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Matrices and Determinants

JEE Mathematics · 271 questions · Page 12 of 28 · Click an option or "Show Solution" to reveal answer

Q111
Let A be a 2 ×\times 2 matrix with det (A) = −- 1 and det ((A + I) (Adj (A) + I)) = 4. Then the sum of the diagonal elements of A can be :
A −-1
B 2
C 1
D −2- \sqrt2
Correct Answer
Option B
Solution
∣(A+I)(adj A+I)∣=4|(A + I)(adj\,A + I)| = 4
⇒∣A adj A+A+adj A+I∣=4\Rightarrow |A\,adj\,A + A + adj\,A + I| = 4
⇒∣(A)I+A+adj A+I∣=4\Rightarrow |(A)I + A + adj\,A + I| = 4
∣A∣=−1⇒∣A+adj A∣=4|A| = - 1 \Rightarrow |A + adj\,A| = 4
A=[abcd] adj A=[a−b−cd]A = \left[ \begin{array}{ll}a & b \\ c & d \end{array} \right]\,adj\,A = \left[ \begin{array}{ll}a & { - b} \\ { - c} & d \end{array} \right]
⇒∣(a+d)00(a+d)∣=4\Rightarrow \left| \begin{array}{ll}{(a + d)} & 0 \\ 0 & {(a + d)} \end{array} \right| = 4
⇒a+d= ± 2\Rightarrow a + d = \, \pm \,2
Q112
 Let A=[111] and B=[92−102112122132−142−152162172], then the value of A′BA is:  \text{ Let } A=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right] \text{ and } B=\left[\begin{array}{ccc} 9^{2} & -10^{2} & 11^{2} \\ 12^{2} & 13^{2} & -14^{2} \\ -15^{2} & 16^{2} & 17^{2} \end{array}\right] \text{, then the value of } A^{\prime} B A \text{ is: }
A 1224
B 1042
C 540
D 539
Correct Answer
Option D
Solution
A′BA=[111][92−102112122132−142−152162172]AA'BA = \left[ \begin{array}{lll}1 & 1 & 1 \end{array} \right]\left[ \begin{array}{lll}{{9^2}} & { - {{10}^2}} & {{{11}^2}} \\ {{{12}^2}} & {{{13}^2}} & { - {{14}^2}} \\ { - {{15}^2}} & {{{16}^2}} & {{{17}^2}} \end{array} \right]A
=[92+122−152−102+132+162112−142+172][111]= \left[ \begin{array}{lll}{{9^2} + {{12}^2} - {{15}^2}} & { - {{10}^2} + {{13}^2} + {{16}^2}} & {{{11}^2} - {{14}^2} + {{17}^2}} \end{array} \right]\left[ \begin{array}{ll}1 \\ 1 \\ 1 \end{array} \right]
=[92+122−152−102+132+162+112−142+172]= \left[ {{9^2} + {{12}^2} - {{15}^2} - {{10}^2} + {{13}^2} + {{16}^2} + {{11}^2} - {{14}^2} + {{17}^2}} \right]
=[(92−102)+(112+122)+(132−142)+(162−152)+172]= [({9^2} - {10^2}) + ({11^2} + {12^2}) + ({13^2} - {14^2}) + ({16^2} - {15^2}) + {17^2}]
=[−19+265+(−27)+31+289]= [ - 19 + 265 + ( - 27) + 31 + 289]
=[585−46]=[539]= [585 - 46] = [539]
Q113
Let A=(12−2−5)A=\left(\begin{array}{cc}1 & 2 \\ -2 & -5\end{array}\right). Let α,β∈R\alpha, \beta \in \mathbb{R} be such that αA2+βA=2I\alpha A^{2}+\beta A=2 I. Then α+β\alpha+\beta is equal to
A −-10
B −-6
C 6
D 10
Correct Answer
Option D
Solution
A2=[12−2−5][12−2−5]=[−3−8821]{A^2} = \left[ \begin{array}{ll}1 & 2 \\ { - 2} & { - 5} \end{array} \right]\left[ \begin{array}{ll}1 & 2 \\ { - 2} & { - 5} \end{array} \right] = \left[ \begin{array}{ll}{ - 3} & { - 8} \\ 8 & {21} \end{array} \right]
αA2+βA=[−3α−8α8α21α]+[β2β−2β−5β]\alpha {A^2} + \beta A = \left[ \begin{array}{ll}{ - 3\alpha } & { - 8\alpha } \\ {8\alpha } & {21\alpha } \end{array} \right] + \left[ \begin{array}{ll}\beta & {2\beta } \\ { - 2\beta } & { - 5\beta } \end{array} \right]
=[−3α+β−8α+2β8α−2β21α−5β]=[2002]= \left[ \begin{array}{ll}{ - 3\alpha + \beta } & { - 8\alpha + 2\beta } \\ {8\alpha - 2\beta } & {21\alpha - 5\beta } \end{array} \right] = \left[ \begin{array}{ll}2 & 0 \\ 0 & 2 \end{array} \right]

On Comparing

8α=2β, −3α+β=2, 21α−5β=28\alpha = 2\beta ,\, - 3\alpha + \beta = 2,\,21\alpha - 5\beta = 2
⇒α=2, β=8\Rightarrow \alpha = 2,\,\beta = 8

So,

α+β=10\alpha + \beta = 10
Q114
Let A=(4−2αβ)A=\left(\begin{array}{rr}4 & -2 \\ \alpha & \beta\end{array}\right). If A2+γA+18I=O\mathrm{A}^{2}+\gamma \mathrm{A}+18 \mathrm{I}=\mathrm{O}, then det⁡(A)\operatorname{det}(\mathrm{A}) is equal to _____________.
A −-18
B 18
C −-50
D 50
Correct Answer
Option B
Solution

Characteristic equation of A is given by

∣A−λI∣=0\left| {A - \lambda I} \right| = 0
∣4−λ−2αβ−λ∣=0\left| \begin{array}{ll}{4 - \lambda } & { - 2} \\ \alpha & {\beta - \lambda } \end{array} \right| = 0
⇒λ2−(4+β)λ+(4β+2α)=0\Rightarrow {\lambda ^2} - (4 + \beta )\lambda + (4\beta + 2\alpha ) = 0

So,

A2−(4+β)A+(4β+2α)I=0{A^2} - (4 + \beta )A + (4\beta + 2\alpha )I = 0
∣A∣=4β+2α=18|A| = 4\beta + 2\alpha = 18
Q115
Let the matrix A=[010001100]A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right] and the matrix B0=A49+2A98B_{0}=A^{49}+2 A^{98}. If Bn=Adj(Bn−1)B_{n}=A d j\left(B_{n-1}\right) for all n≥1n \geq 1, then det⁡(B4)\operatorname{det}\left(B_{4}\right) is equal to :
A 3283^{28}
B 3303^{30}
C 3323^{32}
D 3363^{36}
Correct Answer
Option C
Solution
A=[010001100]A = \left[ \begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right]
⇒A2=[010001100]×[010001100]=[001100010]\Rightarrow {A^2} = \left[ \begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right] \times \left[ \begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right] = \left[ \begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right]
⇒A3=[001100010][010001100]=[100010001]=l\Rightarrow {A^3} = \left[ \begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right]\left[ \begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right] = \left[ \begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] = l

Now

B0=A49+2A98=(A3)16 . A+2(A3)32 . A2{B_0} = {A^{49}} + 2{A^{98}} = {({A^3})^{16}}\,.\,A + 2{({A^3})^{32}}\,.\,{A^2}
B0=A+2A2=[010001100]+[002200020]=[012201120]{B_0} = A + 2{A^2} = \left[ \begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right] + \left[ \begin{array}{lll}0 & 0 & 2 \\ 2 & 0 & 0 \\ 0 & 2 & 0 \end{array} \right] = \left[ \begin{array}{lll}0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0 \end{array} \right]
∣B0∣=9|{B_0}| = 9

Since,

Bn=Adj ∣Bn−1∣⇒∣Bn∣=∣Bn−1∣2{B_n} = Adj\,|{B_{n - 1}}| \Rightarrow |{B_n}| = |{B_{n - 1}}{|^2}

Hence

∣B4∣=∣B3∣2=∣B2∣4=∣B1∣8=∣B0∣16|{B_4}| = |{B_3}{|^2} = |{B_2}{|^4} = |{B_1}{|^8} = |{B_0}{|^{16}}
=∣32∣16=332= |{3^2}{|^{16}} = {3^{32}}
Q116
Let A\mathrm{A} and B\mathrm{B} be any two 3×33 \times 3 symmetric and skew symmetric matrices respectively. Then which of the following is NOT true?
A A4−B4\mathrm{A}^{4}-\mathrm{B}^{4} is a smmetric matrix
B AB−BA\mathrm{AB}-\mathrm{BA} is a symmetric matrix
C B5−A5\mathrm{B}^{5}-\mathrm{A}^{5} is a skew-symmetric matrix
D AB+BA\mathrm{AB}+\mathrm{BA} is a skew-symmetric matrix
Correct Answer
Option C
Solution

(A)

M=A4−B4M = {A^4} - {B^4}
MT=(A4−B4)T=(AT)4−(BT)4{M^T} = {({A^4} - {B^4})^T} = {({A^T})^4} - {({B^T})^4}
=A4−(−B)4=A4−B4=M= {A^4} - {( - B)^4} = {A^4} - {B^4} = M

(B)

M=AB−BAM = AB - BA
MT=(AB−BA)T=(AB)T−(BA)T{M^T} = {(AB - BA)^T} = {(AB)^T} - {(BA)^T}
=BTAT−ATBT= {B^T}{A^T} - {A^T}{B^T}
=−BA−A(−B)= - BA - A( - B)
=AB−BA=M= AB - BA = M

(C)

M=B5−A5M = {B^5} - {A^5}
MT=(BT)5−(AT)5=−(B5+A5)≠−M{M^T} = {({B^T})^5} - {({A^T})^5} = - ({B^5} + {A^5}) \ne - M

(D)

M=AB+BAM = AB + BA
MT=(AB)T+(BA)T{M^T} = {(AB)^T} + {(BA)^T}
=BTAT+ATBT=−BA−AB=−M= {B^T}{A^T} + {A^T}{B^T} = - BA - AB = - M
Q117
Let A and B be two 3×33 \times 3 non-zero real matrices such that AB is a zero matrix. Then
A the system of linear equations AX=0A X=0 has a unique solution
B the system of linear equations AX=0A X=0 has infinitely many solutions
C B is an invertible matrix
D adj⁡(A)\operatorname{adj}(\mathrm{A}) is an invertible matrix
Correct Answer
Option B
Solution

AB is zero matrix

⇒∣A∣=∣B∣=0\Rightarrow |A| = |B| = 0

So neither A nor B is invertible If

∣A∣=0|A| = 0
⇒∣adj A∣=0\Rightarrow |\mathrm{adj}\,A| = 0

so

adj A\mathrm{adj}\,A
AX=0AX = 0

is homogeneous system and

∣A∣=0|A| = 0

So, it is having infinitely many solutions

Q118
Which of the following matrices can NOT be obtained from the matrix [−121−1]\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right] by a single elementary row operation ?
A [011−1]\left[\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right]
B [1−1−12]\left[\begin{array}{cc}1 & -1 \\ -1 & 2\end{array}\right]
C [−12−27]\left[\begin{array}{rr}-1 & 2 \\ -2 & 7\end{array}\right]
D [−12−13]\left[\begin{array}{ll}-1 & 2 \\ -1 & 3\end{array}\right]
Correct Answer
Option C
Solution

Given matrix

A=[−121−1]A = \left[ \begin{array}{ll}{ - 1} & 2 \\ 1 & { - 1} \end{array} \right]

For option A :

R1→R1+R2{R_1} \to {R_1} + {R_2}
A=[011−1]A = \left[ \begin{array}{ll}0 & 1 \\ 1 & { - 1} \end{array} \right]

∴\therefore Option A can be obtained. For option B :

R1↔R2{R_1} \leftrightarrow {R_2}
A=[1−1−12]A = \left[ \begin{array}{ll}1 & { - 1} \\ { - 1} & 2 \end{array} \right]

∴\therefore Option B can be obtained. Option C : Not possible by a single elementary row operation. Option D :

R2→R2+2R1{R_2} \to {R_2} + 2{R_1}
A=[−12−13]A = \left[ \begin{array}{ll}{ - 1} & 2 \\ { - 1} & 3 \end{array} \right]

∴\therefore Option D can be obtained.

Q119
If the system of equations x+y+z=62x+5y+αz=βx+2y+3z=14 \begin{aligned} &x+y+z=6 \\ &2 x+5 y+\alpha z=\beta \\ &x+2 y+3 z=14 \end{aligned} has infinitely many solutions, then α+β\alpha+\beta is equal to
A 8
B 36
C 44
D 48
Correct Answer
Option C
Solution

Given,

x+y+z=6x + y + z = 6

...... (1)

2x+5y+αz=β2x + 5y + \alpha z = \beta

..... (2)

x+2y+3z=14x + 2y + 3z = 14

...... (3) System of equation have infinite many solutions. ∴\therefore

Δx=Δy=Δz=0{\Delta _x} = {\Delta _y} = {\Delta _z} = 0

and

Δ=0\Delta = 0

Now,

Δ=∣11125α123∣=0\Delta = \left| \begin{array}{lll}1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{array} \right| = 0
C1→C1−C3{C_1} \to {C_1} - {C_3}
C2→C2−C3{C_2} \to {C_2} - {C_3}
⇒∣0012−α5−αα−2−13∣=0\Rightarrow \left| \begin{array}{lll}0 & 0 & 1 \\ {2 - \alpha } & {5 - \alpha } & \alpha \\ { - 2} & { - 1} & 3 \end{array} \right| = 0
⇒−2+α+10−2α=0\Rightarrow - 2 + \alpha + 10 - 2\alpha = 0
⇒8−α=0\Rightarrow 8 - \alpha = 0
⇒α=8\Rightarrow \alpha = 8

Now,

x+y+z=6x + y + z = 6
2x+5y+8z=β2x + 5y + 8z = \beta
x+2y+3z=14x + 2y + 3z = 14

∴\therefore

Δx=∣611β581423∣=0{\Delta _x} = \left| \begin{array}{lll}6 & 1 & 1 \\ \beta & 5 & 8 \\ {14} & 2 & 3 \end{array} \right| = 0
C1→C1−6C3{C_1} \to {C_1} - 6{C_3}
C2→C2−C3{C_2} \to {C_2} - {C_3}
⇒∣001β−48−38−4−13∣=0\Rightarrow \left| \begin{array}{lll}0 & 0 & 1 \\ {\beta - 48} & { - 3} & 8 \\ { - 4} & { - 1} & 3 \end{array} \right| = 0
⇒−β+48−12=0\Rightarrow - \beta + 48 - 12 = 0
⇒β=36\Rightarrow \beta = 36

∴\therefore

α+β=8+36=44\alpha + \beta = 8 + 36 = 44
Q120
For the system of linear equations αx+y+z=1,x+αy+z=1,x+y+αz=β\alpha x+y+z=1,x+\alpha y+z=1,x+y+\alpha z=\beta, which one of the following statements is NOT correct?
A It has infinitely many solutions if α=1\alpha=1 and β=1\beta=1
B It has infinitely many solutions if α=2\alpha=2 and β=−1\beta=-1
C x+y+z=34x+y+z=\dfrac{3}{4} if α=2\alpha=2 and β=1\beta=1
D It has no solution if α=−2\alpha=-2 and β=1\beta=1
Correct Answer
Option B
Solution

For infinite solution Δ=Δx=Δy=Δz=0\Delta=\Delta_x=\Delta_y=\Delta_z=0

Δ=∣α111α111α∣=0⇒(α3−3α+2)=0⇒α=1,−2\Delta=\left|\begin{array}{lll} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{array}\right|=0 \Rightarrow\left(\alpha^3-3 \alpha+2\right)=0 \Rightarrow \alpha=1,-2

If β=1\beta=1, then all planes are overlapping ∴\therefore Option (A) is correct.

Option (B) : If α=2⇒Δ≠0\alpha=2 \Rightarrow \Delta \neq 0 ∴\therefore Unique solution exist ∴\therefore Option (B) is incorrect.

Option (C) :

α=2,β=12x+y+z=1x+2y+z=1x+y+2z=1\begin{aligned} & \alpha=2, \beta=1 \\\\ & 2 x+y+z=1 \\\\ & x+2 y+z=1 \\\\ & x+y+2 z=1 \end{aligned}

Adding all three equations,

x+y+z=34x+y+z=\frac{3}{4}

∴\therefore option (C) is correct.

Option (D) : If α=−2\alpha=-2 and β=1\beta=1, then Δ=0,Δx≠0\Delta=0, \Delta_x \neq 0 ∴\therefore No solution.

∴\therefore Option (D) is correct.

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