Mathematics Tricks & Tips of Determinants for NDA
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Tricks & Tips of Determinants for NDA

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Calculating value of determinants

See examples.
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 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 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 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 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 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 2806867778

If `Delta = | ( 1 + a , 1 , 1 ),(1 , 1 + b , 1),(1,1, 1 + c) |`, then consider the following statements
I. If `1/a + 1/b + 1/c = 0`, then `delta = abc`.
II. If `a^(-1) + b^(-1) + c^(-1) = -1`, then `Delta = 0`.
Which of the above statement(s) is/are correct ?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

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

Expanding with respect to `R_1`

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

`- 1( 1 + c - 1) + (1 - 1 - b)`

`=(1+ a) (b + c + bc) - c - b`

`= bc + ab + ac + abc`

`= abc {1/a + 1/b + 1/c } + abc`

If `1/a + 1/b + 1/c = 0`,

then `Delta = abc (0) + abc = abc `

`:.` Statement I is correct.

Now, `1/a + 1/b + 1/c = - 1`

` => Delta = abc (-1) + abc = 0`

`:.` Statement II is correct.
Correct Answer is `=>` (C) Both I and II
Q 2410178910

What is the value of `|( sin 10°,-cos 10°),( sin 80°,cos 80°)|`?
NDA Paper 1 2008
(A)

`0`

(B)

`1`

(C)

`-1`

(D)

`1//2`

Solution:

` |( sin 10°,-cos 10°),( sin 80°,cos 80°)|`

`= sin 10° cos 80° + sin 80° cos 10°`

`= sin (10° + 80°) = sin 90°`

`= 1 [ ∵ sin (A + B) = sin A· cos B + cos A . sin B]`
Correct Answer is `=>` (B) `1`
Q 2514145959

The determinant

`| (a,b,a alpha +b ),(b,c, b alpha +c), (a alpha +b , b alpha +c, 0) | = 0`if `a, b, c` are in
UPSEE 2008
(A)

AP

(B)

GP

(C)

HP

(D)

None of these

Solution:

If three numbers `a. b, c` are in G.P,. then

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

`=> | (a,b, a alpha +b ), (b,c, b alpha +c ), (0,0, -(a alpha^2 + 2 b alpha +c)) | =0`

`(R_3 -> R_3 -alpha R_1 -R_2)`

`=> a { -c (a alpha^2 + 2 b alpha +c) -0}`

`= -b { -b (a alpha^2 +2 b alpha +c) -0} =0`


`=> -ac (a alpha^2 +2 b alpha +c)+b^2 (a alpha^2 +2 b alpha +c)=0`

`=> (b^2 -ac ) (a alpha^2 +2 b alpha +c) =0`

`=>` Either `b^2 -ac =0`, or ` c alpha^2 + a b alpha +c =0`


`:. ` Either `a,b,c` care in GP or `a alpha^2 + 2 b alpha + c = 0`
Correct Answer is `=>` (B) GP
Q 1419534410

If `Delta_1 = |(1 ,1 ,1),(a, b, c),( a^2 , b^2 ,c^2)|, Delta_2 = |(1, bc ,a),(1, ca, b),( 1, ab, c)|`, then

(A)

`Delta_1 + Delta_2 =0`

(B)

`Delta_1 + 2 Delta_2 =0`

(C)

`Delta_1 = Delta_2`

(D)

`Delta_1 = 2 Delta_2`

Solution:

`Delta_1 = |(1, 1 ,1),(a, b, c),(a^2, b^2, c^2)| = (a-b) ( b-c) (c-a)`

and `Delta_2 = |(1, bc,a),( 1, ca, b),( 1, ab, c)|`

`=1/(abc) |( a, abc, a^2),(b, abc, b^2),( c, abc, c^2)|`

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

`=- | (1,1,1),(a,b,c),(a^2 ,b^2,c^2)| = - Delta _1`

`:. Delta_1 + Delta_2 =0`
Correct Answer is `=>` (A) `Delta_1 + Delta_2 =0`
Q 2418434300

If ` | ( x, x^2 , 1 + x^3),(y, y^2 , 1 + y^3),(z ,z^2 , 1 + z^3)| = 0` and `x, y, z` are all
distinct, then `xyz` is equal to
UPSEE 2009
(A)

`-1`

(B)

`1`

(C)

`0`

(D)

`3`

Solution:

Given,

` | ( x, x^2 , 1 + x^3),(y, y^2 , 1 + y^3),(z ,z^2 , 1 + z^3)| = 0`

` => | (x,x^2 + 1),(y,y^2 + 1),( z,z^2 + 1)| + |(x,x^2,x^3),(y,y^2,y^3),(z,z^2,z^3)| = 0`

` => | (x,x^2 + 1),(y,y^2 + 1),( z,z^2 + 1)| + xyz |(1,x,x^2),(1,y,y^2),(1,z,z^2)| = 0`

` => ( 1 + xyz) |(1,x,x^2),(1,y,y^2),(1,z,z^2)| = 0`

` => 1 + xyz = 0 => xyz = -1`
Correct Answer is `=>` (A) `-1`
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 2470280116

If ` | (2,4,0),(0,5,16),(0,0,1+p)| = 20` , then what is the value of
`p`?
NDA Paper 1 2008
(A)

`0`

(B)

`1`

(C)

`2`

(D)

`5`

Solution:

`| (2,4,0),(0,5,16),(0,0,1+p)| = 20`

On expanding along `C_1`,

`2 {5(1 + p)- 0} = 20`

`=> 1 + p = 2`

`:. p = 1`

Alternate Method

In triangular matrix, the determinant of the matrix

= Product of principle diagonal elements

`:. 2.5(1 + p) = 20`

`=> 1 + p = 2`

` :. p = 1`
Correct Answer is `=>` (B) `1`
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`

Minors and Cofactors of a Determinant

See examples
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 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 2826878771

Consider the determinant, ` Delta = | (p ,q , r),(x , y ,z),(l , m ,n) | ; M_(ij)` denotes the minor of an element in ith row and jth column; `C_(ij)` denotes the cofactor of an element in i th row and j th column.
The value of `p · C_(21) + q · C_(22) + r · C_(23)` is



(A)

`0`

(B)

`- Delta`

(C)

`Delta`

(D)

`Delta^2`

Solution:

`p, q, r` are the entries of first row and

`C_(21) , C_(22) , C_(23)` are cofactors of second row.

`:. p . C_(21) +q ·C_(22) + r·C_(23) = 0`
Correct Answer is `=>` (A) `0`
Q 2577578486

Find the determinants of minors and cofactors of the determinant ` | (2,3,4),(7,2,-5),(8,-1,3)|`



Solution:

Here, `M_(11) = | (2,-5),(-1,3)| = 6 - 5 = 1` [delete 1st row and 1st column]

` :. C_(11) =(-1)^(1 + 1) M_(11) = M_(11) = 1`

`M_(12) = | (7, -5),(8,3) | =21 + 40 = 61` [delete 1st row and 2nd column]

`C_(12) = (-1)^(1 + 2) M_(12) = -M_(12) = -61`

`M_(13) = | (7 ,2),(8, -1) | = -7 - 16 = -23` [delete 1st row and 3rd column]

`C_(13) = (-1)^(1 +3) M_(13) = M_(13) = -23`

` M_(21) = | (3, 4),(-1 , 3) |` [delete 2nd row and 1st column]

`= 9 + 4 = 13`

`C_(21) = (-1)^(2+1) M_(21) = - M_(21) = - 13`

`M_(22) = | (2 ,4),(8,3)|` [delete 2nd row and 2nd column]

`= 6 - 32 = -26`

`C_(22) = (-1)^(2+ 2) M_(22) = M_(22) = -26`

`M_(23) = | (2 ,3),(8 , -1) |` [delete 2nd row and 3rd column]

`= - 2 - 24 = -26`

`C_(23) = (-1)^(2+3) M_(23) =- M_(23) =26`

`M_(31) = | (3,4),(2 , -5) |` [delete 3rd row and 1st column]

`= - 15 - 8 = -23`

`C_(31) = (-1)^(3 +1) M_(31) = M_(31) = -23`

`M_(32) = | ( 2, 4),(7 , - 5) |` [delete 3rd row and 2nd column]

`= - 10 - 28 = -38`

`:. C_(32) = (-1)^(3 + 2) M_(32) = - M_(32) = 38`

`M_(33) = |(2,3),(7,2)| = 4 -21 = -17`[delete 3rd row and 3rd column]

`:. C_(33) = (-1)^(3+3) M_(33) = M_(33) = -17`

Hence, determinants of minors and cofactors are

` | ( 1 , 61 , - 23),(13 ,- 26 , - 26),( -23 , -38, - 17)|` and ` | (1, -61 , -23 ),(-13 , -26 ,26),(-23 , 38 , -17 ) |` respectively.
Q 2632423332

Find the minor of the element of second row and third column `(a_(23))` in the following determinant:
` |(2,-3,5),(6,0,4),(1,5,-7)|`

CBSE-12th 2010
Solution:

Given determinant `= |(2,-3,5),(6,0,4),(1,5,-7)|`

Minor of the element `a_(23)` is `M_(23)`

Obtained by deleting III column and II row

`M_(23) = |(2,-3),(1,5) |`

` = 10 - (-3)`

` = 13`

Algebra of determinants

Q 2856167074

What is the value of
` |(cos 15° , sin 15°),(cos 45° , sin 45°) | xx | (cos 45° , cos 15°),(sin 45° , sin 15°)|` ?

(A)

`1/4`

(B)

`sqrt3/2`

(C)

`-1/4`

(D)

`-3/4`

Solution:

`Delta = |(cos 15° , sin 15°),(cos 45° , sin 45°) | xx | (cos 45° , cos 15°),(sin 45° , sin 15°)|`

` = | ( cos 15°·cos 45^o + sin 15^o · sin 45^o cos^2 15^o + sin^2 15^o ) , ( cos^2 45° + sin^2 45° cos 45^o · cos 15^o + sin 45^o · sin^o 15^o) |`

` = | ( cos( 45^o - 15^o ) , 1 ),( 1, cos( 45^o - 15^o )) |`

` = | ( sqrt3/2 , 1),(1 , sqrt3/2) | = 3/4 - 1 = - 1/4`
Correct Answer is `=>` (C) `-1/4`
Q 2354045854

If `D = |(1,a,b),(1,b,c),(1,c,a)|` , then `| (a,b,c),(b,c,a),(1,1,1)|` equals :
BITSAT Mock
(A)

`−D`

(B)

`D`

(C)

0

(D)

None of these

Solution:

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

[interchanging rows and columns]

`= (−1) | (a,1,b),(b,1,c),(c,1,a)|`

[interchanging `C_2` and `C_3`]

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

[interchanging `C_1` and `C_2`]

` = | (1,a,b),(1,b,c),(1,c,a)| = D`
Correct Answer is `=>` (B) `D`
Q 2846778673

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

(A)

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

(B)

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

(C)

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

(D)

None of these

Solution:

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

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

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

Now,

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

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

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

Properties of deteminats and transformation in determinants

See examples.
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 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 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 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 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 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 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 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 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 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 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 2671180926

Using properties of determinants prove the following :
` | (1,1,1),(a,b,c),(a^3,b^3,c^3)| = (a – b) (b – c) (c – a) (a + b + c)`
CBSE-12th 2012
Solution:

` Delta = | (1,1,1),(a,b,c),(a^3,b^3,c^3)| `

Applying `C_1 → C_1 − C_3` and `C_2 → C_2 − C_3`, we have :

` Delta = | ( 1 - 1 , 1- 1 ,1),(a- c, b- c, c),(a^3 - c^3, b^3 - c^3, c^3) |`

` = | ( 0,0,1),( a- c, b- c, c), ( (a-c)(a^2+ac+ c^2), (b-c)(b^2+bc+ c^2),c^3 )|`

` = (c-a )(b-c) | ( 0,0,1),( -1,1, c),(- (a^2+ac+c^2), (b^2+bc+c^2),c^3)|`

Applying `C_1 → C_1 + C_2`, we have:

` Delta = (c-a )(b-c) |( 0,0,1),( 0,1, c),(-(a + b+c),(b^2 + bc +c^2),c^3)|`

` = (a-b) (b-c)(c-a)(a + b+c) |( 0,0,1),( 0,1, c),(-1 , (b^2 + bc+c^2), c^3) |`

Expanding along `C_1`, we have:

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

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

Hence proved.
Q 2826156971

If `T_P , T_q , T_r` are pth, qth and rth terms of an AP, then ` |(T_P , T_q , T_r),(p,q,r),(1,1,1) | ` is equal to

(A)

`1`

(B)

`-1`

(C)

`0`

(D)

`p + q + r`

Solution:

Let a be the first term and d be

the common difference of given AP.

Then,

`T_p = a + (p - 1) d, T = a + (q - 1)d`,

`T_r = a + ( r - 1) d`

`:. T_p - T_r = (p - r)d` and `T_q - T_r = (q - r)d`

Applying `C_1 -> C_2 - C_3` and

`C_2 -> C_2 - C_3` in given determinant,

we have

` Delta | ( (p - r)d , (q - r) d , T_r),(p-r, q-r , r),(0 , 0 ,1) |`

` = (p-r)(q-r) | (d , d , T_r),(1 , 1, r),( 0, 0 ,1) |`

` = 0 [ ∵ C_1 ` and `C_2` are identical ]
Correct Answer is `=>` (C) `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 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 2440778613

What is the value of ` | (1-i , omega^2 , -omega),( omega^2 + i , omega , -i),(1-2i- omega^2 , omega^2 - omega , i - omega)|`,
where `omega` is the cube root of unity?
NDA Paper 1 2009
(A)

`-1`

(B)

`1`

(C)

`2`

(D)

`0`

Solution:

Let `Delta = | (1-i , omega^2 , -omega),( omega^2 + i , omega , -i),(1-2i- omega^2 , omega^2 - omega , i - omega)|`

Applying `R_3 -> R_3 - (R_1 - R_2`),

` = | (1-i , omega^2 , -omega),( omega^2 + i , omega , -i),(0,0,0)| = 0 `

(since, all elements in `R_3` is zero).
Correct Answer is `=>` (D) `0`
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`

Solution:

Given, `A = |(2a,3r,x),(4b,6s,2y),(-2c,-3t,-z)| = lamda | (a,r,x),(b,s,y),(c,t,z)|`

Taking `2` common from `C_1` and `3` from `C_2` from LHS, we get

` 2 xx 3 |(a,r,x),(2b,2s,2y),(-c,-t,-z)| = lamda | (a,r,x),(b,s,y),(c,t,z)|`

Taking `2` common from `R_2` and `-1` from `R_3` in LHS, we get

`-12 |(a,r,x),(b,s,y),(c,t,z)| = lamda | (a,r,x),(b,s,y),(c,t,z)|`

On comparing, we get

` lamda = -12`
Correct Answer is `=>` (B) `-12`
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 2410078810

If `|(y,x,y+z),(z,y,x+y),(x,z,z+x)| = 0` then which one of the following is correct?
NDA Paper 1 2009
(A)

Either `x + y = z` or `x = y`

(B)

Either `x + y =- z` or `x = z`

(C)

Either `x + z = y` or `z = y`

(D)

Either `z + y = x` or `x = y`

Solution:

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

`=> |(x+yz,x+y+z,2(x+y+z)),(z,y,x+y),(x,z,z+x)| = 0`

(`∵ R_1 -> R_1 + R_2 + R_3`)

` => (x + y + z) |(1,1,2),(z,y,x+y),(x,z,z+x)| = 0`

`=> (x + y + z) |(1,0,0),(z,-z+y,x+y-2z),(x,z-x,z-x)| = 0`

(`∵ C_2 -> C_2 - C_1 ,C_3 -> C_3 - 2C_1`)

Expand with respect to `R_1`,

` (x + y + z) |(-z+y,x+y-2z),(z-x,z-x)| = 0`

`=> (x + y + z)(z- x)(-z + y- x - y + 2z) = 0`

`=> (x + y + z) (z- x)^2 = 0`

`=> x + y = - z` or `z = x`
Correct Answer is `=>` (B) Either `x + y =- z` or `x = z`
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 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 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 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 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 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 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 2632345232

Using properties of determinants prove the following:
` |(a,b,c),(a-b,b-c,c-a),(b+c,c+a,a+b) | = a^3 + b^3 +c^3 - 3abc`


CBSE-12th 2009
Solution:

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

Applying ` C_1 -> C_2 + C_2 + C_3`

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

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

`R_3 -> R_3 - 2 R_1`

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

Expanding along `C_1`, we have,

` Delta = (a + b + c) ((b –c) (a + b – 2c) – (c – a) (c + a – 2b))`

`=> Delta =(a + b + c) ((ba + b^2 - 2bc - ca - cb + 2c^2 - ( c^2 + ac - 2bc -ac - a^2 + 2ab))`

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

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

`=> Delta = a^3 + b^3 + c^2 - 3abc = R.H.S.`

Differentiation of deteminants

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)

24

(C)

35

(D)

50

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 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 2876378276

Given two determinants
`Delta_1 = |(x,b,b),(a,x,b),(a,a,x)| ` and ` Delta_1 = |(x,b),(a,x)|`


The `d/(dx) Delta_1 ` is

(A)

`Delta_2`

(B)

`Delta_2//2`

(C)

`3 Delta_2`

(D)

None of these

Solution:

` d/(dx) Delta_1 = | (1,0,0),(a,x,b),(a,a,x) | + | (x,b,b),(0,1,0),(a,a,x)|`

` + |(x,b,b),(a,x,b),(0,0,1) |`

` = | (x,b),(a,x)| + | (x,b),(a,x) | + |(x,b),(a,x)| = 3 Delta_2`
Correct Answer is `=>` (C) `3 Delta_2`
Q 2406256178

`Delta_1 = | (x,b,b),(a,x,b) , (a,a,x) |` and `Delta_2 = | (x,b), (a,x) |` are the given determinants, then
UPSEE 2013
(A)

`Delta_1 = 3 (Delta_2)^2`

(B)

`d/(dx) (Delta_1) =3 Delta_2`

(C)

`d/(dx) (Delta_1) = 3 (Delta_2)^2`

(D)

`Delta_1 = 3 Delta_2^(3/2)`

Solution:

`Delta_1 = x (x^2 -ab) -b (ax -ab) +b (a^2 -ax)`

`=x^3 - 3abx + ab^2 + a^(2b)`

`d/(dx) (Delta_1) = 3x^2 - 3ab = 3 (x^2 -ab) = 3 Delta_2`
Correct Answer is `=>` (B) `d/(dx) (Delta_1) =3 Delta_2`

Solution of System of Linear Equations

See examples
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 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 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 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 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 2410180019

If ` (a_1//x) + ( b _1//y) = c_1 ,(a_2//x) + ( b _2//y) = c_2`
` Delta_1 = |(a_1,b_1),(a_2,b_2)| , Delta_2 = |(b_1,c_1),(b_2,c_2)|` and
` Delta_3 = |(c_1,a_1),(c_2,a_2)|`,
then `(x, y)` is equal to which one of the following?
NDA Paper 1 2008
(A)

`(Delta_2//Delta_1 ,Delta_3//Delta_1)`

(B)

`(Delta_3//Delta_1 ,Delta_2//Delta_1)`

(C)

`(Delta_1//Delta_2 ,Delta_1//Delta_3)`

(D)

`(- Delta_1//Delta_2 ,- Delta_1//Delta_3)`

Solution:

Let ` 1/x = u` and `1/y = v`

`:. a_1u + b_1v = c_1` and `a_2 u + b_2 v = c_2`

Using the methods of cross-multiplication,

` u/(b_1 c_2 - b_2c_1) = v/(c_1a_2 - c_2 a_1) = (-1)/(a_1b_2 - a_2b_1)`

` => (1/x)/(|(b_1,c_1),(b_2,c_2)|) = (1/y)/(|(c_1,a_1),(c_2,a_2)|) = (-1)/(|(a_1,b_1),(a_2,b_2)|)`

` = (1/x)/Delta_2 = (1/y)/Delta_3 = (-1)/Delta_1 `

`:. 1/x = - Delta_2/Delta_1` and ` 1/y = - Delta_3/Delta_1`

` => x = Delta_1/Delta_2` and ` y = - Delta_1/Delta_3`
Correct Answer is `=>` (D) `(- Delta_1//Delta_2 ,- Delta_1//Delta_3)`
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 2434123052

The value of `lambda` and `mu` for which the
simultaneous equation

`x + y + z = 6, x + 2y + 3z = 10` and
`x + 2y + lambda z = mu` have a unique solution are
UPSEE 2011

(This question may have multiple correct answers)

(A) `lambda ne 3`
(B) `mu=3` only
(C) `lambda =3` and `mu=3`
(D) `lambda ne 3` and `mu` can take any value
Solution:

For unique solution `| (1,1,1),(1,2,3),(1,2, lambda)| ne 0`

`=> 1(2lambda - 6) - 1(lambda - 3) + 1(2- 2) ne 0`

`=>lambda-3ne 0`

`=> lambda ne 3`
Correct Answer is `=>` (A)
Q 2477491386

Considers the system of equations
`ax+by+cz=2`
`bx+cy+az=2`
`cx+ay+bz=2`

Where a,b,c are real numbers such that `a+b+c=0`
Then the system
EAMCET 2016
(A)

Has two solutions

(B)

Is inconsistent

(C)

Has unique solutions

(D)

Has infinitely many solutions

Solution:

`\Delta =|( a , b , c ),( b , c , a ),( c , a , b )|`

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

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

`\Delta = 3abc - [(a+b+c)^3 - 3ab(a+b) - 3bc(b+c) - 3ac(a+c) - 6abc]`

`\Delta = 3abc - [0 - 3ab(-c) - 3bc(-a) - 3ac(-b) -6abc] \ \ \ \ \ \( \because a+b+c=0)`

`\Delta = 3abc - 3abc = 0`

No solution `\Rightarrow` inconsistent
Correct Answer is `=>` (B) Is inconsistent
Q 2434123052

The value of `lambda` and `mu` for which the
simultaneous equation

`x + y + z = 6, x + 2y + 3z = 10` and
`x + 2y + lambda z = mu` have a unique solution are
UPSEE 2011

(This question may have multiple correct answers)

(A) `lambda ne 3`
(B) `mu=3` only
(C) `lambda =3` and `mu=3`
(D) `lambda ne 3` and `mu` can take any value
Solution:

For unique solution `| (1,1,1),(1,2,3),(1,2, lambda)| ne 0`

`=> 1(2lambda - 6) - 1(lambda - 3) + 1(2- 2) ne 0`

`=>lambda-3ne 0`

`=> lambda ne 3`
Correct Answer is `=>` (A)
Q 2466045875

For the equations

`x + 2y + 3z = 1, 2x + y + 3z = 2`

and `5x + 5y + 9z = 4`
UPSEE 2010
(A)

there is only one solution

(B)

there exists infinitely many solution

(C)

there is no solution

(D)

None of the above

Solution:

The determinant of the coefficient matrix of
given system of equation is

`|(1,2,3),(2,1,3),(5,5,9)|=1(9-15)-2(18-5)+3 (10-5)`

`=3 ne 0`

Hence, the system of given equation has unique
solution.
Correct Answer is `=>` (A) there is only one solution
Q 2477491386

Considers the system of equations
`ax+by+cz=2`
`bx+cy+az=2`
`cx+ay+bz=2`

Where a,b,c are real numbers such that `a+b+c=0`
Then the system
EAMCET 2016
(A)

Has two solutions

(B)

Is inconsistent

(C)

Has unique solutions

(D)

Has infinitely many solutions

Solution:

`\Delta =|( a , b , c ),( b , c , a ),( c , a , b )|`

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

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

`\Delta = 3abc - [(a+b+c)^3 - 3ab(a+b) - 3bc(b+c) - 3ac(a+c) - 6abc]`

`\Delta = 3abc - [0 - 3ab(-c) - 3bc(-a) - 3ac(-b) -6abc] \ \ \ \ \ \( \because a+b+c=0)`

`\Delta = 3abc - 3abc = 0`

No solution `\Rightarrow` inconsistent
Correct Answer is `=>` (B) Is inconsistent

Area of triangle using determinants

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`
 
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