Mathematics Previous year Matrices Question for NDA
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Previous year Matrices Question for NDA

Previous year Matrices Question for NDA

Set - 1
Q 2713291149

If ` A = [ (alpha , 2),(2 , alpha) ]` and `det (A^3) = 125,` then `alpha` is equal to
NDA Paper 1 2017
(A)

`±1`

(B)

`±2`

(C)

`±3`

(D)

`±5`

Solution:

`A =[(alpha, 2),(2, alpha)]`

`det (A^3) = 125`

`(det A)^3 = 125`

`(alpha^2 -4)^3 = 125 = 5^3`

`alpha^2 =9`

`alpha= pm 3`
Correct Answer is `=>` (C) `±3`
Q 2743391243

If B is a non-singular matrix and A is a square matrix, then the value of `det `(`B^(-1) AB`) is equal to
NDA Paper 1 2017
(A)

det (B)

(B)

det (A)

(C)

det (`B^(-1)`)

(D)

det (`A^(-1)`)

Solution:

` | B^(-1) AB|`

`= |B^(-1) | * |AB| ` (Since `|AB| = |A| * |B|`)

`= 1/(|B|) * |A| * |B|` (since `|B^(-1)| = 1/(|B|)`)

`=|A|`

`Det (A)`
Correct Answer is `=>` (B) det (A)
Q 2713391249

The equations
`x + 2y + 3z = 1`
`2x + y + 3z = 2`
`5x + 5y + 9z = 4`
NDA Paper 1 2017
(A)

have the unique solution

(B)

have infinitely many solutions

(C)

arc inconsistent

(D)

None of the above

Solution:

`Delta = [ ( 1,2, 3 ), ( 2, 1 , 3 ), ( 5 , 5 , 9 ) ]`

`= 3 `

`Delta _1 = [ ( 1, 2 , 3 ), ( 2, 1 ,3 ), ( 4 , 5 , 9 ) ]`

`= 0`

`Delta _2 = [( 1, 1 , 3 ),(2,2,3),(5,4,9)] = -3`

So `Delta != 3, Delta _1 != 0` Equation have unique solution.
Correct Answer is `=>` (A) have the unique solution
Q 2733491342

`A = [ (x + y , y),(x , x - y)] , B = [ (3),(-2) ] ` and ` C = [ (4),(-2) ] ` If `AB = C`, then what is `A^2` equal to?
NDA Paper 1 2017
(A)

` [ (4 , 8),(-4 , - 16)]`

(B)

` [ (4 , -4),(8 , - 16)]`

(C)

` [ (-4 , 8),(4 , 12)]`

(D)

` [ (-4 , -8),(4 , 12)]`

Solution:

`AB = C`

`[(x + y , y ) , ( x , x -y ) ] [ (3), (-2) ] = [ (4), (-2) ] `

`[ (3x + 3 y - 2 y ) , ( 3 x -2x + 2 y ) ] = [ (4), ( -2 ) ]`

`3x + y = 4`

`x +2 y =-2 `

`x =2 , y = -2`

`A = [ (0, (-2 )) , (2, 4 ) ]`

`A^2 = [ ( -4 , -8 ), ( 4 ,12 ) ]`
Correct Answer is `=>` (D) ` [ (-4 , -8),(4 , 12)]`
Q 2733591442

Consider the set A of all matrices of order `3 xx 3` with entries `0` or `1` only. Let B be the subset of A consisting of all matrices whose determinant is `1`. Let C be the set of A consisting of all matrices whose determinant is `-1`. Then which one of the following is correct ?

NDA Paper 1 2017
(A)

`C` is empty

(B)

`B` has as many elements as `C`

(C)

`A = B cup C`

(D)

`B` has thrice as many elements as `C`

Solution:

(1) `C` cannot be empty as

If `A= [(1, 0,0), (0,0,1),(0,1,0)] det (A)=-1`

(2) For every matrix in B there is a corresponding matrix in C obtained by reversing the first and second columns of

the matrix as value of determinant changes sign , if we replace first and second row and vice versa.

`=> n(B) = n(C) `
(3) If `A= [(1,0,1),(1,0,0),(0,1,1)]` then `det |A| =2`

So, `A ne B cup C`

(4) `n(B) ne 3 n(C) ->` Doesn't make any sense.
Correct Answer is `=>` (B) `B` has as many elements as `C`
Q 2763591445

If `A = [ ( cos theta , sin theta),( - sin theta , cos theta)]` , then what is ` A^3` equal to?
NDA Paper 1 2017
(A)

`[ ( cos 3 theta , sin 3 theta),( - sin 3 theta , cos 3 theta)]`

(B)

`[ ( cos^3 theta , sin^3 theta),( - sin^3 theta , cos^3 theta)]`

(C)

`[ ( cos 3 theta , - sin 3 theta),( sin 3 theta , cos 3 theta)]`

(D)

`[ ( cos^3 theta ,- sin^3 theta),( sin^3 theta , cos^3 theta)]`

Solution:

`A = [ ( cos theta , sin theta),( - sin theta , cos theta)]`

`A^2 = [ (cos theta ,sin theta ), ( - sin theta , cos theta )] [ ( cos theta , sin theta ) , ( - sin theta , cos theta ) ] `

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

`A^3 = A .A ^2 = [(cos theta, sin theta),(-sin theta, cos theta)][(cos 2 theta, sin 2 theta),(-sin 2 theta, cos 2 theta)]`

`=> [ (cos theta cos 2 theta+sin theta cos 2 theta, cos theta sin2 theta+ sin theta cos 2 theta), ( - sin theta cos 2 theta- sin 2 theta cos theta , - sin theta sin 2 theta + cos theta cos 2 theta)]`

`=[ ( cos 3 theta , sin 3 theta),( - sin 3 theta, cos 3 theta)]`
Correct Answer is `=>` (A) `[ ( cos 3 theta , sin 3 theta),( - sin 3 theta , cos 3 theta)]`
Q 2783591447

What is the order of
`[ x,y,z] [ (a,h,g),(h,b,f),(g,f,c)] [(x),(y),(z)]`
NDA Paper 1 2017
(A)

`3 xx 1 `

(B)

`1 xx 1`

(C)

`1 xx 3`

(D)

`3 xx 3`

Solution:

`[ x,y,z] [ (a,h,g),(h,b,f),(g,f,c)] [(x),(y),(z)]`

`:. ( 1 xx 3 ) * ( 3 xx 3 ) * ( 3 xx 1 )`

`= ( 1 xx 3 ) * ( 3 xx 1 )`

`= 1 xx 1`
Correct Answer is `=>` (B) `1 xx 1`
Q 2713591449

If `A = [(0,1),(1,0)]` , then the value of `A^4` is
NDA Paper 1 2017
(A)

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

(B)

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

(C)

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

(D)

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

Solution:

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

`A^2 = [ (1, 0 ) , ( 0, 1 ) ] = I`

`A^4 = A^2 * A^2 = I = [ ( 1, 0), ( 0, 1 ) ]`
Correct Answer is `=>` (A) ` [(1,0),(0,1)]`
Q 2731445322

If A is a square matrix of order `3` and `det A = 5`, then what is `det [(2 A )^-1]` equal to ?
NDA Paper 1 2016
(A)

`1/10`

(B)

`2/5`

(C)

`8/5`

(D)

`1/40`

Solution:

`Det | (2A)^(-1) | = Det |(A^(-1)) * (2^(-1)|`

`= (2^(-1))^3 Det (A^(-1))` [ since of matrix of order `3`]

`= 1/(2^3) |A^(-1)| = 1/(2^3 ) * 1/2^3 * 1/(|A|) = 1/40`
Correct Answer is `=>` (D) `1/40`
Q 2711645520

What is `[ (x, y, z) ] [ (a, h , g ),(h,b,f),(g,f,c)]` equal to ?
NDA Paper 1 2016
(A)

`[ (ax + hy +gz , h + b +f , g +f + c )]`

(B)

`[(a,h,g),(hx, by, fz ), (g,f, c ) ]`

(C)

`[(ax + by + gz), (hx + by + fz ), (gx + fy + cz )]`

(D)

`[ (ax + hy + gz , hx + by + fz , gx + fy + cz ) ]`

Solution:

`[ (x, y, z) ] [ (a, h , g ),(h,b,f),(g,f,c)]= [(ax+by +g^2),(hx+by+f^2),(gx+fy +c^2)]`
Correct Answer is `=>` (D) `[ (ax + hy + gz , hx + by + fz , gx + fy + cz ) ]`
Q 2731656522

Let `ax^3 + bx^2 + cx +d = | ( x+1 , 2x, 3x ) , ( 2x +3 , x +1 , x ) , ( 2-x , 3x + 4 , 5 x -1 ) | ` then
What is the value of `a + b + c +d ?
NDA Paper 1 2016
(A)

62

(B)

63

(C)

65

(D)

68

Solution:

Correct Answer is `=>` (B) 63
Q 2721856721

If `m = [ (1, 0), (0,1) ] ` and `n = [ (0,1), ( -1 , 0 ) ]` , then what is the value of the determinant of `m cos theta - n sin theta`?
NDA Paper 1 2016
(A)

-1

(B)

0

(C)

1

(D)

2

Solution:

Det ` (m cos theta - n sin theta )`

Det `([ (cos theta , 0 ), ( 0 , cos theta ) ] - [ ( 0, sin theta ), ( -sin theta , 0) ] )`

`1 cos^2 theta - (0- sin^2 theta) 1`

`=1`
Correct Answer is `=>` (C) 1
Q 2741056823

If `f (x ) = [ (cos x , -sin x , 0 ), ( sin x , cos x , 0 ) , ( 0, 0 , 1) ]` , then which of the following are correct ?

1. `f (theta) xx f(phi) = f( theta + phi )`
2. The value of the determinant of the matrix `f (theta ) xx f (phi ) ` is 1
3. The determinant of f(x) is an even function.

Select the correct answer using the code given below :
NDA Paper 1 2016
(A)

1 and 2 only

(B)

2 and 3 only

(C)

1 and 3 only

(D)

1, 2 and 3

Solution:

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

(i) `f(theta) xx f (phi) = [ (cos theta , - sin theta , 0 ), ( sin theta , cos theta , 0 ), ( 0,0, 1) ] [ (cos phi , - sin phi , 0 ), (sin phi , cos phi , 0 ), ( 0,0,1) ]`

`= [ (cos theta cos phi - sin theta sin phi , - cos theta sin phi - sin theta cos phi , 0 ), ( sin theta cos phi + cos theta sin phi , cos theta cos phi- sin theta sin phi,0), (0,0,1) ]`

`= [ (cos (theta + phi ), - sin (theta + phi) , 0 ), (sin (theta + phi ) , cos ( theta + phi ),0 ), ( 0,0,1) ]`

`= f (theta + phi )`

(ii) ` | f (theta) xx f (phi) | = | f (theta + phi ) |`

`= | (cos (theta + phi ) , - sin (theta + phi) , 0 ), (sin (theta + phi ) , cos (theta + phi ) , 0 ), ( 0,0,1) | `

` = cos^2 (theta + phi ) + sin^2 (theta + phi )`

`=1`

(iii) Det `f(x)= | f(x) | =1`

`f(x) = f (-x) =1` so `fx`is even function
Correct Answer is `=>` (D) 1, 2 and 3
Q 2731167022

If `A = [ (1, -1 ) , (2,3 ) ]` and `B = [ (2,3),(-1,-2) ]` ,then which of the following is/are correct ?

1. `AB(A^-1B^-1)` is a unit matrix.
2. `(AB)^-1 = A^-1 B^-1`

Select the correct answer using the code given below :
NDA Paper 1 2016
(A)

1 only

(B)

2 only

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

`A= [ ( 1,-1 ), (2,3) ] B = [ (2,3), (-1,-2) ]`

1. `AB (A^(-1) B^(-1) )`

`= [ (1,-1), (2,3) ][ ( 2,3), (-1,-2) ]( 1/5 [ (3,1), (-2,1) ] ) 1/1 [ (-2,-3), (1,2) ] `

`= 1/(-5) [ ( -5,0), (0,-5) ] = [ (1,0), (0,1) ] =I`

2. `AB= [ (1,-1),(2,3) ] [ (2,3), (-1,-2) ] = [ (3,5), (1,0) ]`

`(AB)^(-1) = -1/5 [ ( 0,-5), (-1,3) ]`

`A^(-1) * B^(-1) = 1/5 [ (3,1), (-2,1) ] * 1/(-1) [ (-2,3), (1,2) ] = 1/(-5) [ (-5,11),(5,-4)]`

clearly `(AB)^(-1) ne A^(-1) B^(-1)`
Correct Answer is `=>` (A) 1 only
Q 2701667528

For the system of linear equation `2 x + 3y + 5 z = 9 , 7x + 3y - 2 z = 8` and `2 x + 3 y + lamda z = mu`
Under what condition does the above system of equations have infinitely many solution ?
NDA Paper 1 2016
(A)

`lamda = 5 ` and `mu != 9`

(B)

`lamda != 5 ` and ` mu = 7` only

(C)

`lamda != 5`` and ` mu ` has any real value

(D)

`lamda ` has any real value and `mu != 9`

Solution:

Correct Answer is `=>` (C) `lamda != 5`` and ` mu ` has any real value
Q 2127201181

If A is a square matrix, then what is adj `(A^(-1)) - (adj A)^(-1)`
equal to?
NDA Paper 1 2016
(A)

`2 | A|`

(B)

Null matrix

(C)

Unit matrix

(D)

None of these

Solution:

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

`= 0` ` quad [∵ adj (A^(-1) ) = (adj A)^(-1)]`

Hence, it is a null matrix.
Correct Answer is `=>` (B) Null matrix
Q 2177312286

Consider the following in respect of the matrix

` A = [ ( -1 , 1) , (1 , -1) ]`
`1. A^(2) = - A`
`2. A^(3) = 4A`

Which of the above is/are correct ?
NDA Paper 1 2016
(A)

Only `1`

(B)

Only `2`

(C)

Both `1` and `2`

(D)

Neither `1` nor `2`

Solution:

Given, `A = [ ( -1 , 1) , (1 , -1) ] `

1. Now, ` A^(2) = A xx A = [ ( -1 , 1) , (1 , -1) ] [ ( -1 , 1) , (1 , -1) ]`

` = [ ( 1 + 1 ,- 1 -1 ) , (-1 -1 ,1 +1 ) ]`

` A^(2) = [ ( 2 , -2) , (-2 , 2 ) ] = 2 [ ( 1 , -1) , (-1 , 1) ]`

` A^(3) = A^(2) . A = [ ( 2 , -2) , (-2 , 2 ) ] [ ( -1 , 1) , (1 , -1) ]`

` = [ ( -2 -2 ,2 +2 ) , (2 +2 ,-2 -2 ) ] = [ ( -4 , 4) , (4 , -4) ]`

` = 4 [ ( -1 , 1) , (1 , -1) ]`

`A^(3) = 4A`

Hence, only 2 is correct.
Correct Answer is `=>` (B) Only `2`
Q 1608323208

Which one of the following matrices is an

elementary matrix?
NDA Paper 1 2015
(A)

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

(B)

` [(1,5,0),(0,1,0),(0,0,1)]`

(C)

` [(0,2,0),(1,0,0),(0,0,1)]`

(D)

` [(1,0,0),(0,1,0),(0,5,2)]`

Solution:

` [(1,5,0),(0,1,0),(0,0,1)]` is an elementary matrix. Since, the 'value

of determinant of the given matrix is `1` .
Correct Answer is `=>` (B) ` [(1,5,0),(0,1,0),(0,0,1)]`
Q 1638423302

If `A = [(2,7),(1,5)]` then what is `A + 3A ^(- 1)` equal to?

where, `I` is the identity matrix of order `2`.
NDA Paper 1 2015
(A)

`3I`

(B)

`5I`

(C)

`7I`

(D)

None of these

Solution:

We have, `A = [(2,7),(1,5)]`

`:. | A| = 10 - 7 = 3`

Now `A^(-1) = 1/( | A| ) adj A`

`:. | A| = 1/3 [(5,-1),(-7,2)] = 1/3 [(5,-7),(-1,2)]`

`:. A+ 3A^(-1) = [(2,7),(1,5)] + 3 xx 1/3 [(5,-7),(-1,2)] = [(7,0),(0,7)] = 7I .`
Correct Answer is `=>` (C) `7I`
Q 2240367213

If `A = [(1,0,-2) , (2 ,-3, 4) ]` then the matrix `X` for which
`2X + 3A = 0` holds true is
NDA Paper 1 2015
(A)

` [ (- 3/2 , 0, -3),(-3, - 9/2 , -6) ]`

(B)

` [ ( 3/2 , 0, -3),(3, - 9/2 , -6) ]`

(C)

` [ ( 3/2 , 0, 3),(3, 9/2 , 6) ]`

(D)

` [ (- 3/2 , 0, 3),(-3, 9/2 , -6) ]`

Solution:

Given, `A = [(1,0,-2) , (2 ,-3, 4) ]`

We have, `2X + 3A= 0`

`=> X=(-3)/2 A`

` => X =(-3)/2 [(1,0,-2) , (2 ,-3, 4) ] => [ (- 3/2 , 0, 3),(-3, 9/2 , -6) ]`
Correct Answer is `=>` (D) ` [ (- 3/2 , 0, 3),(-3, 9/2 , -6) ]`
Q 2210367219

If `A= [ ( 1,1,-1), (2,-3,4),(3,-2,3) ]` and ` B = [ (-1,-2,-1),(6,12,6),(5,10,5)]`
then which of the following is/ are correct?
1. `A` and `B` commute.
2. `AB` is a null matrix.

Select the correct answer using the code given below
NDA Paper 1 2015
(A)

Only `1`

(B)

Only `2`

(C)

Both `1` and `2`

(D)

Neither `1` nor `2`

Solution:

We have `A= [ ( 1,1,-1), (2,-3,4),(3,-2,3) ]` and ` B = [ (-1,-2,-1),(6,12,6),(5,10,5)]`

` AB = [ (-1+6-5 , -2+12 -10 , -1+6-5),( -2 -18+20 , -4 -36 +40, -2 -18+20), (-3 -12 +15, -6 -24+30 , -3 -12 +15)]`

` = [(0,0,0),(0,0,0),(0,0,0)]`

Hence, `AB ` is a null matix.
Correct Answer is `=>` (B) Only `2`
Q 1668423305

The matrix ` [(0, -4 + i) ,( 4 +i ,0)]` is
NDA Paper 1 2015
(A)

symmetric

(B)

skew-symmetric

(C)

hermitian

(D)

skew-hermitian

Solution:

A square matrix `A` is said to be skew-hermitian, if

`A' = -A` or `a_(ij) = a_(ji) AA i ` and `j`.

Here, `a_(12) = -4 + i` and `a_(21) = 4 + i`

Now, `a_(21) = - (-4 + i) = - (- i - 4) = 4 + i`

Hence, the given matrix is skew-hermitian matrix.
Correct Answer is `=>` (D) skew-hermitian
Q 2220291111

If `A` is an invertible matrix of order `n` and `k` is any positive
real number, then the value of `[det(kA)]^(-1) det (A)` is
NDA Paper 1 2015
(A)

`k^(-n)`

(B)

`k^(-1)`

(C)

`k^(n)`

(D)

`nk`

Solution:

`[ det (kA)]^(-1) det (A)`

` = 1/(det (kA)) xx det (A) quadquadquadquadquad [ :. a^(-m) = 1/a^m]`

` = 1/ (k^(n) det (A)) xx det ( A) = 1/ k^(n) = k^(-n)`
Correct Answer is `=>` (A) `k^(-n)`
Q 1648523403

Consider the following in respect of two
non-singular matrices `A` and `B` of same order

I. det `(A+ B) = det A + det B`

II. `(A+ B)^(-1) =A ^(-1) + B^(-1)`

Which of the above is/are correct?
NDA Paper 1 2015
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

If `A = B + C`, then it is not necessary that

`det (A) = det (B) + det (C)`

Also, `(A+ B)^(-1) =A^(-1) + B^(-1)` is false.
Correct Answer is `=>` (D) Neither I nor II
Q 1628623501

If `X = [(3,-4),(1,-1) ], B=[ (5,2),(-2,1)]` and `A = [(p,q),(r,s)]` satisfy

the equation `AX = B`, then the matrix `A` is equal to
NDA Paper 1 2015
(A)

` [(-7,26),(1,-5) ]`

(B)

` [(7,26),(4,17) ]`

(C)

` [(-7,-4),(26,13) ]`

(D)

` [(-7,26),(-6,23) ]`

Solution:

`:. AX = - B`

`:. [(p,q),(r,s)] [(3,-4),(1,-1) ] = [ (5,2),(-2,1)]`

`=> [(3p+ q, -4p-q),(3r + s , -4r - s)] = [ (5,2),(-2,1)]`

`=> 3p + q = 5` and `-4p - q = 2`

`=> -p = 7 => p = -7`

Now, `q = 5 + 21 = 26, 3r + s = - 2`

Also, `-4r- s = 1`

`=> -r = - 1 => r = 1` and `s = -2 - 3 = - 5`

`:. A= [(-7,26),(1,-5) ]`
Correct Answer is `=>` (A) ` [(-7,26),(1,-5) ]`
Q 1618823700

Let `A= [ (x+y,y) ,(2x,x-y)] , B [(2),(-1)] ` and ` c= [3/2]` if

`AB = C`, then what is `A^2` equal to?
NDA Paper 1 2015
(A)

` [ (6 , -10), (4 , 26)]`

(B)

` [ (-10 , 5), (4 , 24)]`

(C)

` [ (-5 , -6), (-4 , -20)]`

(D)

` [ (-5 , -7), (-5 , 20)]`

Solution:

We have `AB = C`

`:. [(x +y ,y) ,(2x , x-y)] [(2),(-1)] = [3/2]`

` => [( 2x+2y , -y) ,( 4x,-x-y)] =[3/2]`

` => [ (2x+y),(3x-y)] = [3/2]`

` => 2x + y = 3` and `3x + y = 2 => x = 2- 3 = -1`

`:. y = 5`

` :. A^2 = [(x +y ,y) ,(2x , x-y)]^2`

` = [ (4 , 5), (-2 , -6)] [ (4 , 5), (-2 , -6)]`

` = [ (16 - 10 , 20 - 30), (-8 +12 ,-10 + 36 )] = [ (6 , -10), (4 , 26)]`
Correct Answer is `=>` (A) ` [ (6 , -10), (4 , 26)]`
Q 2261401325

If `A` is an orthogonal matrix of order `3` and `B = [(1,2,3),(-3,0,2),(2,5,0)]`
then which following is/are correct?
1.`|AB| =pm 47`
2. `AB=BA`

Select the correct answer using the code given below
NDA Paper 1 2015
(A)

Only `1`

(B)

Only `2`

(C)

Both `1` and `2`

(D)

Neither `1` nor `2`

Solution:

`A` is an orthogonal matrix.

`|AB|= pm |B|`

` :. |B| = |(1,2,3),(-3,0,2),(2,5,0)| = 47`

Now, ` |AB| =pm 47`
Correct Answer is `=>` (A) Only `1`
Q 1618123900

If `E( theta ) = [ (cos theta , sin theta),(- sin theta , cos theta) ]` then `E( alpha) E (beta)` is equal to
NDA Paper 1 2015
(A)

`E(alpha beta)`

(B)

`E(alpha - beta)`

(C)

`E(alpha + beta)`

(D)

`-E(alpha + beta)`

Solution:

Given, ` E( theta ) = [ (cos theta , sin theta),(- sin theta , cos theta) ]`

` :. E (alpha) = [ (cos alpha , sin alpha ),(- sin alpha , cos alpha ) ]`

and ` E (beta) = [ (cos beta , sin beta ),(- sin beta , cos beta ) ]`

` :. E(alpha)E(beta) [ (cos alpha , sin alpha ),(- sin alpha , cos alpha ) ] [ (cos beta , sin beta ),(- sin beta , cos beta ) ]`

` = [( cos2 . cos beta - sin2 . sin beta , cos2 . sin beta - sin2 . cos beta),(-sin 2 . cos beta - sin beta . cos 2 ,- sin 2 . sin beta - cos2 cos beta )]`

` = [ (cos (alpha +beta) , sin(alpha + beta)) ,(-sin (alpha +beta) , cos (alpha + beta))]`

` = E (alpha + beta)`
Correct Answer is `=>` (C) `E(alpha + beta)`
Q 1608180008

The area of a triangle, whose vertices are `(3, 4),
(5, 2)` and the point of intersection of the lines `x = a`
and `y = 5`, is `3` square units. When is the value of `a`?
NDA Paper 1 2015
(A)

`2`

(B)

`3`

(C)

`4`

(D)

`5`

Solution:

We have, ` Delta = 3,` sq units

` :. 1/2 | (5, 2, 1) ,(3,4,1) ,(a,5,1)| = 3`

` 1/2 [5(4- 5)- 2(3- a)+ 1(15- 4a)] = 3`

`=> 1/2 [-5- 2(3- a)+ (15- 4a)] = 3`

`=> 1/2 [-5-6+ 2a + 15- 4a] = 3`

` => 2 - a = pm 3`

` => a= 5` or `-1`

`:. a=5`
Correct Answer is `=>` (B) `3`
Q 1629691511

If `A` and `B` are square matrices of second order such that
`| A | = -1` and `| B | = 3`, then what is `| 3 AB |` equal to?
NDA Paper 1 2014
(A)

`3`

(B)

`-9`

(C)

`-27`

(D)

None of these

Solution:

`A` and `B` are square matrices of order `2`.

We know that, `| kA | = k^n | A |`, where `n` is order of matrix `A`.

`:. | 3AB | = 3^2 | A | | B | quad ( ∵ | AB | = | A || B |)`

` = 9 (-1) (3)`

` = -27 quad ( ∵ | A | = -1, | B | = 3)`
Correct Answer is `=>` (C) `-27`
Q 1782334237

If, `A` and `B` be two matrices such that `AB =A` and `BA =B`.
Then, which of the following statements are correct?
1. `A^2 =A`
2. `B^2 =B`
3. `(AB)^2 = AB`
Select the correct answer using the code given below.
NDA Paper 1 2014
(A)

`1` and `2`

(B)

`2` and `3`

(C)

`1` and `3`

(D)

`1, 2` and `3`

Solution:

1. We have, `AB = A`

`:. A^2 = (AB). (AB) =A. (BA) B`

`= ABB quad (∵ BA = B)`

` = AB`

` = A quad (∵ AB = A)`

Also, `B^2 = (BA). (BA)`

`= B. (AB). A`

`= B. A. A quad (∵ AB = A)`

` = B. A`

` = B quad (∵ BA = B)`

Again `(AB)^2 = (AB).(AB)`

` = A. (BA)B`

` = A. B. B quad (∵ BA = B)`

` = AB`

` = A quad (∵ AB = A)`
Correct Answer is `=>` ()
Q 1783278147

Consider the following statements in respect of the

matrix `A = [ (0,1,2),(-1,0,-3),(-2,3,0)]`

I. The matrix A is skew-symmetric.
II. The matrix A is symmetric.
III. The matrix A is invertible.

Which of the above statements is/ are correct?
NDA Paper 1 2014
(A)

Only 1

(B)

Only 3

(C)

1 and 2

(D)

2 and 3

Solution:

Given matrix

`A = [ (0,1,2),(-1,0,-3),(-2,3,0)]`

Now ` A^T = [ (0,1,2),(-1,0,-3),(-2,3,0)]^T = [ (0,-1,-2),(1,0,3),(2,-3,0)]`

` = - [ (0,1,2),(-1,0,-3),(-2,3,0)] = -A`

`=> A = -A^T`

So, A is skew-symmetric matrix.

and `| A| = [ (0,1,2),(-1,0,-3),(-2,3,0)] = 0 -1 (0- 6) + 2 (- 3)`

` = 6 - 6 =0 `

Since.`| A | = 0` i.e.,` A` is singular matrix.

So, `A` cannot be an invertible matrix.
Correct Answer is `=>` (A) Only 1
Q 1762434335

If the matrix `A` is such that `( (1,3),(0,1) ) A = ( (1,1),(0,-1) )` then
what is A equal to?
NDA Paper 1 2014
(A)

` ( (1,4),(0,-1) )`

(B)

` ( (1,4),(0,1) )`

(C)

` ( (-1,4),(0,-1) )`

(D)

` ( (1,-4),(0,-1) )`

Solution:

` ∵ ( (1,3),(0,1) ) A = ( (1,1),(0,-1) )`

Let ` B = ( (1,3),(0,1) ) ` and ` | B | =1 `

`:. B^(-1) = ( (1,-3),(0,1) ) quad (∵ A^(-1) = 1/(|A|) adj A)`

`:. A = ( (1,-3),(0,1) ) ( (1,1),(0,-1) ) = ( (1,4),(0,-1) )`
Correct Answer is `=>` (A) ` ( (1,4),(0,-1) )`
Q 1713378240

Consider two matrices `A = [(1,2),(2,1),(1,1)]` and `B = [(1,2,-4),(2,1,-4)]`

Which one of the following is correct?
NDA Paper 1 2014
(A)

B is the right inverse of A

(B)

B is the left inverse of A

(C)

B is the both sided inverse of A

(D)

None of the above

Solution:

Given matrices,

`A = [(1,2),(2,1),(1,1)]` and `B = [(1,2,-4),(2,1,-4)]`

1. `AB = [(1,2),(2,1),(1,1)]_(3xx2) [(1,2,-4),(2,1,-4)]_(2xx3)`

` = [(1+4,2+2,-4-8),(2+2,4+1,-8-4),(1+2,2+1,-4-4)]`

` = [(5,4,-12),(4,5,-12),(3,3,-8)]_(3xx3)`

2. `BA = [(1,2,-4),(2,1,-4)]_(2xx3) [(1,2),(2,1),(1,1)]_(3xx2)`

` = [(1+4-4,2+2-4),(2+2-4,4+1-4)]`

` = [(1,0),(0,1)]_(2xx2)`

Now, we observe that `B` is not the right inverse of `A` but `B` is
the left inverse of `A`.
Correct Answer is `=>` (B) B is the left inverse of A
Q 1772434336

Consider the following statements

1. Determinant is a square matrix.

2. Determinant is a number associated with a square
matrix.

Which of the above statements is/are correct?
NDA Paper 1 2014
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

1. We know that, determinant is not a square

matrix, so it is not a true statement.

2. It is true that, determinant is a number associated with a

square matrix.

Hence. Statement `2` is correct.
Correct Answer is `=>` (B) Only 2
Q 1702434338

If `A` is an invertible matrix, then what is det `(A^( -1))` equal
to?
NDA Paper 1 2014
(A)

det `A`

(B)

` 1/(det A)`

(C)

`1`

(D)

None of these

Solution:

` det(A^(-1)) = 1/(det A)`
Correct Answer is `=>` (B) ` 1/(det A)`
Q 1733378242

If `A` is any matrix, then the product `A A` is defined only
when `A` is a matrix of order `m xx n`, where
NDA Paper 1 2014
(A)

`m > n`

(B)

`m < n`

(C)

`m = n`

(D)

`m <= n`

Solution:

Given that, A is any matrix.

Then, the product `A A` is defined only when `A` is a matrix of

order `m xx n` where, `m = n`. i.e., A must be a square matrix.

`A xx A = (m xx n )(m xx n)`

` = (m xx n )(n xx n),` if `m = n`

` = m xx n = n xx n` or `m xx m`

= A is a square matrix
Correct Answer is `=>` (C) `m = n`
Q 1712434339

From the matrix equation `AB = AC`, where `A, B` and `C` are
the square matrices of same order, we can conclude `B = C`
provided
NDA Paper 1 2014
(A)

A is non-singular

(B)

A is singular

(C)

A is symmetric

(D)

A is skew symmetric

Solution:

From the matrix equation `AB = AC`, where `A, B`

and `C` are the square matrices of same order.

We can conclude `B = C` provided `A` is non-singular.
Correct Answer is `=>` (A) A is non-singular
Q 1763378245

The determinant of an odd order skew-symmetric matrix
is always
NDA Paper 1 2014
(A)

zero

(B)

one

(C)

negative

(D)

depends on the matrix

Solution:

We know that, elements of principal diagonals of a skew-symmetric matrix are all zero.

i.e., `a_(ii) =- a_(ii)`

`=> 2 a_(ii) = 0` or `a_(ii) = 0` for fill values of `i`.

e.g., (i) ` A = [(0,-a),(a,0)]_(2xx2)`

` => |A| = |(0,-a),(a,0)| = 0 + a^2 = a^2 != 0`

(ii) `A = [(0,a,b),(-a,0,-c),(-b,c,0)]_(3xx3) => |A| = |(0,a,b),(-a,0,-c),(-b,c,0)|`

` = 0 - a (-bc) + b (-ac)`

` = abc - abc = 0`

Here, we see that, determinant of an even order

skew-symmetric matrix is not zero but an odd order

skew-symmetric matrix is always zero.
Correct Answer is `=>` (A) zero
Q 1712534430

If `A = ( (4, x +2),( 2x-3 , x+1) )` is symmetric, then what is `x` equal
NDA Paper 1 2014
(A)

`2`

(B)

`3`

(C)

`-1`

(D)

`5`

Solution:

` ∵ A = A'`

`=> ( (4, x +2),( 2x-3 , x+1) ) = ( (4, 2x -3),( x + 2 , x+1) )`

` => 2x - 3 = x + 2`

`:. x = 5`
Correct Answer is `=>` (D) `5`
Q 2349178913

Consider the following statements
I. The product of two non-zero matrices can
never be identity matrix.
II. The product of two non-zero matrices can
never be zero matrix.
Which of the above statements is/are correct?
NDA Paper 1 2013
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. We know that, the product of two identity matrix are

always an identity matrix, which is non-zero matrices.

`[(1,0),(0,1)] xx [ (1,0),(0,1) ] = [ (1+0 , 0+0 ),(0 + 0, 1 + 0) ]`

`= [(1,0),(0,1) ] = I = ` Identity matrix

II. The product of two non-zero matrices sometimes be zero
matrix.

`[ (0,c,-b),(-c,0,a),(b,-a,0)] xx [ (a^2,ab,ac),(ab,b^2,bc),(ac,bc,c^2)]`

` = [ (0 + abc - bac , 0 + b^2c - b^2c , 0 + bc^2 - bc^2),( -a^2c + 0 + a^2c , - abc + 0 + abc , -ac^2 + 0 - ac^2) , (a^2b - a^2b + 0 , ab^2 - ab^2 + 0 , abc - abc + 0) ]`

`= [(0,0,0),(0,0,0),(0,0,0)] = 0 =` Zero matrix

So, the statement are correct .
Correct Answer is `=>` (D) Neither I nor II
Q 2480734617

The cofactor of the element `4` in the determinant
` | (1,2,3),(4,5,6),(5,8,9) |` is
NDA Paper 1 2013
(A)

`2`

(B)

`4`

(C)

`6`

(D)

`-6`

Solution:

Let `Delta = | (1,2,3),(4,5,6),(5,8,9) |`

Now, cofactor of element `A = ( -1)^(2 + 1) | (2,3),(8,9) |`

` = - (18 - 24) = 6`
Correct Answer is `=>` (C) `6`
Q 2359178914

Consider the following statements
I. The matrix `[(1,2,1),(a,2a,1),(b,2b,1)]` is singular.
II. The matrix `[(c,2c,1),(a,2a,1),(b,2b,1)]` is non - singular.
Which of the above statements is/are correct?


NDA Paper 1 2013
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. Let `A = [(1,2,1),(a,2a,1),(b,2b,1)]`

Now, `|A| = 1(2a - 2b) - 2(a - b)+ 1(2ab- 2ab)`

`=2a - 2b - 2a + 2b + 0 = 0`

i.e., `A` is a singular matrix.

II. Let `B = [(c,2c,1),(a,2a,1),(b,2b,1)]`

Now, ` | B | = c (2a - 2b) - 2c (a - b)+ 1(2ab - 2ab)`

`= 2ac- 2bc - 2ac + 2bc + 0 = 0`

which is also represent a singular matrix.

So, Statement `I` is correct but Statement `II` is incorrect.
Correct Answer is `=>` (A) Only I
Q 2400834718

If `A` is a square matrix of order `3` with `| A | != 0`, then
which one of the following is correct?
NDA Paper 1 2013
(A)

`|adj A| = | A |`

(B)

`|adj A| = |A|^2`

(C)

`|adj A| = |A|^3`

(D)

`|adj A|^2 = |A|`

Solution:

If `A` is a square matrix of order `n` with `| A | != 0`, then

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

For order `3`, put `n = 3`

`|adj A | = | A |^( 3-1) = | A |^2`
Correct Answer is `=>` (B) `|adj A| = |A|^2`
Q 2369178915

If `A = [ (i ,0),(0,-i)] , B = [ (0,-1),(1,0) ] ` and ` C = [ (0,i),(i,0) ]`, where
`i = sqrt(-1)`, then which one of the following is correct?

where, `I` is the identity matrix.

NDA Paper 1 2013
(A)

`AB = -C`

(B)

`AB = C`

(C)

`A^2 = B^2 = C^2 = I`

(D)

`BA != C`

Solution:

Given that, `A = [ (i ,0),(0,-i)] , B = [ (0,-1),(1,0) ] `

and ` C = [ (0,i),(i,0) ]`

Now , `AB = [ (i ,0),(0,-i)] [ (0,-1),(1,0) ] `

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

` = [ (0 , -i),(-i ,0)] = - [ (0,i),(i,0)] = - C`
Correct Answer is `=>` (A) `AB = -C`
Q 2379178916

If `2A = [ (2,1),(3, 2)]`, then `A^(-1)` is equal to
NDA Paper 1 2013
(A)

` [ (2,-1),(-3, 2)]`

(B)

`1/2 [ (2,-1),(-3, 2)]`

(C)

`1/4 [ (2,-1),(-3, 2)]`

(D)

None of these

Solution:

Given that, `2A = [ (2,1),(3, 2)]`

`=> A = 1/2 [ (2,1),(3, 2)] = [ (1,1//2),(3//2, 1)]`

Now, adj `A = [ (1,-3//2),(-1//2, 1)]^T`

` = [ (1,-1//2),(-3//2, 1)]`

and `| A | = 1 - 3//4 = 1//4`

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

` = 4 [ (1,-1//2),(-3//2, 1)]`

` = [ (4 , -2),(-6 , 4)]`
Correct Answer is `=>` (D) None of these
Q 2389178917

If ` [ (2,3),(4,1)] xx [ (5,-2),(-3,1)] = [ (1,-1),(17,lamda)]`, then `lamda` is equal to
NDA Paper 1 2013
(A)

`7`

(B)

`-7`

(C)

`9`

(D)

`-9`

Solution:

Given that, `[ (2,3),(4,1)] xx [ (5,-2),(-3,1)] = [ (1,-1),(17,lamda)]`

` => [(10-9 , -4+ 3),(20-3 , -8 + 1) ] = [ (1 , -1),(17 , lamda) ]`

` => [(1,-1),(17, -7)] = [ (1, -1),(17 , lamda) ]`

On comparing, we get

`lamda = -7`
Correct Answer is `=>` (B) `-7`
Q 2319180010

If `A` and `B` are two non-singular square matrices
such that `AB = A`, then which one of the following
is correct?
NDA Paper 1 2013
(A)

`B` is an identity matrix

(B)

`B = A^(-1)`

(C)

`B = A^2`

(D)

Determinant of `B` is zero

Solution:

Given that `A` and `B` are two non-singular square

matrices.

So, its inverse i.e., `K^(-1)` and `B^(-1)` must be exist.

We have, `AB = A`

`(A^(-1))` operating in left of both sides, we get

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

` => (A^(-1)A)B = (A^(-1) A) ( ∵A A^(-1) = I` and ` IB = BI = B )`

` => IB = I`

` => B = I =` Identity matrix
Correct Answer is `=>` (A) `B` is an identity matrix
Q 2430167912

Consider the following statements
I. A matrix is not a number.
II. Two determinants of different orders may have
the same value.
Which of the above statements is/are correct?
NDA Paper 1 2013
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

Statement I A matrix is only an arrangement of

numbers, it has no definite value.

e.g., `[7] != 7`

Statement II

Let `Delta_1 = | (1,2,3),(1,1,1),(1,0,0)|_(3 xx 3)`

`=1 (2 -3) = - 3`

and `Delta_2 = | (1,3),(2,3)|_(2 xx 2) = 3 - 6 = -3`

Hence, two determinants of different orders may have the same value.
Correct Answer is `=>` (C) Both I and II
Q 2329180011

The inverse of a diagonal matrix is a
NDA Paper 1 2012
(A)

symmetric matrix

(B)

skew-symmetric matrix

(C)

diagonal matrix

(D)

None of the above

Solution:

We know that, by the property of diagonal matrix,

Let `A =` Diagonal `(a_1,a_2,a_3)`

Then, `A^(-1) =` Inverse of `A`

`=` Diagonal `(a_1 quad ^ (-1) , a_2 quad ^(-1) , a_3 quad ^(-1) )`

`=` Diagonal ` ( 1/a_1 , 1/a_2 , 1/a_3)`

Hence, the inverse of diagonal matrix is also a diagonal matrix.
Correct Answer is `=>` (C) diagonal matrix
Q 2349180013

If `A = [ (3,4),(5,6),(7,8)]` and ` B = [ (3,5,7),(4,6,8) ]` , then
which one of the following is correct?
NDA Paper 1 2012
(A)

B is the inverse of A

(B)

B is the adjoint of A

(C)

B is the transpose of A

(D)

None of the above

Solution:

The transpose of any matrix `A` is obtained by

interchange the row into corresponding column. So, `B` is the

transpose of `A`.
Correct Answer is `=>` (C) B is the transpose of A
Q 2379180016

If the sum of the matrices ` [(x),(y),(z)] , [(y),(y),(z)]` and `[(z),(0),(0)]` is the
matrix ` [ (10),(5),(5)]` , then what is the value of `y`?
NDA Paper 1 2012
(A)

`-5`

(B)

`0`

(C)

`5`

(D)

`10`

Solution:

` [(x),(y),(z)] + [(y),(y),(z)] + [(z),(0),(0)] = [ (10),(5),(5)]`

` = [ (x + y + z),(x+ y +0),(y+ z+ 0)] = [ (10),(5),(5)]`

` => x + y + z = 10` ......(i)

` x + y = 5` .......(ii)

` y + z = 5` ......(iii)

From Eqs. (i) and (iii),

` x + (5) = 10 => x = 5`

On putting the value of `x` in Eq. (ii), we get

` 5 + y = 5`

`=> y = 0`
Correct Answer is `=>` (B) `0`
Q 2319180019

If the matrix `AB` is a zero matrix, then which one
of the following is correct?
NDA Paper 1 2012
(A)

A must be equal to zero matrix or B must be equal to zero matrix

(B)

A must be equal to zero matrix and B must be equal to zero matrix

(C)

It is not necessary that either A is zero matrix or B is zero matrix

(D)

None of the above

Solution:

For the matrix `AB` is a zero matrix. It is not necessary

that either `A` is zero matrix or `B` is zero matrix.

e.g., Let `A = [(1,0),(0,0)]` and `B = [(0,0),(0,-1)]`

`:. AB = 0`, where `A, B != 0`
Correct Answer is `=>` (C) It is not necessary that either A is zero matrix or B is zero matrix
Q 2329280111

If the matrix ` [ (alpha ,2 ,2),(-3 ,0 ,4),(1,-1,1) ]` is not invertible, then
NDA Paper 1 2012
(A)

`alpha = - 5`

(B)

`alpha = 5`

(C)

`alpha = 0`

(D)

`alpha = 1`

Solution:

If the matrix is not invertible, then it is necessary that the

value of the determinant of this matrix must be zero.

Matrix is singlular or ` | (alpha ,2,2),(-3, 0, 4),(1,-1,1) | = 0`

` | (alpha, 2+alpha, 2-alpha),(-3, -3, 7),(1,0,0)| = 0`

Expanding the determinant along the third row, we get

`1 | (2 + alpha, 2 - alpha),( -3 ,7 )| = 0`

` => 14+ 7alpha + 3(2 - alpha) = 0`

` 14 + 7alpha + 6 - 3alpha = 0`

` => 20 + 4alpha = 0`

` => 4alpha = - 20`

` => alpha = -5`
Correct Answer is `=>` (A) `alpha = - 5`
Q 2460167915

If `A= [(1,2),(2,3)]` and `B = [(1,0),(1,0)]`, then what is
determinant of `AB`?
NDA Paper 1 2012
(A)

`0`

(B)

`1`

(C)

`10`

(D)

`20`

Solution:

Given, `A = [(1,2),(2,3)]` and `B = [(1,0),(1,0)]`

Now, `AB = [(1,2),(2,3)] [(1,0),(1,0)] = [ (1+2 , 0),(2+3,0)] = [(3,0),(5,0)]`

Then , ` | AB| = [(3,0),(5,0)] = 3 xx 0 - 5 xx 0 = 0`
Correct Answer is `=>` (A) `0`
Q 2339280112

A square matrix `[a_(ij)]` such that `a_(ij) = 0` for `i != j` and
`a_(ij) = k`, where `k` is a constant for `i = j` is called
NDA Paper 1 2012
(A)

diagonal matrix but not scalar matrix

(B)

scalar matrix

(C)

unit matrix

(D)

None of the above

Solution:

Given,

` [a_(ij)] = { tt ((a_(ij) = 0 , text(for) i != j),(a_(ij) = k , text(for) i = j))` where k is a constant.

`∵ [a_(ij)] = [ (a_(11) ,a_(12),a_(13) ),(a_(21) ,a_(22),a_(23) ),(a_(31) ,a_(32),a_(33) )]_(3xx3)` let order `3 xx 3`.

`= [ (k,0,0),(0,k,0),(0,0,k)]_(3 xx 3)` = Scalar matrix
Correct Answer is `=>` (B) scalar matrix
Q 2359280114

`A` and `B` are two matrices such that `AB = A` and
`BA = B`, then what is the value of `B^2`?
where, `I` is the identity matrix.
NDA Paper 1 2012
(A)

`B`

(B)

`A`

(C)

`I`

(D)

`-I`

Solution:

Given that, `AB = A` .....(i)

and `BA = B` ......(ii)

Now, `B^2 = B·B= (BA)·B` [from Eq. (ii)]

`= B· (AB) = B· A` [from Eq. (i)]

` = B` [from Eq. (ii)]
Correct Answer is `=>` (A) `B`
Q 2379280116

Consider the following statements
I. Every zero matrix is a square matrix
II. A matrix has a numerical value
III. A unit matrix is a diagonal matrix
Which of the above statements is/are correct?
NDA Paper 1 2012
(A)

Only II

(B)

Only III

(C)

Both II and III

(D)

Both I and III

Solution:

I. Every zero matrix is not necessarily square matrix.

II. A matrix does not have a numerical value while every

determinant have a numerical value.

III. A unit matrix is a diagonal matrix and scalar matrix also.
Correct Answer is `=>` (B) Only III
Q 2389280117

If a matrix `A` has inverses `B` and `C`, then which one
of the following is correct ?
NDA Paper 1 2012
(A)

B may not be equal to C

(B)

B should be equal to C

(C)

B and C should be unit matrices

(D)

None of the above

Solution:

We know that, every matrix possesses a unique

inverse.

Hence, `B` and `C` should be equal.
Correct Answer is `=>` (B) B should be equal to C
Q 2329380211

If `A` is a square matrix such that `A^2 = I`, where `I` is
the identity matrix, then what is the value of `A^(-1)`?
NDA Paper 1 2012
(A)

`A+ I`

(B)

Null matrix

(C)

`A`

(D)

Transpose of `A`

Solution:

Given condition, `A^2 = I`

`=> A^(-1) · A^2 = A^(-1) . I`

`=> A^(-1) (A· A) = A^(-1) ( ∵ A·I = A)`

` => (A^(-1) . A)·A = A^(-1)`

`=> I · A = A^(-1) ( ∵ A^(-1) A = I )`

`:. A^(-1) = A`
Correct Answer is `=>` (C) `A`
Q 2339480312

If `A = [ (1,2),(1,1)]` and `B = [(0, -1),(1,2)]` , then what is the
value of `B^(-1) A^(-1)`?
NDA Paper 1 2012
(A)

` [ (1,-3),(-1,2)]`

(B)

` [ (-1,3),(1,-2)]`

(C)

` [ (-1,3),(-1,-2)]`

(D)

` [ (-1,-3),(1,-2)]`

Solution:

Given , `A = [ (1,2),(1,1)]` and `B = [(0, -1),(1,2)]`

Here, ` | A| = | (1,2),(1,1) | = 1 - 2 = -1`

adj `(A) = [ (1,-1),(-2,1)] = [ (1,-2),(-1,1)]`

`:. A^(-1) = (adj A)/(|A|) = -1 [ (1,-2),(-1,1)] = [ (-1,2),(1,-1)]`

here, `| B | = | (0,-1),(1,2)| = 0 - (-1) = 1`

` adj(B) = [ (2,-1),(1,0)] = [ (2,1),(-1,0)]`

` :. B^(-1) = (adj B)/(|B|) = 1 . [ (2,1),(-1,0)] => B^(-1) = [ (2,1),(-1,0)]`

` :. B^(-1) A^(-1) = [ (2,1),(-1,0)] [ (-1,2),(1,-1)] = [ (-2+1,4-1),(1+0,-2+0)]`

` = [ (-1,3),(1,-2)]`
Correct Answer is `=>` (B) ` [ (-1,3),(1,-2)]`
Q 2369480315

The sum and product of matrices A and B exist.
Which of the following implications are
necessarily true?
I. A and B are square matrices of same order.
II. A and B are non-singular matrices.
Select the correct answer using the codes given below.
NDA Paper 1 2012
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. The sum and product of matrices `A` and `B` exist, if `A`

and `B` are square matrices of same order.

It is not necessarily that `A` and `B` are non-singular matrices for

addition and product of two matrices.
Correct Answer is `=>` (A) Only I
Q 2329580411

What is the order of the product

`[xyz] [(a,h,g),(h,b,f),(g,f,c)] [(x),(y),(z)]` ?
NDA Paper 1 2012
(A)

`3 xx 1`

(B)

`1 xx 1`

(C)

`1 xx 3`

(D)

`3 xx 3`

Solution:

Here,

`[xyz]_(1xx3) [(a,h,g),(h,b,f),(g,f,c)]_(3xx3) [(x),(y),(z)]_(3xx1)`

`:.` Order of required matrix `= 1 xx 3 : 3 xx 3 : 3 xx 1`

`= 1 xx 3 : 3 xx 1 = 1 xx 1`
Correct Answer is `=>` (B) `1 xx 1`
Q 2379580416

For what value of `x` does

` [ 1,3,2] [ (1,3,0),(3,0,2),(2,0,1)] [(0),(3),(x)] = [0]` hold ?
NDA Paper 1 2011
(A)

`-1`

(B)

`1`

(C)

`9//8`

(D)

`-9//8`

Solution:

Given, ` [ 1,3,2]_(1xx3) [ (1,3,0),(3,0,2),(2,0,1)]_(3xx3) [(0),(3),(x)]_(3xx1) = [0]_(1xx1)`

` => [ 1 + 9 + 4 quad 3 + 0 + 0 quad 0 + 6 + 2]_(1xx3) [(0),(3),(x)]_(3xx1) = [0]_(1xx1)`

`=> [14 quad 3 quad 8]_(1xx3) [(0),(3),(x)]_(3xx1) =[0]_(1xx1)`

`=> [0 + 9 + 8x] =[0]`

`=> [8x + 9] = [0]`

On comparing, we get

` 8x + 9 = 0 => x => - 9/8`
Correct Answer is `=>` (D) `-9//8`
Q 2349680513

Consider the following statements in respect of
the square matrices `A` and `B` of same order :
I. `A` and `B` are non-zero and `AB = 0`
`=>` Either `| A | = 0` or `| B | = 0`
II. `AB = 0 => A = 0` or `B = 0`
Which of the above statements is/are correct?
NDA Paper 1 2011
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

If `AB = 0`, then we conclude that `A = 0` or `B = 0`. But

remember that it is not a necessary condition that `AB = 0` holds, if

`A = 0` or `B = 0`, because that condition is also possible when

`A != 0` and `B != 0`. If `A` and `B` are non-zero matrix and `AB = 0`

`=> | A | = 0` and `| B | = 0`
Correct Answer is `=>` (D) Neither I nor II
Q 2329780611

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

NDA Paper 1 2011
(A)

`S = {0,2, 3)`

(B)

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

(C)

`S = {1, 2, 3}`

(D)

`S = {2, 3}`

Solution:

Since , the given matrix is

` A = [ ( 2-x , 1 , 1),(1 , 3 - x , o) ,(-1 , -3, -x) ] `

Since, this matrix is singular.

`:. | A | = 0`

`=> A = [ ( 2-x , 1, 1),( 1, 3-x , 0) ,(-1 , -3, -x) ] = 0`

Using `R_2 - R_2 + R_3`,

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

`=> (2- x) (x^2 - 3x) + 1(x) + 1(-x) = 0`

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

` :. x = 2 , 0 , 3`

Hence, solution set, `S = {0, 2, 3}`.
Correct Answer is `=>` (A) `S = {0,2, 3)`
Q 2369780615

Consider the following statements
I. The inverse of a square matrix, if it exists, is
unique.
II. If A and B are singular matrices of order n,
then AB is also a singular matrix of order n.
Which of the statements given above is/are correct?
NDA Paper 1 2011
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

The inverse of a square matrix, if it exists, is unique but

if `A` and `B` are singular matrices of order `n`, then `AB` is not a

singular matrix of order `n`.

Hence, only Statement `I` is correct.
Correct Answer is `=>` (A) Only I
Q 2369880715

If the matrix
` A = [ (alpha , beta),(beta , alpha) ]`
is such that `A^2 = I`, then which one of the
following is correct?
NDA Paper 1 2011
(A)

`alpha = 0, beta = 1` or `alpha = 1, beta = 0`

(B)

`alpha = 0, beta != 1` or `alpha != 1, beta = 1`

(C)

`alpha = 1, beta != 0` or `alpha != 1, beta = 1`

(D)

` alpha != 0 , beta != 0`

Solution:

` ∵ A = [ (alpha , beta),(beta , alpha) ]`

`:. A^2 = [ (alpha , beta),(beta , alpha) ] [ (alpha , beta),(beta , alpha) ] = [ ( alpha^2 + beta^2 , 2 alpha beta ),( 2 alpha beta , alpha^2 + beta^2)]`

Now, ` A^2 = I`

` => [ ( alpha^2 + beta^2 , 2 alpha beta ),( 2 alpha beta ,alpha^2 + beta^2) ] = [ (1,0),(0,1)]`

On comparing, we get

`alpha^2 + beta^2 = 1, alpha beta = 0`

` => alpha = 0, beta = 1` or ` beta = 0 , alpha =1`
Correct Answer is `=>` (A) `alpha = 0, beta = 1` or `alpha = 1, beta = 0`
Q 2329180911

If `A= [ ( alpha , 0),(1 , 1) ]` and `B = [ ( 1 , 0),(2 , 1) ]` such that `A^2 = B ` then
what is the value of `alpha`?
NDA Paper 1 2011
(A)

`-1`

(B)

`1`

(C)

`2`

(D)

`4`

Solution:

`∵ A = [ (alpha , 0),(1 ,1) ]`

` => A^2 = [ (alpha , 0),(1 ,1) ] [ (alpha , 0),(1 ,1) ]`

` => A^2 = [ (alpha^2 , 0),(alpha + 1 ,1) ]`

But ` A^2 = B => [ (alpha^2 , 0),(alpha + 1 ,1) ] = [(1,0),(2,1)]`

On comparing, we get

`alpha^2 = 1` and `alpha + 1 = 2`

`:. alpha = 1`
Correct Answer is `=>` (B) `1`
Q 2369191015

`A = [ (3 , 1),(0,4)] `, then `B = [(1,1),(0,2)]` which of the
following is/are correct'?
I. `AB` is defined.
II. `BA` is defined.
III. `AB =BA`
Select the correct answer using the codes given below.
NDA Paper 1 2011
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

I, II and III

Solution:

`∵ A = [ (3 , 1),(0,4)] `, then `B = [(1,1),(0,2)]`

`:. AB = [ (3 , 1),(0,4)] [(1,1),(0,2)] = [(3,5),(0,8)]`

and `BA = [(1,1),(0,2)] [ (3 , 1),(0,4)] = [(3,5),(0,8)]`

`=> AB = BA`

Hence, all the three statements are correct.
Correct Answer is `=>` (D) I, II and III
Q 2410378219

If `|A| = 8`, where `A` is square matrix of order `3`,
then what is `| adj A |` equal to?
NDA Paper 1 2010
(A)

`16`

(B)

`24`

(C)

`64`

(D)

`512`

Solution:

` ∵ |A| = 8` and `A` is a square matrix of order `3`.

(`∵ | adj(A)| = | A|^(n- 1)` when `A` have order `n`)

`:. | adj(A)| = |A|^(3-1) = 8^(3-1) = 8^2 = 64`
Correct Answer is `=>` (C) `64`
Q 2339291112

If ` [ (1 , -3 , 2),(2 , -8 ,5),(4 ,2 , lamda)]` is not an invertible matrix, then

what is the value of `A`?
NDA Paper 1 2010
(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

Since, tile matrix `[ (1 , -3 , 2),(2 , -8 ,5),(4 ,2 , lamda)]` is not an invertible

matrix, i.e., it should be a singular matrix.

`:. | (1 , -3 , 2),(2 , -8 ,5),(4 ,2 , lamda) | = 0`

`=> 1 (-8 lamda - 10) + 3 (2 lamda - 20) + 2 (4 + 32) = 0`

`=> -8 lamda - 10 + 6 lamda - 60 + 72 = 0`

` => -2 lamda + 2 = 0`

` => lamda = 1`
Correct Answer is `=>` (C) `1`

Set - 2

Q 2319291119

If `A = [(0,1),(-1 ,0) ] , B=[(i , 0),(0 , -i)], C = [(0,-i),(-i , 0) ]` then
which one of the following is not correct ?
NDA Paper 1 2010
(A)

`A^2 = B^2`

(B)

`B^2 = C^2`

(C)

`AB = C`

(D)

`AB = BA`

Solution:

`∵ A = [(0,1),(-1 ,0) ] , B=[(i , 0),(0 , -i)]` and `C = [(0,-i),(-i , 0) ]`

` => AB = [(0,1),(-1 ,0) ] [(i , 0),(0 , -i)]`

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

Hence, `AB = C`
Correct Answer is `=>` (C) `AB = C`
Q 2319391210

Consider the following statements in respect of a
square matrix `A` and its transpose `A^T`.
I. `A + A^T` is always symmetric.
II. `A - A^T` is always anti-symmetric.
Which of the statements given above is/are correct?
NDA Paper 1 2010
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

We know that, `A+ A^T` is always symmetric and `A - A^T`

is always anti-symmetric. (by property of transpose)
Correct Answer is `=>` (C) Both I and II
Q 2349391213

If a matrix `A` is such that `3A^3 + 2A^2 + 5A + I = 0`,
then `A^(-1)` is equal to?
NDA Paper 1 2010
(A)

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

(B)

`3A^2 + 2A + 5I`

(C)

`3A^2 - 2A - 5I`

(D)

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

Solution:

`∵ 3A^3 + 2A^2 + 5A + I = 0`

Operate `A^(-1)` on both sides, we get

`=> 3A^3 A^(-1) + 2A^2K^(-1) + 5A A^(-1) + IA^(-1) = 0`

` => 3A^2 I + 2 A I + 5 I + A^(-1) = 0`

`=> A^(-1) = -(3A^2 + 2A + 5I)`
Correct Answer is `=>` (D) `-(3A^2 + 2A + 5I)`
Q 2379391216

Let `A` and `B` be matrices of order `3 xx 3`. If `AB = 0`,
then which of the following can be concluded?
NDA Paper 1 2010
(A)

`A = 0` and `B = 0`

(B)

`|A| = 0` and `|B| = 0`

(C)

Either `|A| = 0` or `|B| = 0`

(D)

Either `A = 0` or `B = 0`

Solution:

If `AB = 0`, then it may be concluded that either `A = 0` or

`B = 0`.

But, it should be noticed that it is not necessary that either `A = 0`

or `B = 0, AB = 0` can be possible, if `A` and `B` are non-zero.

If `A` and `B` are two matrices and `AB = 0 => | A | = 0` and `| B | = 0`.
Correct Answer is `=>` (B) `|A| = 0` and `|B| = 0`
Q 2319491310

If `A` is a square matrix, then `adj A^T - (adj A)^T` is
equal to
NDA Paper 1 2010
(A)

`2 | A |`

(B)

`2 | A | I`

(C)

Null matrix

(D)

Unit matrix

Solution:

` ∵ (adj A^T) = (adj A)^T`

`=> (adj A^T) - (adj A)^T =` Null matrix
Correct Answer is `=>` (C) Null matrix
Q 2329591411

Let `A = [ (5 , 6 ,1),(2 , -1 , 5) ]`. If there exists a matrix `B`
such that `AB = [(35, 49),(29, 13)]`, then `B` is equal to
NDA Paper 1 2010
(A)

`[ (5 , 1 ,4),(2 , 6 , 3) ]`

(B)

`[ (2 , 6 ,3),(5 , 1 , 4) ]`

(C)

` [ (5,2),(1,6),(4,3)]`

(D)

` [ (2,5),(6,1),(3,4)]`

Solution:

Given that, `A = [ (5 , 6 ,1),(2 , -1 , 5) ]_( 2xx 3)`

and let `B = [ (5,2),(1,6),(4,3)]_(3 xx 2)`

` :. AB = [ (5 , 6 ,1),(2 , -1 , 5) ] [ (5,2),(1,6),(4,3)]`

` = [ ( 25 + 6 + 4 ,10 + 36 + 3),( 10- 1 + 20,4-6+15) ]_(2 xx 2) = [(35, 49),(29, 13)]`
Correct Answer is `=>` (C) ` [ (5,2),(1,6),(4,3)]`
Q 2309591418

Consider the following statements
I. If A' =A, then A is a singular matrix, where A'
is the transpose of A.
II. If A is a square matrix such that `A^3 = I`, then A
is non-singular.
Which of the statements given above is/are correct?
NDA Paper 1 2010
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. It is not necessary that, if `A` is a symmetric matrix,

then it is singular.

II. `A^3 = I => |A^3|= | I | => |A|^3 = 1 => | A | = 1`

Hence, `A` is a non-singular matrix.
Correct Answer is `=>` (B) Only II
Q 2319691519

If `A` is a real skew-symmetric matrix of order `n`
such that `A^2 + I = 0`, where `I` being the identity
matrix of the same order as that of `A`, then what is
the order of `A`?
NDA Paper 1 2010
(A)

3

(B)

Odd

(C)

Prime number

(D)

None of these

Solution:

Let the real symmetric matrix of order `2` be.

`A = [ (0,a),(-a, 0)]`

`A^2 = [ (0,a),(-a,0)] [ (0 ,a),(-a , 0) ] = [(-a^2 , 0),(0 , - a^2) ]`

` => A^2 + I = [ (-a^2 , 0),(0 , - a^2) ] + [ (1,0),(0,1)]`

` = [ ( 1- a^2 , 0),(0, 1 - a^2) ]`

When `a = ± 1`, then `A^2 + I = 0`.

So, the given relation ` A^2 + I = 0` is not always true for even order.

Now, let the real symmetric matrix of order `3` be.

` A = [ (0,a , b),(-a,0,c),(-b , -c ,0) ]`

` A^2 = [ (0,a , b),(-a,0,c),(-b , -c ,0) ] [ (0,a , b),(-a,0,c),(-b , -c ,0) ]`

` = [ ( - a^2 - b^2 , - bc , ac ),(-bc , -a^2 - c^2 ,- ab) ,( ac , -ab , -b^2 - c^2) ]`

`:. A^2 + I = [ ( -a^2 - b^2 + 1 , -bc , -ac ) ,( -bc , -a^2 - c^2 + 1 , -ab ) ,(ac , -ab , -b^2 - c^2 + 1) ] != 0`

So, the given relation `A^2 + I = 0` also is not always true for an odd

order.

Hence, none of the option is true.
Correct Answer is `=>` (D) None of these
Q 2369791615

Let `A = [(1,2),(3,4)] = [a_(ij)]`, where `i , j =1, 2`. If its inverse
matrix is `[b_(ij) ]`, then what is `b_(22)` ?
NDA Paper 1 2010
(A)

`-2`

(B)

`1`

(C)

`3/2`

(D)

`-1/2`

Solution:

`∵ A = [ (1,2),(3,4)] , adj(A) = [ (4,-3),(-2,1)]^T = [ (4,-2),(-3,1)]`

and `|A| = 4 - 6 = -2`

`:. A^(-1) = -1/2 [ (4,-3),(-2,1)]`

` => [b_(ij) ] = - 1/2 [ (4,-3),(-2,1)] => b_(22) = -1/2`
Correct Answer is `=>` (D) `-1/2`
Q 2430878712

Under which one of the following conditions does
the system of equations
`kx + y + z = k - 1`
`x + ky + z = k - 1`
`x + y + kz = k - 1`
have no solution?
NDA Paper 1 2009
(A)

`k = 1`

(B)

`k != -2`

(C)

`k = 1` or `k = -2`

(D)

`k = -2`

Solution:

The given system of equations is

`kx + y + z = k - 1`

`x + ky + z = k - 1`

`x + y + kz = k - 1`

`:. A = [(k,1,1),(1,k,1),(1,1,k)] , B = [(k-1),(k-1),(k-1)]` and ` X = [(x),(y),(z)]`

Now , `|A| = |(k,1,1),(1,k,1),(1,1,k)|`

Expanding along `R_1`,

`= k(k^2 - 1) - 1 (k - 1) + 1 (1 - k)`

`= k^3 - k - k + 1 + 1 - k = k^3 - 3k + 2`

The given system of equations has no solution, if `| A | = 0`.

`:. k^3 - 3k + 2 = 0`

`=> (k-1)^2 (k+2) = 0 => k = 1` or `k = -2`
Correct Answer is `=>` (C) `k = 1` or `k = -2`
Q 2379091816

If `X` and `Y` are the matrices of order `2 xx 2` each and
` 2X - 3Y = [ (-7,0),(7 , -13) ] ` and `3X + 2Y = [ (9,13),(4,13)]`
then `Y` is equal to
NDA Paper 1 2009
(A)

` [ (1,3),(-2,1)]`

(B)

` [ (1,3),(2,1)]`

(C)

` [ (3,2),(-1,5)]`

(D)

` [ (3,2),(1,-5)]`

Solution:

Given, `2X- 3Y = [ (-7,0),(7 , -13) ]` .....(i)

and ` 3X + 2Y = [ (9,13),(4,13)]` .........(ii)

On multiplying Eq. (i) by `3` and Eq. (ii) by `2` and subtracting Eq. (i)

from Eq. (ii), we get

` 13Y = 2 [ (9,13),(4,13)] -3 [ (-7,0),(7 , -13) ]`

` => 13Y = [ (39 , 26 ),(-13 , 65)] `

`:. Y = [ (3,2),(-1 ,5) ]`
Correct Answer is `=>` (C) ` [ (3,2),(-1,5)]`
Q 2309191918

If a matrix `A` is symmetric as well as
anti-symmetric, then which one of the following is
correct?
NDA Paper 1 2009
(A)

A is a diagonal matrix

(B)

A is a null matrix

(C)

A is a unit matrix

(D)

A is n trangular matrix

Solution:

Since `A' =A` and `A'= - A`

`=> A = -A => A = 0`

is a null matrix.
Correct Answer is `=>` (B) A is a null matrix
Q 2430101012

If `A = [ ( 1, -2 , -3),(2,1 ,-2),(3,2,1)]` then which one of the
following is correct?

NDA Paper 1 2009
(A)

A is symmetric matrix

(B)

A is anti-symmetric matrix

(C)

A is singular matrix

(D)

A is non-singular matrix

Solution:

Here, we see that its diagonal elements are not zero,

so it is not anti-symmetric matrix.

Now, `| A | = 1 (1 + 4) + 2(2 + 6)- 3(4- 3)`

`=5 + 16 - 3 = 18 != 0`

Hence, it is non-singular matrix.
Correct Answer is `=>` (D) A is non-singular matrix
Q 2480101017

It `A = [(omega,0),(0,omega)]`, where `omega` is cube root of unity, then
`A^(100)` is equal to
NDA Paper 1 2009
(A)

`A`

(B)

`-A`

(C)

null matrix

(D)

identity matrix

Solution:

Given, `A = [(omega,0),(0,omega)]`

Now, `A^2 = [(omega,0),(0,omega)] [(omega,0),(0,omega)] = [(omega^2,0),(0,omega^2)]`

`A^3 = [(omega^2,0),(0,omega^2)] [(omega,0),(0,omega)] = [(omega^3,0),(0,omega^3)]`

Similarly, ` A^(100) = [(omega^(100),0),(0,omega^(100))] = [((omega^3)^(33).omega ,0),(0,(omega^3)^(33). omega )]`

` = [(1.omega,0),(0,omega.1)] = [(omega,0),(0,omega)] `

` = A`
Correct Answer is `=>` (A) `A`
Q 2410401310

A matrix `X` has `(a +b)` rows and `(a + 2) ` columns
and a matrix `Y` has `(b + 1)` rows and `(a + 3)`
columns. If both `XY` and `YX` exist, then what are
the values of `a, b` respectively'?
NDA Paper 1 2009
(A)

`3, 2`

(B)

`2, 3`

(C)

`2, 4`

(D)

`4, 3`

Solution:

The order of a given matrices are

`[X]_((a+ b)xx(a + 2))` and `[Y]_((b + 1)xx(a + 3)`

As `[XY]` and `[YX]` exist.

`:. a + 2 = b + 1` and `a + 3 = a + b`

`=> a = 2, b = 3`
Correct Answer is `=>` (B) `2, 3`
Q 2470401316

If `A = [(1,2),(3,4)]` and `B = [(a,0),(0,b)]` where `a` and `b` are
positive integers, then which one of the following
is correct ?
NDA Paper 1 2009
(A)

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

(B)

There exists 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:

Let `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,4b)]`

If ` AB = BA`

` => [(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
Q 2480401317

Consider a matrix `M = [ (3,4,0),(2,1,0),(3,1,k)]` and the
following statements
Statement I Inverse of `M` exists.
Statement II `k != 0`
Which one of the following in respect of the above matrix
and statement is correct?
NDA Paper 1 2009
(A)

I implies II but II does not imply I

(B)

II implies I but I does not imply II

(C)

Neither I implies II nor II implies I

(D)

I implies II as well as II implies I

Solution:

` M = [ (3,4,0),(2,1,0),(3,1,k)]`

Now, `| M | = | (3,4,0),(2,1,0),(3,1,k) | = k(3 - 8) = - 5k`

If `k != 0`, then inverse of `M` exists i.e., `M` is non-singular. Thus,

Statement I implies II as well as II implies I.
Correct Answer is `=>` (D) I implies II as well as II implies I
Q 2430501412

Which one of the following is correct in respect of
The matrix `A = [ (0,0,-1),(0,-1,0),(-1,0,0) ]` ?
NDA Paper 1 2009
(A)

`A^(-1)` does not exist

(B)

`A = (-1)I`

(C)

`A` is a unit matrix

(D)

`A^2 = I`

Solution:

`∵ A = [ (0,0,-1),(0,-1,0),(-1,0,0) ]`

`:. | A | = | (0,0,-1),(0,-1,0),(-1,0,0) | = - 1 (-1) = 1 != 0`

So, `A^(- 1)` exists.

Now, `A^2 = [ (0,0,-1),(0,-1,0),(-1,0,0) ] [ (0,0,-1),(0,-1,0),(-1,0,0) ] = [ (1,0,0),(0,1,0),(0,0,1) ]`

` => A^2 = I`
Correct Answer is `=>` (D) `A^2 = I`
Q 2410501419

If `A= [ (3,2),(1,4)]` , then `A (adj A)` is equal to
NDA Paper 1 2009
(A)

` [ (0,10),(10,0)]`

(B)

` [ (10,0),(0,10)]`

(C)

` [ (1,10),(10,1)]`

(D)

` [ (10,1),(1,10)]`

Solution:

` ∵ A = [ (3,2),(1,4)]`

`:. A (adj A) = I_2 | A |`

` = [ (1,0),(0,1)] | (3,2),(1,4) | = [ (1,0),(0,1)] (12 - 2)`

` = [ (1,0),(0,1)] xx 10 = [ (10,0),(0,10)]`
Correct Answer is `=>` (B) ` [ (10,0),(0,10)]`
Q 2450701614

What is the inverse of ` [ ( 0 , 0, 1),(0,1,0),( 1,0,0) ] `?

NDA Paper 1 2009
(A)

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

(B)

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

(C)

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

(D)

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

Solution:

Lat `A = [ ( 0 , 0, 1),(0,1,0),( 1,0,0) ] `

`:. | A | = -1`

Cofactors of `A`

` a_(11) = | (1,0),(0,0)| = 0 , a_(12) = - | (0,0),(1,0)| = 0 , a_(13) = - | (0,1),(1,0)| = -1`

` a_(21) = - | (0,1),(0,0)| = 0 , a_(22) = | (0,1),(1,0)| = -1 , a_(23) = - | (0,0),(1,0)| = 0`

` a_(31) = | (0,1),(1,0)| = -1 , a_(32) = - | (0,1),(0,0)| = 0 , a_(33) = | (0,0),(0,1)| = 0`

`=> adj(A) = [ ( 0 , 0, -1),(0,-1,0),( -1,0,0) ]^7 ` and `adj (A) = [ ( 0 , 0, -1),(0,-1,0),( -1,0,0) ]`

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

` = - 1/1 [ ( 0 , 0, -1),(0,-1,0),( -1,0,0) ] = [ ( 0 , 0, 1),(0,1,0),( 1,0,0) ] `
Correct Answer is `=>` (B) `[ ( 0 , 0, 1),(0,1,0),( 1,0,0) ]`
Q 2410701619

Consider the following statements in respect of
symmetric matrices `A` and `B`.
I. `AB` is symmetric.
II. `A^2 + B^2` is symmetric.
Which of the above statement(s) is/are correct?
NDA Paper 1 2009
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

Given that, `A = A', B = B'`

Now, we have `AB = A'B' = (BA)'`

Therefore, `AB` is not symmetric.

and `A^2 + B^2 = (A')^2 + (B')^2 = (A^2 + B^2)'`

So, `A^2 + B^2` is symmetric.
Correct Answer is `=>` (B) Only II
Q 2460801715


NDA Paper 1 2009

Assertion : ` M = [ (5,10),(4,8)]` is invertible.

Reason : `M` is singular.

(A) Both A and R individually true and R is the correct explanation of A
(B) Both A and R are individually true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Solution:

`M = [ (5,10),(4,8)]`

`|M| = | (5,10),(4,8)| = 40 - 40 = 0`

So, that `M` is not invertible because `M` is a singular matrix.

`R M` is singular matrix.

Therefore, `A` is false and `R` is true.
Correct Answer is `=>` (D)
Q 2410380210

If the matrix `B` is the adjoint of the square matrix
`A` and `alpha` is the value of the determinant of `A`, then
what is `AB` equal to?
where, `I` is identity matrix
NDA Paper 1 2008
(A)

`alpha`

(B)

`(1/alpha)I`

(C)

`I`

(D)

`alpha I`

Solution:

Since, adjoint of the square matrix `A` is `B` and value of

the determinant of `A` is `alpha` i.e., `B = adj A` and `| A | = alpha`.

Then, `AB = | A | I [ ∵ A (adj A )= | A | I]`

`= alpha I`
Correct Answer is `=>` (D) `alpha I`
Q 2410780610

If `| A_(n xx n) | = 3` and `| adj A | = 243`, then what is the
value of `n`?
NDA Paper 1 2008
(A)

`4`

(B)

`5`

(C)

`6`

(D)

`7`

Solution:

`∵ | A_(n xx n)| = 3` and ` |adj A | = 243`

We know that,

`|adj(A)| = | A_(n xx n) |^(n-1)`

`=> 243 = 3^(n- 1) => 3^5 = 3^(n- 1)`

On comparing, we get

`n - 1 = 5 => n = 6`
Correct Answer is `=>` (C) `6`
Q 2400780618

If `A` is matrix of order `3 xx 2` and `B` is matrix of
order `2 xx 3`, then what is `| kAB |` equal to (where, `k` is
any scalar quantity)?
NDA Paper 1 2008
(A)

`k | AB |`

(B)

`k^2 | AB |`

(C)

`k^3 | AB |`

(D)

` | AB |`

Solution:

Since, order of `A` and `B` are `3 xx 2` and `2 xx 3`,

respectively.

`:. | kAB | = k^3 | AB |` (since, order of `AB` is `3 xx 3`)
Correct Answer is `=>` (C) `k^3 | AB |`
Q 2460001815

If `adj (A) = [ (a , 0),(-1 ,b)]` and `ab != 0`, then what is the
value of `| A^(-1) | `?

NDA Paper 1 2008
(A)

1

(B)

ab

(C)

`1/sqrt(ab)`

(D)

`1/(ab)`

Solution:

For `2 xx 2` matrix,

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

`= (ab - 0) = ab = |( a ,0),( -1, b)|`

`:. A^(-1) = (adj A)/(|A|) = 1/(ab) . [(a ,0),(-1,b)]`

`| A^(-1) | = 1/(ab) (ab) = 1`
Correct Answer is `=>` (A) 1
Q 2470112016

If `l + m + n = 0`, then the system of equations
`-2x+y+z = l`
`x-2y+z = m`
`x+y-2z = n`
has
NDA Paper 1 2008
(A)

a trivial solution

(B)

no solution

(C)

a unique solution

(D)

infinitely many solutions

Solution:

Here, `A = [ (-2 , 1 ,1),(1, -2 ,1),(1,1,-2)]`

` B = [ (l),(m),(n)]` and ` x = [ (x),(y),(z)]`

`:. |A| = -2 | (-2 ,1),(1,-2)| -1 |(1,1),(1 , -2)| + 1 | (1,1),(-2,1)|`

` = -2(4-1)-1 (-2 - 1)+ (1 + 2)`

` = -6 + 3 + 3 = 0`

Now `adj A = [ (3,3,3),(3,3,3),(3,3,3)]`

`∵ (adj A) B = [ (3,3,3),(3,3,3),(3,3,3)] [ (l),(m),(n)]`

` = 3 [ (l+m+n),(l+m+n),(l+m+n)] = 3 [ (0),(0),(0)]`

`:. (adj A)· B = 0 (∵ l + m + n = 0)`

So, the given system of equations has an infinitely many

solutions.
Correct Answer is `=>` (D) infinitely many solutions
Q 2410212119

If `A` and `B` are two matrices such that `AB = A` and
`BA = B`, then which one of the following is
correct?
NDA Paper 1 2008
(A)

` (A^T)^2 = A^T`

(B)

` (A^T)^2 = B^T`

(C)

` (A^T)^2 = (A^(-1))^(-1)`

(D)

None of these

Solution:

Given that.

`AB = A` and `BA = B`

(a) `A^T = (AB)^T = B^T A^T`

`=> (A^T)^2 = (B^T A^T) (B^T A^T)`

`= (AB)^T (AB)^T = A^T· A^T = (A^T)^2`

(b) `(A^T)^2 = (B^T A^T) (B^T A^T)`

`= (A^T B^T) (A^T B^T)`

`= (BA)^T (BA)^T = B^TB^T = (B^T )^2`

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

`=> (A^(-1))^(-1) = (B^(-1)A^(-1))^(-1) = AB =A`
Correct Answer is `=>` (D) None of these
Q 2430412312

If `[(1,3),(0,1)] A = [(1,-1),(0,1)]`, then what is the matrix `A`?
NDA Paper 1 2008
(A)

`[(1,-3),(0,1)]`

(B)

`[(2,2),(0,2)]`

(C)

`[(-4,-1),(1,0)]`

(D)

`[(1,-4),(0,1)]`

Solution:

Let `B = [(1,3),(0,1)] ` and `C = [(1,-1),(0,1)]`

Now, `| B | = 1 - 0 = 1`

`:. adj (B) = [(1,0),(-3,1)]^T = [(1,-3),(0,1)]`

` => B^(-1) = (adj (B))/(|B|)`

Then, `B^(-1) = [(1,-3),(0,1)]`

Now, `BA = C`

`=> B^(-1) BA = B^(-1)C`

`:. A = B^(-1) C`

` = [ (1,-3),(0,1)] [ (1,-1),(0,1)] = [(1, -4),(0,1)]`
Correct Answer is `=>` (D) `[(1,-4),(0,1)]`
Q 2480512417

If `X = [(1,-2),(0, 3)]` and `I` is a `2 xx 2` identity matrix,
then `X^2 - 2X + 3I` equals to which one of the
following?
NDA Paper 1 2008
(A)

`- I`

(B)

`-2X`

(C)

`2X`

(D)

`4X`

Solution:

`∵ X = [(1,-2),(0, 3)]`

`:. X^2 = [(1,-2),(0, 3)] [(1,-2),(0, 3)]`

` = [(1,-2-6),(0, 9)] = [(1,-8),(0, 9)]`

`:. X^2 - 2X + 3I = [(1,-8),(0, 9)] - 2 [(1,-2),(0, 3)] + 3 [(1,0),(0, 1)]`

` = [(1,-8),(0, 9)] + [(-2,4),(0, -6)] + [(3,0),(0, 3)]`

` = [ ( 1-2+3 , -8 + 4),(0, 9 - 6 + 3) ]`

` = [ (2,-4),(0,6)] = 2 [(1,-2),(0,3)]`

` =2X`
Correct Answer is `=>` (C) `2X`
Q 2440612513

Under what condition does `A(BC) = (AB)C` hold,
where `A, B` and `C` are three matrices?
NDA Paper 1 2008
(A)

AB and BC both must exist

(B)

Only AB must exists

(C)

Only BC must exists

(D)

Always true

Solution:

To hold the condition `A(BC) = (AB)C`. Hence, `AB` and

`BC` both must exist.
Correct Answer is `=>` (A) AB and BC both must exist
Q 2470712616

If `[(5,0),(0,7)]^(-1) [(x),(-y)] = [(-1),(2)]` then which one of the
following is correct?
NDA Paper 1 2008
(A)

`x = 5, y = 14`

(B)

`x = - 5, y = 14`

(C)

`x = -5, y = -14`

(D)

`x =5, y = -14`

Solution:

We know that,

` [(a,b),(c,d)]^(-1) = 1/(ad - bc) [ (d , -b),(-c ,a)]`

`:. [(5,0),(0,7)]^(-1) = 1/(35) [(7,0),(0,5)]`

Hence, `[(5,0),(0,7)]^(-1) [(x),(-y)] = [(-1),(2)]`(given)

`=> 1/(35) [(7,0),(0,5)] [(x),(-y)] = [(-1),(2)]`

`=> 1/(35) [(7x),(-5y)] = [(-1),(2)]`

On comparing, we get

` (7x)/(35) = -1 ` and ` - (5y)/(35) = 2`

` => x = -5 ` and ` y = - 14`
Correct Answer is `=>` (C) `x = -5, y = -14`
Q 2450891714

Consider the following statements
I. If `det A = 0,` then `det (adj A) = 0.`
II. If A is non-singular, then `det (A^(-1)) = (det A)^(-1)`.
Which of the above statements is/are correct?
NDA Paper 1 2007
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

We know that,

if `det(A) = 0`,then `det(adj A) = 0 (∵ | adj A | = | A |^(n- 1))`

and `det (A^(-1)) = (det A)^(-1)` if `A` is non-singular. (by property).

Then, both I and II are correct.
Correct Answer is `=>` (C) Both I and II
Q 2440091813

Let A be a square matrix of order `n xx n` where `n >= 2`.
If `B` be a matrix obtained from `A` with first and
second rows interchanged, then which one of the
following is correct?
NDA Paper 1 2007
(A)

det (A)= det (B)

(B)

det (A) =- det (B)

(C)

A = B

(D)

A = -B

Solution:

Let A be a square matrix of order `n xx n` where `n >= 2`.and

`B` be a matrix obtained from `A` with first and second rows

interchanged.

e.g., Let ` A = [(a,b),(c,d)] , B = [(c,d),(a,b)]`

`| A | = ad - cb, | B | = bc - ad`

`=> | B| = - (ad - bc)`

`=> |B| = - |A|`

Then, det (A) = - det (B)
Correct Answer is `=>` (B) det (A) =- det (B)
Q 2410191910

What should be the value of k, so that the system
of linear equations `x - y + 2z = 0, k x - y + z = 0`,
`3x + y - 3z = 0` does not possess a unique
solution?
NDA Paper 1 2007
(A)

`0`

(B)

`3`

(C)

`4`

(D)

`5`

Solution:

Given equations are

`x - y + 2 z = 0` ... (i)

`kx - y + z = 0` ... (ii)

`3x + y - 3z = 0` ... (iii)

Thus, system of equations does not posses a unique solution, if

` |A| = |(1,-1,2),(k, -1, 1),(3,1,-3)| = 0`

Expand with respect to `R_1` ,

`1 (3 - 1) + 1· ( -3k - 3) + 2 (k + 3) = 0`

`=> 2 - 3k - 3 + 2k + 6 = 0`

`=> -k + 5 = 0 => k = 5`

Thus, the system does not possess a unique solution, if `k = 5`.
Correct Answer is `=>` (D) `5`
Q 2450012814

The matrix `A = [ (1,2),(2,2)]` satisfies which one of the
following polynomial equations?
NDA Paper 1 2007
(A)

`A^2 + 3A + 2I = 0`

(B)

`A^2 + 3A - 2I = 0`

(C)

`A^2 - 3A - 2I = 0`

(D)

`A^2 - 3A + 2I = 0`

Solution:

Given that, `A = [ (1,2),(2,2)]`

`:. A^2 = [ (1,2),(2,2)] [ (1,2),(2,2)] = [ (1+4 , 2 + 4),(2+4 ,4 + 4)]`

` = [ (5,6),(6,8)]`

Now, `A^2 - 3A - 2I = [ (5,6),(6,8)] - [ (3,6),(6,6)] - [ (2,0),(0,2)]`

` = [ (5 - 3- 2 , 6 - 6 - 0),(6 - 6 - 0 , 8 - 6 - 2)] = [(0,0),(0,0)]`

` => A^2 - 3A - 2I = 0`
Correct Answer is `=>` (C) `A^2 - 3A - 2I = 0`
Q 2470112916

If `A = [(2x,0),(x,x)]` and `A^(-1) = [(1,0),(-1,2) ]` then what Is
the value of `x`?
NDA Paper 1 2007
(A)

` -1/2`

(B)

`1/2`

(C)

`1`

(D)

`2`

Solution:

Given, ` A = [(2x,0),(x,x)]` and ` A^(-1) = [(1,0),(-1,2) ]`

` | A| = 2x^2 - 0 = 2x^2`

`:. adj (A) = [(x,-x),(0,2x)]^T = [(x,0),(-x,2x)]`

` => A^(-1) = (adj (A))/(|A|)`

` => A^(-1) = 1/(2x^2) [(x,0),(-x,2x)]`

` = [ (1/(2x) ,0),(- 1/(2x) , 1/x) ]`

But ` A^(-1) = [(1,0),(-1 ,2)]`

`:. 1/(2x) = 1 => x = 1/2`

Alternate Method

We know that, `A A^(-1) =I`

` :. [(2x,0),(x,x)] [(1,0),(-1 ,2)] = [ (1,0) ,(0,1) ]`

` => [(2x,0),(0,2x)] = [(1,0),(0,1)]`

On comparing, we get

`2x = 1 => x = 1/2`
Correct Answer is `=>` (B) `1/2`
Q 2420223111

Let `A = [a_(ij)]_(m xx m)` be a matrix and `C = [C_(ij)]_(m xx m)` be
another matrix, where `c_(ij)` is the cofactor of `a_(ij)`
Then, what is the value of `| AC |`?
NDA Paper 1 2007
(A)

`| A|^(m- 1)`

(B)

`|A|^m`

(C)

`|A|^(m+ 1)`

(D)

Zero

Solution:

Let `A= [a_(ij)]_(m xx m)` be a matrix and `C = [c_(ij)]_(m xx m)` be

another matrix, where `c_(ij)` is the cofactor of `a_(ij)` i.e., `X= adj (A)`.

Since, `A` is a square matrix .

`:. adj A = (adj A)` both have the same order.

`=> | adj A| = | A|^(m - 1)`

`=> | A | | adj A | = | A |^(m - 1) { A }`

`=> | adj A | = | A |^m`

`:. |AC| = | A |^m`
Correct Answer is `=>` (B) `|A|^m`
Q 2480323217

If `A = [(2,2),(2,2)],` then `A^n` is equal to
NDA Paper 1 2007
(A)

`[ (2^n ,2^n),(2^n,2^n)]`

(B)

`[ (2n ,2n),(2n,2n)]`

(C)

`[ (2^(2n-1) ,2^(2n-1)),(2^(2n-1),2^(2n-1))]`

(D)

`[ (2^(2n+1) ,2^(2n+1)),(2^(2n+1),2^(2n+1))]`

Solution:

`∵ A = [(2,2),(2,2)]`

`:. A^2 = [(2,2),(2,2)] [(2,2),(2,2)] = [(4+4,4+4),(4+4,4+4)]`

` = [(2^3,2^3),(2^3,2^3)]`

`:. A^3 = [(8,8),(8,8)] [(2,2),(2,2)]`

` = [(16+16,16+16),(16+16,16+16)]`

` = [(2^5,2^5),(2^5,2^5)]`

From the above, it is clear that the power of element form an `AP`.

`:. n`th term of `AP = 1 + (n - 1) 2 = 2n - 1`

` A^n = [(2^(2n -1),2^(2n -1)),(2^(2n -1),2^(2n -1))]`
Correct Answer is `=>` (C) `[ (2^(2n-1) ,2^(2n-1)),(2^(2n-1),2^(2n-1))]`
Q 2480423317

If the least number of zeroes in a lower triangular
matrix is `10`, then what is the order of the matrix?
NDA Paper 1 2007
(A)

`3 xx 3`

(B)

`4 xx 4`

(C)

`5 xx 5`

(D)

`10 xx 10`

Solution:

If the least number of zeroes in a lower triangular matrix

is `10`, then order of the matrix is `5 xx 5`.

e.g., ` A = [ (a,0,0,0,0),(b,c,0,0,0),(d,e,f,0,0),(g,h,i,j,0),(k,l,m,n,0)]`

Since, the number of zero's in lower triangular matrix

` = (n(n- 1))/2 = (5 xx 4)/2 = 10`
Correct Answer is `=>` (C) `5 xx 5`
Q 2400623518

If the inverse of ` [ (1,p,q),(0,x,0),(0,0,1)]` is ` [ (1, -p ,-q),(0,1,0),(0,0,1) ]` ,
then what is the value of `x`?
NDA Paper 1 2007
(A)

`1`

(B)

Zero

(C)

`-1`

(D)

`1/p + 1/q`

Solution:

Lat `A = [ (1,p,q),(0,x,0),(0,0,1)]` and ` B = [ (1, -p ,-q),(0,1,0),(0,0,1) ]`

` |B| = 1(1) + p(0) - q(0) = 1`

`C_(11) = 1, C_(12) = 0, C_(13) = 0`

` C_(21) = p, C_(22) = 1, C_(23) = 0`

` C_(31) = q , C_(32) = 0 , C_(33) = 1`

`:. adj (B) = [ ( 1,0,0),(p,1,0),(q,0,1)]^T = [(1,p,q),(0,1,0),(0,0,1)]`

` B^(-1) = (adj (B))/(|B|)`

Thus, ` B^(-1) = [ (1,p,q),(0,1,0),(0,0,1)]`

But `B` is inverse of `A`, therefore `A = B^(-1)`

(according to the question)

` => [ (1,p,q),(0,x,0),(0,0,1)] = [ (1,p,q),(0,1,0),(0,0,1)] => x = 1`

Alternate Method

Given, `B` is the inverse of `A`, then `BA = I`

` [ (1, -p ,-q),(0,1,0),(0,0,1) ] [ (1,p,q),(0,x,0),(0,0,1)] = [ (1,0,0),(0,1,0),(0,0,1)] `

` => [ (1,p-px,0),(0,1,0),(0,0,1)] = [ (1,0,0),(0,1,0),(0,0,1)] `

On comparing, we get

`x = 1`
Correct Answer is `=>` (A) `1`
Q 2420823711

If `AB = [(4,11),(4,5)]` and `A = [(3,2),(1,2)]`, then what is the
value of the determinant of the matrix `B`?
NDA Paper 1 2007
(A)

`4`

(B)

`-6`

(C)

`-1/4`

(D)

`-28`

Solution:

Given that, ` AB = [(4,11),(4,5)]` and `A = [(3,2),(1,2)]`

`| A | = 6 - 2 = 4 , adj (A) = [(2,-1),(-2,3)]^T = [(2,-2),(-1,3)]`

`:. A^(-1) = 1/(6-2) [(2,-2),(-1,3)] = [(1/2,-1/2),(-1/4,3/4)] `

Now, ` AB = [ (4,11),(4,5)]`

` => B = A^(-1) [ (4,11),(4,5)]`

` = [(1/2,-1/2),(-1/4,3/4)] [ (4,11),(4,5)]`

` = [ (2 - 2, (11)/5 - 5/2 ),( -1 + 3 , - (11)/4 + (15)/4 ) ] = [(0,3),(2,1)]`

`:. | B | = [(0,3),(2,1)] = 0 - 6 = -6`
Correct Answer is `=>` (B) `-6`
Q 2460023815


NDA Paper 1 2007

Assertion : If `A = [(2,3),(1,4)] B = [(1,0),(0,1)]` then `(A+ B)^2 = A^2 + B^2 + 2AB`.

Reason : In the above, `AB = BA`.

(A) Both A and R individually true and R is the correct explanation of A
(B) Both A and R are individually true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Solution:

`∵ A = [(2,3),(1,4)] , B = [(1,0),(0,1)] , A + B = [(3,3),(1,5)]`

`=> (A + B)^2 = [ (12,24),(8,24)]`

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

Now, ` A^2 = A.A = [ (2,3),(1,4) ] [ (2,3),(1,4) ] = [ (7,18),(6,19) ]`

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

` B^2 = B.B = [ (1,0),(0,1)] [ (1,0),(0,1)] = [ (1,0),(0,1)]`

` A^2 + b^2 + 2AB = [ (7,18),(7,19) ] + [ (1,0),(0,1)] + [ (4,6),(2,8)] = [ (12,24),(8,28)]`

`:. AB = BA`

`:. (A+B)^2 =A^2 +B^2 +2AB`

Hence, both `A` and `R` are individually true and `R` is the correct

explanation of `A`.
Correct Answer is `=>` (A)
Q 2410123919


NDA Paper 1 2007

Assertion : If `A = [ (cos alpha , sin alpha ),(cos alpha , sin alpha )]` and ` B = [ (cos alpha , cos alpha ),(sin alpha , sin alpha )]`, then `AB != I`.

Reason : The product of two matrices can never be equal to an identity matrix.

(A) Both A and R individually true and R is the correct explanation of A
(B) Both A and R are individually true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Solution:

`∵ A = [ (cos alpha , sin alpha ),(cos alpha , sin alpha )] , B = [ (cos alpha , cos alpha ),(sin alpha , sin alpha )]`

`:. AB = [ (cos alpha , sin alpha ),(cos alpha , sin alpha )] , B = [ (cos alpha , cos alpha ),(sin alpha , sin alpha )]`

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

` = [(1,1),(1,1)]`

`=> AB != I`

Sometimes the product of two matrices can be equal to an

identity matrix.

e.g., `A A^(-1) = I`.

Hence, A is true but R is false
Correct Answer is `=>` (C)
Q 2450234114

If `A` is any `2 xx 2` matrix such that
` [(1,2),(0,3)] A = [(-1,0),(6,3)]`
then A is equal to
NDA Paper 1 2007
(A)

` [(-5,1),(-2,2)]`

(B)

` [(-5,-2),(1,2)]`

(C)

` [(-5,-2),(2,1)]`

(D)

` [(5,2),(-2,-1)]`

Solution:

Let ` [(1,2),(0,3)] = B , | B | = 3`

and `adj (B) = [(3,0),(-2,1)]^T = [(3,-2),(0,1)]`

`:. B^(-1) = 1/3 [(3,-2),(0,1)]`

Now, ` [(1,2),(0,3)] A = [(-1,0),(6,3)] `

` ( ∵ BA = [(-1,0),(6,3)] => A = B^(-1) [(-1,0),(6,3)])`

` => A = 1/3 [(3,-2),(0,1)] [(-1,0),(6,3)]`

` = 1/3 [(-3 - 12,-6),(6,3)]`

` = [(-5,-2),(2,1)]`
Correct Answer is `=>` (C) ` [(-5,-2),(2,1)]`
Q 2430334212

If A is a `3 xx 3` matrix such that `| A | = 4`, then
`A (adj A)` is equal to
NDA Paper 1 2007
(A)

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

(B)

`[(4,0,0),(0,4,0),(0,0,4)]`

(C)

`[(16,0,0),(0,16,0),(0,0,16)]`

(D)

Cannot be determined

Solution:

We know that,

`A (adj A) = | A| I_n = 4 [(1,0,0),(0,1,0),(0,0,1)] = [(4,0,0),(0,4,0),(0,0,4)]`
Correct Answer is `=>` (B) `[(4,0,0),(0,4,0),(0,0,4)]`
Q 2460434315

If `A = [ (x,x^2,1+x^2),(y , y^2 ,1 + y^2),(z ,z^2 ,1 + z^2)]`, where `x`, `y` and `z` are
distinct, then what is the value of `| A |`?
NDA Paper 1 2007
(A)

`0`

(B)

`x^2y - y^2x + xyz`

(C)

`(x - y)(y- z)(z- x)`

(D)

`xyz`

Solution:

` A = [ (x,x^2,1+x^2),(y , y^2 ,1 + y^2),(z ,z^2 ,1 + z^2)]`

` | A| = | (x,x^2,1+x^2),(y , y^2 ,1 + y^2),(z ,z^2 ,1 + z^2)|`

` = | (x - y , (x-y)(x + y) , (x-y)(x + y) ),( y - z , (x-z)(x + z) , (x-z)(x + z) ) ,(z ,z^2 ,1 + z^2)|`

(use operations `R_1 -> R_1 - R_2,R_2 -> R_2 - R_3`)

` = (x - y)(y - z) | (1 , x + y , x + y),(1 , y + z , y + z ),( z , z^2 , 1 + z^2)|`

` = 0 `(since, two rows are identical)
Correct Answer is `=>` (A) `0`
Q 2400534418

Under which of the following condition (s), will
the matrix `A = [ ( 0,0,q),(2,5,1),(8,p,p)]` be singular ?
I. `q =0`
II. `p = 0`
III. `p = 20`
Select the correct answer using the code given below.
NDA Paper 1 2007
(A)

Both I and II

(B)

Only III

(C)

I and III

(D)

Either I or III

Solution:

` A = [ ( 0,0,q),(2,5,1),(8,p,p)]`

I. For `q = 0`,

`A = [ ( 0,0,0),(2,5,1),(8,p,p)] => |A| = 0`

So, A is singular.

II. For `p = 0,`

` A = [ ( 0,0,q),(2,5,5),(8,0,0)]`

`=> |A| = - 40q`

So, `A` is not singular.

III. For `p = 20`,

` A = [ ( 0,0,q),(2,5,1),(8,20,20)]`

` | A| = q | ( 2,5),(8,20)| = 4 - 40 = 0`

So, `A` is singular.

Thus, both I and III are correct.
Correct Answer is `=>` (C) I and III
Q 2420634511

Let `A` be an `m xx n` matrix. Under which one of the
following conditions does `A^(-1)` exists?
NDA Paper 1 2007
(A)

Only `m = n`

(B)

`m = n` and det `(A) != 0`

(C)

`m = n` and del `(A)= 0`

(D)

`m != n`

Solution:

Let `A` be an `m xx n` matrix, then `A^(-1)` will exists, if `m = n`

and del `(A) != 0`.
Correct Answer is `=>` (B) `m = n` and det `(A) != 0`
Q 2450634514

Let `A` and `B` be two matrices of order `'n' xx n`. Let `A`
be non-singular and `B` be singular. Consider the
following statements
I. `AB` is singular.
II. `AB` is non-singular.
III. `A^( -1) B` is singular.
IV. `A^(-1) B` is non-singular.
Which of the above statements is/are correct?
NDA Paper 1 2007
(A)

Only I

(B)

Only II

(C)

Both I and III

(D)

Both II and IV

Solution:

If `A` is non-singular and `B` is singular, then `AB` and `A^(-1)B`

are non-singular because the inverse of `A` will exists when `A` is

non-singular.
Correct Answer is `=>` (D) Both II and IV
 
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