Matrices and Determinants — Page 16 | JEE Mathematics

Q151

If A is a 3

\times

3 matrix and

|A| = 2

, then

|3\,adj\,(|3A|{A^2})|

is equal to :

A

{3^{12}}\,.\,{6^{10}}

B

{3^{11}}\,.\,{6^{10}}

C

{3^{12}}\,.\,{6^{11}}

D

{3^{10}}\,.\,{6^{11}}

Correct Answer

Option B

Solution

Given that $A$ is $3 \times 3$ matrix and $|A|=2$

\begin{aligned} & \text{ Now, | 3adj }\left(|3 A| A^2\right) \text{ | } \\\\ & =3^3\left|\operatorname{adj}\left(|3 A| A^2\right)\right| \\\\ & =3^3\left|\operatorname{adj}\left(54 A^2\right)\right| \\\\ & =3^3\left|54 A^2\right|^2 \\\\ & =3^3 \times\left(54^3\right)^2 \times|A|^4 \\\\ & =3^3 \times(54)^6 \times 2^4 ~~~~~ {[|A|=2 \text{ given }]} \\\\ & =3^3 \times\left(3^3 \times 2\right)^6 \times 2^4 ~~~~~ {\left[\left(a^m\right)^n=a^{m n}\right]} \\\\ & =3^{11} \times 3^{10} \times 2^{10} ~~~~~~~ {\left[(a b)^m=a^m b^m\right]} \\\\ & =(3)^{11} \times(6)^{10} \end{aligned}

Q152

For the system of linear equations

2x - y + 3z = 5

3x + 2y - z = 7

4x + 5y + \alpha z = \beta

, which of the following is NOT correct?

A The system has infinitely many solutions for

\alpha=-6

and

\beta=9

B The system has a unique solution for

\alpha

\ne

-5

and

\beta=8

C The system is inconsistent for

\alpha=-5

and

\beta=8

D The system has infinitely many solutions for

\alpha=-5

and

\beta=9

Correct Answer

Option A

Solution

Given system of linear equation is

\begin{gathered} 2 x-y+3 z=5 \\\\ 3 x+2 y-z=7 \\\\ 4 x+5 y+\alpha z=\beta \\\\ \text{ Now, } \Delta=\left|\begin{array}{ccc} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \alpha \end{array}\right|=7(\alpha+5) \end{gathered}

So, this system of equation has unique solution, if $\alpha \neq-5$ and $\Delta_1=\left|\begin{array}{ccc}5 & -1 & 3 \\ 7 & 2 & -1 \\ \beta & 5 & \alpha\end{array}\right|=17 \alpha-5 \beta+30$ and $\Delta_2=\left|\begin{array}{ccc}2 & 5 & 3 \\ 3 & 7 & -1 \\ 4 & \beta & \alpha\end{array}\right|=-11 \beta+\alpha+104$ and $\Delta_3=\left|\begin{array}{ccc}2 & -1 & 5 \\ 3 & 2 & 7 \\ 4 & 5 & \beta\end{array}\right|=7(\beta-9)$ For infinitely many solutions,

\Delta=\Delta_1=\Delta_2=\Delta_3=0

\begin{aligned} & \text{ So, } \Delta=0 \\\\ & \Rightarrow 7(\alpha+5)=0 \\\\ & \Rightarrow \alpha=-5 \end{aligned}

\begin{aligned} & \Delta_3 =0 \\\\ & \Rightarrow 7(\beta-9) =0 \\\\ & \Rightarrow \beta =9 \end{aligned}

If $\alpha=-5$ and $\beta=8$ , then $\Delta$ equals zero but $\Delta_3$ does not, which would imply the system is inconsistent for $\alpha=-5$ and $\beta=8$ .

Therefore, the option "The system is inconsistent for $\alpha=-5$ and $\beta=8$ " is correct.

Q153

If

A=\left[\begin{array}{cc}1 & 5 \\ \lambda & 10\end{array}\right], \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}

and

\alpha+\beta=-2

, then

4 \alpha^{2}+\beta^{2}+\lambda^{2}

is equal to :

A 12

B 10

C 19

D 14

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{A}=\left[\begin{array}{cc} 1 & 5 \\ \lambda & 10 \end{array}\right] \\\\ & \Rightarrow|\mathrm{A}-x \mathrm{I}|=0 \\\\ & \Rightarrow\left|\begin{array}{cc} 1-x & 5 \\ \lambda & 10-x \end{array}\right|=0 \\\\ & \Rightarrow(1-x)(10-x)-5 \lambda=0 \\\\ & \Rightarrow 10-11 x+x^2-5 \lambda=0 \end{aligned}

Also, $\Rightarrow \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$

\Rightarrow \alpha A^2+\beta A-I=0

and $A^2-11 A+(10-5 \lambda) I=0$ On solving, we get

\alpha=\frac{1}{5}, \beta=-\frac{11}{5}

\begin{aligned} & \text{ So, } 5 \lambda-10=5 \Rightarrow \lambda=3 \\\\ & \therefore 4 \alpha^2+\beta^2+\lambda^2 \\\\ & =4\left(\frac{1}{25}\right)+\left(\frac{121}{25}\right)+9 \\\\ & =\frac{125}{25}+9=14 \end{aligned}

Q154

Let S be the set of all values of

\theta \in[-\pi, \pi]

for which the system of linear equations

x+y+\sqrt{3} z=0

-x+(\tan \theta) y+\sqrt{7} z=0

x+y+(\tan \theta) z=0

has non-trivial solution. Then

\dfrac{120}{\pi} \sum\limits_{\theta \in \mathrm{s}} \theta

is equal to :

A 40

B 30

C 10

D 20

Correct Answer

Option D

Solution

Since, the given system has a non trivial solution,

\text{ So, } \Delta=0

\Rightarrow \Delta=\left|\begin{array}{ccc} 1 & 1 & \sqrt{3} \\ -1 & \tan \theta & \sqrt{7} \\ 1 & 1 & \tan \theta \end{array}\right|=0

\begin{aligned} & \Rightarrow 1\left(\tan ^2 \theta-\sqrt{7}\right)-1(-\tan \theta-\sqrt{7})+\sqrt{3}(-1-\tan \theta)=0 \\\\ & \Rightarrow \tan ^2 \theta-\sqrt{7}+\tan \theta+\sqrt{7}-\sqrt{3}-\sqrt{3} \tan \theta=0 \\\\ & \Rightarrow \tan \theta(\tan \theta-\sqrt{3})+1(\tan \theta-\sqrt{3})=0 \\\\ & \Rightarrow \tan \theta=\sqrt{3} \text{ or } \tan \theta=-1 \\\\ & \therefore \theta=\left\{\frac{\pi}{3}, \frac{-2 \pi}{3}, \frac{-\pi}{4}, \frac{3 \pi}{4}\right\} \\\\ & \text{ So, } \frac{120}{\pi} \sum \theta=\frac{120}{\pi}\left\{\frac{4 \pi-8 \pi-3 \pi+9 \pi}{12}\right\} \\\\ & =\frac{120}{\pi}\left[\frac{2 \pi}{12}\right]=20 \end{aligned}

Q155

Let

A=\left[\begin{array}{ccc}2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2\end{array}\right]

. If

|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 A))|=(16)^{n}

, then

n

is equal to :

A 9

B 8

C 10

D 12

Correct Answer

Option C

Solution

We have,

\begin{aligned} & |\mathrm{A}|=\left|\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{array}\right|=2(4-1)-1(2-0)+0 \\\\ & =6-2=4 \\\\ & \text{ So, }|2 \mathrm{~A}|=2^3|\mathrm{~A}|=8 \times 4=32 \\\\ & \text{ Now, }|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 \mathrm{~A}))|=|2 \mathrm{~A}|^{(n-1)^3} \\\\ & =(32)^{2^3}=32^8 \\\\ & \Rightarrow 16^n=(32)^8=2^8 \times 16^8 \\\\ & \Rightarrow 16^n=16^{2+8} \Rightarrow n=10 \end{aligned}

Concepts : (a) $|k \mathrm{~A}|=k^n|\mathrm{~A}|$ (b) $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{n-1}$

Q156

Let

P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]

and

Q=P A P^{T}

. If

P^{T} Q^{2007} P=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]

, then

2 a+b-3 c-4 d

equal to :

A 2004

B 2006

C 2007

D 2005

Correct Answer

Option D

Solution

\text{ Here, } P=\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{array}\right], A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]

\text{ Here, } \mathrm{PP}^{\mathrm{T}}=\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{array}\right]\left[\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{-1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array}\right]

=\left[\begin{array}{cc} \frac{3}{4}+\frac{1}{4} & -\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \\ \frac{-\sqrt{3}}{4}+\frac{\sqrt{3}}{4} & \frac{1}{4}+\frac{3}{4} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\mathrm{I}

Similarly $P^T P=1$ $\because$

Q=P A P^{T}

Now, $Q^{2007}=\left(P A P^T\right)\left(P A P^T\right) \ldots 2007$ times $=P A^{2007} P^T$

\begin{aligned} & \text{ As, } A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] \\\\ & \Rightarrow A^2=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1+0 & 1+1 \\ 0+0 & 0+1 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right] \end{aligned}

A^3=\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]

. . . .

A^{2007}=\left[\begin{array}{cc} 1 & 2007 \\ 0 & 1 \end{array}\right]

\begin{aligned} & \text{ Hence, } \mathrm{P}^{\mathrm{T}} \mathrm{Q}^{2007} \mathrm{P}=\mathrm{A}^{2007}=\left[\begin{array}{cc} 1 & 2007 \\ 0 & 1 \end{array}\right] \\\\ & \Rightarrow\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} 1 & 2007 \\ 0 & 1 \end{array}\right] \end{aligned}

\begin{aligned} & \Rightarrow a=1, b=2007, c=0, d=1 \\\\ & \therefore 2 a+b-3 c-4 d=2(1)+2007-3(0)-4(1) \\\\ & =2+2007-4=2005 \end{aligned}

Q157

If the system of equations

x+y+a z=b

2 x+5 y+2 z=6

x+2 y+3 z=3

has infinitely many solutions, then

2 a+3 b

is equal to :

A 28

B 25

C 20

D 23

Correct Answer

Option D

Solution

Given system of equations,

\text{ and } \quad \begin{aligned} x+y+a z & =b \\ 2 x+5 y+2 z & =6 \\ x+2 y+3 z & =3 \end{aligned}

Since, given system of equation has infinitely many solutions

\therefore D=0 \text{ and } D_1=D_2=D_3=0

\begin{aligned} & \text{ Here, } D=\left|\begin{array}{lll} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{array}\right|=0 \\\\ & \Rightarrow 1(15-4)-1(6-2)+a(4-5)=0 \\\\ & \Rightarrow -a+11-4=0 \\\\ & \Rightarrow a=7 \end{aligned}

\begin{aligned} & \text{ and } D_1=\left|\begin{array}{ccc} b & 1 & a \\ 6 & 5 & 2 \\ 3 & 2 & 3 \end{array}\right|=0 \\\\ & \Rightarrow b(15-4)-1(18-6)+a(12-15)=0 \\\\ & \Rightarrow 11 b-12-3 a=0 \\\\ & \Rightarrow 11 b-12-21=0 ~~~~~~~(\because a=7)\\\\ & \Rightarrow 11 b-33=0 \\\\ & \Rightarrow b=3 \\\\ & \therefore 2 a+3 b=2(7)+3(3)=14+9=23 \end{aligned}

Q158

Let

\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{2 \times 2}

, where

\mathrm{a}_{\mathrm{ij}} \neq 0

for all

\mathrm{i}, \mathrm{j}

and

\mathrm{A}^{2}=\mathrm{I}

. Let a be the sum of all diagonal elements of

\mathrm{A}

and

\mathrm{b}=|\mathrm{A}|

. Then

3 a^{2}+4 b^{2}

is equal to :

A 4

B 3

C 14

D 7

Correct Answer

Option A

Solution

Given, $A^2=I$ and $b=|A|$ Let

A=\left[\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right]

\begin{aligned} \therefore \quad A^2 & =\left[\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right]\left[\begin{array}{ll} a_1 & b_1 \\ a_2 & b_2 \end{array}\right] \\\\ & =\left[\begin{array}{cc} a_1^2+b_1 a_2 & a_1 b_1+b_1 b_2 \\ a_1 a_2+a_2 b_2 & b_1 a_2+b_2^2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \left(\because A^2=I\right) \end{aligned}

By equality of matrices, we get

\begin{aligned} & a_1 a_2+a_2 b_2=0 \text{ and } a_1 b_1+b_1 b_2=0 \\\\ &\Rightarrow a_2\left(a_1+b_2\right)=0 \text{ and } b_1\left(a_1+b_2\right)=0 \\\\ &\Rightarrow a_1+b_2=0 \text{ (since, } b_1, a_2 \neq 0 \text{ ) }\\\\ &\Rightarrow \text{ Sum of diagonal elements }=0 \\\\ &\Rightarrow a=0 \end{aligned}

\begin{aligned} & \text{ Now, } A^2=I \\\\ & \Rightarrow \left|A^2\right|=1 \\\\ & \Rightarrow |A|^2=1 ~~~~~~~\left(\because\left|A^n\right|=|A|^n\right)\\\\ & \Rightarrow b^2=1 ~~~~~~~~~~~(\because|A|=b)\\\\ & \therefore 3 a^2+4 b^2=3(0)+4(1)=4 \end{aligned}

Q159

Let

P

be a square matrix such that

P^{2}=I-P

. For

\alpha, \beta, \gamma, \delta \in \mathbb{N}

, if

P^{\alpha}+P^{\beta}=\gamma I-29 P

and

P^{\alpha}-P^{\beta}=\delta I-13 P

, then

\alpha+\beta+\gamma-\delta

is equal to :

A 18

B 22

C 24

D 40

Correct Answer

Option C

Solution

We have, $P^2=I-P$

\begin{aligned} \Rightarrow P^4 & =(I-P)^2=I+P^2-2 P \\\\ & =2 I-3 P \text{ [Using Eq. (i)] } \end{aligned}

\begin{aligned} \Rightarrow P^8 & =(2 I-3 P)^2 \\\\ & =4 I+9 P^2-12 P \\\\ & =13 I-21 P \text{ [Using Eq. (i)] } \end{aligned}

and

\begin{aligned} P^6 & =(I-P)(2 I-3 P) \\\\ & =2 I-3 P-2 P+3 P^2 \\\\ & =5 I-8 P \text{ [Using Eq. (i)] } \end{aligned}

$\begin{aligned} & \text{ Now, } P^8+P^6=18 I-29 P \\\\ & \text{ and } P^8-P^6=8 I-13 P \\\\ & \therefore \alpha=8, \beta=6, \gamma=18, \delta=8 \\\\ & \therefore \alpha+\beta+\gamma-\delta=8+6+18-8=24\end{aligned}$

Q160

For the system of equations

x+y+z=6

x+2 y+\alpha z=10

x+3 y+5 z=\beta

, which one of the following is NOT true?

A System has a unique solution for

\alpha=3,\beta\ne14

.

B System has infinitely many solutions for

\alpha=3, \beta=14

.

C System has no solution for

\alpha=3, \beta=24

.

D System has a unique solution for

\alpha=-3, \beta=14

.

Correct Answer

Option A

Solution

Given system of equations,

\begin{aligned} x+y+z & =6 ........(i)\\\\ x+2 y+\alpha z & =10 ........(ii)\\\\ x+3 y+5 z & =\beta ........(iii) \end{aligned}

Here,

\begin{aligned} \Delta & =\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & \alpha \\ 1 & 3 & 5 \end{array}\right| \\\\ & =1(10-3 \alpha)-1(5-\alpha)+1(3-2) \\\\ & =10-3 \alpha-5+\alpha+1 \\\\ & =6-2 \alpha \end{aligned}

For unique solution, $\Delta \neq 0$

\Rightarrow 6-2 \alpha \neq 0 \Rightarrow \alpha \neq 3

When, $\alpha=3$

\begin{aligned} \Delta_1 & =\left|\begin{array}{ccc} 6 & 1 & 1 \\ 10 & 2 & 3 \\ \beta & 3 & 5 \end{array}\right| \\\\ & =\left|\begin{array}{ccc} 0 & 1 & 0 \\ -2 & 2 & 1 \\ \beta-18 & 3 & 2 \end{array}\right|=-1(-4-\beta+18) \\\\ & =\beta-14 \end{aligned}

and $\begin{aligned} \Delta_2 & =\left|\begin{array}{ccc}1 & 6 & 1 \\ 1 & 10 & 3 \\ 1 & \beta & 5\end{array}\right| \\\\ & =1(50-3 \beta)-6(5-3)+1(\beta-10) \\\\ & =50-3 \beta-12+\beta-10 \\\\ & =28-2 \beta=2(14-\beta)\end{aligned}$ and

\begin{aligned} \Delta_3 & =\left|\begin{array}{ccc} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 1 & 3 & \beta \end{array}\right| \\ & =1(2 \beta-30)-1(\beta-10)+6(3-2) \\\\ & =2 \beta-30-\beta+10+6 \\\\ & =\beta-14 \end{aligned}

Thus, at $\beta=14, \Delta_1=\Delta_2=\Delta_3=0$

\Rightarrow \alpha=3, \beta=14

So, system has infinite solutions.