)ᵀ = BᵀAᵀ — the order reverses.
You can verify by size: AB is 3×4, so (AB)ᵀ is 4×3. Now Bᵀ is 4×2 and Aᵀ is 2×3, so BᵀAᵀ is 4×3 ✓. But AᵀBᵀ would be (2×3)(4×2) — inner dims 3 ≠ 4 → not even defined.
Memory trick: Transpose flips the order, just like taking the inverse: (AB)⁻¹ = B⁻¹A⁻¹.