Practice Start Test

Matrices and Determinants

JEE Mathematics · 271 questions · Page 5 of 28 · Click an option or "Show Solution" to reveal answer

Q41
A value of θ∈(0,π3)\theta \in \left( {0,{\pi \over 3}} \right), for which ∣1+cos⁡2θsin⁡2θ4cos⁡6θcos⁡2θ1+sin⁡2θ4cos⁡6θcos⁡2θsin⁡2θ1+4cos⁡6θ∣=0\left| \begin{array}{lll}{1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \\ {{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \\ {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \end{array} \right| = 0, is :
A π18{\pi \over {18}}
B π9{\pi \over {9}}
C 7π24{{7\pi } \over {24}}
D 7π36{{7\pi } \over {36}}
Correct Answer
Option B
Solution
∣1+cos⁡2θsin⁡2θ4cos⁡6θcos⁡2θ1+sin⁡2θ4cos⁡6θcos⁡2θsin⁡2θ1+4cos⁡6θ∣=0\left| \begin{array}{lll}{1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \\ {{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \\ {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \end{array} \right| = 0

R1 →\to R1 - R2, R2 →\to R2 - R3

⇒∣1−1001−1cos⁡2θsin⁡2θ1+4cos⁡6θ∣=0\Rightarrow \left| \begin{array}{lll}1 & { - 1} & 0 \\ 0 & 1 & { - 1} \\ {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \end{array} \right| = 0

C2 →\to C2 + C1

⇒∣10001−1cos⁡2θ11+4cos⁡6θ∣=0\Rightarrow \left| \begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & { - 1} \\ {{{\cos }^2}\theta } & 1 & {1 + 4\cos 6\theta } \end{array} \right| = 0
⇒1+4cos⁡6θ+1=0\Rightarrow 1 + 4\cos 6\theta + 1 = 0
⇒2cos⁡6θ=−1⇒cos⁡6θ=−12\Rightarrow 2\cos 6\theta = - 1 \Rightarrow \cos 6\theta = - {1 \over 2}

=

cos⁡2π3\cos {{2\pi } \over 3}
⇒6θ=2nπ±2π3\Rightarrow 6\theta = 2n\pi \pm {{2\pi } \over 3}
⇒θ=nπ3±π9   n∈1\Rightarrow \theta = {{n\pi } \over 3} \pm {\pi \over 9}\,\,\,n \in 1
⇒θ=π9,2π9,4π9\Rightarrow \theta = {\pi \over 9},{{2\pi } \over 9},{{4\pi } \over 9}
Q42
If the system of equations x + y + z = 5 x + 2y + 3z = 9 x + 3y + az = β\beta has infinitely many solutions, then β\beta −- α\alpha equals -
A 8
B 21
C 18
D 5
Correct Answer
Option A
Solution
D=∣11112313α∣=∣11101202α−1∣D = \left| \begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha \end{array} \right| = \left| \begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 2 & {\alpha - 1} \end{array} \right|
=(α−1)−4=(α−5)= \left( {\alpha - 1} \right) - 4 = \left( {\alpha - 5} \right)

for infinite solutions

D=0⇒α=5D = 0 \Rightarrow \alpha = 5
Dx=0⇒∣511923β35∣=0{D_x} = 0 \Rightarrow \left| \begin{array}{lll}5 & 1 & 1 \\ 9 & 2 & 3 \\ \beta & 3 & 5 \end{array} \right| = 0
⇒∣001−1−13β−15−25∣=0\Rightarrow \left| \begin{array}{lll}0 & 0 & 1 \\ { - 1} & { - 1} & 3 \\ {\beta - 15} & { - 2} & 5 \end{array} \right| = 0
⇒2+β−15=0⇒β−13=0\Rightarrow 2 + \beta - 15 = 0 \Rightarrow \beta - 13 = 0

on

β=13\beta = 13

we get

Dy=Dz=0{D_y} = {D_z} = 0
α=5,β=13\alpha = 5,\beta = 13
Q43
Let A = [2b1bb2+1b1b2]\left[ \begin{array}{lll}2 & b & 1 \\ b & {{b^2} + 1} & b \\ 1 & b & 2 \end{array} \right] where b > 0. Then the minimum value of det⁡(A)b{{\det \left( A \right)} \over b} is -
A 3\sqrt 3
B −- 232\sqrt 3
C −3 - \sqrt 3
D 232\sqrt 3
Correct Answer
Option D
Solution

A =

[2b1bb2+1b1b2]\left[ \begin{array}{lll}2 & b & 1 \\ b & {{b^2} + 1} & b \\ 1 & b & 2 \end{array} \right]

(b > 0)

∣A∣\left| A \right|

= 2(2b2 + 2 −- b2) −- b(2b −- b) + 1(b2 −- b2 −- 1)

∣A∣\left| A \right|

= 2(b2 + 2) −- b2 −- 1

∣A∣\left| A \right|

= b2 + 3

∣A∣b=b+3b⇒b+3b2≥3{{\left| A \right|} \over b} = b + {3 \over b} \Rightarrow {{b + {3 \over b}} \over 2} \ge \sqrt 3
b+3b≥23b + {3 \over b} \ge 2\sqrt 3
Q44
The set of all values of λ\lambda for which the system of linear equations x – 2y – 2z = λ\lambda x x + 2y + z = λ\lambda y – x – y = λ\lambda z has a non-trivial solutions :
A is an empty set
B contains more than two elements
C is a singleton
D contains exactly two elements
Correct Answer
Option C
Solution
∣λ−12212−λ1111∣=0\left| \begin{array}{lll}{\lambda - 1} & 2 & 2 \\ 1 & {2 - \lambda } & 1 \\ 1 & 1 & 1 \end{array} \right| = 0
⇒(λ−1)3=0⇒λ=1\Rightarrow {\left( {\lambda - 1} \right)^3} = 0 \Rightarrow \lambda = 1
Q45
An ordered pair (α\alpha , β\beta ) for which the system of linear equations (1 + α\alpha ) x + β\beta y + z = 2 α\alpha x + (1 + β\beta )y + z = 3 α\alpha x + β\beta y + 2z = 2 has a unique solution, is :
A (–3, 1)
B (1, –3)
C (–4, 2)
D (2, 4)
Correct Answer
Option D
Solution

For unique solution

Δ\Delta

≠\ne 0 ⇒\Rightarrow

∣1+αβ1α1+β1αβ2∣≠0\left| \begin{array}{lll}{1 + \alpha } & \beta & 1 \\ \alpha & {1 + \beta } & 1 \\ \alpha & \beta & 2 \end{array} \right| \ne 0
∣1−1001−1αβ2∣≠0⇒α+β≠−2\left| \begin{array}{lll}1 & { - 1} & 0 \\ 0 & 1 & { - 1} \\ \alpha & \beta & 2 \end{array} \right| \ne 0 \Rightarrow \alpha + \beta \ne - 2
Q46
Let P = [100310931]\left[ \begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{array} \right] and Q = [qij] be two 3 × \times 3 matrices such that Q – P5 = I3. Then q21+q31q32{{{q_{21}} + {q_{31}}} \over {{q_{32}}}} is equal to :
A 15
B 9
C 135
D 10
Correct Answer
Option D
Solution
P=[100310931]P = \left[ \begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{array} \right]
P2=[1003+3109+9+93+31]{P^2} = \left[ \begin{array}{lll}1 & 0 & 0 \\ {3 + 3} & 1 & 0 \\ {9 + 9 + 9} & {3 + 3} & 1 \end{array} \right]
P3=[1003+3+3106.93+3+31]{P^3} = \left[ \begin{array}{lll}1 & 0 & 0 \\ {3 + 3 + 3} & 1 & 0 \\ {6.9} & {3 + 3 + 3} & 1 \end{array} \right]
Pn=[1003n10n(n+1)2323n1]{P^n} = \left[ \begin{array}{lll}1 & 0 & 0 \\ {3n} & 1 & 0 \\ {{{n\left( {n + 1} \right)} \over 2}{3^2}} & {3n} & 1 \end{array} \right]
P5=[1005.31015.95.31]{P^5} = \left[ \begin{array}{lll}1 & 0 & 0 \\ {5.3} & 1 & 0 \\ {15.9} & {5.3} & 1 \end{array} \right]
Q=P5+I3Q = {P^5} + {{\rm I}_3}
Q=[2001520135152]Q = \left[ \begin{array}{lll}2 & 0 & 0 \\ {15} & 2 & 0 \\ {135} & {15} & 2 \end{array} \right]
q21+q31q32=15+13515=10{{{q_{21}} + {q_{31}}} \over {{q_{32}}}} = {{15 + 135} \over {15}} = 10

Aliter

P=(100010001)+(000300930)P = \left( \begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) + \left( \begin{array}{lll}0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{array} \right)
P=I+XP = {\rm I} + X
X=(000300930)X = \left( \begin{array}{lll}0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{array} \right)
X2=(000000900){X^2} = \left( \begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{array} \right)
X3=0{{X_3} = 0}
P5=I+5X+10X2{{P^5} = {\rm I} + 5X + 10{X^2}}
Q=P5+I=2I+5X+10X2{Q = {P^5} + {\rm I} = 2{\rm I} + 5X + 10{X^2}}
Q=(200020002)+(000150015150)+(0000009000)Q = \left( \begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right) + \left( \begin{array}{lll}0 & 0 & 0 \\ {15} & 0 & 0 \\ {15} & {15} & 0 \end{array} \right) + \left( \begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ {90} & 0 & 0 \end{array} \right)
⇒  Q=(2001520135152)\Rightarrow \,\,Q = \left( \begin{array}{lll}2 & 0 & 0 \\ {15} & 2 & 0 \\ {135} & {15} & 2 \end{array} \right)
Q47
If A=[cos⁡θisin⁡θisin⁡θcos⁡θ]A = \left[ \begin{array}{ll}{\cos \theta } & {i\sin \theta } \\ {i\sin \theta } & {\cos \theta } \end{array} \right], (θ=π24)\left( {\theta = {\pi \over {24}}} \right) and A5=[abcd]{A^5} = \left[ \begin{array}{ll}a & b \\ c & d \end{array} \right], where i=−1i = \sqrt { - 1} then which one of the following is not true?
A aa2 - cc2 = 1
B 0≤a2+b2≤10 \le {a^2} + {b^2} \le 1
C a a2 - dd2 = 0
D a2−b2=12{a^2} - {b^2} = {1 \over 2}
Correct Answer
Option D
Solution

∵\because

A=[cos⁡θisin⁡θisin⁡θcos⁡θ]A = \left[ \begin{array}{ll}{\cos \theta } & {i\sin \theta } \\ {i\sin \theta } & {\cos \theta } \end{array} \right]

∴\therefore

An=[cos⁡ nθisin⁡ nθisin⁡ nθcos⁡ nθ],n∈N{A^n} = \left[ \begin{array}{ll}{\cos \,n\theta } & {i\sin \,n\theta } \\ {i\sin \,n\theta } & {\cos \,n\theta } \end{array} \right],n \in N

∴\therefore

A5=[cos⁡ 5θisin⁡ 5θisin⁡ 5θcos⁡ 5θ]=[abcd]{A^5} = \left[ \begin{array}{ll}{\cos \,5\theta } & {i\sin \,5\theta } \\ {i\sin \,5\theta } & {\cos \,5\theta } \end{array} \right] = \left[ \begin{array}{ll}a & b \\ c & d \end{array} \right]

∴\therefore

a=cos⁡5θ, b=isin⁡5θ=c, d=cos⁡5θa = \cos 5\theta ,\,b = i\sin 5\theta = c,\,d = \cos 5\theta

∴\therefore

a2−b2=cos⁡25θ+sin⁡25θ=1{a^2} - {b^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1
a2−c2=cos⁡25θ+sin⁡25θ=1{a^2} - {c^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1
a2−d2=cos⁡25θ−cos⁡25θ=1{a^2} - {d^2} = {\cos ^2}5\theta - {\cos ^2}5\theta = 1
a2+b2=cos⁡25θ−sin⁡25θ=cos⁡10θ=cos⁡10π24{a^2} + {b^2} = {\cos ^2}5\theta - {\sin ^2}5\theta = \cos 10\theta = \cos {{10\pi } \over {24}}

and

0<cos⁡5π12<10 < \cos {{5\pi } \over {12}} < 1
⇒0≤a2+b2≤1\Rightarrow 0 \le {a^2} + {b^2} \le 1
Q48
Let A=(cos⁡α−sin⁡αsin⁡αcos⁡α)A = \left( \begin{array}{ll}{\cos \alpha } & { - \sin \alpha } \\ {\sin \alpha } & {\cos \alpha } \end{array} \right), (α\alpha ∈ \in R) such that A32=(0−110){A^{32}} = \left( \begin{array}{ll}0 & { - 1} \\ 1 & 0 \end{array} \right) then a value of α\alpha is
A 0
B π16{\pi \over {16}}
C π32{\pi \over {32}}
D π64{\pi \over {64}}
Correct Answer
Option D
Solution

From here sin 32α\alpha = 1 and cos 32α\alpha = 0 ∴\therefore 32α\alpha = 2nπ\pi +

π2{\pi \over 2}

⇒\Rightarrow α\alpha =

π64+nπ16{\pi \over {64}} + {{n\pi } \over {16}}

Putting n = 0, α\alpha =

π64{\pi \over {64}}
Q49
Suppose the vectors x1, x2 and x3 are the solutions of the system of linear equations, Ax = b when the vector b on the right side is equal to b1, b2 and b3 respectively. if x1=[111]{x_1} = \left[ \begin{array}{ll}1 \\ 1 \\ 1 \end{array} \right], x2=[021]{x_2} = \left[ \begin{array}{ll}0 \\ 2 \\ 1 \end{array} \right], x3=[001]{x_3} = \left[ \begin{array}{ll}0 \\ 0 \\ 1 \end{array} \right] b1=[100]{b_1} = \left[ \begin{array}{ll}1 \\ 0 \\ 0 \end{array} \right], b2=[020]{b_2} = \left[ \begin{array}{ll}0 \\ 2 \\ 0 \end{array} \right] and b3=[002]{b_3} = \left[ \begin{array}{ll}0 \\ 0 \\ 2 \end{array} \right], then the determinant of A is equal to :
A 32{3 \over 2}
B 4
C 2
D 12{1 \over 2}
Correct Answer
Option C
Solution

Let A =

[a1a2a3a4a5a6a7a8a9]\left[ \begin{array}{lll}{{a_1}} & {{a_2}} & {{a_3}} \\ {{a_4}} & {{a_5}} & {{a_6}} \\ {{a_7}} & {{a_8}} & {{a_9}} \end{array} \right]

For Ax1 = b1 : ⇒\Rightarrow

[a1a2a3a4a5a6a7a8a9][111]=[100]\left[ \begin{array}{lll}{{a_1}} & {{a_2}} & {{a_3}} \\ {{a_4}} & {{a_5}} & {{a_6}} \\ {{a_7}} & {{a_8}} & {{a_9}} \end{array} \right]\left[ \begin{array}{ll}1 \\ 1 \\ 1 \end{array} \right] = \left[ \begin{array}{ll}1 \\ 0 \\ 0 \end{array} \right]

∴\therefore

a1+a2+a3=1{a_1} + {a_2} + {a_3} = 1

....(1)

a4+a5+a6=0{a_4} + {a_5} + {a_6} = 0

......(2)

a7+a8+a9=0{a_7} + {a_8} + {a_9} = 0

.....(3) For Ax2 = b2 :

[a1a2a3a4a5a6a7a8a9][021]=[020]\left[ \begin{array}{lll}{{a_1}} & {{a_2}} & {{a_3}} \\ {{a_4}} & {{a_5}} & {{a_6}} \\ {{a_7}} & {{a_8}} & {{a_9}} \end{array} \right]\left[ \begin{array}{ll}0 \\ 2 \\ 1 \end{array} \right] = \left[ \begin{array}{ll}0 \\ 2 \\ 0 \end{array} \right]

∴\therefore

2a2+a3=02{a_2} + {a_3} = 0

.....(4)

2a5+a6=22{a_5} + {a_6} = 2

....(5)

2a8+a9=02{a_8} + {a_9} = 0

....(6) For Ax3 = b3 :

[a1a2a3a4a5a6a7a8a9][001]=[002]\left[ \begin{array}{lll}{{a_1}} & {{a_2}} & {{a_3}} \\ {{a_4}} & {{a_5}} & {{a_6}} \\ {{a_7}} & {{a_8}} & {{a_9}} \end{array} \right]\left[ \begin{array}{ll}0 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{ll}0 \\ 0 \\ 2 \end{array} \right]

∴\therefore

a3=0{{a_3} = 0}
a6=0{{a_6} = 0}
a9=2{{a_9} = 2}

Putting value of

a3{{a_3}}

in equation (4), we get

a2{{a_2}}

= 0 Putting value of

a6{{a_6}}

in equation (5), we get

a5{{a_5}}

= 1 Putting value of

a9{{a_9}}

in equation (6), we get

a8{{a_8}}

= -1 Putting value of

a2{{a_2}}

and

a3{{a_3}}

in equation (1), we get

a1{{a_1}}

= 1 Putting value of

a5{{a_5}}

and

a6{{a_6}}

in equation (6), we get

a4{{a_4}}

= -1 Putting value of

a5{{a_5}}

and

a6{{a_6}}

in equation (6), we get

a4{{a_4}}

= -1 Putting value of

a8{{a_8}}

and

a9{{a_9}}

in equation (6), we get

a7{{a_7}}

= -1 ∴\therefore A =

[100−110−1−12]\left[ \begin{array}{lll}1 & 0 & 0 \\ { - 1} & 1 & 0 \\ { - 1} & { - 1} & 2 \end{array} \right]

So, |A| = 2(1) = 2

Q50
If the system of linear equations 2x + 2y + 3z = a 3x – y + 5z = b x – 3y + 2z = c where a, b, c are non zero real numbers, has more one solution, then :
A b – c – a = 0
B a + b + c = 0
C b – c + a = 0
D b + c – a = 0
Correct Answer
Option A
Solution

P1 : 2x + 2y + 3z = a P2 : 3x −- y + 5z = b P3 : x −- 3y + 2z = c We find P1 + P3 = P2 ⇒\Rightarrow a + c = b

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