Mathematics Test

Result

Mathematics Test - 47

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The value of $B=\sum_{0 \leq r \leq s \leq n} \sum\left(C_{r}-C_{s}\right)^{2}$is :

A
(n + 1)ⁿ− 2²ⁿ

B
(n + 1)²ⁿC_n− 2ⁿ

C
(n + 1)²ⁿC_n- 2²ⁿ

D
2²ⁿ− 2ⁿ

Solution

Given $A=\sum \sum_{0 \leq r \leq s \leq n} C_{r} C_{s}$
and $\sum_{r=0}^{n} C_{r}^{2}={ }^{2 n} C_{n} \ldots . .(2)$
Given expansion can be written as $\left(\sum_{r=0}^{n} C_{r}\right)^{2}=\left(C_{0}+C_{1}+C_{2}+\ldots .+C_{n}\right)^{2}$
Consider, $\left(C_{0}+C_{1}+C_{2}+\ldots .+C_{n}\right)^{2}$
$\Rightarrow\left(C_{0}+C_{1}+C_{2}+\ldots+C_{n}\right)^{2}=\sum_{r=0}^{n} C_{r}^{2}+2 \sum_{0 \leq r \leq s \leq n} C_{r} C_{s}$
We know that
$C_{0}+C_{1}+C_{2}+\ldots+C_{n}=2^{n}$
$\Rightarrow\left(2^{n}\right)^{2}={ }^{2 n} C_{n}+2 A \quad(b y(1),(2))$
$\Rightarrow 2^{2 n}={ }^{2 n} C_{n}+2 A$
$\Rightarrow 2 A=2^{2 n}-{ }^{2 n} C_{n} \quad \ldots . .(3)$
Now, consider $B=\sum_{0 \leq r \leq n \leq n} \sum\left(C_{r}-C_{s}\right)^{2}$
$=\left(C_{0}-C_{1}\right)^{2}+\left(C_{0}-C_{2}\right)^{2}+\left(C_{0}-C_{3}\right)^{2}+\ldots \ldots+\left(C_{0}-C_{n}\right)^{2}$
$+\left(C_{1}-C_{2}\right)^{2}+\left(C_{1}-C_{3}\right)^{2}+\ldots+\left(C_{1}-C_{n}\right)^{2}$
$+\left(C_{2}-C_{3}\right)^{2}+\left(C_{2}-C_{4}\right)^{2}+\ldots+\left(C_{2}-C_{n}\right)^{2}$
$+\ldots+\left(C_{n-1}-C_{n}\right)^{2}$
$=\left(C_{0}^{2}+C_{1}^{2}-2 C_{0} C_{1}\right)+\left(C_{0}^{2}+C_{2}^{2}-2 C_{0} C_{2}\right)+\left(C_{0}^{2}+C_{3}^{2}-2 C_{0} C_{3}\right)+\ldots+\left(C_{0}^{2}+C_{n}^{2}\right.$
$\left(C_{1}^{2}+C_{2}^{2}-2 C_{1} C_{2}\right)+\left(C_{1}^{2}+C_{3}^{2}-2 C_{1} C_{3}\right)+\ldots+\left(C_{1}^{2}+C_{n}^{2}-2 C_{1} C_{n}\right)+$
$\left(C_{2}^{2}+C_{3}^{2}-2 C_{2} C_{3}\right)+\left(C_{2}^{2}+C_{4}^{2}-2 C_{2} C_{4}\right)+\ldots+\left(C_{2}^{2}+C_{n}^{2}-2 C_{2} C_{n}\right)+$
$\cdots+\left(C_{n-1}^{2}+C_{n}^{2}-2 C_{n-1} C_{n}\right)$
$\Rightarrow B=n\left(C_{0}^{2}+C_{1}^{2}+\ldots+C_{n}^{2}\right)-2 \sum_{0 \leq r \leq s \leq n} C_{r} C_{s}$
$\Rightarrow B=n\left(2 n_{n}\right)-2 A \quad\left(b_{y}(2)\right)$
$\Rightarrow B=n\left(2 n_{n}\right)-2^{2 n}+2 n C_{n}\left(b_{3}(3)\right)$
$\Rightarrow B=(n+1)^{2 n} C_{n}-2^{2 n}$
Hence, the correct option is (C)
Question 2
1 / -0

(A + AB)^T= XA^T, X= ?

A
B^T

B
B

C
I + B^T

D
B^TA^T

Solution

From equation $\Rightarrow(A+A B)^{T}$
$\Rightarrow A^{T}+B^{T} A^{T}$
$\left\{\begin{array}{ll}W e \quad \text { known } & \text { that } \left.(A B)^{T}=B^{T} A^{T}\right\}\end{array}\right.$
$\Rightarrow A^{T}\left(I+B^{T}\right)$
$\Rightarrow A^{T} X$
$\therefore X=I+B^{T}$
Hence, the correct option is (C)
Question 3
1 / -0

If A and B are two square matrices of order n and A and B commute then for any real number K, then ?

A
A − K I, B −K commute.

B
A − K I, B − K I are equal.

C
A − K I, B − K I do not commute.

D
A + KI, B − K I commute.

Solution

$A B=B A \quad$ (GIven)
Consider $(A-K I)(B-K I)$
where, $K$ is any real number
$$
(A-K I)(B-K I)=A B-K(A+B)+K^{2}
$$
Interchanging $A$ and $B$, we get
$$
(B-K I)(A-K I)=B A-K(B+A)+K^{2}
$$
since $A, B$ commute with each other
$$
\begin{aligned}
\therefore(B-K I)(A-K I) &=B A-K(B+A)+K^{2} \\
&=A B-K(A+B)+K^{2}=(A-K I)(B-K I)
\end{aligned}
$$
Hence, option A is true and option C is false.
We cannot say anything about equality of $A-K I$ and $B-K I$, since it is not given that $A$ and $B$ are equal or not. Hence, option $\mathrm{B}$ is false.
Consider $(A+K I)(B-K I)$
$(A+K I)(B-K I)=A B-K(A-B)-K^{2}$
since $A, B$ commute with each other, $\therefore(B-K I)(A+K I)=B A-K(B+A)+K^{2}=A B+K(A-B)+K^{2} \neq(A+K I)(B-K I)$
It is proved that, $A+K I, B-K I$ do not commute with each other. Hence, option D is false.
Hence, the correct option is (A)
Question 4
1 / -0

If
$A=\left[a_{i j}\right]_{3 \times 3}, B=\left[b_{i j}\right]_{2 \times 2}, C=\left[c_{i j}\right]_{3 \times 3}$
are three matrices so that
$|A|=3,|B|=-2,|C|=4,$ then $|A B C|$
is equal to :

A
24

B
-24

C
0

D
Not defined

Solution

If $A=\left[a_{i j}\right]_{3 \times 3}, B=\left[b_{i j}\right]_{2 \times 2}$
then
$A B$ does not exists.
$\therefore$ Number of columns of $A$ iss not same as number of rows of $B$
So, $A B C$ also does not exist. $\therefore|A B C|$ is not defined.
Hence, option D.
Question 5
1 / -0

If A, B are symmetric matrices of the same order then AB − BA is :

A
symmetric matrix

B
skew symmetric matrix

C
Diagonal matrix

D
identity matrix

Solution

Let $A=\left[\begin{array}{ll}a & c \\ c & l\end{array}\right]$
$B=\left[\begin{array}{ll}d & f \\ f & e\end{array}\right]$
$A B=\left[\begin{array}{ll}(a d+c f) & (a f+c e) \\ (c d+l f) & (c f+l e)\end{array}\right]$
$B A=\left[\begin{array}{ll}(d a+f c) & (d c+l f) \\ (f a+c e) & (f c+l e)\end{array}\right]$
$A B-B A=\left[\begin{array}{cc}0 & a f+c e-(d c+l f) \\ -[(a f+c e)-(d c+l f)] & 0\end{array}\right]$
$\therefore$ skew symmetric matrix.
Hence, the correct option is (B)
Question 6
1 / -0

The greatest value of positive integer which divides

(r + 2) (r + 3) (r + 4) (r + 5) ∀r ∈ N is

A
4!

B
2!

C
5!

D
3!

Solution

$\operatorname{Let} P(r)=(r+2)(r+3)+(r+4)(r+5) \forall r \epsilon N$ as $r \epsilon N$ means $P(r)$ is the product of
four consecutive positive integer which means $\mathrm{p}(\mathrm{r})$ is the multiple of $4 \mathrm{I}$
$\therefore$ greatest positive integer which divides $(r+1)(r+2)(r+3)(r+4)$ is $4 !$
Hence, the correct option is (A)
Question 7
1 / -0

Let A and B be 3 × 3 matrices such that A^T= −A, B^T= B , then matrix (λAB + 3BA) is a skew symmertric matrix for :

A
λ = 3

B
λ = −3

C
λ = 3 or λ = −3

D
λ = 3 and λ = −3

Solution

If the matrix $\lambda A B+3 B A$ to be skew-symmetry, Then, it must satisfy $(\lambda A B+3 B A)^{T}=-(\lambda A B+3 B A)$
Now, considering $(\lambda A B+3 B A)^{T}$
$=\lambda(A B)^{T}+3(B A)^{T}$
$=\lambda B^{T} A^{T}+3 A^{T} B^{T}$
Substituting values of $A^{T}=-A$ and $B^{T}=B$, we get $=-\lambda B A-3 A B=-(3 A B+\lambda B A)$
From this, we get $\lambda=3$ clearly.
Hence, option $\mathrm{A}$.
Question 8
1 / -0

If A is a real skew-symmetric matrix such that A²+ I = O, then :

A
A is a square matrix of even order with |A| = ± 1

B
A is a square matrix of odd order with |A| = ± 1

C
A can be a square matrix of any order with |A| = ± 1

D
A is a skew-symmetric matrix of even order with |A| = 1

Solution

Given A is a real skew- - - symmetric matrix. $A^{\prime}=-A$
We know that $|A|=\left|A^{\prime}\right|$ $|k A|=k^{n}|A|$
$\therefore\left|A^{\prime}\right|=(-1)^{n}|A|$
$\Rightarrow|A|=(-1)^{n}|A|$
$\Rightarrow\left(1-(-1)^{n}\right)|A|=O$
$\Rightarrow|A|=0$ or $1-(-1)^{n}=0$ (if
$n$ is even $)$ But given $A^{2}+I=O$ $\Rightarrow A^{2}=-I$
$\Rightarrow|A|^{2}=(-1)^{n}|I|=(-1)^{n} \neq 0$
Hence, A is of even order. $|A|=1$
Hence, option 'D' is correct.
Question 9
1 / -0

If A is a symmetric matrix and nϵN then Aⁿ is :

A
symmetric

B
skew-symmetric

C
a diagonal matrix

D
none of these

Solution

Since, A is symmetric $A^{T}=A$
Now, $\left(A^{2}\right)^{T}=(A A)^{T}=A^{T} A^{T}=A^{2}$
i.e. $\left(A^{2}\right)^{T}=A^{2}$
Hence, $A^{2}$ is symmetric.
$\left(A^{3}\right)^{T}=\left(A^{2} A\right)^{T}=A^{T}\left(A^{2}\right)^{T}=A A^{2}=A^{3}$
i.e. $\left(A^{3}\right)^{T}=A^{3}$
Hence, $A^{3}$ is symmetric.
Hence, for all $n \in N, A^{n}$ is symmetric
Hence, the correct option is (A)
Question 10
1 / -0

If A is a 3^rd order square diagonal matrix of non-positive entries such that A²= I, then :

A
There may exist some diagonal element in A which is zero

B
Value of |A| is −1

C
A⁻¹ does not exist

D
|3adj(2A)| = 1728

Solution

If A is an involutor and diagonal matrix, then $A^{2}=I$. $\operatorname{Let} A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right] .$ Then $A^{2}=I$
$\Rightarrow|A|^{2}=1 \quad \Rightarrow|A|=-1$
¿. - entries are non-positive? Thus $|A|=-1$ $A^{-1}=A$
$|2 A|=-8$
$|3 a d j(2 A)|=27 \cdot|a d j(2 A)|=27(-8)^{2}=1728$
Hence, the correct option is (D)

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mathematics Test - 47

Result

Mathematics Test - 47

Score

-

Rank

-

SHARING IS CARING

Question 1

(n + 1)n− 22n

(n + 1)2nCn− 2n

(n + 1)2nCn- 22n

22n− 2n

Solution

Question 2

BT

B

I + BT

BTAT

Solution

Question 3

A − K I, B −K commute.

A − K I, B − K I are equal.

A − K I, B − K I do not commute.

A + KI, B − K I commute.

Solution

Question 4

24

-24

0

Not defined

Solution

Question 5

symmetric matrix

skew symmetric matrix

Diagonal matrix

identity matrix

Solution

Question 6

4!

2!

5!

3!

Solution

Question 7

λ = 3

λ = −3

λ = 3 or λ = −3

λ = 3 and λ = −3

Solution

Question 8

A is a square matrix of even order with |A| = ± 1

A is a square matrix of odd order with |A| = ± 1

A can be a square matrix of any order with |A| = ± 1

A is a skew-symmetric matrix of even order with |A| = 1

Solution

Question 9

symmetric

skew-symmetric

a diagonal matrix

none of these

Solution

Question 10

There may exist some diagonal element in A which is zero

Value of |A| is −1

A−1 does not exist

|3adj(2A)| = 1728

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

(n + 1)ⁿ− 2²ⁿ

(n + 1)²ⁿC_n− 2ⁿ

(n + 1)²ⁿC_n- 2²ⁿ

2²ⁿ− 2ⁿ

B^T

I + B^T

B^TA^T

A⁻¹ does not exist