**1. Apply orthogonality:** For real vectors, orthogonality means their dot product is zero. Here \((1,a,2)\cdot(2,-1,b)=2-a+2b\).
**2. Rearrange:** \(2-a+2b=0\) gives \(a-2b=2\). This condition is both necessary and sufficient for these two vectors to be orthogonal.
**Answer: A — a-2b=2.**
A: **Correct:** It is the rearranged zero-dot-product equation.
B: **Incorrect:** It gives \(a-2b=-2\), the wrong sign.
C: **Incorrect:** It also gives \(a-2b=-2\).
D: **Incorrect:** It would leave the dot product equal to 2, not zero.
**1. Check continuity at the joining point:** For \(x\le1\), \(f(x)=x^2+2\); for \(x>1\), \(f(x)=x+2\). Both one-sided limits at 1 equal 3, and \(f(1)=3\). Thus the function is continuous there.
**2. Compare one-sided derivatives:** The left derivative is \(2x|_{x=1}=2\), while the right derivative is 1. Since these differ, the derivative at 1 does not exist.
**Answer: A — continuous but not differentiable at x=1.**
A: **Correct:** The values match but the one-sided slopes differ.
B: **Incorrect:** The slopes 2 and 1 prevent differentiability.
C: **Incorrect:** Both limits equal the function value, so continuity does hold.
D: **Incorrect:** Differentiability implies continuity; moreover the derivative here does not exist.
**1. Examine the Jordan block:** \(A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\) has eigenvalue 1 twice, but \((A-I)v=0\) forces the second coordinate to zero. Its eigenspace has dimension one, so it lacks two independent eigenvectors.
**2. Examine the second matrix:** \(B=\begin{pmatrix}2&0&0\\0&0&1\\0&1&0\end{pmatrix}\) is real symmetric. More explicitly its eigenpairs include \(2,(1,0,0)\); \(1,(0,1,1)\); and \(-1,(0,1,-1)\). These three independent eigenvectors give a diagonalization.
**Answer: B — A is not diagonalizable; B is diagonalizable.**
A: **Incorrect:** It reverses both classifications.
B: **Correct:** A is defective, whereas B has a complete eigenbasis.
C: **Incorrect:** A has only one independent eigenvector.
D: **Incorrect:** B is diagonalizable, so this rejects too much.
for a regression problem where each element is of the form
, for
.
is the feature and
is the label of the
data-point.
Let
be two regression models such that:
Which of the following options is correct?
6406536344506.
has the lower mean squared loss.
6406536344507.
has the lower mean squared loss.
6406536344508.
Both models have same mean squared loss.
Published solution
**1. Evaluate both models:** For inputs 1, 2, 3, the labels are 1, 4, 7. Model \(f_1(x)=2x-1\) predicts 1, 3, 5; \(f_2(x)=3x-2\) predicts 1, 4, 7.
**2. Average the squared errors:** \(L_1=(0^2+(-1)^2+(-2)^2)/3=5/3\), while \(L_2=0\). Therefore the second model has lower mean squared loss.
**Answer: B — f₂ has the lower mean squared loss.**
A: **Incorrect:** Its mean squared loss is \(5/3>0\).
B: **Correct:** It exactly matches all three labels, giving zero loss.
C: **Incorrect:** Their losses are unequal.
**1. Find the column dependency:** The columns are \(c_1=(1,2,1)^T\), \(c_2=(1,-1,0)^T\), \(c_3=(2,1,1)^T\). Since \(c_3=c_1+c_2\), the first two columns span the column space.
**2. Prove independence and reject other candidates:** If \(\alpha c_1+\beta c_2=0\), the third coordinate gives \(\alpha=0\), then \(\beta=0\). Thus the space has dimension two. Every vector in it satisfies \(y=3z-x\); this also tests proposed vectors outside the space.
**Answer: A — {(1,2,1), (1,-1,0)}.**
A: **Correct:** It gives the two independent spanning columns.
B: **Incorrect:** \((1,0,0)\) fails \(y=3z-x\), so the pair is not in the column space.
C: **Incorrect:** It includes the dependent third column, so it is a spanning set but not a basis.
D: **Incorrect:** \((1,3,2)\) fails \(3=3(2)-1\); it is outside the space.
**1. Use trace:** The sum of eigenvalues is the trace, so \(m+n=0+3=3\).
**2. Use determinant:** For \(P=\begin{pmatrix}m&2\\1&n\end{pmatrix}\), \(\det P=mn-2=0(3)=0\). Thus \(mn=2\). The parameters are consequently 1 and 2, in either order.
**Answer: B, C — B and C — mn=2 and m+n=3.**
A: **Incorrect:** The determinant condition requires \(mn=2\), not zero.
B: **Correct:** It follows from \(mn-2=0\).
C: **Correct:** It is the trace equation.
D: **Incorrect:** The eigenvalue sum is 3, not 2.
The rank of a matrix is equal to the number of non-zero rows in it.
6406536344523.
If
is a matrix which is obtained by permuting the rows of the
identity
matrix, then the rank of
is
.
6406536344524.
If
is a non-zero column vector in
, then the rank of the matrix
is
.
Published solution
**1. Test the claimed rank formulas:** Rank is subadditive, not generally additive. For \(A=I,B=-I\), \(\operatorname{rank}(A+B)=0\). Counting nonzero rows works after echelon reduction, not for an arbitrary matrix: \(\begin{pmatrix}1&0\\1&0\end{pmatrix}\) has two nonzero rows but rank one.
**2. Establish the two valid claims:** Row permutations are invertible operations, so a permuted \(I_5\) retains rank five. For nonzero real \(u\), every column of \(uu^T\) is a multiple of \(u\), and at least one is nonzero, giving rank one.
**Answer: C, D — C and D.**
A: **Incorrect:** Cancellation can reduce the rank of a sum.
B: **Incorrect:** Nonzero rows can be dependent unless the matrix is in an appropriate echelon form.
C: **Correct:** Permuting identity rows preserves full rank.
D: **Correct:** A nonzero real outer product has a one-dimensional column space.
**1. Compare direction vectors:** The directions \((4,-4,5)\) and \((8,-3,1)\) are not scalar multiples, so the lines are not parallel.
**2. Solve the first two coordinate equations:** Intersection would require \(4t-8s=1\) and \(-4t+3s=-1\). Adding gives \(s=0\), then \(t=1/4\).
**3. Check the third coordinate:** On the first line, \(z=-1+5/4=1/4\); on the second, \(z=5\). They do not match, so the lines do not meet. They are skew.
**Answer: B, D — B and D — nonparallel and nonintersecting.**
A: **Incorrect:** Their direction vectors are not proportional.
B: **Correct:** The directions fail the proportionality test.
C: **Incorrect:** The only parameters satisfying x and y fail the z equation.
D: **Correct:** There is no common point after all three coordinates are checked.
Every square matrix with real entries has at least one real eigenvalue.
6406536344534.
A set of two eigenvectors corresponding to the same eigenvalue of a square
matrix must be linearly dependent.
6406536344535.
Given a square matrix of order
, the set of all eigenvectors for an eigenvalue
, along with the zero vector, forms a subspace of
.
6406536344536.
A square matrix is not invertible if and only if
is one of its eigenvalues.
Published solution
**1. Use counterexamples for A and B:** The real rotation matrix \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) has only the nonreal eigenvalues \(\pm i\). For \(I_2\), the independent vectors \(e_1,e_2\) share eigenvalue 1. Thus A and B are false.
**2. Identify the eigenspace:** Eigenvectors for a real eigenvalue \(\lambda\), together with zero, form \(\ker(A-\lambda I)\), a subspace. Finally, \(A\) is singular exactly when \(Av=0\) has a nonzero solution, which is precisely the eigenvalue-zero condition.
**Answer: C, D — C and D.**
**Convention note:** C uses the usual real-vector-space setting and a real eigenvalue. For complex eigenvalues/eigenvectors, the corresponding eigenspace is considered over the complex field instead.
A: **Incorrect:** Even-order real matrices can have only nonreal conjugate eigenvalues.
B: **Incorrect:** An eigenspace can have dimension greater than one.
C: **Correct:** The set including zero is the nullspace of a linear map.
D: **Correct:** A nontrivial null vector is an eigenvector for eigenvalue zero.
Suppose you are building a credit card fraud detection system and have two approaches:
(i) Manually writing rules such as "If transaction amount is above Rs. 50,000 and the location is
overseas, mark it as fraud.”
(ii) Training a model using millions of past transactions labeled as fraud or not fraud.
Which of the following options is/are correct?
6406536344529.
Approach (i) is not machine learning.
6406536344530.
Approach (ii) is machine learning.
6406536344531.
Both approaches are machine learning.
6406536344532.
Neither approach is machine learning.
Published solution
**1. Distinguish rules from learning:** A programmer-specified fraud threshold is an explicit decision rule. Its parameters are not being learned from examples in the described approach.
**2. Identify supervised learning:** Training on historical transactions with fraud/not-fraud labels learns a decision function from labeled examples. Therefore (i) is not machine learning as described, while (ii) is supervised machine learning.
**Answer: A, B — Approach (i) is not machine learning.; Approach (ii) is machine learning..**
A: **Correct:** The decision rule is manually specified rather than learned.
B: **Correct:** The model learns a classification rule from labeled transactions.
C: **Incorrect:** It incorrectly treats the manual rule as a learning procedure.
D: **Incorrect:** It incorrectly excludes the explicitly trained model.
**1. Write the tangent approximation:** At \(x_0=0\), \(h(x)=f(0)+f^{\prime}(0)x\). Here \(f(x)=(1+2x)^{-5}\), so \(f(0)=1\).
**2. Differentiate using the chain rule:** \(f^{\prime}(x)=-10(1+2x)^{-6}\), hence \(f^{\prime}(0)=-10\). Thus \(a=1,b=-10\), and \(a-b=1-(-10)=11\).
**Answer: 11.**
in the direction of the vector . Your answer should be an integer.
Published solution
**1. Compute the gradient:** For \(f=x^2y-yz^3+z\), \[\nabla f=(2xy,\ x^2-z^3,\ 1-3yz^2).\] At \((1,-2,0)\), this is \((-4,1,1)\).
**2. Check and use the direction:** \(u=(-2,-1,2)/3\) has norm one. The directional derivative is \[D_uf=\nabla f\cdot u=\frac{8-1+2}{3}=3.\]
**Answer: 3.**
Let be the projection matrix that projects vectors in
onto the line
.
Let be the matrix that projects vectors in
onto the nullspace of
.
Based on the above data, answer the given subquestions.
Find
. Your answer should be an integer.
Published solution
**1. Identify a spanning direction:** The line \(y=3x\) is spanned by \(v=(1,3)^T\). Its orthogonal projection matrix is \[P=\frac{vv^T}{v^Tv}=\frac1{10}\begin{pmatrix}1&3\\3&9\end{pmatrix}.\]
**2. Sum the entries:** \(a+b+c+d=(1+3+3+9)/10=16/10\). Therefore \(10(a+b+c+d)=16\).
**Answer: 16.**
Let be the projection matrix that projects vectors in
onto the line
.
Let be the matrix that projects vectors in
onto the nullspace of
.
Based on the above data, answer the given subquestions.
Find
. Your answer should be an integer.
Published solution
**1. Project onto the complementary subspace:** For an orthogonal projector \(P\), its nullspace is orthogonal to its image. Thus the orthogonal projector onto \(\ker P\) is \(A=I-P\). With \(P=\frac1{10}\begin{pmatrix}1&3\\3&9\end{pmatrix}\), \[A=\frac1{10}\begin{pmatrix}9&-3\\-3&1\end{pmatrix}.\]
**2. Evaluate the required sum:** \(10(e+f+g+h)=9-3-3+1=4\). The negative off-diagonal entries are essential.
**Answer: 4.**
Based on the above data, answer the given subquestions.
If
is the smallest eigenvalue of
, enter the value of
. Your answer should be an integer.
Published solution
**1. Form the characteristic polynomial:** For \(A=\begin{pmatrix}1&5\\4&2\end{pmatrix}\), \[\det(A-\lambda I)=(1-\lambda)(2-\lambda)-20=\lambda^2-3\lambda-18.\]
**2. Factor and order the roots:** \((\lambda-6)(\lambda+3)=0\), so the eigenvalues are 6 and \(-3\). The smallest is \(-3\).
**Answer: -3.**
Based on the above data, answer the given subquestions.
If
is an eigenvector of
corresponding to
, find
. Enter your answer correct to two
decimal places.
Published solution
**1. Identify the eigenvalue:** \(\det(A-\lambda I)=(\lambda-6)(\lambda+3)\), so the smallest eigenvalue is \(-3\).
**2. Apply the eigenvector equation:** With \(v=(a,-1)^T\), \(Av=(a-5,4a-2)^T=-3(a,-1)^T\). The first coordinate gives \(a-5=-3a\), hence \(a=5/4=1.25\).
**3. Verify the remaining coordinate:** \(4(5/4)-2=3\), as required by \(-3(-1)=3\).
**Answer: 1.25.**
