Mathematics for Data Science II · Quiz 1 · Jan 2026

**1. Eliminate the first variable:** From the equations, subtract twice the first row from the second, and three times the first row from the third. The last two equations become \(y+(k-2)z=0\) and \(2y+(k-2)z=0\). Subtract twice the former from the latter to obtain \((2-k)z=0\). **2. Separate the parameter cases:** If \(k\ne2\), then \(z=0\), \(y=0\), and \(x=3\), a unique solution. If \(k=2\), the equations reduce to \(y=0\), \(x+z=3\), giving \((x,y,z)=(3-t,0,t)\) for every real \(t\). **3. Evaluate the statements:** S1 and S3 are true. S2 is false because the exceptional case has infinitely many solutions, rather than no solution. **Answer: C — Only S1 and S3 are correct..** A: **Incorrect:** It omits the true statement S3. B: **Incorrect:** S2 is false: the exceptional system is consistent. C: **Correct:** It includes exactly the two true statements, S1 and S3. D: **Incorrect:** It includes false S2 and omits true S3. E: **Incorrect:** S2 is false.

**1. Parameterize the subspace:** The constraints imply \(x=-y-z\) and \(w=y+z\). Hence \[(x,y,z,w)=y(-1,1,0,1)+z(-1,0,1,1).\] The two displayed generators are independent, so \(\dim W=2\). **2. Check the proposed pair:** Both \((-4,1,3,4)\) and \((0,-5,5,0)\) satisfy the constraints. They are not scalar multiples: the first has nonzero first coordinate while the second has first coordinate zero. Thus they form an independent pair in a two-dimensional space and are a basis. **Answer: A — the pair {(-4,1,3,4), (0,-5,5,0)}.** A: **Correct:** Both vectors lie in \(W\) and are independent; their number equals \(\dim W=2\). B: **Incorrect:** It has three vectors in a two-dimensional space. In fact the third is the negative of the sum of the first two. C: **Incorrect:** Its first vector is twice the second plus the third, so the set is dependent. D: **Incorrect:** The second vector is \(-6\) times the first, so this pair spans only a line.

**1. Apply the subspace test:** A nonempty subset must contain the zero vector and be closed under addition and real scalar multiplication. The plane \(x+y+z=0\) is the kernel of a linear map, and \(z=0\) is a coordinate plane. Both pass these tests. **2. Reject the other sets:** The plane \(z=2\) excludes zero. The union in D need not be closed: \(u=(1,0,0)\) lies in \(z=0\), and \(v=(0,1,-1)\) lies in \(x+y+z=0\), but \(u+v=(1,1,-1)\) lies in neither plane. **Answer: B, C — B and C — the two homogeneous planes.** A: **Incorrect:** The zero vector has \(z=0\), so it does not belong to this set. B: **Correct:** The homogeneous linear equation is preserved by addition and scalar multiplication. C: **Correct:** It is \(\operatorname{span}\{(1,0,0),(0,1,0)\}\). D: **Incorrect:** The explicit sum \((1,0,0)+(0,1,-1)=(1,1,-1)\) is outside the union.

**1. Use determinant properties:** If all row sums vanish, then \(A(1,1,1,1)^T=0\). A nonzero null vector makes \(A\) singular, so its determinant is zero. Interchanging two rows multiplies a determinant by \(-1\). **2. Check repeated parameters and scaling:** Equal parameters produce identical rows, hence zero determinant. Multiplying each of four rows by \(-1\) gives \(\det(-A)=(-1)^4\det A=\det A\), so it does not reverse the sign. **3. Interpret the hypotheses carefully:** With distinct real parameters this Vandermonde matrix is nonsingular. A and C are valid conditional determinant rules; their premises are not asserted to occur simultaneously with the stated distinctness. **Answer: A, B, C — A, B and C.** A: **Correct:** Zero row sums give the nonzero null vector \((1,1,1,1)^T\). B: **Correct:** A single row interchange reverses the determinant sign. C: **Correct:** If two parameters were equal, two rows would coincide and the determinant would vanish. D: **Incorrect:** Four sign multipliers have product \((-1)^4=1\), so the determinant is unchanged.

**1. Preserve the solution set:** Every elementary row operation on the augmented matrix is reversible. Thus applying such operations to \([A\mid b]\) produces an equivalent system with exactly the same solutions. **2. Test consistency after reduction:** A row \([0\ 0\ 0\ 0\mid1]\) means \(0=1\), so RREF alone does not guarantee a solution. Conversely, if no such contradictory row exists, free variables can be chosen and pivot variables solved. **3. Delete the last column:** Removing the augmented column from an RREF matrix retains the coefficient pivots and their reduced-column properties. A possible pivot row in the last column becomes a zero row at the bottom. Hence the coefficient block remains in RREF. **Answer: A, C, D — A, C and D.** A: **Correct:** Reversible row operations preserve equivalence. B: **Incorrect:** An inconsistent system also has an RREF; a last-column pivot can encode a contradiction. C: **Correct:** Deleting the augmented column preserves RREF in the coefficient block. D: **Correct:** Absence of a contradictory row is exactly the consistency criterion.

**1. Translate both resource constraints:** Let \(a,b\) count gadgets A and B. Using all hours gives \(2a+3b=8\) and \(3a+2b=7\). **2. Solve for the production quantities:** Subtracting twice the second equation from three times the first yields \(5b=10\), so \(b=2\) and \(a=1\). Both are nonnegative whole numbers. **3. Calculate and check profit:** Assembly uses \(2+6=8\) hours and testing uses \(3+4=7\). Profit is \(5a+4b=5+8=13\) rupees. **Answer: 13.**

**1. Expand the linear combination:** With \(a=(1,2)\) and \(b=(3,-1)\), \(v=ka+lb=(k+3l,2k-l)\). **2. Use equal coordinates:** \(k+3l=2k-l\) gives \(k=4l\). Combining this with \(k+l=5\) gives \(5l=5\), hence \(l=1\) and \(k=4\). **3. Verify:** \(v=4(1,2)+(3,-1)=(7,7)\), which has equal coordinates as required. **Answer: 4.**

**1. Impose a nontrivial linear relation:** Write \(\alpha(-1,c,2)+\beta(0,1,3)+\gamma(-2,0,1)=0\). The first coordinate gives \(\alpha=-2\gamma\); the second gives \(\beta=2c\gamma\). **2. Use the third coordinate:** Substitution gives \(-4\gamma+6c\gamma+\gamma=(6c-3)\gamma=0\). For a nontrivial relation \(\gamma\ne0\), so \(c=1/2\). Therefore the requested value is \(7c=7/2=3.5\). **3. Check dependence:** At \(c=1/2\), the relation \(-2(-1,1/2,2)+(0,1,3)+(-2,0,1)=0\) is nontrivial. **Answer: 3.5.**

**1. Form the price equations:** \[m+2c+3k=140,\qquad2m+c+4k=160,\qquad3m+3c+5k=240.\] **2. Eliminate the muffin and cookie prices:** Add the first two equations and subtract the third: \(2k=60\), giving \(k=30\). The first two equations then become \(m+2c=50\) and \(2m+c=40\), so \(c=20\) and \(m=10\). **3. Add the prices:** \(m+c+k=10+20+30=60\). These values reproduce all three daily totals. **Answer: 60.**

**1. Use the independent first two rows:** Their minor in columns 1 and 3 is \(3(7)-5(-1)=26\ne0\). Rank two therefore requires the third row to be a linear combination \(R_3=\alpha R_1+\beta R_2\). **2. Solve from the fixed columns:** Columns 1 and 3 give \(3\alpha+5\beta=7\) and \(-\alpha+7\beta=15\). Thus \(\beta=2\), \(\alpha=-1\), and \(R_3=-R_1+2R_2\). **3. Read the remaining entries:** \(k_1=-4+2(2)=0\) and \(k_2=-3+2(1)=-1\). This makes the third row dependent while the first two remain independent, so the rank is exactly two. **Answer: 0.**

**1. Use the independent first two rows:** Their minor in columns 1 and 3 is \(3(7)-5(-1)=26\ne0\). Rank two therefore requires the third row to be a linear combination \(R_3=\alpha R_1+\beta R_2\). **2. Solve from the fixed columns:** Columns 1 and 3 give \(3\alpha+5\beta=7\) and \(-\alpha+7\beta=15\). Thus \(\beta=2\), \(\alpha=-1\), and \(R_3=-R_1+2R_2\). **3. Read the remaining entries:** \(k_1=-4+2(2)=0\) and \(k_2=-3+2(1)=-1\). This makes the third row dependent while the first two remain independent, so the rank is exactly two. **Answer: -1.**

**1. Check the operations in C:** Addition and scaling are \((x_1,y_1,4)\oplus(x_2,y_2,4)=(x_1+x_2,y_1+y_2,4)\) and \(c\odot(x,y,4)=(cx,cy,4)\). Both retain the fixed last coordinate. **2. Transfer the vector-space structure:** The bijection \(\phi:W\to\mathbb R^2\), \(\phi(x,y,4)=(x,y)\), preserves these operations. Consequently all vector-space axioms follow from those of \(\mathbb R^2\). The zero vector in this structure is \((0,0,4)\), and the additive inverse of \((x,y,4)\) is \((-x,-y,4)\). **3. Reject the other choices:** Scaling the third coordinate to \(4c\) fails closure for general \(c\). Defining the sum as the first operand fails commutativity when the operands differ. **Answer: C — C — keep the third coordinate fixed at 4 in both operations.** A: **Incorrect:** Scaling by zero gives \((0,0,0)\notin W\); scalar multiplication is not closed. B: **Incorrect:** The proposed addition returns the first operand, so reversing distinct operands changes the sum. C: **Correct:** The operations correspond exactly to ordinary vector operations on the first two coordinates. D: **Incorrect:** Addition fails commutativity, and scaling the last coordinate also fails closure.

**1. Use the specified operations:** The space consists of \((x,y,4)\), with coordinatewise addition and scaling on the first two coordinates while the third stays 4. Its zero vector is \((0,0,4)\). **2. Construct a basis:** Let \(e_1=(1,0,4)\) and \(e_2=(0,1,4)\). Under these operations, \(x\odot e_1\oplus y\odot e_2=(x,y,4)\), so the pair spans. A linear combination is \((0,0,4)\) only when both coefficients are zero, proving independence. **3. Count basis vectors:** This basis has two vectors, so \(\dim W=2\). The fixed third coordinate does not contribute a free parameter. **Answer: 2.**