Practice Start Test

Matrices and Determinants

JEE Mathematics · 271 questions · Page 24 of 28 · Click an option or "Show Solution" to reveal answer

Q231
If A = [−4−131]\left[ \begin{array}{ll}{ - 4} & { - 1} \\ 3 & 1 \end{array} \right], then the determinant of the matrix (A2016 − 2A2015 − A2014) is :
A 2014
B −- 175
C 2016
D −- 25
Correct Answer
Option D
Solution

Given,

A=[−4−131]A = \left[ \begin{array}{ll}{ - 4} & { - 1} \\ 3 & 1 \end{array} \right]
A2=[−4−131][−4−131]{A^2} = \left[ \begin{array}{ll}{ - 4} & { - 1} \\ 3 & 1 \end{array} \right]\left[ \begin{array}{ll}{ - 4} & { - 1} \\ 3 & 1 \end{array} \right]
=[133−9−2]= \left[ \begin{array}{ll}{13} & 3 \\ { - 9} & { - 2} \end{array} \right]

A2 −- 2A −- I

=[133−9−2]−[−8−262]−[1001]= \left[ \begin{array}{ll}{13} & 3 \\ { - 9} & { - 2} \end{array} \right] - \left[ \begin{array}{ll}{ - 8} & { - 2} \\ 6 & 2 \end{array} \right] - \left[ \begin{array}{ll}1 & 0 \\ 0 & 1 \end{array} \right]
=[205−15−5]= \left[ \begin{array}{ll}{20} & 5 \\ { - 15} & { - 5} \end{array} \right]
∣A∣=∣−4−131∣\left| A \right| = \left| \begin{array}{ll}{ - 4} & { - 1} \\ 3 & 1 \end{array} \right|

== −- 4 + 3 == −- 1 Now,

∣A2016−2A2015−A2014∣\left| {{A^{2016}} - 2{A^{2015}} - {A^{2014}}} \right|

==

∣A∣2014∣A2−2A−I∣{\left| A \right|^{2014}}\left| {{A^2} - 2A - {\rm I}} \right|
=(−1)2014∣205−15−5∣= {\left( { - 1} \right)^{2014}}\left| \begin{array}{ll}{20} & 5 \\ { - 15} & { - 5} \end{array} \right|

== 1 ×\times (−- 100 + 75) == −- 25

Q232
If A=[ete−tcos⁡te−tsin⁡tet−e−tcos⁡t−e−tsin⁡t−e−tsin⁡t+e−tcostet2e−tsin⁡t−2e−tcos⁡t]A = \left[ \begin{array}{lll}{{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \\ {{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s}\nolimits} t} \\ {{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \end{array} \right] then A is :
A invertible for all t∈ \in R.
B invertible only if t == π\pi
C not invertible for any t∈ \in R
D invertible only if t == π2{\pi \over 2}.
Correct Answer
Option A
Solution
A=[ete−tcos⁡te−tsin⁡tet−e−tcos⁡t−e−tsin⁡t−e−tsin⁡t+e−tcostet2e−tsin⁡t−2e−tcos⁡t]A = \left[ \begin{array}{lll}{{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \\ {{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s}\nolimits} t} \\ {{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \end{array} \right]
∣A∣=et. e−t.e−t∣1cos⁡tsin⁡t1−cos⁡t−sin⁡t−sin⁡t+cos⁡t12sin⁡t−2cos⁡t∣\left| A \right| = {e^t}.\,{e^{ - t}}.{e^{ - t}}\left| \begin{array}{lll}1 & {\cos t} & {\sin t} \\ 1 & { - \cos t - \sin t} & { - \sin t + \cos t} \\ 1 & {2\sin t} & { - 2\cos t} \end{array} \right|

Apply operations R2 < R2 −-R1, R3 < R3 −- R1, R1 < R1

∣A∣=e−t∣1cos⁡tsin⁡t0−sin⁡t−2cos⁡t−2sin⁡t+cos⁡t02sin⁡t−cos⁡t−2cos⁡t−sin⁡t∣\left| A \right| = {e^{ - t}}\left| \begin{array}{lll}1 & {\cos t} & {\sin t} \\ 0 & { - \sin t - 2\cos t} & { - 2\sin t + \cos t} \\ 0 & {2\sin t - \cos t} & { - 2\cos t - \sin t} \end{array} \right|

Open the determinant by R1

∣A∣=5e−t\left| A \right| = 5{e^{ - t}}

Invertible for all t

∈\in

R

Q233
The total number of matrices A=(02y12xy−12x−y1)A = \left( \begin{array}{lll}0 & {2y} & 1 \\ {2x} & y & { - 1} \\ {2x} & { - y} & 1 \end{array} \right) (x, y ∈ \in R,x ≠ \ne y) for which ATA = 3I3 is :-
A 3
B 4
C 2
D 6
Correct Answer
Option B
Solution

Given ATA = 3I3 ⇒\Rightarrow

[02x2x2yy−y1−11][02y12xy−12x−y1]\left[ \begin{array}{lll}0 & {2x} & {2x} \\ {2y} & y & { - y} \\ 1 & { - 1} & 1 \end{array} \right]\left[ \begin{array}{lll}0 & {2y} & 1 \\ {2x} & y & { - 1} \\ {2x} & { - y} & 1 \end{array} \right]

=

3[100010001]3\left[ \begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]

⇒\Rightarrow

[8x20006y20003]\left[ \begin{array}{lll}{8{x^2}} & 0 & 0 \\ 0 & {6{y^2}} & 0 \\ 0 & 0 & 3 \end{array} \right]

=

[300030003]\left[ \begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array} \right]

∴\therefore 8x2 = 3, 6y2 = 3 ⇒\Rightarrow x =

±38\pm \sqrt {{3 \over 8}}

, y =

±12\pm \sqrt {{1 \over 2}}

Total possible combination of x and y = 2 ×\times 2 = 4

Q234
If A=[2−3−41]A = \left[ \begin{array}{ll}2 & { - 3} \\ { - 4} & 1 \end{array} \right], then adj(3A2 + 12A) is equal to
A [51638472]\left[ \begin{array}{ll}{51} & {63} \\ {84} & {72} \end{array} \right]
B [51846372]\left[ \begin{array}{ll}{51} & {84} \\ {63} & {72} \end{array} \right]
C [72−63−8451]\left[ \begin{array}{ll}{72} & {-63} \\ {-84} & {51} \end{array} \right]
D [72−84−6351]\left[ \begin{array}{ll}{72} & {-84} \\ {-63} & {51} \end{array} \right]
Correct Answer
Option A
Solution

We have,

A=[2−3−41]A = \left[ \begin{array}{ll}2 & { - 3} \\ { - 4} & 1 \end{array} \right]

∴\therefore A2 = A.A =

[2−3−41][2−3−41]\left[ \begin{array}{ll}2 & { - 3} \\ { - 4} & 1 \end{array} \right]\left[ \begin{array}{ll}2 & { - 3} \\ { - 4} & 1 \end{array} \right]

=

[4+12−6−3−8−412+1]\left[ \begin{array}{ll}{4 + 12} & { - 6 - 3} \\ { - 8 - 4} & {12 + 1} \end{array} \right]

=

[16−9−1213]\left[ \begin{array}{ll}{16} & { - 9} \\ { - 12} & {13} \end{array} \right]

Now, 3A2 + 12A =

3[16−9−1213]+12[2−3−41]3\left[ \begin{array}{ll}{16} & { - 9} \\ { - 12} & {13} \end{array} \right] + 12\left[ \begin{array}{ll}2 & { - 3} \\ { - 4} & 1 \end{array} \right]

=

[48−27−3639]+[24−36−4812]\left[ \begin{array}{ll}{48} & { - 27} \\ { - 36} & {39} \end{array} \right] + \left[ \begin{array}{ll}{24} & { - 36} \\ { - 48} & {12} \end{array} \right]

=

[72−63−8451]\left[ \begin{array}{ll}{72} & { - 63} \\ { - 84} & {51} \end{array} \right]

∴\therefore adj(3A2 + 12A) =

[51638472]\left[ \begin{array}{ll}{51} & {63} \\ {84} & {72} \end{array} \right]
Q235
If the system of linear equations x + ay + z = 3 x + 2y + 2z = 6 x + 5y + 3z = b has no solution, then :
A a = −- 1, b = 9
B a = −- 1, b ≠ \ne 9
C a ≠ \ne −- 1, b = 9
D a = 1, b ≠ \ne 9
Correct Answer
Option B
Solution

As the given system of equations has no solution then

Δ\Delta

= 0 and at least one of

Δ\Delta

1,

Δ\Delta

2 and

Δ\Delta

2 should not be zero. ∴\therefore

Δ\Delta

=

∣1a1122153∣=0\left| \begin{array}{lll}1 & a & 1 \\ 1 & 2 & 2 \\ 1 & 5 & 3 \end{array} \right| = 0

⇒\Rightarrow -

aa

- 1 = 0 ⇒\Rightarrow a = - 1

Δ\Delta

2 =

∣1311621b3∣≠0\left| \begin{array}{lll}1 & 3 & 1 \\ 1 & 6 & 2 \\ 1 & b & 3 \end{array} \right| \ne 0

⇒\Rightarrow b ≠\ne 0

Q236
The number of distinct real roots of the equation, ∣cos⁡xsin⁡xsin⁡xsin⁡xcos⁡xsin⁡xsin⁡xsin⁡xcos⁡x∣=0\left| \begin{array}{lll}{\cos x} & {\sin x} & {\sin x} \\ {\sin x} & {\cos x} & {\sin x} \\ {\sin x} & {\sin x} & {\cos x} \end{array} \right| = 0 in the interval [−π4,π4]\left[ { - {\pi \over 4},{\pi \over 4}} \right] is :
A 4
B 3
C 2
D 1
Correct Answer
Option C
Solution

Given,

∣cos⁡xsin⁡xsin⁡xsin⁡xcos⁡xsin⁡xsin⁡xsin⁡xcos⁡x∣=0\left| \begin{array}{lll}{\cos x} & {\sin x} & {\sin x} \\ {\sin x} & {\cos x} & {\sin x} \\ {\sin x} & {\sin x} & {\cos x} \end{array} \right| = 0

R1 →\to R1 −- R3 R1 →\to R2 −- R3

∣cos⁡x−sin⁡x0sin⁡x−cos⁡x0cos⁡x−sin⁡xsin⁡x−cos⁡xsin⁡xsin⁡xcos⁡x∣=0\left| \begin{array}{lll}{\cos x - \sin x} & 0 & {\sin x - \cos x} \\ 0 & {\cos x - \sin x} & {\sin x - \cos x} \\ {\sin x} & {\sin x} & { \cos x} \end{array} \right| = 0

C3 →\to C3 + C2

∣cos⁡x−sin⁡x0sin⁡x−cos⁡x0cos⁡x−sin⁡x0sin⁡xsin⁡xsin⁡x+cos⁡x∣=0\left| \begin{array}{lll}{\cos x - \sin x} & 0 & {\sin x - \cos x} \\ 0 & {\cos x - \sin x} & 0 \\ {\sin x} & {\sin x} & {\sin x + \cos x} \end{array} \right| = 0

Expanding using first column, (cosx −- sinx)(cos −- sinx) (sinx + cos x) + sinx (cosx −- sinx) (sinx −- cosx) = 0 ⇒\Rightarrow (cosx −- sinx)2 (sinx + cosx) ++ sinx (cosx −- sinx)2 = 0 ⇒\Rightarrow (cosx −- sinx)2 (sinx + cosx ++ sinx) = 0 ⇒\Rightarrow (2sinx + cosx )(cosx −- sinx)2 = 0 ∴\therefore cosx = -2sinx or cosx = sinx ⇒\Rightarrow tanx =

−12- {1 \over 2}

or tanx = 1 ∴\therefore x =

−tan⁡−1(12)- {\tan ^{ - 1}}\left( {{1 \over 2}} \right)

,

π4{\pi \over 4}

∴\therefore Number of solutions = 2

Q237
If P = [3212−1232],A=[1101]   \left[ \begin{array}{ll}{{{\sqrt 3 } \over 2}} & {{1 \over 2}} \\ { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \end{array} \right],A = \left[ \begin{array}{ll}1 & 1 \\ 0 & 1 \end{array} \right]\,\,\, Q = PAPT, then PT Q2015 P is :
A [0201500]\left[ \begin{array}{ll}0 & {2015} \\ 0 & 0 \end{array} \right]
B [2015102015]\left[ \begin{array}{ll}{2015} & 1 \\ 0 & {2015} \end{array} \right]
C [2015012015]\left[ \begin{array}{ll}{2015} & 0 \\ 1 & {2015} \end{array} \right]
D [1201501]\left[ \begin{array}{ll}1 & {2015} \\ 0 & 1 \end{array} \right]
Correct Answer
Option D
Solution

P =

[3212−1232]\left[ \begin{array}{ll}{{{\sqrt 3 } \over 2}} & {{1 \over 2}} \\ { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \end{array} \right]

∴\therefore PT =

[32−121232]\left[ \begin{array}{ll}{{{\sqrt 3 } \over 2}} & { - {1 \over 2}} \\ {{1 \over 2}} & {{{\sqrt 3 } \over 2}} \end{array} \right]

As PPT = PTP = I given, Q = PAPT ∴\therefore PTQ = PTP APT ⇒\Rightarrow PTQ = IAPT = APT [ as PTP = I] Now, PT Q2015 P = PTQ .

Q2014 .

P = APT Q2014 P = APT .

Q .

Q2013 .

P = A2PT .

Q2013 .

P . . .

= A2014 .

PTQP = A2014 .

APTP = A2015 As A =

[1101]\left[ \begin{array}{ll}1 & 1 \\ 0 & 1 \end{array} \right]

∴\therefore A2 =

[1101][1101]\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]

A3 =

[1201]\left[ \begin{array}{ll}1 & 2 \\ 0 & 1 \end{array} \right]
[1101]\left[ \begin{array}{ll}1 & 1 \\ 0 & 1 \end{array} \right]

=

[1301]\left[ \begin{array}{ll}1 & 3 \\ 0 & 1 \end{array} \right]

A2015 =

[1201501]\left[ \begin{array}{ll}1 & {2015} \\ 0 & 1 \end{array} \right]
Q238
If S={x∈[0,2π]:∣0cos⁡x−sin⁡xsin⁡x0cos⁡xcos⁡xsin⁡x0∣=0},S = \left\{ {x \in \left[ {0,2\pi } \right]:\left| \begin{array}{lll}0 & {\cos x} & { - \sin x} \\ {\sin x} & 0 & {\cos x} \\ {\cos x} & {\sin x} & 0 \end{array} \right| = 0} \right\}, then ∑x∈Stan⁡(π3+x)\sum\limits_{x \in S} {\tan \left( {{\pi \over 3} + x} \right)} is equal to :
A 4+234 + 2\sqrt 3
B −2+3 - 2 + \sqrt 3
C −2−3 - 2 - \sqrt 3
D −  4−23-\,\,4 - 2\sqrt 3
Correct Answer
Option C
Solution

Given,

∣0cos⁡x−sin⁡xsin⁡x0cos⁡xcos⁡xsin⁡x0∣\left| \begin{array}{lll}0 & {\cos x} & { - \sin x} \\ {\sin x} & 0 & {\cos x} \\ {\cos x} & {\sin x} & 0 \end{array} \right|

= 0 ⇒\Rightarrow

   \,\,\,

0 (0 −- cosx sinx) −- cosx (0 −- cos2x) −- sinx(sin2x) = 0 ⇒\Rightarrow

   \,\,\,

cos3x −- sin3 x = 0 ⇒\Rightarrow

   \,\,\,

tan3x = 1 ⇒\Rightarrow

   \,\,\,

tanx = 1 ∴\therefore

   \,\,\,
∑x ∈  S tan⁡(π3+x)\sum\limits_{x\, \in \,\,S} {\,\tan \left( {{\pi \over 3} + x} \right)}

=

tan⁡π3+tan⁡x1−tan⁡π3tan⁡x{{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}

=

3+11−3{{\sqrt 3 + 1} \over {1 - \sqrt 3 }}

=

(3+1)(1−3)×1+31+3{{\left( {\sqrt 3 + 1} \right)} \over {\left( {1 - \sqrt 3 } \right)}} \times {{1 + \sqrt 3 } \over {1 + \sqrt 3 }}

=

1+3+23−2{{1 + 3 + 2\sqrt 3 } \over { - 2}}

= −- 2 −-

3\sqrt 3
Q239
Let A = [100110111]\left[ \begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{array} \right] and B = A20. Then the sum of the elements of the first column of B is :
A 210
B 211
C 231
D 251
Correct Answer
Option C
Solution

A =

[100110111]\left[ \begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{array} \right]

A2 = A.A =

[100110111]×[100110111]\left[ \begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{array} \right] \times \left[ \begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{array} \right]

=

[100210321]\left[ \begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array} \right]

A3 = A2.A =

[100210321]×[100110111]\left[ \begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array} \right] \times \left[ \begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{array} \right]

=

[100310631]\left[ \begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 0 \\ 6 & 3 & 1 \end{array} \right]

Similarly A4 =

[1004101041]\left[ \begin{array}{lll}1 & 0 & 0 \\ 4 & 1 & 0 \\ {10} & 4 & 1 \end{array} \right]

From this we can say, An =

[100n10n(n+1)2n1]\left[ \begin{array}{lll}1 & 0 & 0 \\ n & 1 & 0 \\ {{{n\left( {n + 1} \right)} \over 2}} & n & 1 \end{array} \right]
∴   \therefore\,\,\,

A20 =

[1002010210201]\left[ \begin{array}{lll}1 & 0 & 0 \\ {20} & 1 & 0 \\ {210} & {20} & 1 \end{array} \right]
∴   \therefore\,\,\,

Sum of the first column = 1 + 20 + 210 = 231

Q240
The system of linear equations x + y + z = 2 2x + 3y + 2z = 5 2x + 3y + (a2 – 1) z = a + 1 then
A has infinitely many solutions for a = 4
B has a unique solution for |a| = 3\sqrt3
C is inconsistent when |a| = 3\sqrt3
D is inconsistent when a = 4
Correct Answer
Option C
Solution
D=∣11123223α2−1∣D = \left| \begin{array}{lll}1 & 1 & 1 \\ 2 & 3 & 2 \\ 2 & 3 & {{\alpha ^2} - 1} \end{array} \right|

D = 3

aa

2 −- 3 −- 6 −- 2

aa

2 + 2 + 4 + 2

aa

2 −- 2 −- 4 D = (

aa

2 −- 3) When D ≠\ne 0 then system of equiation has unique solution. ∴\therefore 3(

aa

2 −- 3) ≠\ne 0 ⇒\Rightarrow

∣a∣\left| a \right|
≠3\ne \sqrt 3

When

3(a2−3)=03({a^2} - 3) = 0

⇒\Rightarrow

∣a∣=3\left| a \right| = \sqrt 3

then D = 0 If D = 0 then two cases possible (1) System of equation has infinite many solution.

(2) System of equation has no solution and inconsistent.

Here D1 =

∣211532a+13a2−1∣=a2−a+1\left| \begin{array}{lll}2 & 1 & 1 \\ 5 & 3 & 2 \\ {a + 1} & 3 & {{a^2} - 1} \end{array} \right| = {a^2} - a + 1

D2 =

∣1212522a+1a2−1∣=a2−3\left| \begin{array}{lll}1 & 2 & 1 \\ 2 & 5 & 2 \\ 2 & {a + 1} & {{a^2} - 1} \end{array} \right| = {a^2} - 3

D3 =

∣11223523a+1∣=a−4\left| \begin{array}{lll}1 & 1 & 2 \\ 2 & 3 & 5 \\ 2 & 3 & {a + 1} \end{array} \right| = a - 4

System of equation will have infinite solution if D1 = D2 = D3 = 0.

And system of equation will have no solution if at last one of D1, D2, D2 is non zero.

At

∣a∣=3\left| a \right| = \sqrt 3

we get D = 0 But D1 = 3 ±\pm

3+1\sqrt 3 + 1

≠\ne 0 and D3 = ±\pm

3−4\sqrt 3 - 4

≠\ne 0 So, system of equations has no solution at

∣a∣=3\left| a \right| = \sqrt 3

then system is in consistent

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