Practice Start Test

Matrices and Determinants

JEE Mathematics · 271 questions · Page 11 of 28 · Click an option or "Show Solution" to reveal answer

Q101
Let A be a 3 ×\times 3 real matrix such that A(110)=(110);A(101)=(−101)A\left( \begin{array}{ll}1 \\ 1 \\ 0 \end{array} \right) = \left( \begin{array}{ll}1 \\ 1 \\ 0 \end{array} \right);A\left( \begin{array}{ll}1 \\ 0 \\ 1 \end{array} \right) = \left( \begin{array}{ll}{ - 1} \\ 0 \\ 1 \end{array} \right) and A(001)=(112)A\left( \begin{array}{ll}0 \\ 0 \\ 1 \end{array} \right) = \left( \begin{array}{ll}1 \\ 1 \\ 2 \end{array} \right). If X=(x1,x2,x3)TX = {({x_1},{x_2},{x_3})^T} and I is an identity matrix of order 3, then the system (A−2I)X=(411)(A - 2I)X = \left( \begin{array}{ll}4 \\ 1 \\ 1 \end{array} \right) has :
A no solution
B infinitely many solutions
C unique solution
D exactly two solutions
Correct Answer
Option B
Solution

Let

A=[abcdefghi]A = \left[ \begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i \end{array} \right]
A=[110]=[110]⇒[abcdefghi][110]=[110]⇒a+b=1d+e=1g+h=0A = \left[ \begin{array}{ll}1 \\ 1 \\ 0 \end{array} \right] = \left[ \begin{array}{ll}1 \\ 1 \\ 0 \end{array} \right] \Rightarrow \left[ \begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i \end{array} \right]\left[ \begin{array}{ll}1 \\ 1 \\ 0 \end{array} \right] = \left[ \begin{array}{ll}1 \\ 1 \\ 0 \end{array} \right] \Rightarrow \begin{array}{ll}{a + b = 1} \\ {d + e = 1} \\ {g + h = 0} \end{array}
A=[101]=[−101]⇒[abcdefghi][101]=[−101]⇒a+c=−1d+f=1g+i=0A = \left[ \begin{array}{ll}1 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{ll}{ - 1} \\ 0 \\ 1 \end{array} \right] \Rightarrow \left[ \begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i \end{array} \right]\left[ \begin{array}{ll}1 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{ll}{ - 1} \\ 0 \\ 1 \end{array} \right] \Rightarrow \begin{array}{ll}{a + c = - 1} \\ {d + f = 1} \\ {g + i = 0} \end{array}
A=[001]=[112]⇒[abcdefghi][001]=[112]⇒c=1f=1i=2A = \left[ \begin{array}{ll}0 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{ll}1 \\ 1 \\ 2 \end{array} \right] \Rightarrow \left[ \begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i \end{array} \right]\left[ \begin{array}{ll}0 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{ll}1 \\ 1 \\ 2 \end{array} \right] \Rightarrow \begin{array}{ll}{c = 1} \\ {f = 1} \\ {i = 2} \end{array}

Solving will get

a=−2, b=3, c=1, d=−1, e=2, f=1, g=−1, h=1, i=2a = - 2,\,b = 3,\,c = 1,\,d = - 1,\,e = 2,\,f = 1,\,g = - 1,\,h = 1,\,i = 2
A=[−231−121−112]⇒A=2I=[−431−101−110]A = \left[ \begin{array}{lll}{ - 2} & 3 & 1 \\ { - 1} & 2 & 1 \\ { - 1} & 1 & 2 \end{array} \right] \Rightarrow A = 2I = \left[ \begin{array}{lll}{ - 4} & 3 & 1 \\ { - 1} & 0 & 1 \\ { - 1} & 1 & 0 \end{array} \right]
(A−2I)x=[411](A - 2I)x = \left[ \begin{array}{ll}4 \\ 1 \\ 1 \end{array} \right]
⇒−4x1+3x2+x3=4\Rightarrow - 4{x_1} + 3{x_2} + {x_3} = 4

..... (i)

−x1+x3=1- {x_1} + {x_3} = 1

...... (ii)

−x1+x2=1- {x_1} + {x_2} = 1

...... (iii) So 3(iii) + (ii) = (i) ∴\therefore Infinite solution

Q102
Let A=[0−220]A = \left[ \begin{array}{ll}0 & { - 2} \\ 2 & 0 \end{array} \right]. If M and N are two matrices given by M=∑k=110A2kM = \sum\limits_{k = 1}^{10} {{A^{2k}}} and N=∑k=110A2k−1N = \sum\limits_{k = 1}^{10} {{A^{2k - 1}}} then MN2 is :
A a non-identity symmetric matrix
B a skew-symmetric matrix
C neither symmetric nor skew-symmetric matrix
D an identity matrix
Correct Answer
Option A
Solution
A=[0−220]A = \left[ \begin{array}{ll}0 & { - 2} \\ 2 & 0 \end{array} \right]
A2=[0−220][0−220]=[−400−4]=−4I{A^2} = \left[ \begin{array}{ll}0 & { - 2} \\ 2 & 0 \end{array} \right]\left[ \begin{array}{ll}0 & { - 2} \\ 2 & 0 \end{array} \right] = \left[ \begin{array}{ll}{ - 4} & 0 \\ 0 & { - 4} \end{array} \right] = - 4I
M=A2+A4+A6+  .....  +  A20M = {A^2} + {A^4} + {A^6} + \,\,.....\,\, + \,\,{A^{20}}
=−4I+16I−64I  += - 4I + 16I - 64I\,\, +

..... upto 10 terms

=−I= - I

[

4−16+644 - 16 + 64

.... + upto 10 terms]

=−I . 4[(−4)10−1−4−1]=45(220−1)I= - I\,.\,4\left[ {{{{{( - 4)}^{10}} - 1} \over { - 4 - 1}}} \right] = {4 \over 5}({2^{20}} - 1)I
N=A1+A3+A5+  ....  +  A19N = {A^1} + {A^3} + {A^5} + \,\,....\,\, + \,\,{A^{19}}
=A−4A+16A  += A - 4A + 16A\,\, +

..... upto 10 terms

=A((−4)10−1−4−1)=−(220−15)A= A\left( {{{{{( - 4)}^{10}} - 1} \over { - 4 - 1}}} \right) = - \left( {{{{2^{20}} - 1} \over 5}} \right)A
N2=(220−1)225A2=−424(220−1)2I{N^2} = {{{{({2^{20}} - 1)}^2}} \over {{2^5}}}{A^2} = {{ - 4} \over {24}}{({2^{20}} - 1)^2}I
MN2=−16125(220−1)3I=KI     (K≠± 1)M{N^2} = {{ - 16} \over {125}}{({2^{20}} - 1)^3}I = KI\,\,\,\,\,(K \ne \pm \,1)
(MN2)T=(KI)T=KI{(M{N^2})^T} = {(KI)^T} = KI

∴\therefore A is correct

Q103
Let the system of linear equations x + y + α\alphaz = 2 3x + y + z = 4 x + 2z = 1 have a unique solution (x∗^ * , y∗^ * , z∗^ * ). If (α\alpha, x∗^ * ), (y∗^ * , α\alpha) and (x∗^ * , −-y∗^ * ) are collinear points, then the sum of absolute values of all possible values of α\alpha is
A 4
B 3
C 2
D 1
Correct Answer
Option C
Solution

Given system of equations

x+y+az=2x + y + az = 2

..... (i)

3x+y+z=43x + y + z = 4

..... (ii)

x+2z=1x + 2z = 1

..... (iii) Solving (i), (ii) and (iii), we get x = 1, y = 1, z = 0 (and for unique solution a ≠\ne −-3) Now, (α\alpha, 1), (1, α\alpha) and (1, −-1) are collinear ∴\therefore

∣α111α11−11∣=0\left| \begin{array}{lll}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & { - 1} & 1 \end{array} \right| = 0
⇒α(α+1)−1(0)+1(−1−α)=0\Rightarrow \alpha (\alpha + 1) - 1(0) + 1( - 1 - \alpha ) = 0
⇒α2−1=0\Rightarrow {\alpha ^2} - 1 = 0

∴\therefore

α= ± 1\alpha = \, \pm \,1

∴\therefore Sum of absolute values of

α=1+1=2\alpha = 1 + 1 = 2
Q104
The number of values of α\alpha for which the system of equations : x + y + z = α\alpha α\alphax + 2α\alphay + 3z = −-1 x + 3α\alphay + 5z = 4 is inconsistent, is
A 0
B 1
C 2
D 3
Correct Answer
Option B
Solution

Δ=∣111α2α313α5∣\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & 2 \alpha & 3 \\ 1 & 3 \alpha & 5\end{array}\right|

=1(10α−9α)−1(5α−3)+1(3α2−2α)=α−5α+3+3α2−2α=3α2−6α+3\begin{aligned} &=1(10 \alpha-9 \alpha)-1(5 \alpha-3)+1\left(3 \alpha^{2}-2 \alpha\right) \\\\ &=\alpha-5 \alpha+3+3 \alpha^{2}-2 \alpha \\\\ &=3 \alpha^{2}-6 \alpha+3 \end{aligned}

For inconsistency Δ=0\Delta=0 i.e. α=1\alpha=1 Now check for α=1\alpha=1

x+y+z=1...(i)x+y+z=1\quad\quad...(i)

x+2y+3z=−1...(ii)x+2 y+3 z=-1\quad\quad...(ii) x+3y+5z=4...(iii)x+3 y+5 z=4\quad\quad...(iii) By (ii) ×2−\times 2- (i) ×1\times 1

x+3y+5z=−3x+3 y+5 z=-3

so equations are inconsistent for α=1\alpha=1

Q105
Let S = {n\sqrt{n} : 1 ≤\le n ≤\le 50 and n is odd}. Let a ∈\in S and A=[10a−110−a01]A = \left[ \begin{array}{lll}1 & 0 & a \\ { - 1} & 1 & 0 \\ { - a} & 0 & 1 \end{array} \right]. If ∑a ∈ Sdet⁡(adj A)=100λ\sum\limits_{a\, \in \,S}^{} {\det (adj\,A) = 100\lambda } , then λ\lambda is equal to :
A 218
B 221
C 663
D 1717
Correct Answer
Option B
Solution

Given,

A=[10a−110−a01]3×3A = {\left[ \begin{array}{lll}1 & 0 & a \\ { - 1} & 1 & 0 \\ { - a} & 0 & 1 \end{array} \right]_{3 \times 3}}

S = {

n\sqrt{n}

: 1 ≤\le n ≤\le 50 and n is odd} ∴\therefore S =

{1,3,5,7,....,49}\left\{ {1,\sqrt 3 ,\sqrt 5 ,\sqrt 7 ,....,\sqrt {49} } \right\}

We know,

∣adj A∣=∣A∣n−1\left| {adj\,A} \right| = {\left| A \right|^{n - 1}}

Here, n = order of matrix. Here, n = 3 ∴\therefore

∣adj A∣=∣A∣3−1=∣A∣2\left| {adj\,A} \right| = {\left| A \right|^{3 - 1}} = {\left| A \right|^2}

Now,

∣A∣=∣10a−110−a01∣\left| A \right| = \left| \begin{array}{lll}1 & 0 & a \\ { - 1} & 1 & 0 \\ { - a} & 0 & 1 \end{array} \right|
=1(1−0)−0+a(0−(−a))= 1(1 - 0) - 0 + a(0 - ( - a))
=a2+1= {a^2} + 1

∴\therefore

∣adj A∣=∣A∣2=(a2+1)2\left| {adj\,A} \right| = {\left| A \right|^2} = {({a^2} + 1)^2}

Now,

∑a ∈ Sdet⁡(adj A)\sum\limits_{a\, \in \,S}^{} {\det (adj\,A)}
=∑a ∈ S(a2+1)2= \sum\limits_{a\, \in \,S}^{} {{{({a^2} + 1)}^2}}

=

(12+1)2+((3)2+1)2+((5)2+1)2+....+((49)2+1)2{\left( {{1^2} + 1} \right)^2} + {\left( {{{\left( {\sqrt 3 } \right)}^2} + 1} \right)^2} + {\left( {{{\left( {\sqrt 5 } \right)}^2} + 1} \right)^2} + .... + {\left( {{{\left( {\sqrt {49} } \right)}^2} + 1} \right)^2}

=

(12+1)2+(3+1)2+(5+1)2+....+(49+1)2{\left( {{1^2} + 1} \right)^2} + {\left( {3 + 1} \right)^2} + {\left( {5 + 1} \right)^2} + .... + {\left( {49 + 1} \right)^2}

=

22+42+62+....+502{2^2} + {4^2} + {6^2} + .... + {50^2}

=

22(12+22+32+....+252){2^2}\left( {{1^2} + {2^2} + {3^2} + .... + {{25}^2}} \right)

=

4.25.26.516=100.2214.{{25.26.51} \over 6} = 100.221

∴\therefore

100K=100.221100K = 100.221
⇒K=221\Rightarrow K = 221
Q106
Let A=[1−2αα2−1]A = \left[ \begin{array}{lll}1 & { - 2} & \alpha \\ \alpha & 2 & { - 1} \end{array} \right] and B=[2α−124−5], α∈CB = \left[ \begin{array}{ll}2 & \alpha \\ { - 1} & 2 \\ 4 & { - 5} \end{array} \right],\,\alpha \in C. Then the absolute value of the sum of all values of α\alpha for which det(AB) = 0 is :
A 3
B 4
C 2
D 5
Correct Answer
Option A
Solution

Given,

A=[1−2αα2−1]A = \left[ \begin{array}{lll}1 & { - 2} & \alpha \\ \alpha & 2 & { - 1} \end{array} \right]

and

B=[2α−124−5]B = \left[ \begin{array}{ll}2 & \alpha \\ { - 1} & 2 \\ 4 & { - 5} \end{array} \right]
AB=[1−2αα2−1][2α−124−5]AB = \left[ \begin{array}{lll}1 & { - 2} & \alpha \\ \alpha & 2 & { - 1} \end{array} \right]\left[ \begin{array}{ll}2 & \alpha \\ { - 1} & 2 \\ 4 & { - 5} \end{array} \right]
=[4+4α−4α−42α−6α2+9]= \left[ \begin{array}{ll}{4 + 4\alpha } & { - 4\alpha - 4} \\ {2\alpha - 6} & {{\alpha ^2} + 9} \end{array} \right]

Given,

∣AB∣=0|AB| = 0

∴\therefore

∣4+4α−4α−42α−6α2+9∣=0\left| \begin{array}{ll}{4 + 4\alpha } & { - 4\alpha - 4} \\ {2\alpha - 6} & {{\alpha ^2} + 9} \end{array} \right| = 0
⇒(4α+4)∣1−12α−6α2+9∣=0\Rightarrow (4\alpha + 4)\left| \begin{array}{ll}1 & { - 1} \\ {2\alpha - 6} & {{\alpha ^2} + 9} \end{array} \right| = 0
⇒(4α+4)(α2+9+2α−6)=0\Rightarrow (4\alpha + 4)({\alpha ^2} + 9 + 2\alpha - 6) = 0
⇒(4α+4)(α2+2α+3)=0\Rightarrow (4\alpha + 4)({\alpha ^2} + 2\alpha + 3) = 0

∴\therefore

α−=−1\alpha - = - 1

or

α2+2α+3=0{\alpha ^2} + 2\alpha + 3 = 0
α1+α2=−2{\alpha _1} + {\alpha _2} = - 2

∴\therefore Sum of all values of

α=−1−2=−3\alpha = - 1 - 2 = - 3

∴\therefore Absolute value of

α=∣−3∣=3\alpha = | - 3| = 3
Q107
Let A and B be two square matrices of order 2. If det (A)=2det\,(A) = 2, det (B)=3det\,(B) = 3 and det⁡((det⁡ 5(det A)B)A2)=2a3b5c\det \left( {(\det \,5(det\,A)B){A^2}} \right) = {2^a}{3^b}{5^c} for some a, b, c, ∈\in N, then a + b + c is equal to :
A 10
B 12
C 13
D 14
Correct Answer
Option B
Solution

Given,

det⁡(A)=2\det (A) = 2

,

det⁡(B)=3\det (B) = 3

and

det⁡((det⁡(5(det⁡A)B))A2)=2a3b5c\det \left( {\left( {\det \left( {5\left( {\det A} \right)B} \right)} \right){A^2}} \right) = {2^a}{3^b}{5^c}
⇒∣det⁡(5(det⁡A)B)A2∣=2a3b5c\Rightarrow \left| {\det \left( {5\left( {\det A} \right)B} \right){A^2}} \right| = {2^a}{3^b}{5^c}
⇒∣∣5(det⁡A)B∣A2∣=2a3b5c\Rightarrow \left| {\left| {5\left( {\det A} \right)\left. B \right|{A^2}} \right.} \right| = {2^a}{3^b}{5^c}
⇒∣∣5∣A∣B∣A2∣=2a3b5c\Rightarrow \left| {\left| {5\left| {A\left| B \right.\left| {{A^2}} \right.} \right.} \right.} \right| = {2^a}{3^b}{5^c}
⇒∣∣5 . 2 . B∣A2∣=2a . 3b . 5c\Rightarrow \left| {\left| {5\,.\,2\,.\,B} \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}
⇒∣∣10B∣A2∣=2a . 3b . 5c\Rightarrow \left| {\left| {10B} \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}
⇒∣102 . ∣B∣A2∣=2a . 3b . 5c\Rightarrow \left| {{{10}^2}\,.\,\left| B \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}

As

∣k . A∣=kn∣A∣\left| {k\,.\,A} \right| = {k^n}|A|
⇒∣100×3A2∣=2a . 3b . 5c\Rightarrow \left| {100 \times 3{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}
⇒(300)2 . ∣A2∣=2a . 3b . 5c\Rightarrow {(300)^2}\,.\,|{A^2}| = {2^a}\,.\,{3^b}\,.\,{5^c}
⇒(300)2 . ∣A∣2=2a . 3b . 5c\Rightarrow {(300)^2}\,.\,|A{|^2} = {2^a}\,.\,{3^b}\,.\,{5^c}
⇒(300)2 . 22=2a . 3b . 5c\Rightarrow {(300)^2}\,.\,{2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}
⇒9×100×100×22=2a . 3b . 5c\Rightarrow 9 \times 100 \times 100 \times {2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}
⇒32×22×52×22×52×22=2a . 3b . 5c\Rightarrow {3^2} \times {2^2} \times {5^2} \times {2^2} \times {5^2} \times {2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}
⇒26 . 32 . 54=2a . 3b . 5c\Rightarrow {2^6}\,.\,{3^2}\,.\,{5^4} = {2^a}\,.\,{3^b}\,.\,{5^c}

Comparing both sides, we get

a=6a = 6

,

b=2b = 2

,

c=4c = 4

∴\therefore

a+b+c=6+2+4=12a + b + c = 6 + 2 + 4 = 12
Q108
The number of θ∈(0,4π)\theta \in(0,4 \pi) for which the system of linear equations 3(sin⁡3θ)x−y+z=23(cos⁡2θ)x+4y+3z=36x+7y+7z=9 \begin{aligned} &3(\sin 3 \theta) x-y+z=2 \\\\ &3(\cos 2 \theta) x+4 y+3 z=3 \\\\ &6 x+7 y+7 z=9 \end{aligned} has no solution, is :
A 6
B 7
C 8
D 9
Correct Answer
Option B
Solution

Given,

3(sin⁡3θ)x−y+z=23(\sin 3\theta )x - y + z = 2
3(cos⁡2θ)x+4y+3z=33(\cos 2\theta )x + 4y + 3z = 3
6x+7y+7z=96x + 7y + 7z = 9

For no solutions determinant of coefficient will be = 0 ∴\therefore

D=∣3sin⁡3θ−113cos⁡2θ43677∣=0D = \left| \begin{array}{lll}{3\sin 3\theta } & { - 1} & 1 \\ {3\cos 2\theta } & 4 & 3 \\ 6 & 7 & 7 \end{array} \right| = 0
⇒3sin⁡3θ(28−21)+1(21cos⁡2θ−18)+1(21cos⁡2θ−24)=0\Rightarrow 3\sin 3\theta (28 - 21) + 1(21\cos 2\theta - 18) + 1(21\cos 2\theta - 24) = 0
⇒21sin⁡3θ+42cos⁡2θ−42=0\Rightarrow 21\sin 3\theta + 42\cos 2\theta - 42 = 0
⇒sin⁡3θ+2cos⁡2θ−2=0\Rightarrow \sin 3\theta + 2\cos 2\theta - 2 = 0
⇒3sin⁡θ−4sin⁡3θ+2(1−2sin⁡2θ)−2=0\Rightarrow 3\sin \theta - 4{\sin ^3}\theta + 2(1 - 2{\sin ^2}\theta ) - 2 = 0
⇒3sin⁡θ−4sin⁡3θ−4sin⁡2θ=0\Rightarrow 3\sin \theta - 4{\sin ^3}\theta - 4{\sin ^2}\theta = 0
⇒4sin⁡3θ+4sin⁡2θ−3sin⁡θ=0\Rightarrow 4{\sin ^3}\theta + 4{\sin ^2}\theta - 3\sin \theta = 0
⇒sin⁡θ(4sin⁡2θ+4sin⁡θ−3)=0\Rightarrow \sin \theta (4{\sin ^2}\theta + 4\sin \theta - 3) = 0

∴\therefore

sin⁡θ=0\sin \theta = 0
⇒θ=π,2π,3π\Rightarrow \theta = \pi ,2\pi ,3\pi

when

θ∈(0,4π)\theta \in (0,4\pi )

or,

4sin⁡2θ+4sin⁡θ−3=04{\sin ^2}\theta + 4\sin \theta - 3 = 0
⇒4sin⁡2θ+6sin⁡θ−2sin⁡θ−3=0\Rightarrow 4{\sin ^2}\theta + 6\sin \theta - 2\sin \theta - 3 = 0
⇒2sin⁡θ(2sin⁡θ+3)−1(2sin⁡θ+3)=0\Rightarrow 2\sin \theta (2\sin \theta + 3) - 1(2\sin \theta + 3) = 0
⇒(2sin⁡θ−1)(2sin⁡θ+3)=0\Rightarrow (2\sin \theta - 1)(2\sin \theta + 3) = 0

∴\therefore

sin⁡θ=12\sin \theta = {1 \over 2}

or,

sin⁡θ=−32\sin \theta = - {3 \over 2}

[not possible as

sin⁡∈[−1,1]\sin \in [ - 1,1]

] ∴\therefore

sin⁡θ=12\sin \theta = {1 \over 2}
⇒θ=π6,5π6,13π6,17π6\Rightarrow \theta = {\pi \over 6},{{5\pi } \over 6},{{13\pi } \over 6},{{17\pi } \over 6}

∴\therefore Possible values of

θ=π,2π,3π,π6,5π6,13π6,17π6\theta = \pi ,2\pi ,3\pi ,{\pi \over 6},{{5\pi } \over 6},{{13\pi } \over 6},{{17\pi } \over 6}

∴\therefore Total 7 values of θ\theta possible.

Q109
The number of real values of λ\lambda, such that the system of linear equations 2x −- 3y + 5z = 9 x + 3y −- z = −-18 3x −- y + (λ\lambda2 −- | λ\lambda |)z = 16 has no solutions, is
A 0
B 1
C 2
D 4
Correct Answer
Option C
Solution
Δ=∣2−3513−13−1λ2−∣λ∣∣=2(3λ2−3∣λ∣−1)+3(λ2−∣λ∣+3)+5(−1−9)\Delta = \left| \begin{array}{lll}2 & { - 3} & 5 \\ 1 & 3 & { - 1} \\ 3 & { - 1} & {{\lambda ^2} - |\lambda |} \end{array} \right| = 2\left( {3{\lambda ^2} - 3|\lambda | - 1} \right) + 3\left( {{\lambda ^2} - |\lambda | + 3} \right) + 5( - 1 - 9)
=9λ2−9∣λ∣−43= 9{\lambda ^2} - 9|\lambda | - 43
=9∣λ∣2−9∣λ∣−43= 9|\lambda {|^2} - 9|\lambda | - 43
Δ=0\Delta = 0

for 2 values of

∣λ∣|\lambda |

out of which one is

−ve-\mathrm{ve}

and other is

+ve+\mathrm{ve}

So, 2 values of λ\lambda satisfy the system of equations to obtain no solution.

Q110
If the system of linear equations. 8x+y+4z=−28x + y + 4z = - 2 x+y+z=0x + y + z = 0 λx−3y=μ\lambda x - 3y = \mu has infinitely many solutions, then the distance of the point (λ,μ,−12)\left( {\lambda ,\mu , - {1 \over 2}} \right) from the plane 8x+y+4z+2=08x + y + 4z + 2 = 0 is :
A 353\sqrt 5
B 4
C 269{{26} \over 9}
D 103{{10} \over 3}
Correct Answer
Option D
Solution
Δ=∣814111λ−30∣\Delta = \left| \begin{array}{lll}8 & 1 & 4 \\ 1 & 1 & 1 \\ \lambda & { - 3} & 0 \end{array} \right|
=8(3)−1(−λ)+4(−3−λ)= 8(3) - 1( - \lambda ) + 4( - 3 - \lambda )
=24+λ−12−4λ= 24 + \lambda - 12 - 4\lambda
=12−3λ= 12 - 3\lambda

So for

λ=4\lambda = 4

, it is having infinitely many solutions.

Δx=∣−214011μ−30∣{\Delta _x} = \left| \begin{array}{lll}{ - 2} & 1 & 4 \\ 0 & 1 & 1 \\ \mu & { - 3} & 0 \end{array} \right|
=−2(3)−1(−μ)+4(−μ)= - 2(3) - 1( - \mu )+4( - \mu )
=−6−3μ=0= - 6 - 3\mu = 0

For

μ=−2\mu = - 2

Distance of

(4,−2,−12)(4, - 2,{{ - 1} \over 2})

from

8x+y+4z+2=08x + y + 4z + 2 = 0
=32−2−2+264+1+16=103= {{32 - 2 - 2 + 2} \over {\sqrt {64 + 1 + 16} }} = {{10} \over 3}

units

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