Practice Start Test

Matrices and Determinants

JEE Mathematics · 271 questions · Page 26 of 28 · Click an option or "Show Solution" to reveal answer

Q251
If A=(1525−2515)A = \left( \begin{array}{ll}{{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \\ {{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \end{array} \right), B=(10i1)B = \left( \begin{array}{ll}1 & 0 \\ i & 1 \end{array} \right), i=−1i = \sqrt { - 1} , and Q = ATBA, then the inverse of the matrix A Q2021 AT is equal to :
A (15−2021202115)\left( \begin{array}{ll}{{1 \over {\sqrt 5 }}} & { - 2021} \\ {2021} & {{1 \over {\sqrt 5 }}} \end{array} \right)
B (10−2021i1)\left( \begin{array}{ll}1 & 0 \\ { - 2021i} & 1 \end{array} \right)
C (102021i1)\left( \begin{array}{ll}1 & 0 \\ {2021i} & 1 \end{array} \right)
D (1−2021i01)\left( \begin{array}{ll}1 & { - 2021i} \\ 0 & 1 \end{array} \right)
Correct Answer
Option B
Solution
AAT=(1525−2515)(15−252515)A{A^T} = \left( \begin{array}{ll}{{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \\ {{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \end{array} \right)\left( \begin{array}{ll}{{1 \over {\sqrt 5 }}} & {{{ - 2} \over {\sqrt 5 }}} \\ {{2 \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \end{array} \right)
AAT=(1001)=IA{A^T} = \left( \begin{array}{ll}1 & 0 \\ 0 & 1 \end{array} \right) = I
Q2=ATBA ATBA=ATBIBA{Q^2} = {A^T}BA\,{A^T}BA = {A^T}BIBA
⇒Q2=ATB2A\Rightarrow {Q^2} = {A^T}{B^2}A
Q3=ATB2AATBA⇒Q3=ATB3A{Q^3} = {A^T}{B^2}A{A^T}BA \Rightarrow {Q^3} = {A^T}{B^3}A

Similarly :

Q2021=ATB2021A{Q^{2021}} = {A^T}{B^{2021}}A

Now,

B2=(10i1)(10i1)=(102i1){B^2} = \left( \begin{array}{ll}1 & 0 \\ i & 1 \end{array} \right)\left( \begin{array}{ll}1 & 0 \\ i & 1 \end{array} \right) = \left( \begin{array}{ll}1 & 0 \\ {2i} & 1 \end{array} \right)
B3=(102i1)(10i1)⇒B3=(103i1){B^3} = \left( \begin{array}{ll}1 & 0 \\ {2i} & 1 \end{array} \right)\left( \begin{array}{ll}1 & 0 \\ i & 1 \end{array} \right) \Rightarrow {B^3} = \left( \begin{array}{ll}1 & 0 \\ {3i} & 1 \end{array} \right)

Similarly

B2021=(102021i1){B^{2021}} = \left( \begin{array}{ll}1 & 0 \\ {2021i} & 1 \end{array} \right)

∴\therefore

AQ2021=AT=AATB2021 AAT=IB2021IA{Q^{2021}} = {A^T} = A{A^T}{B^{2021}}\,A{A^T} = I{B^{2021}}I
⇒AQ2021 AT=B2021=(102021i1)\Rightarrow A{Q^{2021}}\,{A^T} = {B^{2021}} = \left( \begin{array}{ll}1 & 0 \\ {2021i} & 1 \end{array} \right)

∴\therefore

(AQ2021 AT)−1=(102021i1)−1=(10−2021i1){(A{Q^{2021}}\,{A^T})^{ - 1}} = {\left( \begin{array}{ll}1 & 0 \\ {2021i} & 1 \end{array} \right)^{ - 1}} = \left( \begin{array}{ll}1 & 0 \\ { - 2021i} & 1 \end{array} \right)
Q252
If α\alpha + β\beta + γ\gamma = 2π\pi, then the system of equations x + (cos γ\gamma)y + (cos β\beta)z = 0 (cos γ\gamma)x + y + (cos α\alpha)z = 0 (cos β\beta)x + (cos α\alpha)y + z = 0 has :
A no solution
B infinitely many solution
C exactly two solutions
D a unique solution
Correct Answer
Option B
Solution

Given α\alpha + β\beta + γ\gamma = 2π\pi

Δ=∣1cos⁡γcos⁡βcos⁡γ1cos⁡αcos⁡βcos⁡α1∣\Delta = \left| \begin{array}{lll}1 & {\cos \gamma } & {\cos \beta } \\ {\cos \gamma } & 1 & {\cos \alpha } \\ {\cos \beta } & {\cos \alpha } & 1 \end{array} \right|
=1−cos⁡2α−cos⁡γ(cos⁡γ−cos⁡αcos⁡β)+cos⁡β(cos⁡αcos⁡γ−cos⁡β)= 1 - {\cos ^2}\alpha - \cos \gamma (\cos \gamma - \cos \alpha \cos \beta ) + \cos \beta (\cos \alpha \cos \gamma - \cos \beta )
=1−cos⁡2α−cos⁡2β−cos⁡2γ+2cos⁡αcos⁡βcos⁡γ= 1 - {\cos ^2}\alpha - {\cos ^2}\beta - {\cos ^2}\gamma + 2\cos \alpha \cos \beta \cos \gamma
=sin⁡2α−cos⁡2β−cos⁡γ(cos⁡γ−2cos⁡αcos⁡β)= {\sin ^2}\alpha - {\cos ^2}\beta - \cos \gamma (\cos \gamma - 2\cos \alpha \cos \beta )
=−cos⁡(α+β)cos⁡(α−β)−cos⁡γ(cos⁡(2π−(α−β))−2cos⁡αcos⁡β)= - \cos (\alpha + \beta )\cos (\alpha - \beta ) - \cos \gamma (\cos (2\pi - (\alpha - \beta )) - 2\cos \alpha \cos \beta )
=−cos⁡(2π−γ)cos⁡(α−β)−cos⁡γ(cos⁡(α+β)−2cos⁡αcos⁡β)= - \cos (2\pi - \gamma )\cos (\alpha - \beta ) - \cos \gamma (\cos (\alpha + \beta ) - 2\cos \alpha \cos \beta )
=−cos⁡γcos⁡(α−β)+cos⁡γ(cos⁡αcos⁡β+sin⁡αsin⁡β)= - \cos \gamma \cos (\alpha - \beta ) + \cos \gamma (\cos \alpha \cos \beta + \sin \alpha \sin \beta )
=−cos⁡γcos⁡(α−β)+cos⁡γcos⁡(α−β)= - \cos \gamma \cos (\alpha - \beta ) + \cos \gamma \cos (\alpha - \beta )
=0= 0

So, the system of equation has infinitely many solutions.

Q253
The set of all values of λ\lambda for which the system of linear equations: 2x1−2x2+x3=λx12x1−3x2+2x3=λx2−x1+2x2=λx3\begin{array}{ll}{2{x_1} - 2{x_2} + {x_3} = \lambda {x_1}} \\ {2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}} \\ { - {x_1} + 2{x_2} = \lambda {x_3}} \end{array} has a non-trivial solution
A contains two elements
B contains more than two elements
C in an empty set
D is a singleton
Correct Answer
Option A
Solution
2x1−2x2+x3=λx12x1−3x2+2x3=λx2          −x1+2x2=λx3}\left. \begin{array}{ll}{2{x_1} - 2{x_2} + {x^3} = \lambda {x_1}} \\ {2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}} \\ {\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} = \lambda {x_3}} \end{array} \right\}
⇒       (2−λ)x1−2x2+x3=0            2x1−(3+λ)x2+2x3=0                           −x1+2x2−λx3=0\begin{aligned}& \Rightarrow \,\,\,\,\,\,\,\left( {2 - \lambda } \right){x_1} - 2{x_2} + {x_3} = 0 \\ & \,\,\,\,\,\,\,\,\,\,\,\,2{x_1} - \left( {3 + \lambda } \right){x_2} + 2{x_3} = 0 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} - \lambda {x_3} = 0\end{aligned}

For non-trivial solution,

Δ=0\Delta = 0

i.e.

   ∣2−λ−212−(3+λ)2−12−λ∣=0\,\,\,\left| \begin{array}{lll}{2 - \lambda } & { - 2} & 1 \\ 2 & { - \left( {3 + \lambda } \right)} & 2 \\ { - 1} & 2 & { - \lambda } \end{array} \right| = 0
⇒(2−λ)[λ(3+λ)−4]+\Rightarrow \left( {2 - \lambda } \right)\left[ {\lambda \left( {3 + \lambda } \right) - 4} \right] +
         2[−2λ+2]+1[4−(3+λ)]=0\,\,\,\,\,\,\,\,\,2\left[ { - 2\lambda + 2} \right] + 1\left[ {4 - \left( {3 + \lambda } \right)} \right] = 0
⇒λ3+λ2−5λ+3=0\Rightarrow {\lambda ^3} + {\lambda ^2} - 5\lambda + 3 = 0
⇒λ=1,1,3\Rightarrow \lambda = 1,1,3

Hence, λ\lambda has

22

values.

Q254
Let A be a 3 ×\times 3 matrix with det(A) = 4. Let Ri denote the ith row of A. If a matrix B is obtained by performing the operation R2 → \to 2R2 + 5R3 on 2A, then det(B) is equal to :
A 64
B 16
C 128
D 80
Correct Answer
Option A
Solution
A=[R11R12R13R21R22R23R31R32R33]A = \left[ \begin{array}{lll}{{R_{11}}} & {{R_{12}}} & {{R_{13}}} \\ {{R_{21}}} & {{R_{22}}} & {{R_{23}}} \\ {{R_{31}}} & {{R_{32}}} & {{R_{33}}} \end{array} \right]
2A=[2R112R122R132R212R222R232R312R322R33]2A = \left[ \begin{array}{lll}{2{R_{11}}} & {2{R_{12}}} & {2{R_{13}}} \\ {2{R_{21}}} & {2{R_{22}}} & {2{R_{23}}} \\ {2{R_{31}}} & {2{R_{32}}} & {2{R_{33}}} \end{array} \right]
R2→2R2+5R3{R_2} \to 2{R_2} + 5{R_3}
B=[2R112R122R134R21+10R314R22+10R324R23+10R332R312R322R33]B = \left[ \begin{array}{lll}{2{R_{11}}} & {2{R_{12}}} & {2{R_{13}}} \\ {4{R_{21}} + 10{R_{31}}} & {4{R_{22}} + 10{R_{32}}} & {4{R_{23}} + 10{R_{33}}} \\ {2{R_{31}}} & {2{R_{32}}} & {2{R_{33}}} \end{array} \right]
R2→R2−5R3{R_2} \to {R_2} - 5{R_3}
B=[2R112R122R134R214R224R232R312R322R33]B = \left[ \begin{array}{lll}{2{R_{11}}} & {2{R_{12}}} & {2{R_{13}}} \\ {4{R_{21}}} & {4{R_{22}}} & {4{R_{23}}} \\ {2{R_{31}}} & {2{R_{32}}} & {2{R_{33}}} \end{array} \right]
∣B∣=[2R112R122R134R214R224R232R312R322R33]\left| B \right| = \left[ \begin{array}{lll}{2{R_{11}}} & {2{R_{12}}} & {2{R_{13}}} \\ {4{R_{21}}} & {4{R_{22}}} & {4{R_{23}}} \\ {2{R_{31}}} & {2{R_{32}}} & {2{R_{33}}} \end{array} \right]
∣B∣=2×2×4∣R11R12R13R21R22R23R31R32R33∣\left| B \right| = 2 \times 2 \times 4\left| \begin{array}{lll}{{R_{11}}} & {{R_{12}}} & {{R_{13}}} \\ {{R_{21}}} & {{R_{22}}} & {{R_{23}}} \\ {{R_{31}}} & {{R_{32}}} & {{R_{33}}} \end{array} \right|
=16×4= 16 \times 4
=64= 64
Q255
Let S be the set of all λ\lambda ∈ \in R for which the system of linear equations 2x – y + 2z = 2 x – 2y + λ\lambda z = –4 x + λ\lambda y + z = 4 has no solution. Then the set S :
A contains more than two elements.
B contains exactly two elements.
C is a singleton.
D is an empty set.
Correct Answer
Option B
Solution

For no solution :

Δ\Delta

= 0 and

Δ\Delta

1/

Δ\Delta

2/

Δ\Delta

3 ≠\ne 0

Δ\Delta

=

∣2−121−2λ1λ1∣\left| \begin{array}{lll}2 & { - 1} & 2 \\ 1 & { - 2} & \lambda \\ 1 & \lambda & 1 \end{array} \right|

= 0 ⇒\Rightarrow 2(–2 – λ\lambda2) + 1 (1 – λ\lambda) + 2(λ\lambda + 2) = 0 ⇒\Rightarrow –2λ\lambda2 + λ\lambda + 1 = 0 ⇒\Rightarrow λ\lambda = 1,

−12- {1 \over 2}

When λ\lambda = 1 2x – y + 2z = 2 ...(1) x – 2y + z = –4 ...(2) x + y + z = 4 ...(

3) Adding (2) and (3), we get 2x – y + 2z = 0 (contradiction) hence no solution. ∴\therefore λ\lambda = 1 belongs to set S.

When λ\lambda =

−12- {1 \over 2}

2x – y + 2z = 2 ...(1) x – 2y

−12- {1 \over 2}

z = –4 ...(2) x

−12- {1 \over 2}

y + z = 4 ...(3) (1) and (3) contradict each other, hence no solution. ∴\therefore λ\lambda =

−12- {1 \over 2}

belongs to set S.

Q256
Consider the following system of equations : x + 2y −- 3z = a 2x + 6y −- 11z = b x −- 2y + 7z = c, where a, b and c are real constants. Then the system of equations :
A has no solution for all a, b and c
B has a unique solution when 5a = 2b + c
C has infinite number of solutions when 5a = 2b + c
D has a unique solution for all a, b and c
Correct Answer
Option C
Solution
D=∣12−326−111−27∣D = \left| \begin{array}{lll}1 & 2 & { - 3} \\ 2 & 6 & { - 11} \\ 1 & { - 2} & 7 \end{array} \right|

= 20 −- 2(25) −-3(−-10) = 20 −- 50 + 30 = 0

D1=∣a2−3b6−11c−27∣{D_1} = \left| \begin{array}{lll}a & 2 & { - 3} \\ b & 6 & { - 11} \\ c & { - 2} & 7 \end{array} \right|

= 20a −- 2(7b + 11c) −-3(−-2b −- 6c) = 20a −- 14b −- 22c + 6b +18c = 20a −- 8b −- 4c = 4(5a −- 2b −- c)

D2=∣1a−32b−111c7∣{D_2} = \left| \begin{array}{lll}1 & a & { - 3} \\ 2 & b & { - 11} \\ 1 & c & 7 \end{array} \right|

= 7b + 11c −- a(25) −-3(2c −- b) = 7b + 11c −- 25a −- 6c + 3b = −-25a + 10b + 5c = −-5(5a −- 2b −- c)

D3=∣12a26b1−2c∣{D_3} = \left| \begin{array}{lll}1 & 2 & a \\ 2 & 6 & b \\ 1 & { - 2} & c \end{array} \right|

= 6c + 2b −- 2(2c −- b) −- 10a = −-10a + 4b + 2c = −-2(5a −- 2b −- c) for infinite solution

D=D1=D2=D3=0D = {D_1} = {D_2} = {D_3} = 0

⇒\Rightarrow 5a = 2b + c

Q257
The sum of the real roots of the equation ∣x−6−12−3xx−3−32xx+2∣=0\left| \begin{array}{lll}x & { - 6} & { - 1} \\ 2 & { - 3x} & {x - 3} \\ { - 3} & {2x} & {x + 2} \end{array} \right| = 0, is equal to :
A - 4
B 0
C 1
D 6
Correct Answer
Option B
Solution

x(-3x ×\times (x + 2) - 2x(x - 3)) + (– 6) (2(x + 2) + 3 (x – 3)) + (–1) (4x + 3 (–3x)) ⇒\Rightarrow – 5x3 + 30x –30 + 5x = 0 ⇒\Rightarrow x3 – 7x + 6 = 0 ∴\therefore sum of roots = 0

Q258
Let the number 2,b,c be in an A.P. and A = [1112bc4b2c2]\left[ \begin{array}{lll}1 & 1 & 1 \\ 2 & b & c \\ 4 & {{b^2}} & {{c^2}} \end{array} \right]. If det(A) ∈ \in [2, 16], then c lies in the interval :
A [2, 3)
B [4, 6]
C (2 + 23/4, 4)
D [3, 2 + 23/4]
Correct Answer
Option B
Solution

2, b, c are in AP. Let common difference = d ∴\therefore b = 2 + d and c = 2 + 2d |A| =

[1112bc4b2c2]\left[ \begin{array}{lll}1 & 1 & 1 \\ 2 & b & c \\ 4 & {{b^2}} & {{c^2}} \end{array} \right]

C2 = C2 - C1 C3 = C3 - C1 =

∣1002b−2c−24b2−4c2−4∣\left| \begin{array}{lll}1 & 0 & 0 \\ 2 & {b - 2} & {c - 2} \\ 4 & {{b^2} - 4} & {{c^2} - 4} \end{array} \right|

=

(b−2)(c−2)∣1002114b+2c+2∣\left( {b - 2} \right)\left( {c - 2} \right)\left| \begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 1 \\ 4 & {b + 2} & {c + 2} \end{array} \right|

=

(b−2)(c−2)[c+2−b−2]\left( {b - 2} \right)\left( {c - 2} \right)\left[ {c + 2 - b - 2} \right]

=

(b−2)(c−2)(c−b)\left( {b - 2} \right)\left( {c - 2} \right)\left( {c - b} \right)

[ As b = 2 + d and c = 2 + 2d, then b - 2 = 4, c - 2 = 2d and c - b = d] = (d) (2d) (d) = 2d3 Given |A|

∈\in

[2, 16] ∴\therefore 2d3

∈\in

[2, 16] ⇒\Rightarrow d3

∈\in

[1, 8] ⇒\Rightarrow d

∈\in

[1, 2] As c = 2 + 2d then c

∈\in

[4, 6]

Q259
Let A = [aij] be a real matrix of order 3 ×\times 3, such that ai1 + ai2 + ai3 = 1, for i = 1, 2, 3. Then, the sum of all the entries of the matrix A3 is equal to :
A 2
B 1
C 3
D 9
Correct Answer
Option C
Solution
A=[a11a12a13a21a22a23a31a32a33]A = \left[ \begin{array}{lll}{{a_{11}}} & {{a_{12}}} & {{a_{13}}} \\ {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \\ {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \end{array} \right]

Let

x=[111]x = \left[ \begin{array}{ll}1 \\ 1 \\ 1 \end{array} \right]
AX=[a11+a12+a13a21+a22+a23a31+a32+a33]=[111]AX = \left[ \begin{array}{ll}{{a_{11}} + {a_{12}} + {a_{13}}} \\ {{a_{21}} + {a_{22}} + {a_{23}}} \\ {{a_{31}} + {a_{32}} + {a_{33}}} \end{array} \right] = \left[ \begin{array}{ll}1 \\ 1 \\ 1 \end{array} \right]

⇒\Rightarrow AX = X Replace X by AX A2X = AX = X Replace X by AX A3X = AX = X Let

A3=[x1x2x3y1y2y3z1z2z3]{A^3} = \left[ \begin{array}{lll}{{x_1}} & {{x_2}} & {{x_3}} \\ {{y_1}} & {{y_2}} & {{y_3}} \\ {{z_1}} & {{z_2}} & {{z_3}} \end{array} \right]
A3[111]=[x1x2x3y1y2y3z1z2z3]=[111]{A^3}\left[ \begin{array}{ll}1 \\ 1 \\ 1 \end{array} \right] = \left[ \begin{array}{lll}{{x_1}} & {{x_2}} & {{x_3}} \\ {{y_1}} & {{y_2}} & {{y_3}} \\ {{z_1}} & {{z_2}} & {{z_3}} \end{array} \right] = \left[ \begin{array}{ll}1 \\ 1 \\ 1 \end{array} \right]

Sum of all the element = 3

Q260
If α,β≠0,\alpha ,\beta \ne 0, and f(n)=αn+βnf\left( n \right) = {\alpha ^n} + {\beta ^n} and ∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣\left| \begin{array}{lll}3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \\ {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \\ {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \end{array} \right| =K(1−α)2(1−β)2(α−β)2, = K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2},then then KK$ is equal to :
A 11
B −1-1
C αβ\alpha \beta
D 1αβ{1 \over {\alpha \beta }}
Correct Answer
Option A
Solution

Consider

∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣\left| \begin{array}{lll}3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \\ {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \\ {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \end{array} \right|
           =∣1+1+11+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4∣\,\,\,\,\,\,\,\,\,\,\, = \left| \begin{array}{lll}{1 + 1 + 1} & {1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} \\ {1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} \\ {1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} & {1 + {\alpha ^4} + {\beta ^4}} \end{array} \right|
           =∣1111αβ1α2β2∣×∣1111αβ1α2β2∣\,\,\,\,\,\,\,\,\,\,\, = \left| \begin{array}{lll}1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {{\alpha ^2}} & {{\beta ^2}} \end{array} \right| \times \left| \begin{array}{lll}1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {{\alpha ^2}} & {{\beta ^2}} \end{array} \right|
           =∣1111αβ1α2β2∣2\,\,\,\,\,\,\,\,\,\,\, = {\left| \begin{array}{lll}1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {{\alpha ^2}} & {{\beta ^2}} \end{array} \right|^2}
           =[(1−α)(1−β)(α−β)]2\,\,\,\,\,\,\,\,\,\,\, = {\left[ {\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {\alpha - \beta } \right)} \right]^2}

So,

k=1k=1
Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →