Mathematics for Data Science II · Quiz 1 · Sep 2025

**1. Find the row dependence:** The coefficient rows satisfy \(R_3=4R_1-R_2\), because \(4(1,2,1)-(3,0,-1)=(1,8,5)\). The first two rows are independent; their first-two-column minor is \(-6\ne0\). **2. Impose the same relation on the constants:** Consistency requires \(b_3=4b_1-b_2\), equivalently \(-4b_1+b_2+b_3=0\). This is also sufficient: the first two independent equations are solvable, and the third is then their same linear combination. **3. Compare the alternatives:** For example, \(b=(0,1,-1)\) satisfies the derived condition, so it is a consistent right-hand side. It fails each of A, B and C, showing that those are not necessary consistency conditions. **Answer: D — -4b₁ + b₂ + b₃ = 0.** A: **Incorrect:** At the consistent right-hand side \((0,1,-1)\), its left side is 2, not zero. B: **Incorrect:** At \((0,1,-1)\), its left side is \(-2\), not zero. C: **Incorrect:** At \((0,1,-1)\), its left side is 2, not zero. D: **Correct:** It imposes exactly the row relation \(b_3=4b_1-b_2\).

**1. Evaluate S1:** For a square matrix, \(\operatorname{RREF}(A)=I\) means every column has a pivot. Thus \(A\) is invertible and \(Ax=b\) has the unique solution \(x=A^{-1}b\) for every \(b\). S1 is true. **2. Test the converse S2:** Choose \(A=0_{2\times2}\) and \(b=0\). Every \(x\in\mathbb R^2\) solves the system, but \(\operatorname{RREF}(A)=0\ne I\). Existence of a solution does not require invertibility, so S2 is false. **Answer: C — S1 is true and S2 is false..** A: **Incorrect:** S2 fails for a consistent singular system. B: **Incorrect:** S1 is true because an invertible square matrix solves every right-hand side. C: **Correct:** It gives the correct truth values: S1 true, S2 false. D: **Incorrect:** It reverses both established truth values.

**1. Apply the zero and scaling tests:** Positive real numbers exclude zero and are not closed under negative scalars, so A is false. If \(cv=0\) and \(c\ne0\), multiplying by \(1/c\) yields \(v=0\), proving B. **2. Analyze finite real subspaces:** If a subspace contains any nonzero \(v\), it contains all \(tv\) for \(t\in\mathbb R\). These are infinitely many distinct vectors. Therefore a finite subspace can only be \(\{0\}\), proving C. **3. Distinguish spanning from independence:** Removing vectors can destroy spanning: \(\{(1,0),(0,1)\}\) spans \(\mathbb R^2\), but its singleton subset does not. Independence survives passage to a subset, because any relation in the subset extends to the original set by assigning zero coefficients to omitted vectors. Thus E is true and D false. **Answer: B, C, E — B, C and E.** A: **Incorrect:** It excludes zero and, for example, \((-1)(1)=-1\) is outside the set. B: **Correct:** For \(c\ne0\), multiplying \(cv=0\) by the inverse scalar forces \(v=0\). C: **Correct:** A nonzero vector would generate infinitely many distinct real multiples. D: **Incorrect:** A singleton subset of a two-vector basis of \(\mathbb R^2\) fails to span the plane. E: **Correct:** A nontrivial dependence in the subset would also be a dependence in the original set.

**1. Use column dependence:** A zero column or one column proportional to another makes the columns linearly dependent. Either condition forces \(\det A=0\). **2. Use determinant identities:** \(\det(A^T)=\det A\), so D suffices. If \(A^2=0\), then \((\det A)^2=\det(A^2)=0\), forcing \(\det A=0\), so E also suffices. **3. Reject the diagonal-only condition:** The matrix \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\) has zero diagonal but determinant \(-1\). Thus zero diagonal entries alone do not imply singularity. **Answer: A, B, D, E — A, B, D and E.** A: **Correct:** A zero column forces the determinant to vanish. B: **Correct:** Proportional columns are dependent, including when the proportionality scalar is zero. C: **Incorrect:** The displayed two-by-two permutation matrix is an invertible counterexample. D: **Correct:** Transposition preserves the determinant. E: **Correct:** Multiplicativity gives \((\det A)^2=0\).

**1. Add the vectors:** \((1,2)+(2,x)=(3,2+x)\). Being a scalar multiple of \((2,1)\) means \((3,2+x)=t(2,1)\). **2. Compare coordinates:** \(3=2t\) gives \(t=3/2\). Hence \(2+x=3/2\), so \(x=-1/2=-0.5\). **3. Check:** The sum becomes \((3,3/2)=(3/2)(2,1)\). **Answer: -0.5.**

**1. Apply the singularity condition:** A square matrix is not invertible exactly when its determinant is zero. For \[A=\begin{pmatrix}\beta-1&0&-1\\2&\beta+4&2\\\beta+1&\beta+1&\beta+1\end{pmatrix},\] expand along the first row. **2. Factor the determinant:** \[\begin{aligned}\det A&=(\beta-1)\big[(\beta+4)(\beta+1)-2(\beta+1)\big]\\&\quad-\big[2(\beta+1)-(\beta+4)(\beta+1)\big]\\&=(\beta-1)(\beta+1)(\beta+2)+(\beta+1)(\beta+2)\\&=\beta(\beta+1)(\beta+2).\end{aligned}\] **3. Select the minimum root:** The singular values are \(\beta=-2,-1,0\). The smallest is \(-2\). **Answer: -2.**

**1. Set up the monetary equations:** \(2x_1+3x_2=19\) and \(x_1+5x_2=27\), where the variables are the values of individual type-A and type-B coins. **2. Eliminate \(x_1\):** Twice the second equation minus the first gives \(7x_2=35\), so \(x_2=5\). Then \(x_1=27-25=2\). **3. Subtract in the requested order:** \(x_1-x_2=2-5=-3\). The values check both collection totals: \(4+15=19\) and \(2+25=27\). **Answer: -3.**

**1. Count the unconstrained entries:** A real \(3\times3\) matrix has nine independent entries. The trace condition is the single linear equation \(a_{11}+a_{22}+a_{33}=0\). **2. Solve the constraint:** Choose the six off-diagonal entries and \(a_{11},a_{22}\) freely; then \(a_{33}=-a_{11}-a_{22}\). This gives eight free parameters. **3. Confirm by a basis:** If \(E_{ij}\) has a 1 in position \((i,j)\) and zeros elsewhere, a basis consists of the six \(E_{ij}\) with \(i\ne j\), together with \(E_{11}-E_{33}\) and \(E_{22}-E_{33}\). These eight independent matrices span every trace-zero matrix, so \(\dim W=8\). **Answer: 8.**

**1. Bound the rank:** For a \(5\times7\) matrix, \(0\le\operatorname{rank}(A)\le\min(5,7)=5\). **2. Show both extremes occur:** The all-zero matrix has rank zero. The matrix \([I_5\mid0_{5\times2}]\) has five independent rows and rank five. Thus \(n_1=0\) and \(n_2=5\). **3. Compute the difference:** \(n_2-n_1=5-0=5\). **Answer: 5.**

**1. Identify an independent pair:** The first two rows are independent since their first-two-column minor is \(2(1)-3(-3)=11\ne0\). Rank exactly two means the third row must be \(\alpha R_1+\beta R_2\). **2. Use columns 1 and 3:** \(2\alpha+3\beta=4\) and \(\alpha-4\beta=-6\) give \(\beta=16/11\) and \(\alpha=-2/11\). Column 4 checks automatically: \(-4\alpha+5\beta=8\). **3. Read column 2:** \(k=-3\alpha+\beta=6/11+16/11=2\). At this value the third row is dependent while the first two remain independent, establishing rank two. **Answer: 2.**

**1. Compare a decisive coordinate:** The vectors are \(v_1=(1,0,2)\) and \(v_2=(2,3,a)\). Every scalar multiple of \(v_1\) has second coordinate zero, whereas \(v_2\) has second coordinate 3. Thus they cannot be scalar multiples for any \(a\). **2. Verify independence directly:** If \(\alpha v_1+\beta v_2=0\), the second coordinate gives \(3\beta=0\), hence \(\beta=0\). The first coordinate then gives \(\alpha=0\). The pair is independent for every real \(a\), so the stated existence claim is false. **Answer: B — False.** A: **Incorrect:** The second coordinate 3 prevents \(v_2\) from being any multiple of \(v_1\), irrespective of \(a\). B: **Correct:** Only the trivial linear relation is possible for every value of \(a\).