Practice Start Test

Matrices and Determinants

JEE Mathematics · 271 questions · Page 10 of 28 · Click an option or "Show Solution" to reveal answer

Q91
Let [λ\lambda] be the greatest integer less than or equal to λ\lambda. The set of all values of λ\lambda for which the system of linear equations x + y + z = 4, 3x + 2y + 5z = 3, 9x + 4y + (28 + [λ\lambda])z = [λ\lambda] has a solution is :
A R
B (−-∞\infty, −-9) ∪\cup (−-9, ∞\infty)
C [−-9, −-8)
D (−-∞\infty, −-9) ∪\cup [−-8, ∞\infty)
Correct Answer
Option A
Solution
D=∣1113259428+[λ]∣=−24−[λ]+15=−[λ]−9D = \left| \begin{array}{lll}1 & 1 & 1 \\ 3 & 2 & 5 \\ 9 & 4 & {28 + [\lambda ]} \end{array} \right| = - 24 - [\lambda ] + 15 = - [\lambda ] - 9

if

[λ]+9≠0[\lambda ] + 9 \ne 0

then unique solution if

[λ]+9=0[\lambda ] + 9 = 0

then D1 = D2 = D3 = 0 so infinite solutions Hence, λ\lambda can be any red number.

Q92
If the system of linear equations 2x + y −- z = 7 x −- 3y + 2z = 1 x + 4y + δ\deltaz = k, where δ\delta, k ∈\in R has infinitely many solutions, then δ\delta + k is equal to:
A −-3
B 3
C 6
D 9
Correct Answer
Option B
Solution
2x+y−z=72x + y - z = 7
x−3y+2z=1x - 3y + 2z = 1
x+4y+δz=kx + 4y + \delta z = k
Δ=∣21−11−3214δ∣=−7δ−21=0\Delta = \left| \begin{array}{lll}2 & 1 & { - 1} \\ 1 & { - 3} & 2 \\ 1 & 4 & \delta \end{array} \right| = - 7\delta - 21 = 0
δ=−3\delta = - 3
Δ1=∣71−11−32k4−3∣{\Delta _1} = \left| \begin{array}{lll}7 & 1 & { - 1} \\ 1 & { - 3} & 2 \\ k & 4 & { - 3} \end{array} \right|
⇒6−k=0⇒k=6\Rightarrow 6 - k = 0 \Rightarrow k = 6
δ+k=−3+6=3\delta + k = - 3 + 6 = 3
Q93
Let A=[aij]A = [{a_{ij}}] be a square matrix of order 3 such that aij=2j−i{a_{ij}} = {2^{j - i}}, for all i, j = 1, 2, 3. Then, the matrix A2 + A3 + ...... + A10 is equal to :
A (310−32)A\left( {{{{3^{10}} - 3} \over 2}} \right)A
B (310−12)A\left( {{{{3^{10}} - 1} \over 2}} \right)A
C (310+12)A\left( {{{{3^{10}} + 1} \over 2}} \right)A
D (310+32)A\left( {{{{3^{10}} + 3} \over 2}} \right)A
Correct Answer
Option A
Solution

Given,

aij=2j−i{a_{ij}} = {2^{j - i}}

Now,

A=[2021222−120212−22−120]A = \left[ \begin{array}{lll}{{2^0}} & {{2^1}} & {{2^2}} \\ {{2^{ - 1}}} & {{2^0}} & {{2^1}} \\ {{2^{ - 2}}} & {{2^{ - 1}}} & {{2^0}} \end{array} \right]
=[124121214121]= \left[ \begin{array}{lll}1 & 2 & 4 \\ {{1 \over 2}} & 1 & 2 \\ {{1 \over 4}} & {{1 \over 2}} & 1 \end{array} \right]
A2=[124121214121][124121214121]{A^2} = \left[ \begin{array}{lll}1 & 2 & 4 \\ {{1 \over 2}} & 1 & 2 \\ {{1 \over 4}} & {{1 \over 2}} & 1 \end{array} \right]\left[ \begin{array}{lll}1 & 2 & 4 \\ {{1 \over 2}} & 1 & 2 \\ {{1 \over 4}} & {{1 \over 2}} & 1 \end{array} \right]
=[1+1+12+2+24+4+412+12+121+1+12+2+214+14+1412+12+121+1+1]= \left[ \begin{array}{lll}{1 + 1 + 1} & {2 + 2 + 2} & {4 + 4 + 4} \\ {{1 \over 2} + {1 \over 2} + {1 \over 2}} & {1 + 1 + 1} & {2 + 2 + 2} \\ {{1 \over 4} + {1 \over 4} + {1 \over 4}} & {{1 \over 2} + {1 \over 2} + {1 \over 2}} & {1 + 1 + 1} \end{array} \right]
=[3612323634323]= \left[ \begin{array}{lll}3 & 6 & {12} \\ {{3 \over 2}} & 3 & 6 \\ {{3 \over 4}} & {{3 \over 2}} & 3 \end{array} \right]
=3[124121214121]= 3\left[ \begin{array}{lll}1 & 2 & 4 \\ {{1 \over 2}} & 1 & 2 \\ {{1 \over 4}} & {{1 \over 2}} & 1 \end{array} \right]
=3A= 3A

Similarly,

A3=32A{A^3} = {3^2}A
A4=33A{A^4} = {3^3}A

∴\therefore

A2+A3+  ......  +  A10{A^2} + {A^3} + \,\,......\,\, + \,\,{A^{10}}
=3A+32A+33A+  ......  +  39A= 3A + {3^2}A + {3^3}A + \,\,......\,\, + \,\,{3^9}A
=A(3+32+33+  ......  +  39)= A(3 + {3^2} + {3^3} + \,\,......\,\, + \,\,{3^9})
=A(3(39−1)3−1)=3(39−1)2A= A\left( {{{3({3^9} - 1)} \over {3 - 1}}} \right) = {{3({3^9} - 1)} \over 2}A

=

(310−32)A\left( {{{{3^{10}} - 3} \over 2}} \right)A
Q94
If the system of equations α\alphax + y + z = 5, x + 2y + 3z = 4, x + 3y + 5z = β\beta has infinitely many solutions, then the ordered pair (α\alpha, β\beta) is equal to :
A (1, −-3)
B (−-1, 3)
C (1, 3)
D (−-1, −-3)
Correct Answer
Option C
Solution

Given system of equations

αx+y+z=5\alpha x + y + z = 5
x+2y+3z=4x + 2y + 3z = 4

, has infinite solution

x+3y+5z=βx + 3y + 5z = \beta

∴\therefore

Δ=∣α11123135∣=0⇒α(1)−1(2)+1(1)=0\Delta = \left| \begin{array}{lll}\alpha & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{array} \right| = 0 \Rightarrow \alpha (1) - 1(2) + 1(1) = 0
⇒α=1\Rightarrow \alpha = 1

and

Δ1=∣511423β35∣=0{\Delta _1} = \left| \begin{array}{lll}5 & 1 & 1 \\ 4 & 2 & 3 \\ \beta & 3 & 5 \end{array} \right| = 0
⇒5(1)−1(20−3β)+1(12−2β)=0\Rightarrow 5(1) - 1(20 - 3\beta ) + 1(12 - 2\beta ) = 0
⇒β=3\Rightarrow \beta = 3

and

Δ2=∣1511431β5∣=0⇒(20−3β)−5(2)+1(β−4)=0{\Delta _2} = \left| \begin{array}{lll}1 & 5 & 1 \\ 1 & 4 & 3 \\ 1 & \beta & 5 \end{array} \right| = 0 \Rightarrow (20 - 3\beta ) - 5(2) + 1(\beta - 4) = 0
⇒−2β+6=0\Rightarrow - 2\beta + 6 = 0
⇒β=3\Rightarrow \beta = 3

Similarly,

⇒Δ3=∣11512413β∣=0⇒β=3\Rightarrow {\Delta _3} = \left| \begin{array}{lll}1 & 1 & 5 \\ 1 & 2 & 4 \\ 1 & 3 & \beta \end{array} \right| = 0 \Rightarrow \beta = 3

∴\therefore (α\alpha, β\beta) = (1, 3)

Q95
Let A=(2−102)A = \left( \begin{array}{ll}2 & { - 1} \\ 0 & 2 \end{array} \right). If B=I−5C1(adj A)+5C2(adj A)2−  .....  −5C5(adj A)5B = I - {}^5{C_1}(adj\,A) + {}^5{C_2}{(adj\,A)^2} - \,\,.....\,\, - {}^5{C_5}{(adj\,A)^5}, then the sum of all elements of the matrix B is
A −-5
B −-6
C −-7
D −-8
Correct Answer
Option C
Solution

Given

A=[2−102]A = \left[ \begin{array}{ll}2 & { - 1} \\ 0 & 2 \end{array} \right]

and

B=I−5C1(adj A)+5C2(adj A)2−5C3(adj A)3+5C4(adj A)4−5C5(adj A)5B = I - {5_{{C_1}}}(adj\,A) + {5_{{C_2}}}{(adj\,A)^2} - {5_{{C_3}}}{(adj\,A)^3} + {5_{{C_4}}}{(adj\,A)^4} - {5_{{C_5}}}{(adj\,A)^5}
=(I−(adj A))5= {\left( {I - (adj\,A)} \right)^5}

Cofactor of

A=[(−1)1+1 . 2(−1)1+2 . 0(−1)2+1 . (−1)(−1)2+2 . 2]A = \left[ \begin{array}{ll}{{{( - 1)}^{1 + 1}}\,.\,2} & {{{( - 1)}^{1 + 2}}\,.\,0} \\ {{{( - 1)}^{2 + 1}}\,.\,( - 1)} & {{{( - 1)}^{2 + 2}}\,.\,2} \end{array} \right]
=[2012]= \left[ \begin{array}{ll}2 & 0 \\ 1 & 2 \end{array} \right]

Transpose of cofactor of

A=[2102]A = \left[ \begin{array}{ll}2 & 1 \\ 0 & 2 \end{array} \right]

∴\therefore

adj A=[2102]adj\,A = \left[ \begin{array}{ll}2 & 1 \\ 0 & 2 \end{array} \right]

Now,

I−adj AI - adj\,A
=[1001]−[2102]= \left[ \begin{array}{ll}1 & 0 \\ 0 & 1 \end{array} \right] - \left[ \begin{array}{ll}2 & 1 \\ 0 & 2 \end{array} \right]
=[−1−10−1]= \left[ \begin{array}{ll}{ - 1} & { - 1} \\ 0 & { - 1} \end{array} \right]

Now let,

P=I−adj A=[−1−10−1]P = I - adj\,A = \left[ \begin{array}{ll}{ - 1} & { - 1} \\ 0 & { - 1} \end{array} \right]

∴\therefore

P2=[−1−10−1][−1−10−1]{P^2} = \left[ \begin{array}{ll}{ - 1} & { - 1} \\ 0 & { - 1} \end{array} \right]\left[ \begin{array}{ll}{ - 1} & { - 1} \\ 0 & { - 1} \end{array} \right]
=[1201]= \left[ \begin{array}{ll}1 & 2 \\ 0 & 1 \end{array} \right]
P4=P2 . P2=[1201][1201]=[1401]{P^4} = {P^2}\,.\,{P^2} = \left[ \begin{array}{ll}1 & 2 \\ 0 & 1 \end{array} \right]\left[ \begin{array}{ll}1 & 2 \\ 0 & 1 \end{array} \right] = \left[ \begin{array}{ll}1 & 4 \\ 0 & 1 \end{array} \right]
P5=P4 . P=[1401][−1−10−1]=[−1−50−1]{P^5} = {P^4}\,.\,P = \left[ \begin{array}{ll}1 & 4 \\ 0 & 1 \end{array} \right]\left[ \begin{array}{ll}{ - 1} & { - 1} \\ 0 & { - 1} \end{array} \right] = \left[ \begin{array}{ll}{ - 1} & { - 5} \\ 0 & { - 1} \end{array} \right]

∴\therefore

B=[−1−50−1]B = \left[ \begin{array}{ll}{ - 1} & { - 5} \\ 0 & { - 1} \end{array} \right]

Now sum of elements

=−1−5−1+0=−7= - 1 - 5 - 1 + 0 = - 7
Q96
If the system of linear equations 2x+3y−z=−22x + 3y - z = - 2 x+y+z=4x + y + z = 4 x−y+∣λ∣z=4λ−4x - y + |\lambda |z = 4\lambda - 4 where, λ\lambda ∈\in R, has no solution, then
A λ\lambda = 7
B λ\lambda = −-7
C λ\lambda = 8
D λ\lambda2 = 1
Correct Answer
Option B
Solution
Δ=∣23−11111−1∣λ∣∣=0⇒∣λ∣=7\Delta = \left| \begin{array}{lll}2 & 3 & { - 1} \\ 1 & 1 & 1 \\ 1 & { - 1} & {|\lambda |} \end{array} \right| = 0 \Rightarrow |\lambda | = 7

But at

λ=7, Dx=Dy=Dz=0\lambda = 7,\,{D_x} = {D_y} = {D_z} = 0
P1:2x+3y−z=−2{P_1}:2x + 3y - z = - 2
P2:x+y+z=4{P_2}:x + y + z = 4
P3:x−y+∣λ∣z=4λ−4{P_3}:x - y + |\lambda |z = 4\lambda - 4

So clearly

5P2−2P1=P35{P_2} - 2{P_1} = {P_3}

, so at

λ=7\lambda = 7

, system of equation is having infinite solutions. So

λ=−7\lambda = - 7

is correct answer.

Q97
Let A be a matrix of order 3 ×\times 3 and det (A) = 2. Then det (det (A) adj (5 adj (A3))) is equal to _____________.
A 512 ×\times 106
B 256 ×\times 106
C 1024 ×\times 106
D 256 ×\times 1011
Correct Answer
Option A
Solution
∣A∣=2|A| = 2
∣∣A∣=adj (5 adj A3)∣||A| = adj\,(5\,adj\,{A^3})|
=∣25∣A∣adj (adj A3)∣= |25|A|adj\,(adj\,{A^3})|
=253∣A∣3 . ∣adj A3∣2= {25^3}|A{|^3}\,.\,|adj\,{A^3}{|^2}
=253 . 23 . ∣A3∣4= {25^3}\,.\,{2^3}\,.\,|{A^3}{|^4}
=253 . 23 . 212=106 . 512= {25^3}\,.\,{2^3}\,.\,{2^{12}} = {10^6}\,.\,512
Q98
Let A and B be two 3 ×\times 3 matrices such that AB=IAB = I and ∣A∣=18|A| = {1 \over 8}. Then ∣adj (B adj(2A))∣|adj\,(B\,adj(2A))| is equal to
A 16
B 32
C 64
D 128
Correct Answer
Option C
Solution

A and B are two matrices of order 3 ×\times 3. and

AB=IAB = I

,

∣A∣=18|A| = {1 \over 8}

Now,

∣A∣∣B∣=1|A||B| = 1
∣B∣=8|B| = 8

∴\therefore

∣adj(B(adj(2A))∣=∣B(adj(2A))∣2|adj(B(adj(2A))| = |B(adj(2A)){|^2}
=∣B∣2∣adj(2A)∣2= |B{|^2}|adj(2A){|^2}
=26∣2A∣2×2= {2^6}|2A{|^{2 \times 2}}
=26. 212. 1212=64= {2^6}.\,{2^{12}}.\,{1 \over {{2^{12}}}} = 64
Q99
Let the system of linear equations x+2y+z=2x + 2y + z = 2, αx+3y−z=α\alpha x + 3y - z = \alpha , −αx+y+2z=−α - \alpha x + y + 2z = - \alpha be inconsistent. Then α\alpha is equal to :
A 52{5 \over 2}
B −-52{5 \over 2}
C 72{7 \over 2}
D −-72{7 \over 2}
Correct Answer
Option D
Solution
x+2y+z=2x + 2y + z = 2
αx+3y−z=α\alpha x + 3y - z = \alpha
−αx+y+2z=−α- \alpha x + y + 2z = - \alpha
Δ=∣121α3−1−α12∣=1(6+1)−2(2α−α)+1(α+3α)\Delta = \left| \begin{array}{lll}1 & 2 & 1 \\ \alpha & 3 & { - 1} \\ { - \alpha } & 1 & 2 \end{array} \right| = 1(6 + 1) - 2(2\alpha - \alpha ) + 1(\alpha + 3\alpha )
=7+2α= 7 + 2\alpha
Δ=0⇒α=−72\Delta = 0 \Rightarrow \alpha = - {7 \over 2}
Δ1=∣221α3−1−α12∣=14+2α≠0{\Delta _1} = \left| \begin{array}{lll}2 & 2 & 1 \\ \alpha & 3 & { - 1} \\ { - \alpha } & 1 & 2 \end{array} \right| = 14 + 2\alpha \ne 0

for

α=−72\alpha = - {7 \over 2}

∴\therefore For no solution

α=−72\alpha = - {7 \over 2}
Q100
Let A be a 3 ×\times 3 invertible matrix. If |adj (24A)| = |adj (3 adj (2A))|, then |A|2 is equal to :
A 66
B 212
C 26
D 1
Correct Answer
Option C
Solution

We know,

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

Now,

∣adj 24A∣=∣adj 3(adj 2A)∣|adj\,24A| = |adj\,3(adj\,2A)|
⇒∣24A∣3−1=∣3 adj 2A∣3−1\Rightarrow |24A{|^{3 - 1}} = |3\,adj\,2A{|^{3 - 1}}
⇒∣24A∣2=∣3 adj 2A∣2\Rightarrow |24A{|^2} = |3\,adj\,2A{|^2}

Also, we know,

∣KA∣=Kn∣A∣|KA| = {K^n}|A|
⇒((24)2)2∣A∣2=((3)3)2∣adj 2A∣2\Rightarrow {\left( {{{(24)}^2}} \right)^2}|A{|^2} = {\left( {{{(3)}^3}} \right)^2}|adj\,2A{|^2}
⇒(24)6∣A∣2=36 . (∣2A∣3−1)2\Rightarrow {(24)^6}|A{|^2} = {3^6}\,.\,{\left( {|2A{|^{3 - 1}}} \right)^2}
⇒(24)6∣A∣2=36 . ∣2A∣4\Rightarrow {(24)^6}|A{|^2} = {3^6}\,.\,|2A{|^4}
⇒(24)6∣A∣2=36 . (23)4 . ∣A∣4\Rightarrow {(24)^6}|A{|^2} = {3^6}\,.\,{\left( {{2^3}} \right)^4}\,.\,|A{|^4}
⇒36 . 86 . ∣A∣2=36 . 84 . ∣A∣4\Rightarrow {3^6}\,.\,{8^6}\,.\,|A{|^2} = {3^6}\,.\,{8^4}\,.\,|A{|^4}
⇒82=∣A∣2\Rightarrow {8^2} = |A{|^2}
⇒∣A∣2=64\Rightarrow |A{|^2} = 64

= 26

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