Solution

Solution

Product of eigenvalues = Determinant of matrix

Sum of eigenvalues = Trace

Therefore,

=3(0+1)-1(0+5)+1(-2+20)

=3-5+18

=16

Solution

Solution

Solution

A diagonal matrix is defined as a square matrix in which all off-diagonal entries are zero. (Note that a diagonal matrix is necessarily symmetric.) Entries on the main diagonal may or may not be zero. Thus a Zero matrix is a diagonal matrix .

If the diagonal values are ‘1’ then its identity matrix, which is a diagonal matrix.

If all entries on the main diagonal are equal scalars, then the diagonal matrix is called a scalar matrix.

Solution

The given system can be represented as

The augmented matrix

Solution

Solution

= -2/27

Solution

R is a partial order relation if R is reflexive, antisymmetric and transitive.

To be reflexive,there must be all (x,x) type ordered pairs. Thus there must be (0,0),(1,1),(2,2) must be present. As (0,0),(1,1) are already present (2,2) must be added to be reflexive.

Its already a transitive relation as there are no (a,b),(b,c) type of ordered pairs, if such pairs exist then (a,c) must be present.

For anti symmetric, there must not be any (x,y) (y,x) type ordered pairs.

There is already (0,1) ordered pair, if we add (1,0) then it will not be antisymmetric. Hence R will not be a partial order.

Thus just adding (2,2) is sufficient to make R a partial order

Solution

a * b = max (a, b)

for instance, if we take

1*10 = max (1,10) = 10

-1*5 = max (-1,5) = 5

So, for every value of a, b result belongs to A. so it satisfies closure property.

∴ It is algebraic structure.

Semi group: To be semi group, it should satisfy (a*b)*c = a*(b*c)

▪ (a*b)*c = a*(b*c)

▪ max (a, b) * c = a* max (b, c)

▪ max (a, b, c) = max (a, b, c)

∴ It satisfies for all values of a, b, c. It is semi group.

Monoid: To be monoid, it should satisfy a × e = a

For every value of a, e should be constant. But e cannot be constant as it changes for some

values. For instance, if we assume e = 50, Monoid satisfies for all values of a < 50 but for a >

50 it doesn’t satisfy. So, it is not monoid but it is a semi group.

Solution

Solution

“is a child of” is irreflexive, asymmetric, intransitive. - No person is a child of him/herself - For all people, if x is a child of y, then y is not a child of x. - if x is a child of y and y a child of z, then for no person is x a child of z.

Solution

Therefore, f is continuous at all real numbers greater than 5.

Hence, f is continuous at every real number and therefore, it is a continuous function.

Solution

Solution

It can be seen that alternatively we must get 0’s for infinite series.

And the numbers {2, 0} give an idea of 1+1, 1-1 kind.

Now, we know that

Solution

T(n)=T(n-1)+n rewrite as

T(n)=n+T(n-1)

now expand T(n-1)

=n+n-1+T(n-2)

expand T(n-1)

=n+(n-1)+(n-2)+T(n-3)

expand T(n-3)

=n+(n-1)+(n-2)+(n-3)+T(n-4)

=n+(n-1)+(n-2)+(n-3)+...

This is sum of first ‘n’ numbers,

i.e., n(n+1)/2

Solution

Now multiply a few rows of the options to get the given matrix.

Method 2:

Follow an LU decomposition method, we get

Solution

Solution

Euler's Formula

For any polyhedron that doesn't intersect itself, let F= Number of Faces, V= the Number of Vertices, E=the Number of Edges

Then F + V − E = 2

Given that dodecahedron has twice the number of faces of a cube. We know the cube has 6 faces, then dodecahedron has 12 faces i.e. F=12.

It is given that V= 20

F + V − E = 2

12+20-E=2

E= 30

Solution

There are already 5 persons , when you join the group its 6 in total.

Since all 8 sweets are identical and 6 people are there then number of ways= C(n+r−1,r−1)

n=8, r=6.

Number of ways = 13C5= 1287

Solution

Rolle’s Theorem:

Suppose that a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then if f(a) = f(b), then there exists at least one point c in the open interval (a, b) for which f’(c) = 0.

f(x) = x^2 - 2x - 8 is continuous on [-1,3]

Also f(x) is differentiable on (-1,3).

This f’(x) = 2x-2

f’(c) = 0 at

2c-2 = 0

c = 1

Note: Given f(x) is a polynomial, thus its considered as continuous and differentiable by default in general

Solution

Note: For planar graphs, its m ≤ 3n - 6, m ≤ 2n - 4 ( can be rememberable)

For the formal proof :

For graph G with f faces, it follows from the handshaking lemma for planar graphs that 2m ≥ 3f because the degree of each face of a simple graph is at least 3), so f ≤ 2/3 m.

Combining this with Euler's formula

Since n - m + f = 2

We get m - n + 2 ≤ 2/3 m

Hence m ≤ 3n - 6.

Solution

A matching of graph G is a subgraph of G such that every edge shares no vertex with any other edge. That is, each vertex in matching M has degree one

We can get maximum of 3 matchings, Thus answer is 3.

Note: There can be many other combinations, but nothing can be more than size 3.

Solution

A bipartite graph has two sets of vertices. The edges are nothing but product of number of vertices in each set. This will be maximu, when number of vertices are equal in each set.

If there are n vertices , then n/2 vertices can be there in each set, and mx number of edges = n/2 *n/2 = n^2/4

Here n= 16 , thus n^2/4 = 64

Solution

The connectivity k(kn) of the complete graph kn is n-1.

Connectivity is the number of edges to be removed such that the number of components will be increased by one after removal.

Solution

Cardinality of a set is the number of elements in that set

It is asked that the number of prime numbers under 100,

Let S be that set then

S= {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}. The cardinality is 25

Solution

Let S be a subset of A in the poset (A, R). If there exists an element a in A such that sRa for all s in S, then a is called an upper bound.

The upper bound if {d} is {d,h,i,j,k,l,m}

I.e. 7

Solution

An item z is a least upper bound (LUB) for items x and y, if z is an upper bound of x and y, and no other upper bound d of x and y exists with d .

For {d,k,f} the LUB is {k}.

Solution

There are 5 elements in A.

The powerset of a set with n elements consists of 2^n elements

Thus, P(A) has 2^5 = 32 number of elements

Solution

We know that P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

= P(A) + P(B) − P(A)P(B) (since A,B are independent.)

Thus 1/4 + P(B) − (1/4)P(B) = 2P(B) − 1/4

P(B) = 2/5

Solution

Let D1 be the event for choosing a Dassault plane in the first selection, D2 be the event that choosing the second Dassault plane.

Solution

A Hamiltonian cycle is a closed loop on a graph where every node (vertex) is visited exactly once.

The total number of hamiltonian cycles in a complete graph are (n-1)!/2,

For n=6, (6-1)!/2 = 60

Solution

If ‘H’ is a subgroup of finite group (G, *) then O(H) is the divisor of O(G).

∵ O(H) = order of subgroup H.

Given that the order of the group (size of the group) is 180.

Its divisors are {1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180}

Among them the largest is 180.

In other cases, the largest divisor of 180 is 180.

Thus ‘G’ itself is the largest subgroup.