Free practice final
Linear Algebra practice final (free, with answers)
A typical Linear Algebra final covers solving systems by row reduction, matrix algebra and determinants, vector spaces, rank and linear transformations, eigenvalues and diagonalization, and orthogonal projections and least squares. Work each problem by hand before opening the answer, then check your steps against the explanation.
Topics
- Systems and Row Reduction
- Matrix Algebra and Inverses
- Determinants
- Vector Spaces and Subspaces
- Linear Independence and Bases
- Rank and Nullity
- Linear Transformations
- Eigenvalues and Diagonalization
- Orthogonality and Least Squares
15 questions
Try each one before you open its answer.
1. For which values of h and k does the system x + 2y − z = 1, 2x + 5y + z = 4, x + 3y + hz = k have no solution?
- A h = 2 and k = 3
- B h = 2 and k ≠ 3
- C h ≠ 2, for any value of k
- D h = 3 and k ≠ 2
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Answer: B. h = 2 and k ≠ 3
Subtracting 2 times row 1 from row 2 gives y + 3z = 2, and subtracting row 1 from row 3 gives y + (h + 1)z = k − 1. Subtracting those two leaves (h − 2)z = k − 3. When h = 2 that row reads 0 = k − 3, a contradiction whenever k ≠ 3; when h = 2 and k = 3 there are infinitely many solutions, and when h ≠ 2 there is exactly one.
2. Solve the system x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.
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Answer: (x, y, z) = (1, 2, 3)
Adding the first and third equations eliminates z and gives 2x + 3y = 8, and subtracting the first equation from the second gives x − 2y = −3. Solving those two gives y = 2 and x = 1, and then z = 6 − 1 − 2 = 3. Substituting (1, 2, 3) into all three equations confirms it.
3. Let A and B be any two n × n matrices. Which statement is always true?
- A AB = BA
- B (A + B)² = A² + 2AB + B²
- C If AB = 0, then A = 0 or B = 0
- D (AB)ᵀ = BᵀAᵀ
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Answer: D. (AB)ᵀ = BᵀAᵀ
Transposing a product reverses the order of the factors, so (AB)ᵀ = BᵀAᵀ always. Matrix multiplication is not commutative, so AB = BA can fail, and so can (A + B)² = A² + 2AB + B², since the true expansion is A² + AB + BA + B². A product of nonzero matrices can also be zero: with A = B = [[0, 1], [0, 0]], AB = 0.
4. Find the inverse of A = [[1, 2, 1], [0, 1, 1], [1, 2, 2]].
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Answer: A⁻¹ = [[0, −2, 1], [1, 1, −1], [−1, 0, 1]]
Row reduce [A | I]: subtract row 1 from row 3, subtract the new row 3 from rows 1 and 2, then subtract 2 times row 2 from row 1. When the left half becomes I, the right half is A⁻¹. Multiplying A by the result gives I, which confirms it.
5. A is a 3 × 3 matrix with det A = 4. What is det(2A⁻¹)?
- A 1/2
- B 2
- C 32
- D 1/32
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Answer: B. 2
Multiplying a 3 × 3 matrix by 2 multiplies each of its 3 rows by 2, which multiplies the determinant by 2³ = 8. Also det(A⁻¹) = 1 / det A = 1/4, so det(2A⁻¹) = 8 · (1/4) = 2.
6. For which values of k is the matrix A = [[1, 2, 3], [0, k, 1], [2, 4, k]] not invertible?
- A k = 0 or k = 6
- B k = 6 only
- C k = −6 or k = 0
- D k = 2 or k = 3
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Answer: A. k = 0 or k = 6
A square matrix fails to be invertible exactly when its determinant is 0. Expanding along the first column, det A = 1 · (k² − 4) + 2 · (2 − 3k) = k² − 6k = k(k − 6), which is zero when k = 0 or k = 6.
7. Which of these sets is a subspace of R³?
- A All (x, y, z) with x + y + z = 1
- B All (x, y, z) with xy = 0
- C All (x, y, z) with x = 2z
- D All (x, y, z) with x ≥ 0
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Answer: C. All (x, y, z) with x = 2z
The set where x = 2z is the null space of the 1 × 3 matrix [[1, 0, −2]], so it contains the zero vector and is closed under addition and scalar multiplication. The plane x + y + z = 1 does not contain the zero vector, the set xy = 0 is not closed under addition because (1, 0, 0) + (0, 1, 0) = (1, 1, 0), and the set x ≥ 0 is not closed under multiplication by −1.
8. Which statement about vectors in R³ is true?
- A Any set of four vectors in R³ is linearly dependent
- B Any set of three vectors in R³ is a basis for R³
- C A set that contains the zero vector can still be linearly independent
- D If no vector in a set of three is a scalar multiple of another, the set is linearly independent
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Answer: A. Any set of four vectors in R³ is linearly dependent
R³ has dimension 3, so a linearly independent set in R³ has at most three vectors, and any four vectors must be dependent. Three vectors form a basis only if they are independent, and any set containing the zero vector is dependent. The vectors (1, 0, 0), (0, 1, 0) and (1, 1, 0) are pairwise non-parallel yet dependent, because the third is the sum of the first two.
9. A is a 6 × 9 matrix, and the null space of A has dimension 4. Which statement is true?
- A rank A = 2, and the columns of A do not span R⁶
- B rank A = 5, and the columns of A span R⁶
- C rank A = 4, and the columns of A span R⁹
- D rank A = 5, and the columns of A do not span R⁶
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Answer: D. rank A = 5, and the columns of A do not span R⁶
The Rank–Nullity Theorem says rank A + dim Nul A equals the number of columns, so rank A = 9 − 4 = 5. The columns are vectors in R⁶, and they span R⁶ only if the column space has dimension 6, so they do not.
10. Find the rank of A = [[1, 2, 0, 3], [2, 4, 1, 8], [3, 6, 1, 11]] and the dimension of its null space.
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Answer: Rank 2; the null space has dimension 2
Subtracting multiples of row 1 gives rows (0, 0, 1, 2) and (0, 0, 1, 2), and subtracting those leaves a zero row, so the echelon form has pivots only in columns 1 and 3. Two pivots means rank 2, and the 4 − 2 = 2 free variables give a null space of dimension 2.
11. The linear map T: R² → R² first rotates each vector 90° counterclockwise about the origin and then reflects the result across the x-axis. What is the standard matrix of T?
- A [[0, −1], [1, 0]]
- B [[0, 1], [1, 0]]
- C [[0, −1], [−1, 0]]
- D [[0, 1], [−1, 0]]
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Answer: C. [[0, −1], [−1, 0]]
The columns of the standard matrix are T(e₁) and T(e₂). Rotating e₁ = (1, 0) gives (0, 1), which reflects to (0, −1), and rotating e₂ = (0, 1) gives (−1, 0), which the reflection leaves unchanged. So the matrix is [[0, −1], [−1, 0]], the reflection matrix times the rotation matrix, with the first step on the right.
12. Find the eigenvalues of A = [[4, 1], [2, 3]] and an eigenvector for the larger eigenvalue.
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Answer: λ = 2 and λ = 5; for λ = 5, any nonzero multiple of (1, 1)
The characteristic polynomial is det(A − λI) = (4 − λ)(3 − λ) − 2 = λ² − 7λ + 10 = (λ − 2)(λ − 5). For λ = 5, A − 5I = [[−1, 1], [2, −2]], whose null space is spanned by (1, 1), and indeed A(1, 1) = (5, 5).
13. Which of these matrices is not diagonalizable?
- A [[2, 0], [0, 2]]
- B [[1, 2], [2, 1]]
- C [[1, 1], [0, 2]]
- D [[2, 1], [0, 2]]
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Answer: D. [[2, 1], [0, 2]]
The matrix [[2, 1], [0, 2]] has the single eigenvalue 2, but A − 2I = [[0, 1], [0, 0]] has rank 1, so there is only one independent eigenvector, too few for a basis of R². The others are diagonalizable: [[2, 0], [0, 2]] is already diagonal, [[1, 2], [2, 1]] is symmetric, and [[1, 1], [0, 2]] has two distinct eigenvalues.
14. What is the orthogonal projection of y = (3, 0, 3) onto the line spanned by u = (1, 2, 2)?
- A (3, 6, 6)
- B (1, 2, 2)
- C (9, 18, 18)
- D (2, −2, 1)
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Answer: B. (1, 2, 2)
The projection is ((y · u) / (u · u)) u. Here y · u = 3 + 0 + 6 = 9 and u · u = 1 + 4 + 4 = 9, so the projection is 1 · u = (1, 2, 2). The leftover part y − (1, 2, 2) = (2, −2, 1) is orthogonal to u, as it should be.
15. Find the least-squares line y = b₀ + b₁x for the data points (0, 1), (1, 3), (2, 4) and (3, 4).
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Answer: y = 1.5 + x (b₀ = 3/2 and b₁ = 1)
With design matrix X whose rows are (1, xᵢ), the normal equations XᵀX b = Xᵀy are 4b₀ + 6b₁ = 12 and 6b₀ + 14b₁ = 23. Solving them gives b₁ = 1 and b₀ = 3/2.