Practice Start Test

Matrices and Determinants

JEE Mathematics · 271 questions · Page 13 of 28 · Click an option or "Show Solution" to reveal answer

Q121
If A=12[13−31]A = {1 \over 2}\left[ \begin{array}{ll}1 & {\sqrt 3 } \\ { - \sqrt 3 } & 1 \end{array} \right], then :
A A30−A25=2I\mathrm{A^{30}-A^{25}=2I}
B A30+A25−A=I\mathrm{A^{30}+A^{25}-A=I}
C A30=A25\mathrm{A^{30}=A^{25}}
D A30+A25+A=I\mathrm{A^{30}+A^{25}+A=I}
Correct Answer
Option B
Solution
A=12[13−31]A = {1 \over 2}\left[ \begin{array}{ll}1 & {\sqrt 3 } \\ { - \sqrt 3 } & 1 \end{array} \right]

Let θ=π3\theta=\dfrac{\pi}{3}

A2=[cos⁡θsin⁡θ−sin⁡θcos⁡θ][cos⁡θsin⁡θ−sin⁡θcos⁡θ]=[cos⁡2θsin⁡2θ−sin⁡2θcos⁡2θ]A3=[cos⁡2θsin⁡2θ−sin⁡2θcos⁡2θ][cos⁡θsin⁡θ−sin⁡θcos⁡θ]=[cos⁡3θsin⁡3θ−sin⁡3θcos⁡3θ]\begin{aligned} A^2 & =\left[\begin{array}{cc} \cos \theta & \sin \theta \\\\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & \sin \theta \\\\ -\sin \theta & \cos \theta \end{array}\right] \\\\ & =\left[\begin{array}{cc} \cos 2 \theta & \sin 2 \theta \\\\ -\sin 2 \theta & \cos 2 \theta \end{array}\right] \\\\ A^3 & =\left[\begin{array}{cc} \cos 2 \theta & \sin 2 \theta \\\\ -\sin 2 \theta & \cos 2 \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & \sin \theta \\\\ -\sin \theta & \cos \theta \end{array}\right] \\\\ & =\left[\begin{array}{cc} \cos 3 \theta & \sin 3 \theta \\\\ -\sin 3 \theta & \cos 3 \theta \end{array}\right] \end{aligned}

∴A30=[cos⁡30θsin⁡30θ−sin⁡30θcos⁡30θ]=[1001]A25=[cos⁡25θsin⁡25θ−sin⁡25θcos⁡25θ]=12[13−31]=A∴A30+A25−A=I\begin{aligned} & \therefore \quad A^{30}=\left[\begin{array}{cc}\cos 30 \theta & \sin 30 \theta \\\\ -\sin 30 \theta & \cos 30 \theta\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\ & A^{25}=\left[\begin{array}{cc}\cos 25 \theta & \sin 25 \theta \\\\ -\sin 25 \theta & \cos 25 \theta\end{array}\right]=\dfrac{1}{2}\left[\begin{array}{cc}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{array}\right]=A \\\\ & \therefore \quad A^{30}+A^{25}-A=I\end{aligned}

Q122
Let SS denote the set of all real values of λ\lambda such that the system of equations λx+y+z=1\lambda x+y+z=1 x+λy+z=1x+\lambda y+z=1 x+y+λz=1x+y+\lambda z=1 is inconsistent, then ∑λ∈S(∣λ∣2+∣λ∣)\sum\limits_{\lambda \in S}\left(|\lambda|^{2}+|\lambda|\right) is equal to
A 12
B 2
C 4
D 6
Correct Answer
Option D
Solution

∣λ111λ111λ∣=0\left|\begin{array}{lll}\lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda\end{array}\right|=0

λ(λ2−1)−1(λ−1)+1(1−λ)=0⇒λ3−λ−λ+1+1−λ=0⇒λ3−3λ+2=0⇒(λ−1)(λ2+λ−2)=0\begin{aligned} & \lambda\left(\lambda^{2}-1\right)-1(\lambda-1)+1(1-\lambda)=0 \\\\ & \Rightarrow \lambda^{3}-\lambda-\lambda+1+1-\lambda=0 \\\\ & \Rightarrow \lambda^{3}-3 \lambda+2=0 \\\\ & \Rightarrow (\lambda-1)\left(\lambda^{2}+\lambda-2\right)=0 \end{aligned}

⇒\Rightarrow λ=1,−2\lambda=1,-2 For λ=1⇒∞\lambda=1 \Rightarrow \infty solution λ=−2⇒\lambda=-2 \Rightarrow no solution ∑λ∈S∣λ∣2+∣λ∣=6\sum\limits_{\lambda \in S}|\lambda|^{2}+|\lambda|=6

Q123
For the system of linear equations x+y+z=6x+y+z=6 αx+βy+7z=3\alpha x+\beta y+7 z=3 x+2y+3z=14x+2 y+3 z=14 which of the following is NOT true ?
A If α=β=7\alpha=\beta=7, then the system has no solution
B For every point (α,β)≠(7,7)(\alpha, \beta) \neq(7,7) on the line x−2y+7=0x-2 y+7=0, the system has infinitely many solutions
C There is a unique point (α,β)(\alpha, \beta) on the line x+2y+18=0x+2 y+18=0 for which the system has infinitely many solutions
D If α=β\alpha=\beta and α≠7\alpha \neq 7, then the system has a unique solution
Correct Answer
Option B
Solution

Δ=∣111αβ7123∣\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & \beta & 7 \\ 1 & 2 & 3\end{array}\right|

=1(3β−14)−1(3α−7)+1(2α−β)=3β−14+7−3α+2α−β=2β−α−7\begin{aligned} & =1(3 \beta-14)-1(3 \alpha-7)+1(2 \alpha-\beta) \\\\ & =3 \beta-14+7-3 \alpha+2 \alpha-\beta \\\\ & =2 \beta-\alpha-7 \end{aligned}

So, for α=β≠7,Δ≠0\alpha=\beta \neq 7, \Delta \neq 0 so unique solution. α=β=7\alpha=\beta=7, equation (i) and (ii) represent 2 parallel planes so no solution.

If α−2β+7=0\alpha-2 \beta+7=0, but (α,β)≠(7,7)(\alpha, \beta) \neq(7,7), then no solution.

Q124
Let A=(10004−1012−3)A = \left( \begin{array}{lll}1 & 0 & 0 \\ 0 & 4 & { - 1} \\ 0 & {12} & { - 3} \end{array} \right). Then the sum of the diagonal elements of the matrix (A+I)11{(A + I)^{11}} is equal to :
A 4094
B 2050
C 6144
D 4097
Correct Answer
Option D
Solution

A2=[10004−1012−3][10004−1012−3]A^{2}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right]

=[10004−1012−3]=A=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{array}\right]=A
⇒A3=A4=.......=A\Rightarrow \mathrm{A}_{3}=\mathrm{A}_{4}=.......=\mathrm{A}

Now,

(A+I)11=11C0A11+11C1A10+…11C11I=A(11C0+11C1…11C10)+I=A(211−1)+I\begin{aligned} (A+I)^{11} & ={ }^{11} C_{0} A^{11}+{ }^{11} C_{1} A^{10}+\ldots{ }^{11} C_{11} I \\\\ & =A\left({ }^{11} C_{0}+{ }^{11} C_{1} \ldots{ }^{11} C_{10}\right)+I \\\\ & =A\left(2^{11}-1\right)+I \end{aligned}

Trace of

(A+I)11=211+4(211−1)+1−3(211−1)+1=2×211−4+3+2=212+1=4097\begin{aligned} (A+I)^{11} & =2^{11}+4\left(2^{11}-1\right)+1-3\left(2^{11}-1\right)+1 \\\\ & =2 \times 2^{11}-4+3+2 \\\\ & =2^{12}+1 \\\\ & =4097 \end{aligned}
Q125
For α,β∈R\alpha, \beta \in \mathbb{R}, suppose the system of linear equations x−y+z=52x+2y+αz=83x−y+4z=β \begin{aligned} & x-y+z=5 \\ & 2 x+2 y+\alpha z=8 \\ & 3 x-y+4 z=\beta \end{aligned} has infinitely many solutions. Then α\alpha and β\beta are the roots of :
A x2+18x+56=0x^2+18 x+56=0
B x2−10x+16=0x^2-10 x+16=0
C x2+14x+24=0x^2+14 x+24=0
D x2−18x+56=0x^2-18 x+56=0
Correct Answer
Option D
Solution
Δ=∣1−1122α3−14∣=0\Delta = \left| \begin{array}{lll}1 & { - 1} & 1 \\ 2 & 2 & \alpha \\ 3 & { - 1} & 4 \end{array} \right| = 0
⇒α=4\Rightarrow \alpha = 4
Δ3=0{\Delta _3} = 0
=∣1−152283−1β∣=0= \left| \begin{array}{lll}1 & { - 1} & 5 \\ 2 & 2 & 8 \\ 3 & { - 1} & \beta \end{array} \right| = 0
⇒β=14\Rightarrow \beta = 14

∴\therefore

x2−18x+56=0{x^2} - 18x + 56 = 0
Q126
Let A=[110310−310110]A = \left[ \begin{array}{ll}{{1 \over {\sqrt {10} }}} & {{3 \over {\sqrt {10} }}} \\ {{{ - 3} \over {\sqrt {10} }}} & {{1 \over {\sqrt {10} }}} \end{array} \right] and B=[1−i01]B = \left[ \begin{array}{ll}1 & { - i} \\ 0 & 1 \end{array} \right], where i=−1i = \sqrt { - 1} . If M=ATBA\mathrm{M=A^T B A}, then the inverse of the matrix AM2023AT\mathrm{AM^{2023}A^T} is
A [1−2023i01]\left[ \begin{array}{ll}1 & { - 2023i} \\ 0 & 1 \end{array} \right]
B [102023i1]\left[ \begin{array}{ll}1 & 0 \\ {2023i} & 1 \end{array} \right]
C [12023i01]\left[ \begin{array}{ll}1 & {2023i} \\ 0 & 1 \end{array} \right]
D [10−2023i1]\left[ \begin{array}{ll}1 & 0 \\ { - 2023i} & 1 \end{array} \right]
Correct Answer
Option C
Solution
AAT=[110310−310110][110−310310110]=[1001]=I B2=[1−i01][1−i01]=[1−2i01]B3=[1−3i01]\begin{aligned} & \mathrm{AA}^{\mathrm{T}}=\left[\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}\right]\left[\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] = I \\\\\ & \mathrm{B}^2=\left[\begin{array}{cc} 1 & -\mathrm{i} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -\mathrm{i} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & -2 \mathrm{i} \\ 0 & 1 \end{array}\right] \\\\ & \mathrm{B}^3=\left[\begin{array}{cc} 1 & -3 \mathrm{i} \\ 0 & 1 \end{array}\right] \end{aligned}

Similarly,

B2023=[1−2023i01]M=ATBAM2=M⋅M=ATBAATBA=ATB2 AM3=M2⋅M=ATB2AATBA=ATB3 A\begin{aligned} & \mathrm{B}^{2023}=\left[\begin{array}{cc} 1 & -2023 \mathrm{i} \\ 0 & 1 \end{array}\right] \\\\ & \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA} \\\\ & \mathrm{M}^2=\mathrm{M} \cdot \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA} \mathrm{A}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{~A} \\\\ & \mathrm{M}^3=\mathrm{M}^2 \cdot \mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^2 \mathrm{AA}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^3 \mathrm{~A} \end{aligned}

Similarly,

M2023=ATB2023 AAM2023 AT=AATB2023AAT=I.B2023.I=B2023=[1−2023i01]\begin{aligned} & \mathrm{M}^{2023}= \mathrm{A}^{\mathrm{T}} \mathrm{B}^{2023} \mathrm{~A} \\\\ & \mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}=\mathrm{AA}^{\mathrm{T}} \mathrm{B}^{2023} \mathrm{AA}^{\mathrm{T}} = \mathrm{I}.\mathrm{B}^{2023}.\mathrm{I}=\mathrm{B}^{2023} \\\\ & =\left[\begin{array}{cc} 1 & -2023 \mathrm{i} \\ 0 & 1 \end{array}\right] \end{aligned}

∴\therefore

 Inverse of (AM2023 AT) is [12023i01]\text{ Inverse of }\left(\mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}\right) \text{ is }\left[\begin{array}{cc} 1 & 2023 i \\ 0 & 1 \end{array}\right]
Q127
If PP is a 3×33 \times 3 real matrix such that PT=aP+(a−1)IP^T=a P+(a-1) I, where a>1a>1, then :
A ∣AdjP∣=1|A d j P|=1
B ∣AdjP∣>1|A d j P|>1
C ∣AdjP∣=12|A d j P|=\dfrac{1}{2}
D PP is a singular matrix
Correct Answer
Option A
Solution
P=[a1b1c1a2b2c2a3b3c3]P = \left[ \begin{array}{lll}{{a_1}} & {{b_1}} & {{c_1}} \\ {{a_2}} & {{b_2}} & {{c_2}} \\ {{a_3}} & {{b_3}} & {{c_3}} \end{array} \right]

Given :

PT=aP+(a−1)I{P^T} = aP + (a - 1)I
[a1a2a3b1b2b3c1c2c3]=[aa1+a−1ab1ac1aa2ab2+a−1ac2aa3ab3ac3+a−1]\left[ \begin{array}{lll}{{a_1}} & {{a_2}} & {{a_3}} \\ {{b_1}} & {{b_2}} & {{b_3}} \\ {{c_1}} & {{c_2}} & {{c_3}} \end{array} \right] = \left[ \begin{array}{lll}{a{a_1} + a - 1} & {a{b_1}} & {a{c_1}} \\ {a{a_2}} & {a{b_2} + a - 1} & {a{c_2}} \\ {a{a_3}} & {a{b_3}} & {a{c_3} + a - 1} \end{array} \right]
⇒a1=aa1+a−1⇒a1(1−a)=a−1⇒a1=−1\Rightarrow {a_1} = a{a_1} + a - 1 \Rightarrow {a_1}(1 - a) = a - 1 \Rightarrow {a_1} = - 1

Similarly,

a1=b2=c3=−1{a_1} = {b_2} = {c_3} = - 1

Now,

a2=ab1b1=aa2]→a2=a2a2⇒a2=0⇒b1=0\left. \begin{array}{ll}{a_2} = a{b_1} \\ {b_1} = a{a_2} \end{array} \right] \to {a_2} = {a^2}{a_2} \Rightarrow {a_2} = 0 \Rightarrow {b_1} = 0
c1=aa3{c_1} = a{a_3}

Similarly, all other elements will also be 0

a2=a3=b1=b3=c1=c2=0{a_2} = {a_3} = {b_1} = {b_3} = {c_1} = {c_2} = 0

∴\therefore

P=[−1000−1000−1]P = \left[ \begin{array}{lll}{ - 1} & 0 & 0 \\ 0 & { - 1} & 0 \\ 0 & 0 & { - 1} \end{array} \right]
∣P∣=−1|P| = - 1
∣Adj(P)∣n×n=∣A∣(n−1)|Adj(P){|_{n \times n}} = |A{|^{(n - 1)}}
⇒∣Adj(P)∣=(−1)2=1\Rightarrow |Adj(P)| = {( - 1)^2} = 1
Q128
Let the system of linear equations x+y+kz=2x+y+kz=2 2x+3y−z=12x+3y-z=1 3x+4y+2z=k3x+4y+2z=k have infinitely many solutions. Then the system (k+1)x+(2k−1)y=7(k+1)x+(2k-1)y=7 (2k+1)x+(k+5)y=10(2k+1)x+(k+5)y=10 has :
A unique solution satisfying x−y=1x-y=1
B infinitely many solutions
C no solution
D unique solution satisfying x+y=1x+y=1
Correct Answer
Option D
Solution
x+y+kz=2x + y + kz = 2

............(i)

2x+3y−z=12x + 3y - z = 1

..........(ii)

3x+4y+2z=k3x + 4y + 2z = k

......(iii) (1) + (2)

3x+4y+z(k−1)=33x + 4y + z(k - 1) = 3

Comparing with (3)

k=3k = 3

Now,

4x+5y=74x + 5y = 7
⇒3x+3y=3\Rightarrow 3x + 3y = 3
7x+8y=107x + 8y = 10

as

47≠58{4 \over 7} \ne {5 \over 8}

∴\therefore Unique solution satisfying

x+y=1x + y = 1
Q129
Let A=(mnpq),d=∣A∣≠0A=\left(\begin{array}{cc}\mathrm{m} & \mathrm{n} \\ \mathrm{p} & \mathrm{q}\end{array}\right), \mathrm{d}=|\mathrm{A}| \neq 0 and ∣A−d(AdjA)∣=0\mathrm{|A-d(A d j A)|=0}. Then
A 1+d2=m2+q21+\mathrm{d}^{2}=\mathrm{m}^{2}+\mathrm{q}^{2}
B 1+d2=(m+q)21+d^{2}=(m+q)^{2}
C (1+d)2=m2+q2(1+d)^{2}=m^{2}+q^{2}
D (1+d)2=(m+q)2(1+d)^{2}=(m+q)^{2}
Correct Answer
Option D
Solution
∣A−d(q−n−pm)∣=0\left| {A - d\left( \begin{array}{ll}q & { - n} \\ { - p} & m \end{array} \right)} \right| = 0
∣m−qdn(1+d)p(1+d)q−md∣=0\left| \begin{array}{ll}{m - qd} & {n(1 + d)} \\ {p(1 + d)} & {q - md} \end{array} \right| = 0
(m−qd)(q−md)=np(1+d)2(m - qd)(q - md) = np{(1 + d)^2}
mq−(q2+m2)d+qmd2=np(1+d2)+2npdmq - ({q^2} + {m^2})d + qm{d^2} = np(1 + {d^2}) + 2npd
d2(mq−np)+1(mq−np)=(2np+m2+q2)d{d^2}(mq - np) + 1(mq - np) = (2np + {m^2} + {q^2})d
(d2+1)(mq−np)=(2np+m+a)d({d^2} + 1)(mq - np) = (2np + m + a)d
d2+1=2np+m2+q2{d^2} + 1 = 2np + {m^2} + {q^2}
2d=2mq−2np2d = 2mq - 2np
⇒(1+d)2=(m+q)2\Rightarrow {(1 + d)^2} = {(m + q)^2}
Q130
The set of all values of t∈R\mathrm{t\in \mathbb{R}}, for which the matrix [ete−t(sin⁡t−2cos⁡t)e−t(−2sin⁡t−cos⁡t)ete−t(2sin⁡t+cos⁡t)e−t(sin⁡t−2cos⁡t)ete−tcos⁡te−tsin⁡t]\left[ \begin{array}{lll}{{e^t}} & {{e^{ - t}}(\sin t - 2\cos t)} & {{e^{ - t}}( - 2\sin t - \cos t)} \\ {{e^t}} & {{e^{ - t}}(2\sin t + \cos t)} & {{e^{ - t}}(\sin t - 2\cos t)} \\ {{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \end{array} \right] is invertible, is :
A {kπ,k∈Z}\left\{ {k\pi ,k \in \mathbb{Z}} \right\}
B R\mathbb{R}
C {(2k+1)π2,k∈Z}\left\{ {(2k + 1){\pi \over 2},k \in \mathbb{Z}} \right\}
D {kπ+π4,k∈Z}\left\{ {k\pi + {\pi \over 4},k \in \mathbb{Z}} \right\}
Correct Answer
Option B
Solution

If the matrix is invertible then its determinant should not be zero. So,

∣ete−t(sin⁡t−2cos⁡t)e−t(−2sin⁡t−cos⁡t)ete−t(2sin⁡t+cos⁡t)e−t(sin⁡t−2cos⁡t)ete−tcos⁡te−tsin⁡t∣≠0\left|\begin{array}{ccc} e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\ e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\ e^t & e^{-t} \cos t & e^{-t} \sin t \end{array}\right| \neq 0
⇒et×e−t×e−t∣1sin⁡t−2cos⁡t−2sin⁡t−cos⁡t12sin⁡t+cos⁡tsin⁡t−2cos⁡t1cos⁡tsin⁡t∣≠0\Rightarrow e^t \times e^{-t} \times e^{-t}\left|\begin{array}{ccc} 1 & \sin t-2 \cos t & -2 \sin t-\cos t \\ 1 & 2 \sin t+\cos t & \sin t-2 \cos t \\ 1 & \cos t & \sin t \end{array}\right| \neq 0

Applying, R1→R1−R2R_1 \rightarrow R_1-R_2 then R2→R2−R3R_2 \rightarrow R_2-R_3,

e−t∣0−sin⁡t−3cos⁡t−3sin⁡t+cos⁡t02sin⁡t−2cos⁡t1cos⁡tsin⁡t∣≠0⇒e−t(2sin⁡tcos⁡t+6cos⁡2t+6sin⁡2t−2sin⁡tcos⁡t)≠0⇒6e−t≠0, for ∀t∈R.\begin{aligned} & e^{-t}\left|\begin{array}{ccc} 0 & -\sin t-3 \cos t & -3 \sin t+\cos t \\ 0 & 2 \sin t & -2 \cos t \\ 1 & \cos t & \sin t \end{array}\right| \neq 0 \\\\ & \Rightarrow e^{-t}\left(2 \sin t \cos t+6 \cos ^2 t+6 \sin ^2 t-2 \sin t \cos t\right) \neq 0 \\\\ & \Rightarrow 6e^{-t} \neq 0, \text{ for } \forall t \in R . \end{aligned}
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