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Matrices and Determinants

JEE Mathematics · 271 questions · Page 8 of 28 · Click an option or "Show Solution" to reveal answer

Q71
The system of equations kx + y + z = 1, x + ky + z = k and x + y + zk = k2 has no solution if k is equal to :
A 0
B −-1
C −-2
D 1
Correct Answer
Option C
Solution
D=∣k111k111k∣=0D = \left| \begin{array}{lll}k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{array} \right| = 0
⇒k(k2−1)−(k−1)+(1−k)=0\Rightarrow k({k^2} - 1) - (k - 1) + (1 - k) = 0
⇒(k−1)(k2+k−1−1)=0\Rightarrow (k - 1)({k^2} + k - 1 - 1) = 0
⇒(k−1)(k2+k−2)=0\Rightarrow (k - 1)({k^2} + k - 2) = 0
⇒(k−1)(k−1)(k+2)=0\Rightarrow (k - 1)(k - 1)(k + 2) = 0
⇒k=1,k=−2\Rightarrow k = 1,k = -2

for k = 1 equation identical so k = −-2 for no solution.

Q72
Let A=[23a0]A = \left[ \begin{array}{ll}2 & 3 \\ a & 0 \end{array} \right], a∈\inR be written as P + Q where P is a symmetric matrix and Q is skew symmetric matrix. If det(Q) = 9, then the modulus of the sum of all possible values of determinant of P is equal to :
A 36
B 24
C 45
D 18
Correct Answer
Option A
Solution
A=[23a0]A = \left[ \begin{array}{ll}2 & 3 \\ a & 0 \end{array} \right]

,

AT=[2a30]{A^T} = \left[ \begin{array}{ll}2 & a \\ 3 & 0 \end{array} \right]
A=A+AT2+A−AT2A = {{A + {A^T}} \over 2} + {{A - {A^T}} \over 2}

Let

P=A+AT2P = {{A + {A^T}} \over 2}

and

Q=A−AT2Q = {{A - {A^T}} \over 2}
Q=(03−a2a−320)Q = \left( \begin{array}{ll}0 & {{{3 - a} \over 2}} \\ {{{a - 3} \over 2}} & 0 \end{array} \right)

Det (Q) = 9

0−(3−a2)(a−32)=90 - \left( {{{3 - a} \over 2}} \right)\left( {{{a - 3} \over 2}} \right) = 9
⇒(a−32)2=9⇒(a−3)2=36\Rightarrow {\left( {{{a - 3} \over 2}} \right)^2} = 9 \Rightarrow {(a - 3)^2} = 36
a−3=± 6⇒a=9,−3a - 3 = \pm \,6 \Rightarrow a = 9, - 3
P=[2a+32a+320]P = \left[ \begin{array}{ll}2 & {{{a + 3} \over 2}} \\ {{{a + 3} \over 2}} & 0 \end{array} \right]
P=[2660]P = \left[ \begin{array}{ll}2 & 6 \\ 6 & 0 \end{array} \right]

or

[2000]\left[ \begin{array}{ll}2 & 0 \\ 0 & 0 \end{array} \right]

| P | = - 36 or 0 ∴\therefore | −-36 + 0 | = 36

Q73
If A=(0sin⁡αsin⁡α0)A = \left( \begin{array}{ll}0 & {\sin \alpha } \\ {\sin \alpha } & 0 \end{array} \right) and det⁡(A2−12I)=0\det \left( {{A^2} - {1 \over 2}I} \right) = 0, then a possible value of α\alpha is :
A π4{\pi \over 4}
B π6{\pi \over 6}
C π2{\pi \over 2}
D π3{\pi \over 3}
Correct Answer
Option A
Solution
A2=[0sin⁡αsin⁡α0][0sin⁡αsin⁡α0]=[sin⁡2α00sin⁡2α]{A^2} = \left[ \begin{array}{ll}0 & {\sin \alpha } \\ {\sin \alpha } & 0 \end{array} \right]\left[ \begin{array}{ll}0 & {\sin \alpha } \\ {\sin \alpha } & 0 \end{array} \right] = \left[ \begin{array}{ll}{{{\sin }^2}\alpha } & 0 \\ 0 & {{{\sin }^2}\alpha } \end{array} \right]
A2−12I=[sin⁡2α00sin⁡2α]−[120012]=[sin⁡2α−1200sin⁡2α−12]{A^2} - {1 \over 2}I = \left[ \begin{array}{ll}{{{\sin }^2}\alpha } & 0 \\ 0 & {{{\sin }^2}\alpha } \end{array} \right] - \left[ \begin{array}{ll}{{1 \over 2}} & 0 \\ 0 & {{1 \over 2}} \end{array} \right] = \left[ \begin{array}{ll}{{{\sin }^2}\alpha - {1 \over 2}} & 0 \\ 0 & {{{\sin }^2}\alpha - {1 \over 2}} \end{array} \right]

Given,

∣A2−12I∣=0\left| {{A^2} - {1 \over 2}I} \right| = 0
⇒∣sin⁡2α−1200sin⁡2α−12∣=0\Rightarrow \left| \begin{array}{ll}{{{\sin }^2}\alpha - {1 \over 2}} & 0 \\ 0 & {{{\sin }^2}\alpha - {1 \over 2}} \end{array} \right| = 0
⇒(sin⁡2α−12)2=0\Rightarrow {\left( {{{\sin }^2}\alpha - {1 \over 2}} \right)^2} = 0
⇒sin⁡2α=12⇒sin⁡α=12,−12\Rightarrow {\sin ^2}\alpha = {1 \over 2} \Rightarrow \sin \alpha = {1 \over {\sqrt 2 }}, - {1 \over {\sqrt 2 }}

∴\therefore

α=π4\alpha = {\pi \over 4}
Q74
If x, y, z are in arithmetic progression with common difference d, x ≠\ne 3d, and the determinant of the matrix [342x452y5kz]\left[ \begin{array}{lll}3 & {4\sqrt 2 } & x \\ 4 & {5\sqrt 2 } & y \\ 5 & k & z \end{array} \right] is zero, then the value of k2 is :
A 72
B 12
C 36
D 6
Correct Answer
Option A
Solution
∣342x452y5kz∣=0\left| \begin{array}{lll}3 & {4\sqrt 2 } & x \\ 4 & {5\sqrt 2 } & y \\ 5 & k & z \end{array} \right| = 0
R1→R1+R3−2R2{R_1} \to {R_1} + {R_3} - 2{R_2}

⇒\Rightarrow

∣042−k−1020452y5kz∣=0\left| \begin{array}{lll}0 & {4\sqrt 2 - k - 10\sqrt 2 } & 0 \\ 4 & {5\sqrt 2 } & y \\ 5 & k & z \end{array} \right| = 0

{ ∵\because 2y = x + z}

⇒(k−62)(4z−5y)=0\Rightarrow (k - 6\sqrt 2 )(4z - 5y) = 0

⇒\Rightarrow k =

626\sqrt 2

or 4z = 5y (Not possible ∵\because x, y, z in A.P.) So, k2 = 72 ∴\therefore Option (A)

Q75
The solutions of the equation ∣1+sin⁡2xsin⁡2xsin⁡2xcos⁡2x1+cos⁡2xcos⁡2x4sin⁡2x4sin⁡2x1+4sin⁡2x∣=0,(0<x<π)\left| \begin{array}{lll}{1 + {{\sin }^2}x} & {{{\sin }^2}x} & {{{\sin }^2}x} \\ {{{\cos }^2}x} & {1 + {{\cos }^2}x} & {{{\cos }^2}x} \\ {4\sin 2x} & {4\sin 2x} & {1 + 4\sin 2x} \end{array} \right| = 0,(0 < x < \pi ), are
A π12,π6{\pi \over {12}},{\pi \over 6}
B π6,5π6{\pi \over 6},{{5\pi } \over 6}
C 5π12,7π12{{5\pi } \over {12}},{{7\pi } \over {12}}
D 7π12,11π12{{7\pi } \over {12}},{{11\pi } \over {12}}
Correct Answer
Option D
Solution

By using C1 →\to C1 −- C2 and C3 →\to C3 −- C2 we get

∣1sin⁡2x0−11+cos⁡2x−104sin⁡2x1∣=0\left| \begin{array}{lll}1 & {{{\sin }^2}x} & 0 \\ { - 1} & {1 + {{\cos }^2}x} & { - 1} \\ 0 & {4\sin 2x} & 1 \end{array} \right| = 0

Expanding by R1 we get

1(1+cos⁡2x+4sin⁡2x)−sin⁡2x(−1)=01(1 + {\cos ^2}x + 4\sin 2x) - {\sin ^2}x( - 1) = 0
⇒2+4sin⁡2x=0\Rightarrow 2 + 4\sin 2x = 0
⇒sin⁡2x=−12\Rightarrow \sin 2x = {{ - 1} \over 2}
⇒2x=nπ+(−1)n(−π6),n∈Z\Rightarrow 2x = n\pi + {( - 1)^n}\left( {{{ - \pi } \over 6}} \right),n \in Z

∴\therefore

2x=7π6,11π62x = {{7\pi } \over 6},{{11\pi } \over 6}
⇒x=7π12,11π2\Rightarrow x = {{7\pi } \over {12}},{{11\pi } \over 2}
Q76
Let α\alpha, β\beta, γ\gamma be the real roots of the equation, x3 + ax2 + bx + c = 0, (a, b, c ∈\in R and a, b ≠\ne 0). If the system of equations (in u, v, w) given by α\alphau + β\betav + γ\gammaw = 0, β\betau + γ\gammav + α\alphaw = 0; γ\gammau + α\alphav + β\betaw = 0 has non-trivial solution, then the value of a2b{{{a^2}} \over b} is
A 5
B 3
C 1
D 0
Correct Answer
Option B
Solution

x3 + ax2 + bx + c = 0 Roots are α\alpha, β\beta, γ\gamma. For non-trivial solutions,

∣αβγβγαγαβ∣=0\left| \begin{array}{lll}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array} \right| = 0

⇒\Rightarrow

α3+β3+γ3−3αβγ=0{\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = 0

⇒\Rightarrow

(α+β+γ)[(α+β+α)2−3(∑αβ)]=0(\alpha + \beta + \gamma )\left[ {{{\left( {\alpha + \beta + \alpha } \right)}^2} - 3\left( {\sum {\alpha \beta } } \right)} \right] = 0

⇒\Rightarrow

(−a)[a2−3b]=0( - a)[{a^2} - 3b] = 0

⇒\Rightarrow

a2=3b{a^2} = 3b

(∵\because a ≠\ne 0)

⇒a2b=3\Rightarrow {{{a^2}} \over b} = 3
Q77
Let A+2B=[1206−33−531]A + 2B = \left[ \begin{array}{lll}1 & 2 & 0 \\ 6 & { - 3} & 3 \\ { - 5} & 3 & 1 \end{array} \right] and 2A−B=[2−152−16012]2A - B = \left[ \begin{array}{lll}2 & { - 1} & 5 \\ 2 & { - 1} & 6 \\ 0 & 1 & 2 \end{array} \right]. If Tr(A) denotes the sum of all diagonal elements of the matrix A, then Tr(A) −- Tr(B) has value equal to
A 1
B 2
C 0
D 3
Correct Answer
Option B
Solution
A=15((A+2B)+2(2A−B))A = {1 \over 5}((A + 2B) + 2(2A - B))
=15([1206−33−531]+[4−2104−212024])= {1 \over 5}\left( {\left[ \begin{array}{lll}1 & 2 & 0 \\ 6 & { - 3} & 3 \\ { - 5} & 3 & 1 \end{array} \right] + \left[ \begin{array}{lll}4 & { - 2} & {10} \\ 4 & { - 2} & {12} \\ 0 & 2 & 4 \end{array} \right]} \right)
=15[501010−515−555]⇒tr(A)=1= {1 \over 5}\left[ \begin{array}{lll}5 & 0 & {10} \\ {10} & { - 5} & {15} \\ { - 5} & 5 & 5 \end{array} \right] \Rightarrow tr(A) = 1

Similarly,

B=15(2(A+2B)−(2A−B))B = {1 \over 5}(2(A + 2B) - (2A - B))
=15([24012−66−1062]−[2−152−16012])= {1 \over 5}\left( {\left[ \begin{array}{lll}2 & 4 & 0 \\ {12} & { - 6} & 6 \\ { - 10} & 6 & 2 \end{array} \right] - \left[ \begin{array}{lll}2 & { - 1} & 5 \\ 2 & { - 1} & 6 \\ 0 & 1 & 2 \end{array} \right]} \right)
=15[06−510−50−1050]⇒tr(B)=−1= {1 \over 5}\left[ \begin{array}{lll}0 & 6 & { - 5} \\ {10} & { - 5} & 0 \\ { - 10} & 5 & 0 \end{array} \right] \Rightarrow tr(B) = - 1
Tr(A)−Tr(B)=1−(−1)=2Tr(A) - Tr(B) = 1 - ( - 1) = 2
Q78
Let the system of linear equations 4x + λ\lambday + 2z = 0 2x −- y + z = 0 μ\mux + 2y + 3z = 0, λ\lambda, μ\mu∈\inR. has a non-trivial solution. Then which of the following is true?
A μ\mu = 6, λ\lambda∈\inR
B λ\lambda = 3, μ\mu∈\inR
C μ\mu = −-6, λ\lambda∈\inR
D λ\lambda = 2, μ\mu∈\inR
Correct Answer
Option A
Solution

Given, system of linear equations 4x + λ\lambday + 2z = 0 2x −- y + z = 0 μ\mux + 2y + 3z = 0 For non-trivial solution,

Δ\Delta

= 0

∣4λ22−11μ23∣=0\left| \begin{array}{lll}4 & \lambda & 2 \\ 2 & { - 1} & 1 \\ \mu & 2 & 3 \end{array} \right| = 0
⇒4(−3−2)−λ(6−μ)+2(4+μ)=0\Rightarrow 4( - 3 - 2) - \lambda (6 - \mu ) + 2(4 + \mu ) = 0
⇒−λ(6−μ)−2(6−μ)=0\Rightarrow - \lambda (6 - \mu ) - 2(6 - \mu ) = 0
⇒(6−μ)(λ+2)=0\Rightarrow (6 - \mu )(\lambda + 2) = 0
⇒λ=−2\Rightarrow \lambda = - 2

and

μ∈R\mu \in R

or μ\mu = 6 and

λ∈R\lambda \in R

.

Q79
The values of λ\lambda and μ\mu such that the system of equations x+y+z=6x + y + z = 6, 3x+5y+5z=263x + 5y + 5z = 26, x+2y+λz=μx + 2y + \lambda z = \mu has no solution, are :
A λ\lambda = 3, μ\mu = 5
B λ\lambda = 3, μ\mu ≠\ne 10
C λ\lambda ≠\ne 2, μ\mu = 10
D λ\lambda = 2, μ\mu ≠\ne 10
Correct Answer
Option D
Solution
x+y+z=6x + y + z = 6

..... (i)

3x+5y+5z=263x + 5y + 5z = 26

.... (ii)

x+2y+λz=μx + 2y + \lambda z = \mu

..... (iii)

5×(i)−(ii)⇒2x=4⇒x=25 \times (i) - (ii) \Rightarrow 2x = 4 \Rightarrow x = 2

∴\therefore from (i) and (iii)

y+z=4y + z = 4

..... (iv)

2y+λz=μ−22y + \lambda z = \mu - 2

.....(v)

(v)−2×(iv)(v) - 2 \times (iv)
⇒(λ−2)z=μ−10\Rightarrow (\lambda - 2)z = \mu - 10
⇒z=μ−10λ−2\Rightarrow z = {{\mu - 10} \over {\lambda - 2}}

&

y=4−μ−10λ−2y = 4 - {{\mu - 10} \over {\lambda - 2}}

∴\therefore For no solution λ\lambda = 2 and μ\mu ≠\ne 10.

Q80
The values of a and b, for which the system of equations 2x + 3y + 6z = 8 x + 2y + az = 5 3x + 5y + 9z = b has no solution, are :
A a = 3, b ≠\ne 13
B a ≠\ne 3, b ≠\ne 13
C a ≠\ne 3, b = 3
D a = 3, b = 13
Correct Answer
Option A
Solution
D=∣23612a359∣=3−aD = \left| \begin{array}{lll}2 & 3 & 6 \\ 1 & 2 & a \\ 3 & 5 & 9 \end{array} \right| = 3 - a
D=∣23812535b∣=b−13D = \left| \begin{array}{lll}2 & 3 & 8 \\ 1 & 2 & 5 \\ 3 & 5 & b \end{array} \right| = b - 13

If a = 3, b ≠\ne 13, no solution.

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