Mathematics Must Do Problems of Determinants for NDA

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Must Do Problems of Determinants for NDA

Q 1419523410

If the value of the determinant `|(a, 1, 1),(1, b, 1),(1 ,1 ,c)|` is positive, then

(A)

`abc > 1`

(B)

`abc > - 8`

(C)

`abc < - 8`

(D)

`abc > - 2`

Solution:

Let `Delta' = | (a, 1, 1),(1, b, 1),(1 ,1, c)|= abc+ 2- a-b-c > 0`

or `abc+2 > a+b+c` .....................(i)

where `AM > GM`

`=> (a+ b + c)/3 > (abc ^(1//3))`

`:. a+ b+ c > 3 (abc)^(1//3)` .........(ii)

From Eqs. (i) and (ii),

then `abc+ 2 > 3(abc)^(1//3)`

let `(abc)^(1//3) = x`,

then `x ^3 + 2 > 3x`

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

`:. x + 2 > 0`

`x > -2`

`=> x ^3 > -8`

`abc > -8`

Correct Answer is `=>` (B) `abc > - 8`

Q 2511780620

The value of determinat `| (1+a^2-b^2 , 2ab , -2b) , (2ab , 1-a^2+b^2 , 2a) , (2b , -2a , 1-a^2-b^2)|` is
WBJEE 2013

(A)

`0`

(B)

`(1+a^2+b^2)`

(C)

`(1+a^2+b^2)^2`

(D)

`(1+a^2+b^2)^3`

Solution:

Let `Delta = | (1+a^2-b^2 , 2ab , -2b) , (2ab , 1-a^2+b^2 , 2a) , (2b , -2a , 1-a^2-b^2)|`

Apply `C_1 -> C_1 - bC_3` and `C_2 -> aC_3 + C_2`

`Delta = |(1+a^2-b^2+2b^2 , 2ab-2ab , -2b) , (2ab-2ab , 1-a^2+b^2+2a^2 , 2a) , (2b-b+a^2b+b^3 , -2a+a-a^3-ab^2 , 1-a^2-b^2)|`

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

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

` = (1+a^2+b^2)^2{(1-a^2-b^2+2a^2)+2b^2}`

` = (1+a^2+b^2)^2(1+a^2+b^2) = (1+a^2+b^2)^3`

Correct Answer is `=>` (D) `(1+a^2+b^2)^3`

Q 2816767679

Consider the system of linear equations
`a_1 x + b_1 y + c_1 z + d_1 = 0, a_2 x + b_2 y + c_2 z + d_2 = 0` and `a_1 x + b_3 y + c_3 z + d_3 = 0`
Let us denote by `Delta (a, b, c)` the determinant
` | (a_1 , b_1 , c_1 ),(a_2 , b_2 , c_2) , ( a_3 , b_3 , c_3) |`
If `delta (a, b, c) != 0`, then the value of `x` has a unique solution of the above equation, then

(A)

`(Delta (b ,c , d))/( Delta (a ,b , c))`

(B)

`- (Delta (b ,c , d))/( Delta (a ,b , c))`

(C)

`(Delta (a ,c , d))/( Delta (a ,b , c))`

(D)

`- (Delta (a ,c , d))/( Delta (a ,b , c))`

Solution:

We know that `, x = ( | (d_1 , b_1 , c_1),(d_2 , b_2 , c_2),( d_3 , b_3 , c_3) |)/(| ( a_1 , b_1 , c_1) , (a_2 , b_2 ,c_2) , (a_3 , b_3 ,c_3)|)`

` = (Delta (d, b , c))/(Delta (a , b , c) ) = (Delta (b ,c , d))/( Delta (a ,b , c))`

Correct Answer is `=>` (A) `(Delta (b ,c , d))/( Delta (a ,b , c))`

Q 2866367275

(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 , 1/y = v`

`a_1 u + b_1 v = c_1` and `a_2 u + b_2 v = c_2`

Using the method of cross-multiplication,

` u/(b_1 c_2 - b_2 c_1) = v/(c_1 a_2 -c_2 a_1) = (-1)/(a_1 b_2 - a_2 b_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 1580167017

Let the determinant of a `3 xx 3` matrix `A` be `6`, then `B` is a matrix defined by `B = 5A^ 2` .Then determinant of `B` is :
BITSAT 2005

(A)

`180`

(B)

`100`

(C)

`80`

(D)

None of these

Solution:

Given that `det(A)=6` .........................(i)

Now, `B = 5A^2`

`=> det (B) = det (5A^2)`

`= 5^3 det (A^2) = 5det (A)^2`

`=5^3 (6)^2` from (i)

`=> det (B) =4500`

Correct Answer is `=>` (D) None of these

Q 2410478319

Let `A` be an `n xx n` martrix. If det `(lamda A) = lamda^s det (A)`,
then what is the value of `s`?
NDA Paper 1 2010

(A)

`0`

(B)

`1`

(C)

`-1`

(D)

`n`

Solution:

If `A` is an `n xx n` matrix, then

`det (lamda A) = lamda^n det (A)`

But `det (lamda A) = lamda^s det (A)`

On comparing, we get

` s = n`

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

Q 2456101074

If the determinant ` Delta = | (3,-2, sin 3 theta),(-7,8,cos2 theta),(-11,14,2)| = 0`
then the value of `sin theta` is
UPSEE 2014

(A)

` 1/3 ` or `1`

(B)

`1/sqrt(2)` or `sqrt(3)/2`

(C)

`0` or `1/2`

(D)

None of these

Solution:

Applying `R_2 -> R_2 + 4R_1` and `R_3 -> R_3 + 7R_1` we

get

` | (3, -2, sin 3theta),(5, 0, cos 2 theta + 4 sin 3 theta),(10,0,2 + 7 sin 3 theta) | = 0`

`=> 2 [5(2 + 7 sin 3 theta ) - 10 (cos 2 theta + 4 sin 3 theta )] = 0`

`=> 2 + 7 sin 3 theta - 2 cos 2 theta - 8 sin 3 theta = 0 `

` => 2 - 2 cos 2 theta - sin 3 theta = 0`

`=> sin theta (4 sin^2 theta + 4 sin theta - 3 ) = 0`

`=> sin theta = 0` or `(2 sin theta - 1) = 0` or `(2 sin theta + 3) = 0`

`=> sin theta = 0` or `sin theta = 1/2`

Correct Answer is `=>` (C) `0` or `1/2`

Q 2826167071

If ` Delta = | (a_1, b_1 , c_1),(a_2, b_2 , c_2),(a_3, b_3 , c_3)|` and `A_1 , B_1 , C_1` denote the cofactors of `a_1 , b_1 ,c_1 ` respectively, then the value of the determinant ` | ( A_1, B_1 , C_1),( A_2, B_2 , C_2),( A_3, B_3 , C_3)|` is

(A)

`Delta`

(B)

`Delta^2`

(C)

`Delta^3`

(D)

`0`

Correct Answer is `=>` (B) `Delta^2`

Q 1479823716

Given that `q^2 - pr < 0, p > 0` the value of

`|(p, q, px+qy),(q ,r ,qx+ry),( px+qy, qx+ ry, 0)|` is

(A)

zero

(B)

positive

(C)

negative

(D)

`q^2 + pr`

Solution:

Applying `C_3 -> C _3 - x C_1 - yC_2`, Then we get

`| (p, q ,0),( q, r, 0),( px + qy ,qx + ry, - {px ^2 + 2qxy + ry^2})|`

`= - (px^2 + 2qxy + ry^2) (pr - q^2)`

`= (px^2 + 2qxy + ry^2) (q^2- pr)`

`= 1/p {(px+ qy)^2 + y^2 (pr-q^2)} (q^2 - pr) < 0`

Correct Answer is `=>` (C) negative

Q 2816656570

If A, B and C are the angles of a triangle, then the value of `Delta = | (-1 , cos C , cos B),( cos C , -1 , cos A),(cos B , cos A , -1) |` is

(A)

cos A cos B cos C

(B)

sin A sin B sin C

(C)

`0`

(D)

None of these

Solution:

In a triangle, `A + B + C = pi`

`:. cos (A + B) = cos ( pi - C) = -cos C`

`=> cos A cos B + cos C = sin A sin B`

and `sin (A+ B)= sin C`

Expanding the given determinant,

`Delta = - ( 1 - cos^2 A)`

`+ cos C [cos C + cos A cos B]`

`+ cos B [cos B + cos A cos C]`

`= - sin^2 A + cos C (sin A sin B)`

`+ cos B (sin A sin C)`

`= - sin^2 A + sin A sin (B +C)`

`= - sin^2 A + sin ^2 A= 0`

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

Q 2611134929

Using properties of determinants, prove that:
` | (1 + a , 1 , 1 ),(1 , 1 + b , 1 ),(1 , 1 ,1 + c) | = abc + bc + ca + ab`
CBSE-12th 2014

Solution:

Consider the detrminant

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

Taking abc common outside, we have

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

Apply the transformation, ` C_1 -> C_1 + C_2 + C_3`

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

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

Apply the transformations ` R_2 -> R_2 - R_1 ` and ` R_3 -> R_3 - R_1 `

` => Delta = abc ( 1 + 1/a + 1/b + 1/c) | ( 1 , 1/b , 1/c ),( 0,1,0) ,( 0,0,1) |`

Expanding along C , we have

` Delta = abc ( 1 + 1/a + 1/b + 1/c) xx 1 xx | (1 ,0),(0,1) |`

` => Delta = abc ( 1 + 1/a + 1/b + 1/c) = abc + ab + bc + ca`

Q 2520267111

If `f(x) = | (1,x, x+1), (2x, x (x-1) , (x+1)x ), (3x (x-1), x (x-1)(x-2), (x+1) x (x-1) ) | `

Then, `f(100)` is equal to
WBJEE 2015

(A)

`0`

(B)

`1`

(C)

`100`

(D)

`10`

Solution:

We have, `f(x) = | (1,x, x+1), (2x, x (x-1) , (x+1)x ), (3x (x-1), x (x-1)(x-2), (x+1) x (x-1) ) | `

Taking common `x, (x-1)` from `R_2` and `R_3`

respectively, we get

`f(x) = x (x-1) | (1,x, x+1), (2,x-1,x+1), (3x, x(x-2) , (x+1)x) |`

Taking common `(x+1)` from `C_3` , we get

`f(x) = x (x-1) (x+1) | (1,x,1), (2,x-1,1), (3x, x (x-2) ,x) |`

Taking common x from `R_3` , we get

`f(x) = x^2 (x-1)(x+1) | (1,x,1),(2,x-1,1), (3,x-2,1) |`

Applying `R_1 -> R_1 -R_2, R_2 -> R_2 -R_3`, we get

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

[`·: R_1` is identical with `R_2` ]

`=> f (100) = 0`

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

Q 1543491343

If `Delta_1=|( x , a , b),( b , x , a),( a, b, x)|` and `Delta_2=|( x , b),(a , x)|` are the given determinants, then:
1976

(A)

`Delta_1=2( Delta_2)^2`

(B)

` frac{d}{dx} ( Delta_1)=3 Delta_2`

(C)

` \frac{d}{dx} (\Delta_1)=3(\Delta_2)^2`

(D)

`\Delta_1=3(\Delta_2)^{3/2}`

Solution:

We have, `Delta_1=|( x , a , b),( b , x , a),( a, b , x)| `

`\therefore \ frac{d}{dx}\Delta_1 = |(1,0,0),(b,x,a),(a,b,x )| +|(x,a,b),(0,1,0),(a,b,x)| +|(x,a,b),(b,x,a),(0,0,1 )| `

`\quad =(x^2-ab)+(x^2-ab)+(x^2-ab)=3(x^2-ab)`

and `\Delta_2=|( x , b),(a , x)| =x^2-ab`

Correct Answer is `=>` (B) ` frac{d}{dx} ( Delta_1)=3 Delta_2`

Q 2259512414

If `1/a + 1/b + 1/c = 0`, then
`| (1 + a,1 ,1),(1,1 + b,1),(1,1,1 + c) |` is equal to :
BITSAT Mock

(A)

`−abc`

(B)

`abc`

(C)

`0`

(D)

None of these

Solution:

The given determinant

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

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

[Taking `a, b, c` from `R_1, R_2` and `R_3` respectively]

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

[Performing `R_1 →R_1 + R_2 + R_3` and
taking `(1// a + 1// b + 1 //c + 1)` common]

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

`[C_1 →(C_1 − C_3)` and `C_2 →(C_2 − C_3)]`

` = (abc) (1/a + 1/b+ 1/c+ 1) .(1) |(0,1),(-1,-1)|`

[Expanding by 1st row]

` = (abc) (1/a + 1/b+ 1/c+ 1) .1`

`= abc (0 + 1) = abc`

[`∵ (1/a + 1/b+ 1/c+ 1) = 0` is given ]

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

Q 2580291117

If `f: [0, pi/ 2) -> R` is defined as

`f(theta) = | (1, tan theta , 1), (-tan theta ,1 , tan theta), (-1, - tan theta , 1) |`. Then, the

range of `f` is
WBJEE 2015

(A)

`(2,oo)`

(B)

`(-oo ,-2)`

(C)

`[2,oo )`

(D)

`(-oo, 2]`

Solution:

We have, `f(theta) = | (1, tan theta , 1), (-tan theta ,1 , tan theta), (-1, - tan theta , 1) |`.

`= 1 (1+ tan^2 theta) - tan theta`

`(- tan theta + tan theta) +1 (tan^2 theta +1)`

`=2 (1+tan^2 theta) -0 = 2 sec^2 theta`

`:.` Range of `f = (2,oo)`

Correct Answer is `=>` (A) `(2,oo)`

Q 2561480325

If `I = ((1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1))` and `P = ((1 , 0 , 0) , (0 , -1 , 0) , (0 , 0 , -2))` then the matrix `P^3+2P^2` is equal to
WBJEE 2013

(A)

`P`

(B)

`I-P`

(C)

`2I+P`

(D)

`2I-P`

Solution:

`I = ((1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1))` and `P = ((1 , 0 , 0) , (0 , -1 , 0) , (0 , 0 , -2))`

The characteristic equation of `P` is

`=> |P-lamdaI| = 0`

`=> | (1-lamda , 0 , 0) , (0 , -1-lamda , 0) , (0 , 0 , -2-lamda)| = 0`

`=> (1-lamda){(1+lamda)(2+lamda)} = 0`

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

`=> 2-2lamda^2+lamda-lamda^3 = 0`

`=> lamda^3+2lamda^2-lamda-2 = 0`
We know that, Caylay Hamilton theorem states that 'Every square matrix satisfy its characteristic
equation'.

`therefore P^3+2P^2-P-2I = 0`

`=> P^3+2P^2 = P+2I`

Correct Answer is `=>` (C) `2I+P`

Q 2816123979

If `Delta = | (1, a , a^2),( cos ( n-1) x , cos n x , cos ( n +1) x ),( sin ( n -1) x , sin nx , sin ( n + 1) x ) | ` then `Delta` is

(A)

independent

(B)

independent of a

(C)

independent of n

(D)

None of these

Solution:

We have

`Delta = | (1, a, a^2),( cos ( n -1) x , cos n x , cos (n +1) x ), ( sin (n -1) x , sin nx , sin ( n +1) x)|`

Since `cos (n -1) x + cos ( n +1) x`

`= 2 cos n x * cos x`

and `sin ( n -1)x + sin ( n +1) x `

` =2 sin n x * cos x`

Applying `C_1 -> C_1 - 2 cos x * C_2 + C_3`,

we get

`Delta = | (1-2a cos x +a^2, a , a^2),( 0, cos n x , cos (n +1) x ),( 0, sin x , sin (n +1) x ) |`

`= ( 1- 2 a cos x + a^2 ) [ cos nx * sin (n +1) x - sin n x * cos ( n +1) x ]`

`= (1 - 2 a cos x + a^2 ) sin x`

`:. Delta ` is independent of n

Correct Answer is `=>` (C) independent of n