Mathematics Previous year question of Determinants for NDA

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Previous year question of Determinants for NDA

Previous year Determinant question for NDA

Q 2763391245

If `a != b != c`, then one value of `x` which satisfies the equation
` | ( 0 , x - a , x - b),( x + a , 0 , x - c), ( x + b , x + c , 0) | = 0` is given by
NDA Paper 1 2017

(A)

`a`

(B)

`b`

(C)

`c`

(D)

`0`

Solution:

`-(x-a) (0-(x-c)(x+b))+(x-b) (x+0)(x+c)=0`

`(x-a) (x+b) (x-c) +(x+0) (x-b) (x+c) =0`

`x=0` satisfies the equation.

Correct Answer is `=>` (D) `0`

Q 2763491345

What is the value of the determinant
` | (1,1,1),(1,1+xyz,1),(1,1,1+xyz) | ` ?
NDA Paper 1 2017

(A)

` 1 + x + y + z`

(B)

`2 xyz`

(C)

`x^2y^2z^2`

(D)

`2x^2y^2z^2`

Solution:

` | (1,1,1),(1,1+xyz,1),(1,1,1+xyz) | `

`R _1 -> R_2 - R_1`

`R_2 -> R_3 -R_2`

`| (0, xyz , 0 ), ( 0 , -xyz , xyz ), ( 1, 1, 1 + xyz) |`

`= x yz ( x yz )`

`= x^2 y^2 z^2 `

Correct Answer is `=>` (C) `x^2y^2z^2`

Q 2773491346

If ` | (x , y ,0),(0 , x ,y),(y , 0 ,x) | = 0` then which one of the following is correct?
NDA Paper 1 2017

(A)

`x/y ` is one of the cube roots of unity

(B)

`x` is one of the cube roots of unity

(C)

`y` is one of the cube roots of unity

(D)

`x/y ` is one of the cube roots of `-1`

Solution:

` | (x , y ,0),(0 , x ,y),(y , 0 ,x) | = 0 `

`x (x^2 ) -y ( 0 - y^2 ) = 0`

`=> x^3 + y^3 = 0`

`:. x^3/y^3 =-1`

`=> x/y =(-1)^(1/3)`

Correct Answer is `=>` (D) `x/y ` is one of the cube roots of `-1`

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 c ?
NDA Paper 1 2016

(A)

-1

(B)

(C)

(D)

Solution:

`ax^3 + bx^2 +cx +d = | (x+1,2x,3x), (2x+3,x+1,x), (2-x , 3x +4, 5x -1) |`

Different with respect to `x` & put `x=0`

`c = | (1,2,3), (3,1,0), (2,4,-1) | + | (1,0,0),(2,1,1),(2,4,-1) | + | (1,0,0 ), ( 3,1,0 ), ( -1, 3 ,5 ) |`

`=1 (-1) -2 (-3) +3 (10) + (-5) +1 (5)`

`= -1-5+30 =24`

Correct Answer is `=>` (B) 24

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)

(B)

(C)

(D)

Correct Answer is `=>` (B) 63

Q 2781156927

Which of the following are correct in respect of the system of equation `x + y + z = 8 , x -y + 2z = 6` and `3x - y + 5 z = k ?`

1. The have no solution ,if `k = 25`
2. They have infinitely many solution , if k =20
3. They have unique solution , if k = 25.

Select the correct answer using 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:

`Delta = | (1,1,1), (1,-1,2), (3,-1,5) | `

`Delta = 1 (-3) -1 (-1) +1 (2) =0`

`Delta_1 = | (8,1,1), (6,-1,2), (k,-1,5) |`

`Delta_1 = 8 (-3) -1 (30-2k) +1 (-6+k)`

`Delta_1 =3 [- 8 -10 -2 + k]`

`Delta_1 = 3 [-20+k]`

`Delta_1 =0` (for `k=20 `)

`Delta_2 = | (1, theta ,1 ), (1,6,2),(3,k,5) |`

`= -k +20`

Similarly

`Delta_2 =0` for `(k=20)`

Similarly `Delta_3 = 0 ` (for `k=20`)

So `Delta = 0 , ` & `Delta_1 =Delta_2 =Delta_3 =0` (for `k=20`)

` :.` system has infinite no. of solution.

Hence `2` is corect

1. If `k= 15`

`Delta = 0 , Delta_1 ne 0 , Delta_2 ne 0, Delta_3 ne 0`

`:.` No. solution for `k=15`

Correct Answer is `=>` (A) 1 and 2 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 = 9`

(C)

`lamda = 9` and ` mu =5 `

(D)

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

Solution:

for infinite solution `Delta = 0` & `Delta_1 =0`

`Delta = | (2,3,5), (7,3,-2), (2,3,lambda) | = 0`

`=> 2 (3 lambda + 6) -3 (7 lambda + 4) + 5 (21 -6) =0`

`lambda = 5`

`Delta_1 = | (9,3,5), (8,3,-2), (mu,3,5) | =0`

`=> 9 (35+6) -3 (40 +2 mu) + 5 (24 -3 mu) =0`

`=> mu = 9`

Correct Answer is `=>` (B) `lamda = 5 ` and ` mu = 9`

Q 2701667528

(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`

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

Q 2106380278

Consider the function `f(x) = |(x^(3), sinx , cos x) , (6 ,-1 , 0) , (p ,p^(2) ,p^(3))|`
where `p` is a constant.

What is the value of `f'(0)?`
NDA Paper 1 2016

(A)

`p^(3)`

(B)

`3p^(3)`

(C)

`6p^(3)`

(D)

` -6p^(3)`

Solution:

Given `f(x) = |(x^(3), sinx , cos x) , (6 ,-1 , 0) , (p ,p^(2) ,p^(3))|`

where, p is a constant.

On differentiating both sides w.r.t. x, we get

` f'(x) = |(3x^(3), cosx , -sin x) , (6 ,-1 , 0) , (p ,p^(2) ,p^(3))| + |(x^(3), sinx , cos x) , (0 ,0 , 0) , (p ,p^(2) ,p^(3))|`

` +|(x^(3), sinx , cos x) , (6 ,-1 , 0) , (0,0,0)|`

` => f'(x) = |(3x^(2), cosx , -sin x) , (6 ,-1 , 0) , (p ,p^(2) ,p^(3))| + 0 + 0`

` => f'(0) = |(0, cos0 , -sin 0) , (6 ,-1 , 0) , (p ,p^(2) ,p^(3))|`

` => f'(0) = |(0, 0 , 0) , (6 ,-1 , 0) , (p ,p^(2) ,p^(3))| = (-1) ( 6 p^(3) - 0) = - 6p^(3)`

Correct Answer is `=>` (D) ` -6p^(3)`

Q 2166480375

Consider the function `f(x) = |(x^(3), sinx , cos x) , (6 ,-1 , 0) , (p ,p^(2) ,p^(3))|`
where p is a constant.

What is the value of p for which `f' '(0) = 0?`
NDA Paper 1 2016

(A)

` - 1/6` or `0`

(B)

` - 1` or `0`

(C)

` - 1/6` or `1`

(D)

` - 1` or `1`

Solution:

From question 49, `f'(x) = |(x^(3), sinx , cos x) , (6 ,-1 , 0) , (p ,p^(2) ,p^(3))|`

Again, differentiating both sides, w.r.t. x, we get

=>

` f''(x) = |(6x, -sinx , -cos x) , (6 ,-1 , 0) , (p ,p^(2) ,p^(3))| + |(3x^(2), cosx , -sin x) , (0 ,0 , 0) , (p ,p^(2) ,p^(3))|`

`+ |(3x^(2), cosx , -sin x) , (6 ,-1 , 0) , (0,0,0)|`

` => f''(x) = |(6x, -sinx , -cos x) , (6 ,-1 , 0) , (p ,p^(2) ,p^(3))| + 0+ 0`

Since, we have `f" (0) = 0`

` => f''(0) = |(0, -sinx , cos x) , (6 ,-1 , 0) , (p ,p^(2) ,p^(3))| = 0`

` => |(0,0,1) , (6 ,-1 , 0) , (p ,p^(2) ,p^(3))| = 0`

` => 6p^(2) + p = 0`

`=> p(6p + 1) =0`

` => p = 0` or `6 p + 1 = 0`

` => p =0` or `p = - 1/6`

`p = - 1/6` or `0`

Correct Answer is `=>` (A) ` - 1/6` or `0`

Q 2107412388

Which of the following determinants have value 'zero'?

`1. | ( 41,1,5) , ( 79,7,9),(29,5,3)|`

` 2. | ( 1,a,a+b) , ( 1,b,c+a),(1,c,a+b)|`

`3. | ( 0,c,b) , ( -c,0,a),(-b,-a,0)|`

Select the correct answer using the code given below.
NDA Paper 1 2016

(A)

`1` and `2`

(B)

`2` and `3`

(C)

`1` and `3`

(D)

`1, 2` and `3`

Solution:

l.Now, ` | ( 41,1,5) , ( 79,7,9),(29,5,3)| = | ( 1,1,5) , ( 7,7,9),(5,5,3)| =0 quad [C_(1) -> C_(1) -8C_(3)]`

[`∵` two columns of determinant are same, then value of
determinant is zero.]

`2. | ( 1,a,a+b) , ( 1,b,c+a),(1,c,a+b)| = | ( 1,a,a+b+c) , ( 1,b, a+b+c ),(1,c,a+b+c)|`

` [ C_1 -> C_2 + C_3]`

` = (a+ b +c) | ( 1,a,1) , ( 1,b,1),(1,c,1)| = 0`

[`∵` two columns cf determinant are same, then value of
determinant is zero.]

`3. | ( 0,c,b) , ( -c,0,a),(-b,-a,0)|`

`= 0(0 + a)- c(0 + ab) + b(ac- 0)`

[expanding with respect to `R_(1)`]

`=0- c(ab) + b(ac)= -abc+ abc =0`

Hence, `1, 2` and `3` determinants have value 'zero'.

Correct Answer is `=>` (D) `1, 2` and `3`

Q 2167712685

The system of linear equations `kx + y + z = 1`,
`x + ky + z = 1` and `x + y + kz = 1` has a unique solution
under which one of the following conditions?
NDA Paper 1 2016

(A)

`k != 1` and `k != - 2`

(B)

`k != 1` and `k != 2`

(C)

`k != -1` and `k != - 2`

(D)

`k != -1` and `k != 2`

Solution:

Given linear equations

`kx + y + z = 1,`

`x + ky + z = 1` and `x + y + kz = 1`

For unique solution, ` | ( k,1,1) ,(1,k,1) ,(1,1,k)| != 0`

` => k(k^(2) - 1) -1 (k -1) +1 (1 - k) != 0`

`=> k(k + 1) (k -1)- (k- 1)- (k- 1) != 0`

`=> k(k- 1) (k + 1) - 2(k- 1) != 0`

`=> (k - 1 )[k^(2) + k - 2] != 0`

` => (k -1) (k- 1) (k + 2) != 0`

` => (k - 1)^(2) (k + 2) != 0`

` => k != 1` and `k != -2`

Correct Answer is `=>` (A) `k != 1` and `k != - 2`

Q 2200267118

Consider the following statements in respect of the
determinant `|(cos^(2) \ \alpha/2 , sin ^(2)\ \ alpha/2) ,(sin ^(2)\ \ beta/2 , cos ^(2) \ \beta /2)|`
where `alpha , beta` are complementary angles.
1. The value of the determinant is ` 1/sqrt(2) cos((alpha - beta )/2)`
2. The maximum value of the determinant is ` 1/sqrt(2)`.

Which of the above statement(s) is/are correct?

NDA Paper 1 2015

(A)

Only `1`

(B)

Only `2`

(C)

Both `1` and `2`

(D)

Neither `1` nor `2`

Solution:

1 . We have, ` Delta =``|(cos^(2) \ \alpha/2 , sin ^(2)\ \ alpha/2) ,(sin ^(2)\ \ beta/2 , cos ^(2) \ \beta /2)|`

` = cos^(2) \ alpha/2 quad cos^(2) \ beta /2 - sin ^(2) \ alpha/2 quad sin ^(2) \ beta/2`

` = cos (alpha/2) quad cos (beta /2) + sin (alpha/2) quad sin (beta/2) `

` = cos (alpha/2) quad cos (beta /2) - sin (alpha/2) quad sin (beta/2) `

` = cos ((alpha - beta )/2 ) cos 45^(0)`

` = 1/sqrt(2) cos ((alpha - beta )/2 )`

2. The maximum value of `cos \ (alpha - beta )/2` is `1`.

`:.` The maximum value of the determinant is ` 1/sqrt(2)`

Hence, both statements are correct.

Correct Answer is `=>` (C) Both `1` and `2`

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 1638823702

The value of ` |(1,1,1),(1 ,1+x,1),(1,1,1+y)|` is

NDA Paper 1 2015

(A)

`x + y`

(B)

`x - y`

(C)

`xy`

(D)

`1 +x+ y `

Solution:

` |(1,1,1),(1 ,1+x,1),(1,1,1+y)|`

` = |(1,1,1),(0 ,x,0),(0,0,y)|`

`= 1 (xy - 0) = xy` [emplaning along first column]

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

Q 2211601520

If `a, b` and `c` are real numbers, then the value of the
determinant
` |(1-a,a-b-c ,b+c),(1-b, b-c-a,c+a),(1-c,c-a-b,a+b) |` is
NDA Paper 1 2015

(A)

`0`

(B)

`(a- b)(b- c)(c- a)`

(C)

`(a+ b + c)^2`

(D)

`(a+ b + c)^3`

Solution:

We have,` Delta = |(1-a,a-b-c ,b+c),(1-b, b-c-a,c+a),(1-c,c-a-b,a+b)|`

Applying `C_2 -> C_2 + C_3` , we get

` Delta = |(1-a,a ,b+c),(1-b, b,c+a),(1-c,c,a+b)|`

Now, applying `C_1 -> C_1 + C_2, C_3 -> C_3 + C_2` and taking

common `a + b + c` from `C_3` , we get

` Delta = (a+b+c) | (1,a,1),(1,b,1),(1,c,1)|`

`∵ C_1` and `C_3` are identical.

Hence, the determinant is `0`.

Correct Answer is `=>` (A) `0`

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 1659278114

`A(3, 4)` and `B(5,- 2)` are two points and `P` is a point such
that `PA =PB`. If the area of `Delta PAB` is `10` sq units, then
what are the coordinates of `P`?
NDA Paper 1 2014

(A)

Only `(1, 0)`

(B)

Only `(7, 2)`

(C)

Either `( 1, 0)` or `(7, 2)`

(D)

Neither `(1, 0)` nor `(7, 2)`

Solution:

We have, `A(3, 4)` and `8(5, -2)`

Let `P(x, y)`

Given that, `PA = PB`

`=> PA^2 = PB^2`

`=> (x - 3)^2 + (y - 4)^2 = (x - 5)^2 + (y + 2)^2`

`=> x^2 - 6x + 9 + y^2 - 8y + 16`

`= x^2 - 10 x + 25 + y^2 + 4y + 4`

`=> 4x - 12y = 4`

`=> x - 3y = 1` .....(1)

∵ Area of `Delta PAB = 10`

` :. 1/2 | (x ,y ,1),(3,4,1),(5 ,-2,1) | = pm 10`

` => x(4 + 2)- y (3- 5) + 1(-6-20) = pm 20`

`=> 6x + 2y - 26 = pm 20`

`=> 6x + 2y -26=20`

or `6x + 2y -26= -20`

`=> 6x + 2y = 46` ......(2)

or `6x + 2y = 6 ` .........(3)

On solving Eqs. (i) and (ii), we get

`x =7, y = 2`

Similarly, solving Eqs. (i) and (iii), we get

`x = 1, y = 0`

Hence, coordinates of Pare `(7, 2)` or `(1, 0)`.

Correct Answer is `=>` (C) Either `( 1, 0)` or `(7, 2)`

Q 1772334236

If `a != b != c` are all positive, then the value of the
determinant ` | ( a,b,c),( b, c, a),( c,a,b) |` is
NDA Paper 1 2014

(A)

non - negative

(B)

non - positive

(C)

negative

(D)

positive

Solution:

` | ( a,b,c),( b, c, a),( c,a,b) | = | ( a+b+c, b, c) ,(a+b+c ,c ,a) ,(a+b+c ,a, b) |`

`(∵ C_1 -> C_1 + C_2 + C_3)`

` = (a + b + c) | ( 1,b,c),( 1, c, a),( 1,a,b) |`

[on taking `(a + b + c)` common from `C_1`]

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

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

` = (a+ b + c) [-(a^2 + b^2 + c^ 2 - ab - bc - ca)]`

` = - 1/2 (a+ b +c) [(a- b)^2 + (b- c )^2 + (c - a)^2]`

`=` Negative value

Correct Answer is `=>` (C) negative

Q 1722434331

If ` | ( 6i, -3i, 1),(4, 3i, -1),(20, 3,i) | = x + iy ,` where` i = sqrt(-1),`

then what is `x` equal to?

NDA Paper 1 2014

(A)

`3`

(B)

`2`

(C)

`1`

(D)

`0`

Solution:

` | ( 6i, -3i, 1),(4, 3i, -1),(20, 3,i) | `

` = 6i [3i^ 2 + 3] + 3i [4i + 20] + 1 [12- 60 i]`

` = 6i [-3 + 3] + 12i^2 + 60i + 12- 60i`

` = - 12 + 12 = 0 = x + iy`

`:. x = 0`

Correct Answer is `=>` (D) `0`

Q 1723378241

One of the roots of ` |(x+a, b, c),(a, x +b, c),(a ,b ,x+c) | =0` is
NDA Paper 1 2014

(A)

`abc`

(B)

`a+ b + c`

(C)

`-(a + b + c)`

(D)

`-abc`

Solution:

Given that,

` |(x+a, b, c),(a, x +b, c),(a ,b ,x+c) | =0`

Use operation `C_1 -> C_1 + C_2 + C_3` ,

`=> |(x+a+b+c, b, c),(x+a+b+c, x +b, c),(x+a+b+c ,b ,x+c) | =0`

` => (x + a + b + c) |(1, b, c),(1, x +b, c),(1 ,b ,x+c) | =0`

Again use operation,

`R_2 -> R_2 - R_1, R_3 -> R_3 - R_1`

` => (x + a + b + c) |(1, b, c),(0, x , 0),(0 ,0,x) | =0`

Now, expanded along `C_1`,

`(x + a + b + c ) . 1 . (x^2) = 0`

`=> x . (x + a+ b + c) = 0`

`=> x + a + b + c = 0`

`=> x = - (a + b + c)`

Correct Answer is `=>` (C) `-(a + b + c)`

Q 1783378247

If any two adjacent rows or columns of a determinant are
interchanged in position, the value of the determinant
NDA Paper 1 2014

(A)

Becomes zero

(B)

Remains the same

(C)

Changes its sign

(D)

ls doubled

Solution:

If any two adjacent rows or column of a

determinant are interchanged in position. The value of the

determinant changes its sign.

e.g.,

(i) Let ` Delta = |(1,2),(2,3)| = 3-4 = -1`

If `R_1 ↔R_2`

` Delta' = |(2,3),(1,2)| = 4- 3 = 1 = - Delta`

` => Delta' = - Delta`

(ii) ` Delta = |(1,2),(2,3)| = 3-4 = -1`

` Delta' = |(2,1),(3,2)| = 4- 3 = 1 = - Delta`

` => Delta' = - Delta`

Correct Answer is `=>` (C) Changes its sign

Q 1732534432

If ` | (a , b ,0),(0,a ,b),(b ,0 ,a) | = 0` then which one of the following is

correct?
NDA Paper 1 2014

(A)

`a/b` is one of the cube roots of unity

(B)

`a/b` is one of the cube roots of `-1`

(C)

`a` is one of the cube roots of unity

(D)

`b` is one of the cube roots of unity

Solution:

`| (a , b ,0),(0,a ,b),(b ,0 ,a) |`

` = a [a^2 - 0] - b [ - b^2 ] + 0`

` = a^3 + b^3 = 0`

` => a^3 = -b^3`

` => (a/b)^3 = -1`

Hence, `a/b` is one of the cube roots of `-1`.

Correct Answer is `=>` (B) `a/b` is one of the cube roots of `-1`

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 2420145011

What is the value of the determinant

`|(1,bc, a (b+ c)),(1 , ca , b(c + a)) ,(1 , ab , c(a + b)) |` ?
NDA Paper 1 2013

(A)

`0`

(B)

`abc`

(C)

`ab + bc + ca`

(D)

`abc(a + b +c)`

Solution:

Let ` Delta = |(1,bc, a (b+ c)),(1 , ca , b(c + a)) ,(1 , ab , c(a + b)) |`

Use operation `R_2 -> R_2 - R_1` and `R_3 -> R_3 - R_1`

` Delta = |(1,bc, ab + ac),(0 , c(a-b) , (b-a) c) ,(0 , b(a - c) , b(c - a)) |`

Expanding along `C_1`

` Delta = bc (a- b) (c -a)- bc (a - c) (b - a)`

` = bc (a- b) (c -a)- bc (a- b) (c -a)= 0`

Alternate Method

Use operation `C_3 -> C_2 + C_3`

` Delta = | ( 1, bc ,ab+ac+bc),(1 , ca ,bc+ba+ca),(1 , ab , ca+bc+ab)|`

Taking common `(ab +bc+ ca)` in `C_3`, we get,

` Delta = (ab + bc + ca) | ( 1, bc , 1),(1 , ca , 1),(1 ,ab, 1)|` (since, `C_1` and `C_3` are identicals)

` = (ab + bc + ca) xx 0 = 0`

Correct Answer is `=>` (A) `0`

Q 2420245111

What is the value of the minor of the element `9` in
the determinant ` | ( 10, 19 , 2),(0,13,1),(9, 24 , 2) |` ?
NDA Paper 1 2013

(A)

`- 9`

(B)

`-7`

(C)

`7`

(D)

`0`

Solution:

Let ` Delta = | ( 10, 19 , 2),(0,13,1),(9, 24 , 2) |`

` :. ` Minor of element `9 = | (19 , 2),(13 , 1)| = 19 - 26 = -7`

Correct Answer is `=>` (B) `-7`

Q 2480345217

The roots of the equation ` | ( 1 , t - 1 ,1 ),(t - 1 , 1 , 1),(1 , 1 , t - 1) | = 0` are
NDA Paper 1 2013

(A)

`1, 2`

(B)

`-1 ,2`

(C)

`1,-2`

(D)

`-1, -2`

Solution:

Given that,

` | ( 1 , t - 1 ,1 ),(t - 1 , 1 , 1),(1 , 1 , t - 1) | = 0`

Applying operation `C_1 -> C_1 + C_2 + C_3`,

` | ( t + 1 , t - 1,1),( t + 1 ,1 ,1 ),( t + 1 ,1 , t -1) | = 0 => (t + 1) | ( 1 , t-1, 1),(1 , 1,1),(1,1,t-1) | = 0`

Applying operation `R_2 -> R_2 - R_1` and ` R_3 -> R_3 - R_1`,

` ( t + 1) | (1, t-1, 1),(0,2-t, 0),(0,2-t, t-2)| = 0`

Expanding along `C_1`,

` (t + 1) | (2-t,0),(2-t, t-2) | = 0`

`=> (t+1)(2-t)(t-2) = 0 => (t + 1) (t -2)^2 = 0`

`:. t = -1 , 2`

Correct Answer is `=>` (B) `-1 ,2`

Q 2400445318

The value of the determinant ` | (m , n ,p),(p,m,n),(n,p,m) |`
NDA Paper 1 2013

(A)

is a perfect cube

(B)

is a perfect square

(C)

has linear factor

(D)

is zero

Solution:

Let `Delta = | (m , n ,p),(p,m,n),(n,p,m) |`

Applying operation `C_1 -> C_1 + C_2 + C_3`,

` Delta = | ( (m+n+p), n, p),((m+n+p), m, n),((m+n+p) , p, m)| =( m+n+p) | (1 , n ,p),(1, m ,n),(1 ,p ,m) |`

Applying operation `R_2 -> R_2 - R_1R_3 -> R_3 - R_1`,

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

`= (m+ n + p)[(m-n)(m- p)-(n- p)(p- n)]`

`= (m+ n + p)(m^2-mn- p m+ pn+ p^2 + n^2- 2pn)`

`= (m+ n + p)(m^2+ n^2 + p^2 -mn-np- p m)`

` = 1/2 (m+ n+ p)[(m-n)^2+ (n- p)^2+ (p- m)^2] =` A linear factor

Correct Answer is `=>` (C) has linear factor

Q 2420167911

If `D` is determinant of order `3` and `D'` is the
determinant obtained by replacing the elements of
`D` by their cofactors, then which one of the
following is correct?
NDA Paper 1 2013

(A)

`D' = D^2`

(B)

`D' = D^3`

(C)

`D' = 2D^2`

(D)

`D' = 3D^3`

Solution:

Given that, `D` is determinant of order `3` and `D'` is the

determinant obtained by replacing the elements of `D` by their

cofactors.

`:. D' =` Cofactor of `D`

` => |D'| = | text(Cofactor of ) D |`

`=> |D'| = |adj D |`

`=> |D'| = |D|^(3-1)`

`=> |D'| = |D|^2`

`:. D' = D^2`

Correct Answer is `=>` (A) `D' = D^2`

Q 2450167914

The value of the determinant ` | (x^2,1,y^2 + z^2),(y^2,1,z^2 + x^2),(z^2,1,x^2+y^2)|` is
NDA Paper 1 2012

(A)

`0`

(B)

`x^2 + y^2 + z^2`

(C)

`x^2 + y^2 + z^2 + 1`

(D)

None of these

Solution:

Let `Delta = | (x^2,1,y^2 + z^2),(y^2,1,z^2 + x^2),(z^2,1,x^2+y^2) |`

Use operation `C_3 -> C_2 + C_1`,

`Delta = | (x^2,1,x^2 + y^2 + z^2),(y^2,1,x^2 + y^2 + z^2),(z^2,1,x^2 + y^2 + z^2) |`

Taking `(x^2 + y^2 + z^2)` common from `C_3`.

` Delta = (x^2 + y^2 + z^2) | (x^2,1,1),(y^1,1,1),(z^2,1,1) |`

Since, `C_2` is identical as ` C_3`.

`:. Delta = (x^2 + y^2 + y^2) · 0`

` = 0`

Correct Answer is `=>` (A) `0`

Q 2410167919

What is the value of ` | (-a^2 ,ab , ac),(ab , -b^2 , bc),(ac, bc , -c^2) |` ?
NDA Paper 1 2012

(A)

`4abc`

(B)

`4a^2bc`

(C)

`4a^2b^2c^2`

(D)

`- 4a^2b^2c^2`

Solution:

Let ` Delta = | (-a^2 ,ab , ac),(ab , -b^2 , bc),(ac, bc , -c^2) |`

Taking common `a, b` and `c` from rows `R_1 , R_2` and `R_3` ,respectively,

` Delta = abc |(-a,b,c),(a,-b,c),(a,b,-c) |`

Again, taking common `a, b` and `c` from columns `C_1 , C_2` and `C_3`, respectively,

` Delta = a^2b^2c^2 | (-1,1,1),(1,-1,1),(1,1,-1)|`

` = a^2b^2c^2 [ -1 (1 -1)- 1 (-1-1) + 1 (1 + 1)]`

`= a^2b^2c^2 (0 + 2 + 2) = 4a^2b^2c^2`

Correct Answer is `=>` (C) `4a^2b^2c^2`

Q 2410178010

If two rows of a determinant are identical, then
what is the value of the determinant?
NDA Paper 1 2012

(A)

`0`

(B)

`1`

(C)

`-1`

(D)

Can be any real value

Solution:

By property of. determinant, if two row/ column of a

determinant are identical to each other, then the value of

determinant should be zero.

e.g., ` |(a,b,c),(a,b,c),(x,y,z)| = |(a,a,x),(b,b,y),(c,c,z)| = 0`

Correct Answer is `=>` (A) `0`

Q 2440178013

If ` | (8,-5,1),(5,x,1),(6,3,1)| = 2` then what is the value of `x`?
NDA Paper 1 2012

(A)

`4`

(B)

`5`

(C)

`6`

(D)

`8`

Solution:

Given that, `| (8,-5,1),(5,x,1),(6,3,1)| = 20`

Using operations `R_2 -> R_2 - R_1` and `R_3 -> R_3 - R_1`,

`| (8,-5,1),(-3,x+5,0),(-2,8,0)| = 2`

Expanding along `C_3`,

`1·(-24 + 2x + 10) = 2 => 2x - 14 = 2`

`=> 2x = 16 => x = 8`

Correct Answer is `=>` (D) `8`

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 2460178015

The roots of the equation ` | (x,alpha,1),(beta,x,1),(beta, gamma ,1)| = 0` are
independent of

NDA Paper 1 2011

(A)

`alpha`

(B)

`beta`

(C)

`gamma`

(D)

`alpha,beta` and `gamma`

Solution:

Given, ` | (x,alpha,1),(beta,x,1),(beta, gamma ,1)| = 0`

(use operations `R_2 -> R_2 - R_1` and `R_3 -> R_3 - R_1`)

` => | (x,alpha,1),(beta-x,x-alpha,1),(beta-x, gamma-alpha ,1)| = 0`

Expand with respect to `C_3`,

`(beta - x)(gamma - alpha)- (x - alpha)(beta - x) = 0`

`=> (beta- x) {(-alpha + gamma)- (x- alpha)} = 0`

` => (beta - x) (-alpha + gamma - x + alpha ) = 0`

` => (beta - x)(gamma - x) = 0`

` x = beta , gamma`

So, the roots of the given equation are independent of `alpha .`

Correct Answer is `=>` (A) `alpha`

Q 2410178019

What is the value of the determinant
` | (a - b,b+c,a),(b-c,c+a,b),(c-a,a+b,c)|` ?
NDA Paper 1 2011

(A)

`a^3 + b^3 + c^3`

(B)

`3bc`

(C)

`a^3 + b^3 + c^3 - 3abc`

(D)

`0`

Solution:

` | (a - b,b+c,a),(b-c,c+a,b),(c-a,a+b,c)|`

` = | (a - b,b+c,a+ b+ c),(b-c,c+a,a+ b+ c),(c-a,a+b,a+ b+ c)|` (use operation `C_3 -> C_3 + C_2`)

` = (a + b + c ) | (a - b,b+c,1),(b-c,c+a,1),(c-a,a+b,1)|`

[take `(a + b + c)` common from `C_3`]

` = (a + b + c ) | (a - b,b+c,1),(2b-a-c,a-b,0),(b+c-2a,a-c,0)|`

(use `R_2 -> R_2 - R_1` and `R_3 -> R_3 - R_1`)

Expand with respect to `C_3`.

`=(a+ b + c) {(a- c) (2b- a- c)- (a- b) (b + c- 2a)}`

`= (a+ b +c) (2ab - a^2 - ac- 2bc + ac + c^2`

`- ab- ac + 2a^2 + b^2 + bc- 2ab)`

`=(a+ b +c) (a^2 + b^2 + c^2 - ab - bc - ca)`

`= (a^3 + b^3 + c^3 - 3abc)`

Correct Answer is `=>` (C) `a^3 + b^3 + c^3 - 3abc`

Q 2420278111

If `|(p,-q,0),(0,p,q),(q,0,p)| = 0`then which one of the following is correct?
NDA Paper 1 2011

(A)

p is one of the cube roots of unity

(B)

q is one of the cube roots of unity

(C)

p/q is one of the cube roots of unity

(D)

None of the above

Solution:

` |(p,-q,0),(0,p,q),(q,0,p)| = 0`

Expand with respect to `R_1`,

`p(p^2 - 0) + q (0 - q^2) + 0 = 0 => p^3 - q^3 = 0`

`=> (p-q)(p^2 + q^2 + pq)= 0`

`=> p - q = 0` and `p^2 + q^2 + pq = 0`

`=> p = q` and `p^2/q^2 + 1 + (pq)/q^2 = 0`

`=> ( p/q) = 1` and `( p/q)^2 + ( p/q) + 1 = 0`

We conclude that `( p/q)` is one of the cube roots of unity.

Correct Answer is `=>` (C) p/q is one of the cube roots of unity

Q 2410278119

If `a^(-1) + b^(-1) + c^(-1) = 0` such that

`|(1+a,1,1),(1,1+b,1),(1,1,1+c)| = lamda` then what is `lamda` equal to?
NDA Paper 1 2011

(A)

`-abc`

(B)

`abc`

(C)

`0`

(D)

`1`

Solution:

Given, `a^(-1) + b^(-1) + c^(-1) = 0` ......(i)

`|(1+a,1,1),(1,1+b,1),(1,1,1+c)| = lamda`

Expand with respect to `R_1`,

`(1 + a) {(1 + b) (1 +c) - 1}- 1 {1 + c- 1} + 1 {1- 1- b} = lamda`.

`=> (1 + a) {b + c + bc} - c - b = lamda`.

`=> b + c + bc + ab + ac + abc - c - b = lamda`.

`=> bc + ab + ac + abc = lamda`

`=> abc (1/a + 1/b + 1/c) + abc = lamda`

`=> abc {(a^(-1) + b^(-1) + c^(-1)) + 1} = lamda`

`=> abc (0 + 1) = lamda` [from Eq. (i)]

`:. lamda = abc`

Correct Answer is `=>` (B) `abc`

Q 2420378211

What is the value of the determinant
`|(x+1 ,x+2,x+4),(x+3 ,x+5,x+8),(x+7 ,x+10,x+14)|`?
NDA Paper 1 2011

(A)

`x + 2`

(B)

`x^2 + 2`

(C)

`2`

(D)

`-2`

Solution:

`|(x+1 ,x+2,x+4),(x+3 ,x+5,x+8),(x+7 ,x+10,x+14)|`

(use operations `C_2 -> C_2 - C_1 ; C_3 -> C_3 - C_1`)

` = |(x+1 ,1,3),(x+3 ,2,5),(x+7 ,3,7)| = |(x+1 ,1,3),(2 ,1,2),(6 ,2,4)|`

(use operations `R_3 -> R_3 - R_1 ; R_2 -> R_2 - R_1`)

`= (x + 1)(0) - 1(8 -12) + 3(4- 6)`

`= 4 - 6 = -2`

Correct Answer is `=>` (D) `-2`

Q 2450378214

If `5` and `7` are the roots of the equation
`|(x,4,5),(7,x,7),(5,8,x)| =0`, then what is the third root?
NDA Paper 1 2011

(A)

`-12`

(B)

`9`

(C)

`13`

(D)

`14`

Solution:

` |(x,4,5),(7,x,7),(5,8,x)| =0`

Expand with respect to `R_1`,

`x(x^2 - 56) - 4(7x - 35) + 5(56 - 5x) = 0`

`=> x^3 - 56x - 28x + 140 + 280 - 25x = 0`

` => x^3 - 109x + 420 = 0`

`=> (x- 5) (x- 7) (x + 12) = 0`

` :. x = -12`

Correct Answer is `=>` (A) `-12`

Q 2440478313

If ` |(a,b,c),(l,m,n),(p,q,r)| = 2`,then what is the value of
` |(6a,3b,15c),(2l,m,5n),(2p,q,5r)|` ?
NDA Paper 1 2010

(A)

`10`

(B)

`20`

(C)

`40`

(D)

`60`

Solution:

`∵ |(a,b,c),(l,m,n),(p,q,r)| = 2`

`:. |(6a,3b,15c),(2l,m,5n),(2p,q,5r)| = 30 |(a,b,c),(l,m,n),(p,q,r)|`

` = 30 xx 2 = 60`

Correct Answer is `=>` (D) `60`

Q 2410678510

If `a, b, c` are non-zero real numbers and
` |(1+a,1,1),(1,1+b,1),(1,1,1+c)| = 0` then what is the value

of ` 1/a+ 1/b = 1/c` ?
NDA Paper 1 2009

(A)

`2`

(B)

`1`

(C)

`- 1`

(D)

`0`

Solution:

Given, ` |(1+a,1,1),(1,1+b,1),(1,1,1+c)| = 0`

Applying `C_2 -> C_2 - C_1` and `C_3 -> C_3 - C_1`,

`=> |(1+a,-a,-a),(1,b,0),(1,0,c)| = 0`

Expanding along `R_3`,

`1(ab) + c(b + ab +a) = 0`

`=> ab +bc + ca + abc = 0`

`=> 1/a + 1/b + 1/c = -1`

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

Q 2470678516

`A = |(2a,3r,x),(4b,6s,2y),(-2c,-3t,-z)| = lamda | (a,r,x),(b,s,y),(c,t,z)|`, then
what is the value of `lamda` ?
NDA Paper 1 2009

(A)

`12`

(B)

`-12`

(C)

`7`

(D)

`-7`

Correct Answer is `=>` (B) `-12`