Matrices and Determinants — Page 14 | JEE Mathematics

Q131

Let

\alpha

and

\beta

be real numbers. Consider a 3

\times

3 matrix A such that

A^2=3A+\alpha I

. If

A^4=21A+\beta I

, then

A

\alpha=1

B

\alpha=4

C

\beta=8

D

\beta=-8

Correct Answer

Option D

Solution

$\mathrm{A}^{2}=3 \mathrm{~A}+\alpha \mathrm{I}$ $A^{3}=3 A^{2}+\alpha A$ $\mathrm{A}^{3}=3(3 \mathrm{~A}+\alpha \mathrm{I})+\alpha \mathrm{A}$ $\mathrm{A}^{3}=9 \mathrm{~A}+\alpha \mathrm{A}+3 \alpha \mathrm{I}$ $\mathrm{A}^{4}=(9+\alpha) \mathrm{A}^{2}+3 \alpha \mathrm{A}$ $=(9+\alpha)(3 \mathrm{~A}+\alpha \mathrm{I})+3 \alpha \mathrm{A}$ $=\mathrm{A}(27+6 \alpha)+\alpha(9+\alpha)$ $\Rightarrow 27+6 \alpha=21 \Rightarrow \alpha=-1$ $\Rightarrow \beta=\alpha(9+\alpha)=-8$

Q132

Consider the following system of equations

\alpha x+2y+z=1

2\alpha x+3y+z=1

3x+\alpha y+2z=\beta

for some

\alpha,\beta\in \mathbb{R}

. Then which of the following is NOT correct.

A It has a solution for all

\alpha\ne-1

and

\beta=2

B It has no solution if

\alpha=-1

and

\beta\ne2

C It has no solution for

\alpha=-1

and for all

\beta \in \mathbb{R}

D It has no solution for

\alpha=3

and for all

\beta\ne2

Correct Answer

Option C

Solution

$D=\left|\begin{array}{ccc}\alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & 2\end{array}\right|=0 \Rightarrow \alpha=-1,3$ $D_{x}=\left|\begin{array}{ccc}2 & 1 & 1 \\ 3 & 1 & 1 \\ \alpha & 2 & \beta\end{array}\right|=0 \Rightarrow \beta=2$ $D_{y}=\left|\begin{array}{ccc}\alpha & 1 & 1 \\ 2 \alpha & 1 & 1 \\ 3 & 2 & \beta\end{array}\right|=0$ $D_{z}=\left|\begin{array}{ccc}\alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & \beta\end{array}\right|=0$ $\beta=2, \alpha=-1$ $\alpha=-1, \beta=2$ Infinite solution

Q133

Let A, B, C be 3

\times

3 matrices such that A is symmetric and B and C are skew-symmetric. Consider the statements (S1) A

^{13}

B

^{26}

-

B

^{26}

A

^{13}

is symmetric (S2) A

^{26}

C

^{13}

-

C

^{13}

A

^{26}

is symmetric Then,

A Only S2 is true

B Only S1 is true

C Both S1 and S2 are false

D Both S1 and S2 are true

Correct Answer

Option A

Solution

$A^{T}=A, B^{T}=-B, C^{T}=-C$

\begin{aligned} P & =A^{13} B^{26}-B^{26} A^{13} \\\\ P^{T} & =\left(A^{13} B^{26}-B^{26} A^{13}\right)^{T}=\left(A^{13} B^{26}\right)^{T}-\left(B^{26} A^{B}\right)^{T} \\\\ & =\left(B^{26}\right)^{T}\left(A^{13}\right)^{T}-\left(A^{13}\right)^{T}\left(B^{26}\right)^{T} \\\\ & =\left(B^{T}\right)^{26}\left(A^{T}\right)^{13}-\left(A^{T}\right)^{13}\left(A^{T}\right)^{26} \\\\ & =B^{26} A^{13}-A^{13} B^{26}=-\left(A^{13} B^{26}-B^{26} A^{13}\right)=-P \end{aligned}

$P$ is skew-symmetric matrix $\Rightarrow S_{1}$ is false $Q=A^{26} C^{13}-C^{13} A^{26}=Q^{T}=\left(A^{26} C^{13}-C^{13} A^{26}\right)^{T}$ $Q=\left(A^{26} C^{13}\right)^{T}-\left(C^{13} A^{26}\right)^{T}=\left(C^{13}\right)^{T}\left(A^{26}\right)^{T}-\left(A^{26}\right)^{T}\left(C^{13}\right)^{T}$ $=\left(C^{T}\right)^{13}\left(A^{T}\right)^{26}-\left(A^{T}\right)^{26}\left(C^{T}\right)^{13}=-C^{13} A^{26}+A^{26} C^{13}$ $=A^{26} C^{13}+C^{13} A^{26}$ $\Rightarrow Q^{T}=Q \Rightarrow Q$ is symmetric matrix $\Rightarrow S_{2}$ is true.

Q134

Let

x,y,z > 1

and

A = \left[ \begin{array}{lll}1 & {{{\log }_x}y} & {{{\log }_x}z} \\ {{{\log }_y}x} & 2 & {{{\log }_y}z} \\ {{{\log }_z}x} & {{{\log }_z}y} & 3 \end{array} \right]

. Then

\mathrm{|adj~(adj~A^2)|}

is equal to

A

6^4

B

2^8

C

4^8

D

2^4

Correct Answer

Option B

Solution

\begin{aligned} & |A|=\frac{1}{\log x \log y \log z}\left|\begin{array}{ccc} \log x & \log y & \log z \\ \log x & 2 \log y & \log z \\ \log x & \log y & 3 \log z \end{array}\right|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 3 \end{array}\right|=2 \\\\ & \Rightarrow\left|\operatorname{adj}\left(\operatorname{adj} A^2\right)\right|=\left|\operatorname{adj}\left(A^2\right)\right|^2=\left(\left|A^2\right|^2\right)^2=|A|^8=2^8 \end{aligned}

Q135

Let S

_1

and S

_2

be respectively the sets of all

a \in \mathbb{R} - \{ 0\}

for which the system of linear equations

ax + 2ay - 3az = 1

(2a + 1)x + (2a + 3)y + (a + 1)z = 2

(3a + 5)x + (a + 5)y + (a + 2)z = 3

has unique solution and infinitely many solutions. Then

A

\mathrm{n({S_1}) = 2}

and S

_2

is an infinite set

B

\mathrm{{S_1} = \Phi}

and

\mathrm{{S_2} = \mathbb{R} - \{ 0\}}

C

\mathrm{{S_1} = \mathbb{R} - \{ 0\}}

and

\mathrm{{S_2} = \Phi}

D S

_1

is an infinite set and n(S

_2

) = 2

Correct Answer

Option C

Solution

Given system of equations

\begin{aligned} & a x+2 a y-3 a z=1 \\\\ & (2 a+1) x+(2 a+3) y+(a+1) z=2 \\\\ & (3 a+5) x+(a+5) y+(a+2) z=3 \\\\ & \text{ Let } A=\left|\begin{array}{ccc} a & 2 a & -3 a \\\\ 2 a+1 & 2 a+3 & a+1 \\\\ 3 a+5 & a+5 & a+2 \end{array}\right| \\\\ & =a\left|\begin{array}{ccc} 1 & 0 & 0 \\\\ 2 a+1 & 1-2 a & 7 a+4 \\\\ 3 a+5 & -5 a-5 & 10 a+17 \end{array}\right| \\\\ & =a\left(15 a^{2}+31 a+37\right) \\\\ & \text{ Now } A=0 \\\\ & \Rightarrow a=0 \end{aligned}

So, $S_{1}=R-\{0\}$ and at $a=0$ System has infinite solution but $a \in R-\{0\}$ $\therefore S_{2}=\Phi$

Q136

Let A be a 3

\times

3 matrix such that

\mathrm{|adj(adj(adj~A))|=12^4}

. Then

\mathrm{|A^{-1}~adj~A|}

is equal to

A 12

B 2

\sqrt3

C 1

D

\sqrt6

Correct Answer

Option B

Solution

$|A|^{(n-1)^{3}}=12^{4}$

\begin{aligned} &|A|^{8}=12^{4} \\\\ &|A|=\sqrt{12} \\\\ &\left|A^{-1} \operatorname{adj} A\right|=\left|A^{-1}\right| \cdot|A|^{2} \\\\ &=|A| = 2\sqrt3 \end{aligned}

Q137

If the system of equations

x+2y+3z=3

4x+3y-4z=4

8x+4y-\lambda z=9+\mu

has infinitely many solutions, then the ordered pair (

\lambda,\mu

) is equal to :

A

\left( {{{72} \over 5},{{21} \over 5}} \right)

B

\left( { - {{72} \over 5}, - {{21} \over 5}} \right)

C

\left( { - {{72} \over 5},{{21} \over 5}} \right)

D

\left( {{{72} \over 5}, - {{21} \over 5}} \right)

Correct Answer

Option D

Solution

For infinite many solution, $\Delta=0$ and $\Delta_x=0$

\begin{aligned} & \Delta=\left|\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 3 & -4 \\ 8 & 4 & -\lambda \end{array}\right|=0 \\\\ & \Rightarrow 1(-3 \lambda+16)-2(-4 \lambda+32)+3(16-24)=0 \\\\ & \Rightarrow 16-3 \lambda+8 \lambda-64-24=0 \Rightarrow 5 \lambda=72 \\\\ & \therefore \lambda=\frac{72}{5} \\\\ & \Delta_x=\left|\begin{array}{ccc} 3 & 2 & 3 \\ 4 & 3 & -4 \\ 9+\mu & 4 & -\lambda \end{array}\right|=0 \end{aligned}

\begin{aligned} & \Rightarrow \quad 3(-3 \lambda+16)-2(-4 \lambda+36+4 \mu)+3(16-27-3 \mu)=0 \\\\ & \Rightarrow-9 \lambda+48+8 \lambda-72-8 \mu-33-9 \mu=0 \\\\ & \Rightarrow-\lambda-17 \mu=57 \\\\ & \Rightarrow-17 \mu=57+\lambda \\\\ & \therefore -\mu=\frac{57+\frac{72}{5}}{17}\\\\ & \Rightarrow \mu=\frac{-357}{85}=\frac{-21}{5} \end{aligned}

\text{ Thus, }(\lambda, \mu) \equiv\left(\frac{72}{5}, \frac{-21}{5}\right)

Q138

If A and B are two non-zero n

\times

n matrices such that

\mathrm{A^2+B=A^2B}

, then :

A

\mathrm{A^2B=I}

B

\mathrm{A^2=I}

or

\mathrm{B=I}

C

\mathrm{A^2B=BA^2}

D

\mathrm{AB=I}

Correct Answer

Option C

Solution

Given : $A^{2}+B=A^{2} B\quad...(i)$ $\Rightarrow A^{2}+B-I=A^{2} B-I$ $\Rightarrow A^{2} B-A^{2}-B+I=I$ $\Rightarrow A^{2}(B-I)-I(B-I)=I$ $\Rightarrow\left(A^{2}-I\right)(B-I)=I$ $\therefore A^{2}-I$ is the inverse matrix of $B-I$ and vice versa.

So, $(B-I)\left(A^{2}-I\right)=I$ $\Rightarrow B A^{2}-B-A^{2}+I=I$ $\therefore A^{2}+B=B A^{2} \quad...(ii)$ So, by (i) and (ii) $A^{2} B=B A^{2}$

Q139

Let

\alpha

be a root of the equation

(a - c){x^2} + (b - a)x + (c - b) = 0

where a, b, c are distinct real numbers such that the matrix

\left[ \begin{array}{lll}{{\alpha ^2}} & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c \end{array} \right]

is singular. Then, the value of

{{{{(a - c)}^2}} \over {(b - a)(c - b)}} + {{{{(b - a)}^2}} \over {(a - c)(c - b)}} + {{{{(c - b)}^2}} \over {(a - c)(b - a)}}

is

A 3

B 6

C 12

D 9

Correct Answer

Option A

Solution

\begin{aligned} & \Delta=0=\left|\begin{array}{ccc} \alpha^2 & \alpha & 1 \\\\ 1 & 1 & 1 \\\\ \mathrm{a} & \mathrm{b} & \mathrm{c} \end{array}\right| \\\\ & \Rightarrow \alpha^2(\mathrm{c}-\mathrm{b})-\alpha(\mathrm{c}-\mathrm{a})+(\mathrm{b}-\mathrm{a})=0 \end{aligned}

It is singular when $\alpha=1$

\begin{aligned} & \frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)} \\\\ & \frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)} \\\\ & =3 \frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3 \end{aligned}

Q140

Let the determinant of a square matrix A of order

m

be

m-n

, where

m

and

n

satisfy

4 m+n=22

and

17 m+4 n=93

. If

\operatorname{det}(n \operatorname{adj}(\operatorname{adj}(m A)))=3^{a} 5^{b} 6^{c}

then

a+b+c

is equal to :

A 96

B 84

C 109

D 101

Correct Answer

Option A

Solution

Given that $|A|=m-n$ , and let's solve the system of linear equations to find the values of $m$ and $n$ : $4m + n = 22$ ......

(1) $17m + 4n = 93$ .......

(2) We can multiply equation (1) by 4 to make the coefficients of $n$ in both equations equal: $16m + 4n = 88$ .........

(3) Now, subtract equation (3) from equation (2): $(17m + 4n) - (16m + 4n) = 93 - 88$ $m = 5$ Now, we can find the value of $n$ by substituting $m$ into equation (1): $4(5) + n = 22$ $20 + n = 22$ $n = 2$ Since the order of matrix A is $m$ , the order is 5.

Now, let's find the determinant of $n \operatorname{adj}(\operatorname{adj}(mA))$ : We know that $\operatorname{det}(A) = m-n = 3$ .

\begin{aligned} & \therefore \operatorname{det}(n \operatorname{adj}(\operatorname{adj}(m A))) \\\\ & =|2 \operatorname{adj}(\operatorname{adj}(5 A))| \\\\ & =2^5|5 A|^{16} \\\\ & =2^5 5^{80}|A|^{16}=2^5 \cdot 3^{16} \cdot 5^{80} \\\\ & =3^{11} 5^{80} 6^5 \end{aligned}

So, $a+b+c=96$