Let $$A = \left( {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right).$$ and $$10$$ $$B = \left( {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alph

Given that $$10B$$ $$\,\,\, = \left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right]$$ $$ \Rightarrow B = {1 \over {10}}\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right]$$ Also since, $$B = {A^{ - 1}} \Rightarrow AB = I$$ $$ \Rightarrow {1 \over {10}}\left[ {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right]\left[ {\matrix{ 4 & 2 &am

Matrices and Determinants — Page 22 | JEE Mathematics

Q211

Let the system of equations :

\begin{aligned} & 2 x+3 y+5 z=9 \\ & 7 x+3 y-2 z=8 \\ & 12 x+3 y-(4+\lambda) z=16-\mu \end{aligned}

have infinitely many solutions. Then the radius of the circle centred at

(\lambda, \mu)

and touching the line

4 x=3 y

is :

A

\dfrac{7}{5}

B

\dfrac{21}{5}

C 7

D

\dfrac{17}{5}

Correct Answer

Option A

Solution

\begin{aligned} & \Delta=\left|\begin{array}{ccc} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(4+\lambda) \end{array}\right| \\ & =2(-12-3 \lambda+6)-3(-28-7 \lambda+24)+5(21-36) \\ & =-12-6 \lambda+12+21 \lambda-75 \\ & =15 \lambda-75 \\ & \Rightarrow 15 \lambda-75=0 \\ & \Rightarrow \lambda=5 \\ & \Delta_1=\left|\begin{array}{ccc} 9 & 3 & 5 \\ 8 & 3 & -2 \\ 16-\mu & 3 & -9 \end{array}\right| \\ & =9(-27+6)-3(-72+32-2 \mu)+5(24-48+3 \mu) \\ & =-189+120+6 \mu-120+15 \mu \\ & =21 \mu-189=0 \\ & \Rightarrow \mu=9 \end{aligned}

\begin{aligned} & \therefore r=\left|\frac{4(5)-3(9)}{\sqrt{(4)^2+(3)^2}}\right| \\ & r=\frac{7}{5} \end{aligned}

Q212

If the system of equations

\begin{aligned} & 2 x+\lambda y+3 z=5 \\ & 3 x+2 y-z=7 \\ & 4 x+5 y+\mu z=9 \end{aligned}

has infinitely many solutions, then

\left(\lambda^2+\mu^2\right)

is equal to :

A 30

B 26

C 22

D 18

Correct Answer

Option B

Solution

\begin{aligned} &\begin{aligned} & 2 x+\lambda y+3 z=5 \\ & 3 x+2 y-z=7 \\ & 4 x+5 y+\mu z=9 \end{aligned}\\ &\text{ For infinite solutions } \Rightarrow \Delta=0=\Delta_1=\Delta_2=\Delta_3 \end{aligned}

\begin{aligned} & \Delta=\left|\begin{array}{lll} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{array}\right|=0 \\ & \Rightarrow-4 \lambda-3 \lambda \mu+4 \mu+31=0 \\ & \Delta_1=\left|\begin{array}{ccc} 5 & \lambda & 3 \\ 7 & 2 & -1 \\ 9 & 5 & \mu \end{array}\right|=0 \Rightarrow-9 \lambda-7 \lambda \mu+10 \mu+76=0 \\ & \Delta_2=\left|\begin{array}{ccc} 2 & 3 & 5 \\ 3 & -1 & 7 \\ 4 & \mu & 9 \end{array}\right|=0 \Rightarrow \mu+5=0 \Rightarrow \mu=-5 \\ & \Delta_3=\left|\begin{array}{lll} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{array}\right|=0 \Rightarrow \lambda+1=0 \Rightarrow \lambda=-1 \end{aligned}

\begin{aligned} &\therefore \text{ For infinite solution } \mu=-5 \text{ and } \lambda=-1\\ &\begin{aligned} \text{ Now } \mu^2+\lambda^2 & =25+1 \\ & =26 \end{aligned} \end{aligned}

Q213

Let

A

be a

3 \times 3

real matrix such that

A^2(A-2 I)-4(A-I)=O

, where

I

and

O

are the identity and null matrices, respectively. If

A^5=\alpha A^2+\beta A+\gamma I

, where

\alpha, \beta

, and

\gamma

are real constants, then

\alpha+\beta+\gamma

is equal to :

A 76

B 12

C 4

D 20

Correct Answer

Option B

Solution

\begin{aligned} & A^2(A-2 I)-4(A-I)=0 \\ & A^3-2 A^2-4 A+4 I=0 \end{aligned}

Multiply by $A$

\begin{aligned} & A^4=2 A^3+4 A^2-4 A \\ & A^4=2\left(2 A^2+4 A-4 I\right)+4 A^2-4 A \\ & A^4=8 A^2+4 A-8 I \end{aligned}

Multiply again by $A$

\begin{aligned} & \Rightarrow A^5=8 A^3+4 A^2-8 A \\ & \Rightarrow A^5=8\left(2 A^2+4 A-4 I\right)+4 A^2-8 A \\ & \Rightarrow A^5=20 A^2+24 A-32 I \end{aligned}

Comparing with $A^5=\alpha A^2+\beta A+\gamma I$

\begin{aligned} & \alpha=20, \beta=24, \gamma=-32 \\ & \therefore \alpha+\beta+\gamma=20+24-32 \\ & =44-32 \\ & =12 \end{aligned}

Q214

Let

A

be a matrix of order

3 \times 3

and

|A|=5

. If

|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))|=2^\alpha \cdot 3^\beta \cdot 5^\gamma, \alpha, \beta, \gamma \in N

, then

\alpha+\beta+\gamma

is equal to

A 26

B 27

C 25

D 28

Correct Answer

Option B

Solution

To find the expression $|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))|$ , we break it down as follows: Recognize that: $|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))| = 2^3 |3A (\operatorname{adj}(2A))|^2$ Apply properties of determinants: $= 2^3 (3^3)^2 |A|^2 \left|\operatorname{adj}(2A)\right|^2$ Further simplify using $|\operatorname{adj}(B)| = |B|^{n-1}$ for a $3 \times 3$ matrix: $= 2^3 \cdot 3^6 \cdot 5^2 \cdot (|2A|^2)^2$ Simplify $|2A|$ : $= 2^3 \cdot 3^6 \cdot 5^2 \cdot (2^3)^4 \cdot |A|^4$ Continue to simplify: $= 2^3 \cdot 3^6 \cdot 5^2 \cdot (2^3)^4 \cdot 5^4$ Expand and combine powers: $= 2^{15} \cdot 3^6 \cdot 5^6$ Therefore, $\alpha = 15$ , $\beta = 6$ , and $\gamma = 6$ .

So, $\alpha + \beta + \gamma = 15 + 6 + 6 = 27$ .

Q215

Let the matrix

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

satisfy

A^n=A^{n-2}+A^2-I

for

n \geqslant 3

. Then the sum of all the elements of

\mathrm{A}^{50}

is :

A 44

B 39

C 52

D 53

Correct Answer

Option D

Solution

\begin{aligned} &\begin{aligned} & A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right] \\ & A^2=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right] \\ & A^3=A+A^2-I \\ & A^3=\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right] \\ & A^4=A^2+A^2-I=2 A^2-I \\ & A^4=\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 0 & 1 \end{array}\right] \text{ and } A^5=\left[\begin{array}{lll} 1 & 0 & 0 \\ 3 & 0 & 1 \\ 2 & 1 & 0 \end{array}\right] \\ & A^{50}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{array}\right] \end{aligned}\\ &\text{ Sum of elements }=53 \end{aligned}

Q216

If A =

\left[ \begin{array}{lll}1 & {\sin \theta } & 1 \\ { - \sin \theta } & 1 & {\sin \theta } \\ { - 1} & { - \sin \theta } & 1 \end{array} \right]

; then for all

\theta

\in

\left( {{{3\pi } \over 4},{{5\pi } \over 4}} \right)

, det (A) lies in the interval :

A

\left( {{3 \over 2},3} \right]

B

\left( {0,{3 \over 2}} \right]

C

\left[ {{5 \over 2},4} \right)

D

\left( {1,{5 \over 2}} \right]

Correct Answer

Option A

Solution

\left| A \right| = \left| \begin{array}{lll}1 & {\sin \theta } & 1 \\ { - \sin \theta } & 1 & {\sin \theta } \\ { - 1} & { - \sin \theta } & 1 \end{array} \right|

= 2(1 + sin2 $\theta$ ) $\theta$

\in

\left( {{{3\pi } \over 4},{{5\pi } \over 4}} \right) \Rightarrow {1 \over {\sqrt 2 }} < \sin \theta < {1 \over {\sqrt 2 }}

$\Rightarrow$ 0 $\le$ sin2 $\theta$ <

{1 \over 2}

$\therefore$

\left| A \right| \in \left[ {2,3} \right)

Q217

If the system of linear equations x

-

4y + 7z = g 3y

-

5z = h

-

2x + 5y

-

9z = k is consistent, then :

A g + 2h + k = 0

B g + h + 2k = 0

C 2g + h + k = 0

D g + h + k = 0

Correct Answer

Option C

Solution

x $-$

4y + 7z = g

3y

$-$

5z = h

$-$

2x + 5y

$-$

9z = k

D = \left| \begin{array}{lll}1 & { - 4} & 7 \\ 0 & 3 & { - 5} \\ { - 2} & 5 & { - 9} \end{array} \right|

D = 1\left( { - 27 + 25} \right) - 2\left( {20 - 21} \right)

D = - 2 + 2 = 0

If system is consistent then

{D_1} = {D_2} = {D_3} = 0

\left| \begin{array}{lll}1 & { - 4} & g \\ 0 & 3 & h \\ { - 2} & 5 & k \end{array} \right| = 0

1\left( {3k - 5h} \right) - 2\left( { - 4h - 3g} \right) = 0

3k - 5h + 8h + 6g = 0

6g + 3h + 3k = 0

2g + h + k = 0

Q218

If

A

and

B

are square matrices of size

n\, \times \,n

such that

{A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right),

then which of the following will be always true?

A

A=B

B

AB=BA

C either of

A

or

B

is a zero matrix

D either of

A

or

B

is identity matrix

Correct Answer

Option B

Solution

{A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)

{A^2} - {B^2} = {A^2} + AB - BA - {B^2}

\Rightarrow AB = BA

Q219

Let

A = \left( \begin{array}{lll}1 & { - 1} & 1 \\ 2 & 1 & { - 3} \\ 1 & 1 & 1 \end{array} \right).

and

10

B = \left( \begin{array}{lll}4 & 2 & 2 \\ { - 5} & 0 & \alpha \\ 1 & { - 2} & 3 \end{array} \right)

. if

B

is the inverse of matrix

A

, then

\alpha

is

A

5

B

-1

C

2

D

-2

Correct Answer

Option A

Solution

Given that

10B

\,\,\, = \left[ \begin{array}{lll}4 & 2 & 2 \\ { - 5} & 0 & \alpha \\ 1 & { - 2} & 3 \end{array} \right]

\Rightarrow B = {1 \over {10}}\left[ \begin{array}{lll}4 & 2 & 2 \\ { - 5} & 0 & \alpha \\ 1 & { - 2} & 3 \end{array} \right]

Also since,

B = {A^{ - 1}} \Rightarrow AB = I

\Rightarrow {1 \over {10}}\left[ \begin{array}{lll}1 & { - 1} & 1 \\ 2 & 1 & { - 3} \\ 1 & 1 & 1 \end{array} \right]\left[ \begin{array}{lll}4 & 2 & 2 \\ { - 5} & 0 & \alpha \\ 1 & { - 2} & 3 \end{array} \right] = \left[ \begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]

\Rightarrow {1 \over {10}}\left[ \begin{array}{lll}{10} & 0 & {5 - 2} \\ 0 & {10} & { - 5 + \alpha } \\ 0 & 0 & {5 + \alpha } \end{array} \right] = \left[ \begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]

\Rightarrow {{5 - \alpha } \over {10}} = 0

\Rightarrow \alpha = 5

Q220

Let

A = \left| \begin{array}{lll}5 & {5\alpha } & \alpha \\ 0 & \alpha & {5\alpha } \\ 0 & 0 & 5 \end{array} \right|.

If

\,\,\left| {{A^2}} \right| = 25,

then

\,\left| \alpha \right|

equals

A

1/5

B

5

C

{5^2}

D

1

Correct Answer

Option A

Solution

\left| {{A^2}} \right| = 25 \Rightarrow {\left| A \right|^2} = 25

\Rightarrow {\left( {25\alpha } \right)^2} = 25 \Rightarrow \left| \alpha \right| = {1 \over 5}