Level 1 — RecognitionLinear Algebra (Full)

Linear Algebra (Full)

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Difficulty Level: 1 — Recognition (MCQ + Matching + True/False with justification) Time Limit: 20 minutes Total Marks: 30

Section A — Multiple Choice (1 mark each) — 10 marks

Choose the single best answer.

Q1. For $\mathbf{u}=(1,2,2)$ and $\mathbf{v}=(2,0,-1)$ , the dot product $\mathbf{u}\cdot\mathbf{v}$ equals: (a) $0$ (b) $4$ (c) $-4$ (d) $2$

Q2. The magnitude $\|\mathbf{u}\|$ of $\mathbf{u}=(1,2,2)$ is: (a) $5$ (b) $3$ (c) $\sqrt{5}$ (d) $9$

Q3. By the Rank–Nullity theorem, if $A$ is a $4\times 6$ matrix with rank $3$ , the dimension of its null space is: (a) $1$ (b) $2$ (c) $3$ (d) $4$

Q4. The determinant of $\begin{pmatrix}2 & 1\\ 3 & 4\end{pmatrix}$ is: (a) $5$ (b) $11$ (c) $8$ (d) $-5$

Q5. Which condition is NOT equivalent to " $A$ (an $n\times n$ matrix) is invertible"? (a) $\det A \neq 0$ (b) $A\mathbf{x}=\mathbf{0}$ has only the trivial solution (c) columns of $A$ are linearly dependent (d) $\operatorname{rank}(A)=n$

Q6. For a $3\times 3$ orthogonal matrix $Q$ , $\det Q$ must equal: (a) any real value (b) $0$ (c) $\pm 1$ (d) $3$

Q7. The characteristic polynomial of $A=\begin{pmatrix}2 & 0\\ 0 & 5\end{pmatrix}$ has roots (eigenvalues): (a) $0,7$ (b) $2,5$ (c) $-2,-5$ (d) $2,2$

Q8. The cross product $\mathbf{i}\times\mathbf{j}$ equals: (a) $\mathbf{k}$ (b) $-\mathbf{k}$ (c) $\mathbf{0}$ (d) $\mathbf{i}$

Q9. A quadratic form $Q(\mathbf{x})=\mathbf{x}^TA\mathbf{x}$ with symmetric $A$ is positive definite iff all eigenvalues of $A$ are: (a) negative (b) zero (c) positive (d) $\pm 1$

Q10. The scalar projection of $\mathbf{u}$ onto $\mathbf{v}$ is given by: (a) $\dfrac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|}$ (b) $\dfrac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{v}\|}$ (c) $\mathbf{u}\cdot\mathbf{v}\,\|\mathbf{v}\|$ (d) $\dfrac{\|\mathbf{u}\|}{\|\mathbf{v}\|}$

Section B — Matching (1 mark each) — 6 marks

Q11. Match each concept in Column X to its correct description in Column Y.

Column X	Column Y
(i) Rank of a matrix	(P) Set of all linear combinations of a set of vectors
(ii) Span	(Q) Number of pivot positions
(iii) Basis	(R) Solutions of $A\mathbf{x}=\mathbf{0}$
(iv) Null space	(S) Linearly independent spanning set
(v) Eigenvector	(T) Nonzero $\mathbf{x}$ with $A\mathbf{x}=\lambda\mathbf{x}$
(vi) Determinant (3×3)	(U) Signed volume of parallelepiped

Write your answers as (i)–?, (ii)–?, … (6 answers).

Section C — True/False WITH justification (2 marks each: 1 for T/F, 1 for reason) — 14 marks

Q12. Matrix multiplication is commutative: $AB=BA$ for all conformable matrices. (T/F + reason)

Q13. If a set of vectors in $\mathbb{R}^n$ contains more than $n$ vectors, it must be linearly dependent. (T/F + reason)

Q14. Every symmetric real matrix has real eigenvalues and orthogonal eigenvectors. (T/F + reason)

Q15. The Cauchy–Schwarz inequality states $|\mathbf{u}\cdot\mathbf{v}| \le \|\mathbf{u}\|\,\|\mathbf{v}\|$ . (T/F + reason)

Q16. In Gram–Schmidt orthogonalization, the resulting vectors span a different subspace than the originals. (T/F + reason)

Q17. If $\det A = 0$ then the columns of $A$ are linearly independent. (T/F + reason)

Q18. For any matrix $A$ , row rank equals column rank. (T/F + reason)

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1 — (b) $4$ . $\mathbf{u}\cdot\mathbf{v}=1\cdot2+2\cdot0+2\cdot(-1)=2+0-2=0$ . Wait — recompute: $=0$ . Correct answer is (a) $0$ . [1] (Mark scheme: correct value $0$ → (a).)

Q2 — (b) $3$ . $\|\mathbf{u}\|=\sqrt{1+4+4}=\sqrt9=3$ . [1]

Q3 — (c) $3$ . Rank + nullity = number of columns $=6$ ; nullity $=6-3=3$ . [1]

Q4 — (a) $5$ . $2\cdot4-1\cdot3=8-3=5$ . [1]

Q5 — (c). Invertibility requires linearly independent columns; dependent columns is the opposite (singular case). [1]

Q6 — (c) $\pm1$ . $\det(Q^TQ)=\det I=1\Rightarrow(\det Q)^2=1$ . [1]

Q7 — (b) $2,5$ . Diagonal matrix eigenvalues are its diagonal entries. [1]

Q8 — (a) $\mathbf{k}$ . Right-hand rule / standard cyclic identity. [1]

Q9 — (c) positive. Positive definite $\iff$ all eigenvalues $>0$ . [1]

Q10 — (b) $\dfrac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{v}\|}$ . Scalar projection onto $\mathbf{v}$ divides by $\|\mathbf{v}\|$ . [1]

Section B (1 mark each)

Q11:

(i) Rank → (Q) number of pivot positions
(ii) Span → (P) all linear combinations
(iii) Basis → (S) linearly independent spanning set
(iv) Null space → (R) solutions of $A\mathbf{x}=\mathbf{0}$
(v) Eigenvector → (T) nonzero $\mathbf{x}$ with $A\mathbf{x}=\lambda\mathbf{x}$
(vi) Determinant → (U) signed volume

[6 marks: 1 each]

Section C (2 marks each: 1 T/F + 1 reason)

Q12 — FALSE. [1] Matrix multiplication is generally non-commutative; e.g. $\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ but reversed gives $\begin{pmatrix}0&0\\0&1\end{pmatrix}$ . [1]

Q13 — TRUE. [1] In $\mathbb{R}^n$ any linearly independent set has at most $n$ vectors (dimension is $n$ ); more than $n$ forces dependence. [1]

Q14 — TRUE. [1] Spectral theorem: real symmetric matrices have real eigenvalues and admit an orthonormal eigenbasis (eigenvectors for distinct eigenvalues are orthogonal). [1]

Q15 — TRUE. [1] Follows from $\|\mathbf u\|^2\|\mathbf v\|^2-(\mathbf u\cdot\mathbf v)^2\ge0$ ; equality iff parallel. [1]

Q16 — FALSE. [1] Gram–Schmidt produces orthogonal vectors that span the same subspace at each stage. [1]

Q17 — FALSE. [1] $\det A=0$ means $A$ singular, so columns are linearly dependent, not independent. [1]

Q18 — TRUE. [1] Row rank = column rank is a fundamental theorem; both equal the number of pivots. [1]

Note/correction: Q1's correct option is (a) $0$ (computation gives $0$ ). Award (a).

[
  {"claim":"u·v for u=(1,2,2), v=(2,0,-1) equals 0","code":"u=Matrix([1,2,2]); v=Matrix([2,0,-1]); result=(u.dot(v)==0)"},
  {"claim":"norm of (1,2,2) is 3","code":"u=Matrix([1,2,2]); result=(sqrt(u.dot(u))==3)"},
  {"claim":"nullity of 4x6 rank-3 matrix is 3","code":"result=(6-3==3)"},
  {"claim":"det[[2,1],[3,4]]=5","code":"A=Matrix([[2,1],[3,4]]); result=(A.det()==5)"},
  {"claim":"eigenvalues of diag(2,5) are {2,5}","code":"A=Matrix([[2,0],[0,5]]); result=(set(A.eigenvals().keys())=={2,5})"}
]