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Must Do Problem For NDA

Must Do Problem For NDA

Q 2886878777

A and B are two matrices of same order `3 xx 3`, where
`A = [( 1,2,3),(2,3,4),(5,6,8)]` and ` B = [ (3,2,5),(2,3,8),(7,2,9)]`
The value of `adj (adj A)` equals

(A)

`-A`

(B)

`4A`

(C)

`8A`

(D)

`16A`

Solution:

`adj (adj A)`

` = | A |^(n - 2) A = | A | A = - A`
Correct Answer is `=>` (A) `-A`
Q 2886878777

A and B are two matrices of same order `3 xx 3`, where
`A = [( 1,2,3),(2,3,4),(5,6,8)]` and ` B = [ (3,2,5),(2,38),(7,29)]`
The value of `| adj ( B) |` equals

(A)

`24`

(B)

`24^2`

(C)

`24^3`

(D)

`8^2`

Solution:

` | adj B | = | B |^(n-1) = |B|^2 = 24^2`
Correct Answer is `=>` (B) `24^2`
Q 2886878777

A and B are two matrices of same order `3 xx 3`, where
`A = [( 1,2,3),(2,3,4),(5,6,8)]` and ` B = [ (3,2,5),(2,38),(7,29)]`
The value of `| (adj (adj (adj (adj A)))) |` equals

(A)

`2^4`

(B)

`2^9`

(C)

`1`

(D)

`2^(19)`

Solution:

`| (adj (adj (adj (adj A)))) |`

`= | adj (adj ( - A)) | = | - A |^((n-1)^2) = |-A|^4`

`= (|A|)^4 = (-1)^4 = 1`
Correct Answer is `=>` (C) `1`
Q 2846778673

`A (theta) = [ (sin theta , i cos theta),(i cos theta , sin theta) ]` , where `i = sqrt(- 1)`
If `B(theta) = A (pi/2 - theta)`. then `AB` equals

(A)

`[ (0, i),(i,0) ]`

(B)

`[ (0, -i),(- i, 0) ]`

(C)

`[ (0, 1),(1 , 0) ]`

(D)

None of these

Solution:

We have, `B(theta) = A ( pi/2 - theta)`

` = [ ( sin ( pi/2 - theta) , i cos ( pi/2 - theta)),( i cos ( pi/2 - theta) , sin ( pi/2 - theta) ) ]`

` = [ ( cos theta , i sin theta),( i sin theta , cos theta) ]`

Now,

` AB = [ ( sin theta , i cos theta),( i sin theta , sin theta) ] [ ( cos theta , i sin theta),( i sin theta , cos theta) ]`

` = [ ( sin theta cos theta + i^2 sin theta cos theta , i sin^2 theta + i cos^2 theta ),( i cos^2 theta + i sin^2 theta , i^2 cos theta sin theta + sin theta cos theta ) ]`

` = [ (0 , i),(i , 0) ]`
Correct Answer is `=>` (A) `[ (0, i),(i,0) ]`
Q 2846278173

Let A be a `2 xx 2` matrix with non-zero entries and let `A^2 = I`, where `I` is `2 xx 2` identity matrix, then consider the following statements
I. Sum of diagonal elements of `A` is `0`.
II. Determinant of matrix `A` is `1`.
Which of the above statement(s) is/are correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

Let `A = [ (a,b) ,( c , b) ]`

Then, ` A^2 = [ (a,b) ,( c , b) ] [ (a,b) ,( c , b) ] `

` = [ (a^2 + bc , ab + bd ),(ac + cd , bc + d^2) ]`

Given, `A^2 = I`

` => [ (a^2 + bc , ab + bd ),(ac + cd , bc + d^2) ] = [ (1,0),(0,1) ]`

`=> b( a + d) = 0` and `c (a + d) = 0`

`=> a + d = 0 [ ∵ b != 0 , c != 0 ]`

So, sum of diagonal elements of `A` is `0`.

So, Statement I is correct.

Now, `|A| = ad - bc = - a^2 - bc`

`= - (a^2 + bc) = - 1 [∵ a^2 + bc = 1]`

So, Statement II is incorrect.
Correct Answer is `=>` (A) Only I
Q 2846178073

If A and B are square matrices such that `B = A^(-1) BA = 0`, then consider the following statements
I. `AB + BA = 0` II. `A^2 - B^2 = (A + B) (A - B)`
Which of the above statement(s) is/are correct ?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

`B = A^(-1) B A`

`=> AB = A A^(-1) BA => AB = I (BA)`

`=> AB = ( BA) => AB - BA = 0`

`:.` Statement I is wrong.

Now, `(A+ B)(A- B)`

`= A ( A B) + B(A - B)`

`= A^2 - AB + BA - B^2`

`= A^2 - AB + AB - B^2 [ ∵ AB = BA]`

`= A^2 - B^2`

`:.` Statement II is correct.
Correct Answer is `=>` (B) Only II
Q 2806567478

If `A = [ ( cos x , sin x , 0),(- sin x , cos x , 0) , ( 0 , 0 ,1) ] = f (x)` , then `A^(-1)` equale

(A)

`f(-x)`

(B)

`- f(x)`

(C)

`- f (- x)`

(D)

`f(x)`

Solution:

` | A| = | ( cos x , sin x , 0),(- sin x , cos x , 0) , ( 0 , 0 ,1) | `

`= cos^2 x + sin^2 x = 1 != 0`

Now, `c_(11) = cos x , c_(12) = sin x , c_(13) = 0`

` c_(21) = - sin x, c_(22) = cos x, c_(23) = 0`

` c_(31) = 0, c_(32) = 0 , c_(33) = 1`

`:. A^(-1) = 1/(|A|) adj (A) = adj (A)`

` = [ ( cos x , - sin x , 0),( sin x , cos x , 0) , ( 0 , 0 ,1) ] `

` = [ ( cos (-x) , sin (-x) , 0) ,(- sin (-x) , cos (-x) , 0) , ( 0 , 0 ,1) ] = f(-x)`
Correct Answer is `=>` (A) `f(-x)`
Q 2816567470

If a matrix `A` is such that `3 A^3 + 2 A^2 + 5A + I = O`, then what is `A^(-1)` equal to ?

(A)

`- (3A^2 + 2A + 5)`

(B)

`3A^2 + 2A + 5l`

(C)

`3A^2 - 2A - 5l`

(D)

`- (3A^2 + 2A + 5l)`

Solution:

`∵ 3 A^3 + 2 A^2 + 5 A+ I = O`

`=> 3 A^3 A^(-1) + 2 A^2 A^(-1)`

`+ 5 A A^(-1) + I A^(-1) = 0`

`=> 3A^2 + 2A + 5 + A^(-1) = O`

`=> A^(-1) = - (3A^2 + 2A + 5)`
Correct Answer is `=>` (A) `- (3A^2 + 2A + 5)`
Q 2876467376

If ` [ ( 1, - tan theta),(tan theta , 1) ] [ ( 1 , tan theta),(- tan theta , 1) ]^(-1) = [ (a , -b),(b ,a) ]` , then

(A)

`a= 1, b = 1`

(B)

`a = cos 2 theta, b = sin 2 theta`

(C)

`a = sin 2 theta, b = cos 2 theta`

(D)

None of these

Solution:

` [ ( 1, - tan theta),(tan theta , 1) ] [ ( 1 , tan theta),(- tan theta , 1) ]^(-1) = [ (a , -b),(b ,a) ]`

`=> [ ( 1, - tan theta),(tan theta , 1) ] 1/(1 + tan theta) [ ( 1, - tan theta),(tan theta , 1) ] = [ (a , -b),(b ,a) ]`

` => 1/(1 + tan theta)`

` [ ( 1 - tan^2 theta , - 2 tan theta), (2 tan theta , 1 - tan^2 theta ) ] = [ (a , -b),(b ,a) ]`

` => [ ( (1 - tan^2 theta)/(1 + tan^2 theta) (- 2 tan theta)/(1 + tan^2 theta) ) , ((2 tan theta)/(1 + tan^2 theta) (1 - tan^2 theta)/(1 + tan^2 theta)) ] = [ (a , -b),(b ,a) ]`

` => [ ( cos 2 theta ,- sin 2 theta ) , ( sin 2 theta , cos 2 theta) ] = [ (a , -b),(b ,a) ]`

`=> a = cos 2 theta , b = sin 2 theta`
Correct Answer is `=>` (B) `a = cos 2 theta, b = sin 2 theta`
Q 2886367277

If the matrix ` A = [ (2-x , 1 , 1),(1 , 3-x, 0),( -1, -3,-x)]` is singular, then what is the solution set S ?

(A)

`S = {0, 2, 3}`

(B)

`S = {-1, 2, 3}`

(C)

`S = {1, 2, 3}`

(D)

`S = {2, 3}`

Solution:

For the singular matrix,

`| (2-x , 1 , 1),(1 , 3-x, 0),( -1, -3,-x)| = 0`

`=> (2 - x) [ x (x - 3) ] - [-x]`

`+ [-3 + (3-x) ] = 0`

`=> x (x - 3) (x - 2) = 0 => x = 0, 2, 3`

So, the solution set is, `S = { 0, 2, 3}`.
Correct Answer is `=>` (A) `S = {0, 2, 3}`
Q 2826712671

If `A =[(1,2),(3, 4)] ` and `b = [( a, 0),( 0, b) ]` where a, bare natural numbers, then which one of the following is correct?

(A)

There exist more than one but finite number of B's such that AB = BA

(B)

There exist exactly one B such that AB = BA

(C)

There exist infinitely many B's such that AB = BA

(D)

There cannot exist any B such that AB = BA

Solution:

`because A = [(1,2),(3, 4)]` and `B = [(a,0),( 0,b)]`

`:. AB = [(1,2),( 3,4)] [(a,0),(0,b)] = [ (a , 2b),( 3a, 4b) ]`

and `BA = [ (a, 0),( 0, b)] [ (1,2),(3,4) ] = [ (a, 2a),( 3b , 4 b)]`

If `AB = BA`, then `[ (a, 2b),(3a ,4b) ] = [ (a, 2a), ( 3b , 4b) ] => a =b`

From the above it is clear that there exist infinitely many B' s such that AB = BA.
Correct Answer is `=>` (C) There exist infinitely many B's such that AB = BA
 
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