Mathematics for Data Science II · Quiz 1 · May 2026

**1. Apply the row-swap rule.** Swapping two rows changes the sign of the determinant, so the intermediate determinant is \(-\det(A)=-2\). **2. Apply the scaling rule.** Multiplying one row by \(-3\) multiplies the determinant by \(-3\): \[ \det(B)=(-3)(-2)=\boxed{6}. \] The two sign changes cancel. A: Incorrect. The value \(3\) misses the original determinant factor \(\det(A)=2\). B: Correct. The combined multiplier is \((-1)(-3)=3\), giving \(3\cdot2=6\). C: Incorrect. Both the sign and magnitude are wrong; the final determinant is \(6\). D: Incorrect. The value \(-6\) accounts for scaling but misses the sign reversal caused by swapping rows.

**1. Form the product.** For \(M=\begin{bmatrix}1&-1&2\end{bmatrix}\), \[ M^TM=\begin{bmatrix}1&-1&2\\-1&1&-2\\2&-2&4\end{bmatrix}. \] Its second and third rows are \(-1\) and \(2\) times its first row. Therefore \(\operatorname{rank}(M^TM)=1\), and its determinant is \(0\). Also, \(\operatorname{rank}(M)=1\). **2. Check RREF.** The only nonzero row of \(M\) starts with a leading \(1\). There are no other entries in that pivot column to clear, so \(M\) is already in RREF. Non-pivot entries \(-1\) and \(2\) need not be zero. A: Incorrect. Although \(\operatorname{rank}(M)=1\), the claimed nonzero determinant is false: \(\det(M^TM)=0\). B: Correct. The displayed product has one independent nonzero row, hence rank \(1\) and determinant \(0\). C: Correct. The row \([1,-1,2]\) already satisfies every RREF condition. D: Incorrect. The RREF remains \([1,-1,2]\), not \([1,0,0]\). Row operations cannot independently erase entries in this single row. E: Incorrect. Both matrices have rank \(1\), so their ranks are equal.

**1. Count the degrees of freedom.** A symmetric matrix \(\begin{bmatrix}a&b\\b&c\end{bmatrix}\) is represented by \((a,b,c)\), so \(\dim V=3\). **2. Find the span of the two given matrices.** A linear combination is \[ \alpha\begin{bmatrix}2&1\\1&3\end{bmatrix}+\beta I =\begin{bmatrix}2\alpha+\beta&\alpha\\\alpha&3\alpha+\beta\end{bmatrix}. \] Thus every matrix in their span satisfies \(c=a+b\); conversely, this relation lets us choose \(\alpha=b\), \(\beta=a-2b\). **3. Extend to a basis.** The third matrix must be symmetric and must violate \(c=a+b\). Exactly the second and third options meet these requirements. A: Incorrect. \(\begin{bmatrix}1&0\\1&1\end{bmatrix}\) is not symmetric, so it is not in \(V\). B: Correct. \(\begin{bmatrix}0&1\\1&0\end{bmatrix}\) is symmetric and has \(c=0\ne a+b=1\); it supplies the third independent direction. C: Correct. \(\begin{bmatrix}2&0\\0&3\end{bmatrix}\) is symmetric and has \(3\ne2+0\); it lies outside the existing span. D: Incorrect. \(\begin{bmatrix}0&1\\1&1\end{bmatrix}=\begin{bmatrix}2&1\\1&3\end{bmatrix}-2I\), so it adds no independent vector. E: Incorrect. \(\begin{bmatrix}1&-1\\-1&0\end{bmatrix}=-\begin{bmatrix}2&1\\1&3\end{bmatrix}+3I\), so it is already in the span.

**1. Collect the known vectors.** \[ 2(1,1,1)+3(1,-1,3)=(5,-1,11). \] **2. Isolate the unknown vector.** \[ 4(x,y,z)=(7,5,3)-(5,-1,11)=(2,6,-8), \quad (x,y,z)=\left(\tfrac12,\tfrac32,-2\right). \] **3. Add the components.** \[ x+y+z=\tfrac12+\tfrac32-2=\boxed{0}. \]

**1. Take determinants** of the identity \(A^T=A^2\): \[ \det(A)=\det(A^T)=\det(A^2)=(\det A)^2. \] **2. Use invertibility.** Writing \(d=\det A\), we obtain \(d(d-1)=0\). Since \(A\) is invertible, \(d\ne0\). Therefore \(\boxed{\det A=1}\).

**1. Use the defining equation.** For \(B=\begin{bmatrix}a&b\\c&d\end{bmatrix}\), the condition \(B^T=-B\) gives \[ a=-a,\qquad d=-d,\qquad c=-b. \] Over the real numbers, \(a=d=0\). Hence \[ S=\left\{\begin{bmatrix}0&t\\-t&0\end{bmatrix}:t\in\mathbb R\right\} =\operatorname{span}\left\{\begin{bmatrix}0&1\\-1&0\end{bmatrix}\right\}. \] **2. Identify the dimension.** This span is a subspace and has one nonzero basis matrix. Thus \(\boxed{\dim S=1}\).

**1. Construct the matrix** using \(a_{ij}=\max(i,j)\): \[ A=\begin{bmatrix}1&2&3\\2&2&3\\3&3&3\end{bmatrix}. \] **2. Test whether its rows are independent.** Expanding the determinant along the first row, \[ \det A=1(6-9)-2(6-9)+3(6-6)=3\ne0. \] A \(3\times3\) matrix with nonzero determinant has three pivots, so \(\boxed{\operatorname{rank}A=3}\).

**1. Put the vectors into a square matrix.** Three vectors in \(\mathbb R^3\) are linearly dependent exactly when \[ \det\begin{bmatrix}k&8&-5\\0&k-6&4\\0&3&k-2\end{bmatrix}=0. \] **2. Expand along the first column.** \[ k\big[(k-6)(k-2)-12\big] =k(k^2-8k)=k^2(k-8)=0. \] **3. Count distinct values.** The possibilities are \(k=0\) and \(k=8\). The repeated factor \(k^2\) still represents only one value, so the answer is \(\boxed{2}\).

**1. Compare the given matrices entry by entry.** They satisfy \[ M_3=M_1-M_2,\qquad M_4=2M_3-M_2=2M_1-3M_2. \] The second identity proves \(2M_3-M_2-M_4=0\). **2. Check uniqueness.** In \(aM_1+bM_2=0\), the \((1,1)\) entry gives \(a=0\), and the \((1,2)\) entry then gives \(b=0\). Therefore the zero combination has the unique coefficient pair \((0,0)\). Similarly, \[ aM_1+bM_3=(a+b)M_1-bM_2=M_2 \] forces \(-b=1\) and \(a+b=0\), giving the unique pair \((a,b)=(1,-1)\). A: Incorrect. Since \(M_3=M_1-M_2\), the expression equals \(2M_3\ne0\). B: Correct. Direct substitution of \(M_4=2M_3-M_2\) makes the expression zero. C: Incorrect. The expression is \(5M_2-3M_1\ne0\); its \((1,1)\) entry is \(-3\). D: Correct. The only pair satisfying \(aM_1+bM_2=0\) is \((0,0)\), because \(M_1,M_2\) are independent. E: Incorrect. There is exactly one pair, \((1,-1)\), satisfying \(aM_1+bM_3=M_2\), rather than more than one.

**1. Check the first pair.** If \(aM_1+bM_2=0\), the \((1,1)\) and \((1,2)\) entries give \(a=b=0\). Thus \(\{M_1,M_2\}\) is linearly independent. **2. Use the matrix relations.** \[ M_3=M_1-M_2,\qquad M_4=2M_3-M_2. \] The first relation makes \(\{M_1,M_2,M_3\}\) dependent. However, neither \(M_3,M_4\) nor \(M_2,M_4\) are scalar multiples: \(M_4\) has different diagonal/off-diagonal ratios from \(M_3\), while \(M_2\) has zero diagonal and \(M_4\) does not. Both pairs are independent. A: Correct. Only the trivial combination of \(M_1\) and \(M_2\) gives the zero matrix. B: Incorrect. The nontrivial relation \(M_1-M_2-M_3=0\) proves dependence. C: Incorrect. In \(aM_3+bM_4=0\), the first two entries give \(a+2b=0\) and \(-a-3b=0\), hence \(a=b=0\). D: Incorrect. In \(aM_2+bM_4=0\), the \((1,1)\) entry gives \(2b=0\), and then \(aM_2=0\) gives \(a=0\).

**1. Interpret the column space.** If the columns of \(A\) are \(a_1,\ldots,a_n\), then \[ Ax=x_1a_1+\cdots+x_na_n. \] Therefore \(b\in C(A)\) means precisely that some \(x\in\mathbb R^n\) satisfies \(Ax=b\). **2. Separate existence from uniqueness.** This proves \[ b\in C(A)\quad\Longleftrightarrow\quad Ax=b\text{ is consistent}. \] Uniqueness additionally requires \(\ker A=\{0\}\), which is not assumed. For example, \(A=\begin{bmatrix}1&0\end{bmatrix}\), \(b=\begin{bmatrix}1\end{bmatrix}\) has all solutions \(x=(1,t)^T\). A: Correct. Membership in the column space guarantees at least one coefficient vector \(x\) with \(Ax=b\). B: Incorrect. Existence does not ensure uniqueness; nonzero null-space vectors can be added to a solution. C: Correct. If \(Ax=b\), then \(b\) is the linear combination of the columns of \(A\) with coefficients from \(x\). D: Correct. A consistent system is one that has at least one solution, exactly what \(b\in C(A)\) guarantees. E: Incorrect. \(b\in C(A)\) implies consistency. Inconsistency occurs when \(b\notin C(A)\).

**1. Find when uniqueness can fail.** For \[ A=\begin{bmatrix}3&a\\4&1\end{bmatrix},\qquad b=\begin{bmatrix}1\\4\end{bmatrix}, \] a nonzero determinant would give a unique solution. Hence no solution is possible only if \[ \det A=3-4a=0\quad\Rightarrow\quad a=\frac34. \] **2. Confirm inconsistency.** At \(a=\tfrac34\), reduce the augmented matrix with \(R_2\leftarrow R_2-\tfrac43R_1\): \[ \left[\begin{array}{cc|c}3&\tfrac34&1\\4&1&4\end{array}\right] \longrightarrow \left[\begin{array}{cc|c}3&\tfrac34&1\\0&0&\tfrac83\end{array}\right]. \] The second row demands \(0=\tfrac83\), which is impossible. Thus \(\boxed{a=0.75}\).