Mathematics Tricks & Tips of matrices For NDA

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Problem based on different Types of Matrices

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 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 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 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 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 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 2160156015

(i) Show that the matrix `A = [ (1 , -1 , 5),( -1 , 2 ,1) ,(5 ,1 ,3)]` is a
symmetric matrix.
(ii) Show that the matrix `A = [ (0 , 1 , -1),( -1 , 0 ,1) ,(1 ,- 1 , 0)]` is
a skew-symmetric matrix.
Class 12 Exercise Q.No. 0

Solution:

(i) For a symmetric matrix `a_(ij) = a_(ji)`

Now, `A = [(1 , -1 , 5),( -1 , 2 ,1) ,(5 ,1 ,3) ], A ' = [ (1 , -1 , 5),( -1 , 2 ,1) ,(5 ,1 ,3) ]`

` a_(21) = - 1 = a_(12) , a_(31) = 5 = a_(13)`

`a_(32) = 1 a_(23) , a_(11) , a_(22) , a_(33)` are `1, 2, 3` respectively.

Hence, `a_(ji) = a_(ij) :.` A is a symmetric matrix.

(ii) For a skew symmetric matrix `a_(ji) = - a_(ij)`

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

`a_(21) = -1 , a_(12) = 1`,

`:. -a_(12) = - 1 ` or `a_(21) = -a_(12)`

`a_(31) = 1 , a_(13) =- 1`,

`:. - a_(13) = 1` or `a_(31) = -a_(13)`

`a_(32) = - 1, a_(23) = 1`,

`:. - a_(32) = - 1` or `a_(32) = - a_(23)`

`a_(11) = 0 , a_(22) = 0, a_(33) = 0`

Hence, A skew - symmetric matrix.

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 1863101045

Choose the correct answer in the following questions:

If `A`, `B` are symmetric matrices of same order then `AB- BA` is a
Class 12 Exercise 3.3 Q.No. 11

(A)

Skew-symmetric matrix

(B)

Symmetric matrix

(C)

Zero matrix

(D)

Identity matrix

Solution:

Now` A'= B, B'= B`

`(AB-BA)' = (AB)'- (BA)'`

`= B'A' -A'B' = BA-AB`

`=-(AB-BA)`

`AB-BA` is a skew-symmetric matrix

Hence, option (1) is correct.

Correct Answer is `=>` (A) Skew-symmetric matrix

Q 1822091831

For the matrix, `A = [(1,5),(6,7)]`, verify that

(i) `(A+ A')` is a symmetric matrix.

(ii) `(A-A')` is a skew-symmetric matrix.
Class 12 Exercise 3.3 Q.No. 8

Solution:

`A = [(1,5),(6,7)] =>A' = [(1,6),(5,7)]`

(i) `A+A' = [(1,5), (6,7) ] + [(1,6),(5,7)] = [ (2,11), (11,14) ]`

Let `Z= [(2,11), (11,14)] => Z' = [(2,11), (11,14) ]`

`:. Z' =Z`

Hence, `Z` or `A+ A' ` is a symmetric matrix.

(ii) `A-A' = [(1,5),(6,7)] - [(1,6),(5,7)] = [(0,-1),(1,0)]`

Let `Z = [(0,-1),(1,0)] => Z' = [(0,1),(-1,0)]`

`=> Z' = - [(0,-1),(1,10)]= -Z`

Hence, `Z` or `A -A'` is a skew-symmetric matrix.

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 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 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 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 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 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`

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 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)`

Algebra Of Matrices

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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 `=>` ()

Transport and Adjoint of a Matrix

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 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 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 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 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 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 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 2640767613

If `A = tt ((cos alpha , sin alpha ),(- sin alpha , cos alpha ))` , find satisfying `0 < pi/2` when `A + A^T = sqrt 2 I_2 ;` where `A^T` is transpose of `A`.
CBSE-12th 2016

Solution:

Consider the given matrix

` A = [ (cos alpha , sin alpha ),(- sin alpha , cos alpha ) ] 0 < alpha < pi/2`

` A + A^T = sqrt 2 I_2`

` [ (cos alpha , sin alpha ),(- sin alpha , cos alpha ) ] + [ (cos alpha , -sin alpha ),( sin alpha , cos alpha ) ] = sqrt2 = [ (1,0) ,(0,1) ]`

` [ ( 2 cos alpha , 0 ),(0, 2 cos alpha) ] = [ ( sqrt2 ,0),(0, sqrt2)]`

` 2 cos alpha = sqrt2`

` cos alpha = sqrt2/2 = 1/sqrt2`

` alpha = pi/4`

Q 2447645583

Suppose `A` and `B` are two square matrices of same order. If `A,B` are symmetric matrices, then `AB-BA` is
EAMCET 2016

(A)

A symmetric matrix

(B)

A skew symmetric

(C)

A scalar matrix

(D)

A triangular matrix

Solution:

`(AB-BA) ^{ T }`

` = (AB ) ^{ T }- (BA) ^{ T }`

`= B ^{ T } A ^{ T }- A ^{ T } B^{ T }`

` =BA-AB`

` =-(AB-BA)`

This shows that `(AB-BA)` is a skew-symmetric matrix

Correct Answer is `=>` (B) A skew symmetric

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`

Correct Answer is `=>` (C) `6`

Inverse of a Matrix

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 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 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 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 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 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 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 2460001815

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

NDA Paper 1 2008

(A)

(B)

(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 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 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 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 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 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 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

Solution of a System of Linear Equations

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 2640867713

For what values of k, the system of linear equations
`x + y + z = 2`
`2x + y – z = 3`
`3x + 2y + kz = 4`
has a unique solution?
CBSE-12th 2016

Solution:

for unique solution `|A| != 0`

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

`C_2 -> C_2 -> C_1 ; C_3 -> C_3 - C_1`

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

expansion along `R_1`

`- (k - 3) - 3 != 0`

` - k + 3 - 3 != 0`

` k != 0`

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 2562123935

The system of linear equations
`lamda x + y + z = 3`
`x - y - 2z = 6`
`- x + y + z = mu` has
WBJEE 2012

(A)

infinite number of solutions for `lamda != -1` and all `mu`.

(B)

infinite number of solutions for `lamda = -1` and `mu = 3`

(C)

no solution for `lamda != -1`

(D)

unique solution for `lamda = -1` and `mu = 3`

Solution:

Augmented matrix `[A : B] = [(1,-1,-2,6),(-1,1,1,mu),(lamda ,1,1,3)]`

`~ [ (1,-1,-2,6),(0,0,-1,mu+6),(lamda +1,0,-1,9)]`

Applying `R_2 -> R_2 + R_1, R_3 -> R_3 + R_1`

` ~ [ (1,-1,-2,6),(0,0,-1,mu+6),(lamda +1,0,0,3-mu)]`

Infinite number of solutions for `lamda = - 1` and

`mu = 3`.

Correct Answer is `=>` (B) infinite number of solutions for `lamda = -1` and `mu = 3`

Q 2366145975

Find the value of `k` for which the system of
equations `kx + 2y = 5` and `3x + y = 1` has no
solution?
NDA Paper 1 2011

(A)

`0`

(B)

`3`

(C)

`6`

(D)

`15`

Solution:

Here, `a_1 = k, a_2 = 3,b_1 = 2, b_2 = 1,c_1 = 5, and c_2 = 1`
For no solution,

`a_1/a_2 = b_1/b_2 ne c_1/c_2`

`=> k/3 = 2/1 => k = 6`

Correct Answer is `=>` (C) `6`

Q 2826767671

If `l + m + n = 0`, then the system of equations `-2x + y + z = l, x - 2y + z = m, x + y- 2z = n` has

(A)

a trivial solution

(B)

no solution

(C)

a unique solution

(D)

infinitely many solutions

Solution:

`Delta = [ (-2 , 1 ,1 ),(1, -2, 1),(1 , 1, -2) ]`

`= -2 (4-1)-1 (-2-1)+ 1+ 2 = 0`

`Delta = |(l , 1,1),(m , -2, 1),(n,1,-2) | = | ( l+m +n , 0 ,0),(m , -2 , 1),(n , 1 ,-2) |`

` [ R_1 -> R_1 + R_2 + R_3 ]`

`= 0 [ ∵ l + m + n = 0 ]`

Similarly, `Delta_2 = Delta_3 = 0`

Hence, the given system of equations

has infinitely many solutions.

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

Q 2502145038

Consider the system of equations `x + y + z = 0, alphax + betay + gamma z = 0 , alpha^2 x + beta^2 y + gamma^2 z = 0` Then, the system of equations has
WBJEE 2013

(This question may have multiple correct answers)

(A) a unique solution for all values of `alpha,beta` and `gamma`.

(B) infinite number of solutions, if any two of `alpha,beta. gamma` are equal.

(D) more than one, but finite· number of solutions depending on values of `alpha, beta` and `gamma`.

Solution:

Given system of equations is

`x + y + z = 0,`
`alphax + betay + gamma z = 0 ,`

` alpha^2 x + beta^2 y + gamma^2 z = 0`

The coefficient matrix `A [(1 ,1 ,1) , (alpha , beta , gamma) , ( alpha^2 , beta^2 , gamma^2)]`

Now `|A| = | [(1 ,1 ,1) , (alpha , beta , gamma) , ( alpha^2 , beta^2 , gamma^2)] |`

` = (alpha-beta)(beta-gamma)(gamma-alpha)`

(i) The system of equations has a unique solution, if `alpha,beta` and `gamma` are distinct
i.e., `|A| ne 0`
(ii) The system of equations has infinite number of solutions, if any two of `alpha, beta` and `gamma` are
equal.
i.e., ` |A| = 0`

Correct Answer is `=>` (B)

Q 2640867713

For what values of k, the system of linear equations
`x + y + z = 2`
`2x + y – z = 3`
`3x + 2y + kz = 4`
has a unique solution?
CBSE-12th 2016

Q 2435391262

The values of `x, y` and `z` for the system of
equations `x + 2y + 3z = 6, 3x - 2y + z = 2` and
`4x + 2y + z = 7` are respectively
UPSEE 2014

(A)

1, 1, 1

(B)

1, 2, 3

(C)

1, 3, 2

(D)

2, 3, 1

Solution:

The given system of equations is

`x + 2y + 3z = 6` ....(i)

`3x - 2y + z = 2` ....(ii)

and `4x + 2y + z = 7` ......(iii)

Here, `A = [ (1,2,3),(3 ,-2,1),(4,2,1)] , B = [(6),(2),(7)] ` and ` X = [ (x),(y),(z)]`

` :. |A| = | (1,2,3),(3 ,-2,1),(4,2,1)| `

`= 1 (-2 -2)-2(3-4) + 3(6+ 8)`

`= - 4 + 2 + 42 = 40`

Now, `adj A = [ (-4,4,8),(1,-11,8),(14,6,-8)]`

`:. = A^(-1) = 1/(|A|) adj A = 1/(40) [ (-4,4,8),(1,-11,8),(14,6,-8)]`

Now, ` X = A^(-1)B = 1/(40) [ (-4,4,8),(1,-11,8),(14,6,-8)] [(6),(2),(7)]`

` = 1/(40) [ (-24,+8,+56) ,(6,-22,+56),(84,+12,-56)] = 1/(40) [(40),(40),(40)]`

` => [(x),(y),(z)] = [(1),(1),(1)]`

` => x = 1, y = 1 , z = 1`

Correct Answer is `=>` (A) 1, 1, 1