**1. Evaluate the function and gradient:** For \(f(x,y)=\sqrt{x+y}\), \(f(4,5)=3\), and \(f_x=f_y=1/(2\sqrt{x+y})\). Thus both partial derivatives at \((4,5)\) equal \(1/6\).
**2. Form the first-order approximation:** \[L(x,y)=f(4,5)+\nabla f(4,5)\cdot(x-4,y-5)=3+\frac{x-4}{6}+\frac{y-5}{6}.\] It agrees with both the value and first derivatives at the base point.
**Answer: B — L(x,y)=3+(x-4)/6+(y-5)/6.**
A: **Incorrect:** It uses the wrong function value 9 and reverses the base coordinates.
B: **Correct:** It has the correct base value and both tangent coefficients.
C: **Incorrect:** Its coefficients are twice the correct partial derivatives.
D: **Incorrect:** Its constant and partial-derivative coefficients are incorrect.
**1. Find the gradient at the current dimensions:** For \(C=x^2+y^2\), \(\nabla C=(2x,2y)\), so \(\nabla C(1,2)=(2,4)\).
**2. Dot with the unit direction:** The supplied vector \(v=(3,4)/5\) has norm one. Thus \[D_vC(1,2)=(2,4)\cdot(3/5,4/5)=\frac{6+16}{5}=\frac{22}{5}.\] The numerical rate is in hundreds of rupees per unit movement in the dimension plane.
**Answer: D — 22/5.**
A: **Incorrect:** 5 is the current cost value, not the directional derivative.
B: **Incorrect:** 4 is only the y-partial derivative.
C: **Incorrect:** It omits the factor 2 in the gradient.
D: **Correct:** It is the gradient dotted with the normalized direction.
**1. Choose a direction on the line:** The line \(x+y=0\) is spanned by \(v=(1,-1)^T\). For orthogonal projection, \(P=vv^T/(v^Tv)\).
**2. Calculate and check:** \[P=\frac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}.\] It fixes \((1,-1)^T\), annihilates the perpendicular vector \((1,1)^T\), and satisfies \(P^2=P\).
**Answer: A — P=(1/2)[[1,-1],[-1,1]].**
A: **Correct:** It is the orthogonal projector onto the specified line.
B: **Incorrect:** It projects onto the perpendicular line y=x.
C: **Incorrect:** It is -2 times the required projector and does not fix vectors on the target line.
D: **Incorrect:** It is not symmetric or idempotent and does not have the stated image.
be a real symmetric matrix. Consider the following statements.
(1)
is orthogonally diagonalizable.
(2)
is not diagonalizable.
(3) All the eigenvalues of
are real.
(4) The eigenvalues of
may be imaginary.
(5) Eigenvectors corresponding to distinct eigenvalues of
are linearly independent.
(6) Eigenvectors corresponding to distinct eigenvalues of
are linearly dependent.
Which of the above statements are true?
6406537022680.
(1), (3), (5)
6406537022681.
(2), (4), (6)
6406537022682.
(2), (3), (5)
6406537022683.
(1), (4), (6)
Published solution
**1. Apply the spectral theorem:** A real symmetric matrix has real eigenvalues and an orthonormal eigenbasis. Hence it is orthogonally diagonalizable, proving statements (1) and (3) and rejecting (2) and the intended nonreal interpretation of (4).
**2. Assess eigenvectors for distinct eigenvalues:** If \(Mu=\lambda u\), \(Mv=\mu v\), symmetry gives \(\lambda u^Tv=u^TMv=\mu u^Tv\). For \(\lambda\ne\mu\), \(u^Tv=0\), so the eigenvectors are independent. Thus (5) is true and (6) false.
**Answer: A — (1), (3), (5).**
A: **Correct:** It contains exactly (1), (3) and (5).
B: **Incorrect:** All three selected statements contradict the spectral theorem or independence.
C: **Incorrect:** It incorrectly includes (2), despite the guaranteed diagonalization.
D: **Incorrect:** It incorrectly includes the nonreal-eigenvalue and dependence claims.
**1. Interpret multiplication by rows:** Row i of \(AB\) equals \(\sum_j a_{ij}r_j\), where \(r_j\) are the rows of B. Thus the second row is \(r_1+3r_2-2r_3+4r_4\). This proves every product row lies in the row space of B.
**2. Distinguish columns and the other row-space claim:** Column 2 of \(AB\) is \(Ac_2\); it is not determined by a single row of A. Nor must the product row space lie in that of A: for example, an A whose rows are multiples of \(a=(1,3,-2,4)\), followed by B=diag(1,0,0,0), gives nonzero rows parallel to \(e_1\), which is not parallel to a.
**Answer: A, D — A and D.**
A: **Correct:** It is the row-combination formula with the supplied coefficients.
B: **Incorrect:** The second column is Ac₂, not the displayed combination of B’s columns.
C: **Incorrect:** The row-space containment is not generally in A; the counterexample shows why.
D: **Correct:** Every row of AB is a linear combination of rows of B.
repeated twice. Which of the following is/are
true?
6406537022688.
is invertible.
6406537022689.
Each of the entries on the main diagonal of
is equal to
.
6406537022690.
has two linearly independent eigenvectors.
6406537022691.
is not invertible, where
is the identity matrix.
Published solution
**1. Use the repeated eigenvalues:** The determinant of P is \(3\cdot3=9\ne0\), so P is invertible. Since 3 is an eigenvalue, \(P-3I\) has a nonzero null vector and is singular.
**2. Avoid conclusions that repetition does not justify:** The matrix \(\begin{pmatrix}4&1\\-1&2\end{pmatrix}\) has characteristic polynomial \((\lambda-3)^2\) but diagonal entries 4 and 2. A Jordan block \(\begin{pmatrix}3&1\\0&3\end{pmatrix}\) has only one independent eigenvector. Thus B and C are not guaranteed.
**Answer: A, D — A and D.**
A: **Correct:** No eigenvalue is zero, so the matrix is invertible.
B: **Incorrect:** Eigenvalues do not have to equal individual diagonal entries for a general matrix.
C: **Incorrect:** A repeated eigenvalue need not have a two-dimensional eigenspace.
D: **Correct:** Subtracting an eigenvalue times the identity makes the matrix singular.
matrix with rank 4. Which of the following is/are true?
6406537022692.
Columns of
are linearly independent.
6406537022693.
Null space of
contains only the zero vector.
6406537022694.
Row space of
is
.
6406537022695.
Left null space of
has dimension 2.
Published solution
**1. Apply rank–nullity to the domain:** A \(6\times4\) matrix maps \(\mathbb R^4\) into \(\mathbb R^6\). Rank 4 means all four columns are independent and \(\dim\ker M=4-4=0\).
**2. Identify the other spaces:** Rows lie in \(\mathbb R^4\), so the row space is actually \(\mathbb R^4\), not \(\mathbb R^6\). The left nullspace is \(\ker M^T\), with dimension \(6-4=2\).
**Answer: A, B, D — A, B and D.**
A: **Correct:** The rank equals the total number of columns.
B: **Correct:** Zero nullity means only the zero vector solves Mx=0.
C: **Incorrect:** Rows have four coordinates; the claim uses the wrong ambient space.
D: **Correct:** The transpose has a six-dimensional domain and rank four.
**1. Use the point-direction equation:** The line is \((x,y)=(2,1)+\alpha(-3,2)\), \(\alpha\in\mathbb R\). This is exactly A.
**2. Eliminate the parameter:** A normal vector is \((2,3)\), since \((2,3)\cdot(-3,2)=0\). Thus \(2(x-2)+3(y-1)=0\), or \(2x+3y=7\), which is B. C swaps the point and direction; D does not even contain \((2,1)\).
**Answer: A, B — A and B.**
A: **Correct:** It uses the specified base point and direction.
B: **Correct:** It is the equivalent normal equation.
C: **Incorrect:** Its direction (2,1) is not parallel to (-3,2).
D: **Incorrect:** At the required point, 2y-x=0, not 7.
**1. Apply Cauchy–Schwarz:** \(|u^Tv|\le\|u\|\|v\|=2(3)=6\). This is an upper bound independent of the five-dimensional coordinates.
**2. Show the bound is attainable:** Take \(u=(2,0,0,0,0)\) and \(v=(3,0,0,0,0)\). Their norms are 2 and 3, and \(|u^Tv|=6\). Therefore the maximum, not just a bound, is 6.
**Answer: 6.**
**1. Form the eigenvector equations:** \(A(3,1)^T=(3,1)^T\) gives \(3a+b=3\), \(3c+d=1\). The eigenpair \(4,(2,1)^T\) gives \(2a+b=8\), \(2c+d=4\).
**2. Solve for the entries:** Subtracting yields \(a=-5,c=-3\), then \(b=18,d=10\). Thus \(A=\begin{pmatrix}-5&18\\-3&10\end{pmatrix}\).
**3. Sum and check:** \(a+b+c+d=-5+18-3+10=20\). Its trace is 5 and determinant 4, consistent with eigenvalues 1 and 4.
**Answer: 20.**
Consider the following dataset consisting of two features
and
:
We want to reduce the dimensionality of the dataset from
to
using the following encoder-
decoder pairs.
The reconstruction loss is defined as the mean squared distance between the reconstructed input
and the original input. Based on the above information, answer the given sub-questions.
Compute the reconstruction loss for Pair 1.
**NOTE:** Your answer should be an integer
Published solution
**1. Encode and decode each sample:** Pair 1 maps \((x_1,x_2)\) to their average and reconstructs it as \((u,u)\). Thus \((2,4)\mapsto3\mapsto(3,3)\), and \((6,8)\mapsto7\mapsto(7,7)\).
**2. Calculate squared Euclidean errors:** The two squared distances are \((3-2)^2+(3-4)^2=2\) and \((7-6)^2+(7-8)^2=2\).
**3. Average over data points:** The specified reconstruction loss is \((2+2)/2=2\). Do not divide by feature count as well; the definition averages squared distances across samples.
**Answer: 2.**
Consider the following dataset consisting of two features
and
:
We want to reduce the dimensionality of the dataset from
to
using the following encoder-
decoder pairs.
The reconstruction loss is defined as the mean squared distance between the reconstructed input
and the original input. Based on the above information, answer the given sub-questions.
Compute the reconstruction loss for Pair 2.
**NOTE:** Your answer should be an integer
Published solution
**1. Apply Pair 2:** The encoder gives \(u=x_1-x_2=-2\) for both \((2,4)\) and \((6,8)\). Both reconstructions are \((-2,-2)\).
**2. Compute both squared distances:** For the first sample the error is \((-2-2)^2+(-2-4)^2=16+36=52\). For the second it is \((-2-6)^2+(-2-8)^2=64+100=164\).
**3. Average over samples:** \(L_2=(52+164)/2=108\). This loss includes the errors in both coordinates.
**Answer: 108.**
Consider the following dataset consisting of two features
and
:
We want to reduce the dimensionality of the dataset from
to
using the following encoder-
decoder pairs.
The reconstruction loss is defined as the mean squared distance between the reconstructed input
and the original input. Based on the above information, answer the given sub-questions.
Which encoder-decoder pair performs better?
6406537022704.
Pair-1
6406537022705.
Pair-2
Published solution
**1. Compare on the same loss definition:** On the two samples \((2,4),(6,8)\), Pair 1 reconstructs \((3,3),(7,7)\), producing squared distances 2 and 2. Pair 2 reconstructs \((-2,-2)\) twice, producing squared distances 52 and 164.
**2. Select the smaller average:** \(L_1=2\), whereas \(L_2=108\). Lower reconstruction error means Pair 1 performs better on this dataset.
**Answer: A — Pair-1.**
A: **Correct:** Its reconstruction loss is 2, the smaller value.
B: **Incorrect:** Its loss is 108, much larger on the given samples.
