Matrices and Determinants — Page 21 | JEE Mathematics

Q201

Let

A=\left[a_{i j}\right]

be a

3 \times 3

matrix such that

A\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right], A\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]

and

A\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]

, then

a_{23}

equals :

A 2

B

-

1

C 1

D 0

Correct Answer

Option B

Solution

\begin{aligned} & \text{ Let } 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] \\ & A\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] \Rightarrow\left[\begin{array}{l} a_{12} \\ a_{22} \\ a_{32} \end{array}\right]=\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right] \Rightarrow \begin{array}{l} a_{22}=0 ; a_{12}=0 \\ a_{32}=1 \end{array} \end{aligned}

\begin{aligned} A\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right] \Rightarrow \begin{array}{l} 4 a_{11}+a_{12}+3 a_{13}=0 \\ 4 a_{21}+a_{22}+3 a_{23}=1 \Rightarrow 4 a_{21}+3 a_{23}=1 \\ 4 a_{31}+a_{32}+3 a_{33}=0 \end{array} \\ A\left[\begin{array}{l} 2 \\ 1 \\ 2 \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right] \Rightarrow \begin{array}{l} 2 a_{11}+a_{12}+2 a_{13}=1 \\ 2 a_{21}+a_{22}+2 a_{23}=0 \Rightarrow a_{21}+a_{23}=0 \\ 2 a_{31}+a_{32}+2 a_{33}=0 \end{array} \\ -4 a_{23}+3 a_{23}=1 \Rightarrow a_{23}=-1 \end{aligned}

Q202

If the system of equations

\begin{aligned} & 2 x-y+z=4 \\ & 5 x+\lambda y+3 z=12 \\ & 100 x-47 y+\mu z=212 \end{aligned}

has infinitely many solutions, then

\mu-2 \lambda

is equal to

A 56

B 59

C 57

D 55

Correct Answer

Option C

Solution

\begin{aligned} & \Delta=0 \Rightarrow\left|\begin{array}{ccc} 2 & -1 & 1 \\ 5 & \lambda & 3 \\ 100 & -47 & \mu \end{array}\right|=0 \\ & 2(\lambda \mu+141)+(5 \mu-300)-235-100 \lambda=0 \ldots (1)\\ & \Delta_3=0 \Rightarrow\left|\begin{array}{ccc} 2 & -1 & 4 \\ 5 & \lambda & 12 \\ 100 & -47 & 212 \end{array}\right|=0 \\ & 6 \lambda=-12 \Rightarrow \lambda=-2 \\ & \text{ Put } \lambda=2 \text{ in }(1) \\ & 2(-2 \mu+141)+5 \mu-300-235+200=0 \\ & \mu=53 \\ & \therefore 57 \end{aligned}

Q203

Let α be a solution of

x^2 + x + 1 = 0

, and for some a and b in

R, \begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}

. If

\dfrac{4}{\alpha^4} + \dfrac{m}{\alpha^a} + \dfrac{n}{\alpha^b} = 3

, then m + n is equal to _______

A 11

B 3

C 8

D 7

Correct Answer

Option A

Solution

Let $\alpha$ be a solution of the equation $x^2 + x + 1 = 0$ .

This implies that $\alpha$ is a cube root of unity, denoted as $\omega$ , where $\alpha^2 + \alpha + 1 = 0$ .

We are given matrix equations with parameters $a$ and $b$ as follows: $\begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$ Solving these equations: $\begin{aligned} & \left[4 - a - 2b, 64 - a - 14b, 52 + 2a - 8b\right] = \left[0, 0, 0\right] \\ & \Rightarrow a + 2b = 4 \\ & \Rightarrow a + 14b = 64 \end{aligned}$ From these, solve for $a$ and $b$ : $\begin{aligned} & 12b = 60 \Rightarrow b = 5 \\ & a = -6 \end{aligned}$ We now substitute $a$ and $b$ into the given function: $\dfrac{4}{\alpha^4} + \dfrac{m}{\alpha^a} + \dfrac{n}{\alpha^b} = 3$ We substitute $\alpha = \omega$ and simplify: $\begin{aligned} & \dfrac{4}{\omega} + \dfrac{m}{1} + \dfrac{n}{\omega^2} = 3 \\ & \Rightarrow 4\omega^2 + m + n\omega = 3 \end{aligned}$ Substitute $\omega = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$ and $\omega^2 = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i$ : $\begin{aligned} & 4\left(-\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i\right) + m + n\left(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right) = 3 \\ & \Rightarrow -2 + m - \dfrac{n}{2} = 3 \\ & \text{and } \dfrac{-4\sqrt{3}}{2} + \dfrac{n\sqrt{3}}{2} = 0 \end{aligned}$ From the above, solve for $n$ and $m$ : $\begin{aligned} & n = 4 \\ & m = 7 \end{aligned}$ Thus, $m + n = 11$ .

Q204

Let

A = \begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix}

. If

\det(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n

,

m, n \in \mathbb{N}

, then

m + n

is equal to

A 22

B 20

C 24

D 26

Correct Answer

Option C

Solution

$|A|=\left|\begin{array}{ccc}2 & 2+p & 2+p+q \\ 4 & 6+2 p & 8+3 p+2 q \\ 6 & 12+3 p & 20+6 p+3 q\end{array}\right|$

\begin{aligned} &\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_2-\mathrm{C}_1 \times \frac{\mathrm{q}}{2}\\ &\text{ Then } \mathrm{C}_3 \rightarrow \mathrm{C}_2-\mathrm{C}_1 \mathrm{X}\left(1+\frac{\mathrm{p}}{2}\right)\\ &\Rightarrow|\mathrm{A}|=\left|\begin{array}{ccc} 2 & 0 & 0 \\ 4 & 2 & 2+\mathrm{p} \\ 6 & 6 & 8+3 \mathrm{p} \end{array}\right| \end{aligned}

\begin{aligned} & \Rightarrow|\mathrm{A}|=2(16+6 \mathrm{p}-12-6 \mathrm{p})=8=2^3 \\ & |\operatorname{adj}(\operatorname{adj}(3 \mathrm{~A}))|=|3 \mathrm{~A}|^{(3-1)^2}=|3 \mathrm{~A}|^4 \\ & =\left(3^3|\mathrm{~A}|\right)^4=\left(3^3 \times 2^3\right)^4=2^{12} \times 3^{12} \\ & \Rightarrow \mathrm{~m}+\mathrm{n}=24 \end{aligned}

Q205

Let

\mathrm{A}=\left[\begin{array}{cc}\alpha & -1 \\ 6 & \beta\end{array}\right], \alpha>0

, such that

\operatorname{det}(\mathrm{A})=0

and

\alpha+\beta=1

. If I denotes

2 \times 2

identity matrix, then the matrix

(I+A)^8

is :

A

\left[\begin{array}{cc}257 & -64 \\ 514 & -127\end{array}\right]

B

\left[\begin{array}{cc}766 & -255 \\ 1530 & -509\end{array}\right]

C

\left[\begin{array}{cc}1025 & -511 \\ 2024 & -1024\end{array}\right]

D

\left[\begin{array}{ll}4 & -1 \\ 6 & -1\end{array}\right]

Correct Answer

Option B

Solution

Let $|A|=0 \Rightarrow \alpha \beta-(-6)=0 \Rightarrow \alpha \beta=-6$ and $\alpha+\beta=1 \Rightarrow \alpha \beta$ are roots of the equation

\begin{aligned} & x^2-x-6=0 \Rightarrow x=3,-2 . \text{ Since } \alpha>0 \\ & \Rightarrow \quad \alpha=3, \beta=-2 \\ & \Rightarrow \quad A=\left[\begin{array}{ll} 3 & -1 \\ 6 & -2 \end{array}\right] \Rightarrow I+A=\left[\begin{array}{ll} 4 & -1 \\ 6 & -1 \end{array}\right] \\ & (I+A)^2=\left[\begin{array}{ll} 10 & -3 \\ 18 & -5 \end{array}\right] \Rightarrow(I+A)^4=\left[\begin{array}{ll} 46 & -15 \\ 90 & -29 \end{array}\right] \\ & \Rightarrow \quad(I+A)^8=\left[\begin{array}{cc} 766 & -255 \\ 1530 & -509 \end{array}\right] \end{aligned}

Q206

Let

a \in R

and

A

be a matrix of order

3 \times 3

such that

\operatorname{det}(A)=-4

and

A+I=\left[\begin{array}{lll}1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2\end{array}\right]

, where

I

is the identity matrix of order

3 \times 3

. If

\operatorname{det}((a+1) \operatorname{adj}((a-1) A))

is

2^{\mathrm{m}} 3^{\mathrm{n}}, \mathrm{m}

,

\mathrm{n} \in\{0,1,2, \ldots, 20\}

, then

\mathrm{m}+\mathrm{n}

is equal to :

A 14

B 17

C 15

D 16

Correct Answer

Option D

Solution

To solve for the value of $m+n$ , we first establish the matrix $A$ and determine the value of $a$ .

The condition given is: $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$ Since $I$ is the identity matrix of size $3 \times 3$ , we have: $A = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} - I = \begin{bmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{bmatrix}$ We know $\operatorname{det}(A) = -4$ .

Calculating the determinant: $\det(A) = 0(0 \times 1 - 0 \times 1) - a(2 \times 1 - 0 \times a) + 1(2 \times 1 - 0 \times 1) = -2a + 2$ Setting $-2a + 2 = -4$ , we solve for $a$ : $-2a + 2 = -4$ $-2a = -6$ $a = 3$ With $a = 3$ , the problem is to find $\operatorname{det}((a+1) \operatorname{adj}((a-1) A))$ .

With $a = 3$ : Calculate $(a+1) \operatorname{adj}((a-1) A)$ : $= 4 \operatorname{adj}(3A)$ Calculate the determinant: $\det(4 \operatorname{adj}(3A)) = 4^3 \times \det(\operatorname{adj}(3A))$ Using the property $\operatorname{det}(\operatorname{adj}(A)) = \det(A)^{n-1}$ for a $n \times n$ matrix: $\operatorname{adj}(3A) = (3A)^{2-1} = (3 \times \det(A))^2$ $= (3^3 \times (-4))^2$ Therefore, calculate $\det(3A)^2$ : $\det((3A)^2) = (3^3 \times (-4))^2 = 3^6 \times 4^2$ Substitute back: $4^3 \times \det((3A)^2) = 4^3 \times 3^6 \times 4^2 = 2^6 \times 3^6 \times 2^4$ Simplifying gives: $= 2^{10} \times 3^6$ Finally, we have powers $m = 10$ and $n = 6$ .

Therefore: $m+n = 10 + 6 = 16$

Q207

If the system of linear equations

\begin{aligned} & 3 x+y+\beta z=3 \\ & 2 x+\alpha y-z=-3 \\ & x+2 y+z=4 \end{aligned}

has infinitely many solutions, then the value of

22 \beta-9 \alpha

is :

A 31

B 37

C 43

D 49

Correct Answer

Option A

Solution

\begin{aligned} &\begin{aligned} & 3 x+y+\beta z=3 \\ & 2 x+\alpha y-z=-3 \\ & x+2 y+z=4 \end{aligned}\\ &\text{ has infinite solution }\\ &\begin{aligned} & \Rightarrow \Delta=0, \Delta_1=\Delta_2=\Delta_3 \\ & \Delta=0 \Rightarrow\left|\begin{array}{ccc} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{array}\right|=0 \\ & \Delta_2=0 \Rightarrow\left|\begin{array}{ccc} 3 & 3 & \beta \\ 2 & -3 & -1 \\ 1 & 4 & 1 \end{array}\right|=0 \\ & \Rightarrow 3(-3+4)-3(2+1)+\beta(8+3)=0 \\ & \Rightarrow 3-9+11 \beta=0 \\ & \Rightarrow \quad \beta=\frac{6}{11} \\ & \Delta_3=0 \Rightarrow\left|\begin{array}{ccc} 3 & 1 & 3 \\ 2 & \alpha & -3 \\ 1 & 2 & 4 \end{array}\right|=0 \\ & \Rightarrow \quad 3(4 \alpha+6)-1(8+3)+3(4-\alpha)=0 \\ & \quad 12 \alpha+18-11+12-3 \alpha=0 \\ & \quad 9 \alpha=-19 \\ & \quad \alpha=\frac{-19}{9} \end{aligned} \end{aligned}

$\therefore \quad 22 \beta-9 \alpha=31$

Q208

Let A =

\left( \begin{array}{lll}0 & {2q} & r \\ p & q & { - r} \\ p & { - q} & r \end{array} \right).

If AAT = I3, then

\left| p \right|

is :

A

{1 \over {\sqrt 2 }}

B

{1 \over {\sqrt 5 }}

C

{1 \over {\sqrt 6 }}

D

{1 \over {\sqrt 3 }}

Correct Answer

Option A

Solution

A is orthogonal matrix $\Rightarrow$ 02 + p2 + p2 = 1 $\Rightarrow$

\left| p \right| = {1 \over {\sqrt 2 }}

Q209

Let the system of equations x + 5y - z = 1 4x + 3y - 3z = 7 24x + y + λz = μ λ, μ ∈ ℝ, have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy 7 ≤ x + y + z ≤ 77, is :

A 4

B 5

C 3

D 6

Correct Answer

Option C

Solution

\begin{aligned} &\text{ For infinitely many solution }\\ &\begin{aligned} & \Delta=0 \\ & \left|\begin{array}{ccc} 1 & 5 & -1 \\ 4 & 3 & -3 \\ 24 & 1 & \lambda \end{array}\right|=0 \\ & \Rightarrow 1(3 \lambda+3)-5(4 \lambda+72)-1(4-72)=0 \\ & \Rightarrow-17 \lambda+3-4 \times 72-4=0 \\ & \Rightarrow 17 \lambda=-289 \\ & \Rightarrow \lambda=-17 \\ & \Delta 1=0 \end{aligned} \end{aligned}

\begin{aligned} & \Rightarrow\left|\begin{array}{ccc} 1 & 5 & -1 \\ 7 & 3 & -3 \\ \mu & 1 & -17 \end{array}\right|=0 \\ & \Rightarrow 1(-51+3)-5(-119+3 \mu)-1(7-3 \mu)=0 \\ & \Rightarrow-48+595-15 \mu-7+3 \mu=0 \\ & \Rightarrow 12 \mu=540 \\ & \mu=45 \\ & x+5 y-z=1 \\ & 4 x+3 y-3 z=7 \\ & 24 x+y-17 z=45 \\ & \text{ Let } z=1 \\ & x+5 y=1+\lambda] \times 4 \\ & 4 x+3 y=7+3 \lambda \end{aligned}

$\dfrac{\underset{-}{4 \mathrm{x}}+20 \mathrm{y}=\underset{-}{4+4 \lambda}}{-17 \mathrm{y}=3-\lambda}$

\begin{aligned} \begin{aligned} \mathrm{y} & =\frac{\lambda-3}{17}, \mathrm{x}=1+\lambda-\frac{5 \lambda-15}{17} \\ & =\frac{32-12 \lambda}{17} \\ 7 & \leq \frac{\lambda-3}{17}+\frac{32+12 \lambda}{17}+\lambda \leq 77 \\ 7 & \leq \frac{30 \lambda+29}{17} \leq 77 \\ 3 & \leq \lambda \leq 42 \\ \lambda & =3,20,37 \end{aligned} \end{aligned}

Q210

Let

A

be a

3 \times 3

matrix such that

|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} \mathrm{A}))|=81

. If

S=\left\{n \in \mathbb{Z}:(|\operatorname{adj}(\operatorname{adj} A)|)^{\dfrac{(n-1)^2}{2}}=|A|^{\left(3 n^2-5 n-4\right)}\right\}

, then

\sum\limits_{n \in S}\left|A^{\left(n^2+n\right)}\right|

is equal to :

A 820

B 866

C 750

D 732

Correct Answer

Option D

Solution

\begin{aligned} & |\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))|=81 \\ & =|A|^{(n-1)^3}=(3)^4 \Rightarrow|A|^8=3^4 \Rightarrow|A|=3^{1 / 2} \\ & |\operatorname{adj}(\operatorname{adj} A)|^{\frac{(n-1)^2}{2}}=|A|^{\left(3 n^2-5 n-4\right)} \\ & {\left[|A|^{(n-1)^2}\right]^{\frac{(n-1)^2}{2}}=|A|^{3 n^2-5 n-4}} \\ & |A|^{2(n-1)^2}=|A|^{3 n^2-5 n-4} \\ & \Rightarrow 2(n-1)^2=3 n^2-5 n-4 \\ & \quad n^2-n-6=0 \\ & \Rightarrow n=-2,3 \\ & \sum_{x \leftarrow 5}\left|A^{n^2+n}\right|=\left|A^2\right|+\left|A^{12}\right| \\ & =3+3^6=732 \end{aligned}