Matrices and Determinants — Page 19 | JEE Mathematics

Q181

If the system of equations

x+4 y-z=\lambda, 7 x+9 y+\mu z=-3,5 x+y+2 z=-1

has infinitely many solutions, then

(2 \mu+3 \lambda)

is equal to :

A

-2

B 2

C 3

D

-3

Correct Answer

Option D

Solution

\begin{aligned} & x+4 y-z=\lambda \\ & 7 x+9 y+\mu z=-3 \\ & 5 x+y+2 z=-1 \\ & {\left[\begin{array}{ccc} 1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} \lambda \\ -3 \\ -1 \end{array}\right]} \\ & A=\left[\begin{array}{lll} 1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2 \end{array}\right], B=\left[\begin{array}{c} \lambda \\ -3 \\ -1 \end{array}\right] \\ & A X=B \\ & X=A^{-1} B \\ & \quad=\frac{\operatorname{adj} A}{|A|} B \end{aligned}

If

|A|=0

and

(\operatorname{adj} A) \cdot B=0

, system has infinitely many solutions.

\begin{aligned} & |A|=18-\mu-4(14-5 \mu)-1(7-45)=0 \\ & \Rightarrow 18-\mu-56+20 \mu+38=0 \\ & \Rightarrow 19 \mu=0 \\ & \Rightarrow \mu=0 \\ & \text{ Also adjA }=\left[\begin{array}{ccc} 18 & -9 & 9 \\ -14 & 7 & -7 \\ -38 & 19 & -19 \end{array}\right] \\ & (\text{ adj } A) \cdot B=0 \\ & {\left[\begin{array}{ccc} 18 & -9 & 9 \\ -14 & 7 & -7 \\ -38 & 19 & -19 \end{array}\right]\left[\begin{array}{c} \lambda \\ -3 \\ -1 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]} \\ & 18 \lambda+27-9=0 \\ & \Rightarrow 18 \lambda=-18 \\ & \Rightarrow \lambda=-1 \\ & \Rightarrow 2 \mu+3 \lambda=3(-1)=-3 \end{aligned}

Q182

Let

A=\left[\begin{array}{lll}2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b\end{array}\right]

. If

A^3=4 A^2-A-21 I

, where

I

is the identity matrix of order

3 \times 3

, then

2 a+3 b

is equal to

A

-10

B

-12

C

-13

D

-9

Correct Answer

Option C

Solution

\begin{aligned} & |A-\lambda I|=0 \\ & \left|\begin{array}{ccc} 2-\lambda & a & 0 \\ 1 & 3-\lambda & 1 \\ 0 & 5 & b-\lambda \end{array}\right|=0 \\ & (2-\lambda)[(3-\lambda)(b-\lambda)-5]-a[b-\lambda-0]+0=0 \\ & (2-\lambda)\left[3 b-3 \lambda-b \lambda+\lambda^2-5\right]-a b+a \lambda=0 \\ & \lambda^3-(b+5) \lambda^2+(1-a+5 b) \lambda+(10-6 b+a b)=0 \\ & A^3-(b+5) A^2+(1-a+5 b) A+(10-6 b+a b) I=0 \\ & \Rightarrow \mathrm{b}+5=4,1-a+5 b=1,10-6 b+a b=21 \\ & \Rightarrow a=-5, b=-1 \\ & \Rightarrow 2 a+3 b=-13 \end{aligned}

Q183

Let A and B be two square matrices of order 3 such that

\mathrm{|A|=3}

and

\mathrm{|B|=2}

. Then

|\mathrm{A}^{\mathrm{T}} \mathrm{A}(\operatorname{adj}(2 \mathrm{~A}))^{-1}(\operatorname{adj}(4 \mathrm{~B}))(\operatorname{adj}(\mathrm{AB}))^{-1} \mathrm{AA}^{\mathrm{T}}|

is equal to :

A 32

B 81

C 64

D 108

Correct Answer

Option C

Solution

\begin{aligned} & |A|=3 \\ & |B|=2 \\ & \left.\left|A^T\right||A| \mid(\operatorname{adj}(2 A))^{-1}\|\operatorname{adj}(4 B)\|(\operatorname{adj}(A B))^{-1}\right)|A|\left|A^T\right| \\ & 3 \cdot 3 \frac{1}{64 \cdot 9}(64)^2 \cdot 4 \cdot \frac{1}{9 \cdot 4} 3 \cdot 3 \\ & =64 \end{aligned}

Q184

For some

a, b,

let

f(x)=\left|\begin{array}{ccc}\mathrm{a}+\dfrac{\sin x}{x} & 1 & \mathrm{~b} \\ \mathrm{a} & 1+\dfrac{\sin x}{x} & \mathrm{~b} \\ \mathrm{a} & 1 & \mathrm{~b}+\dfrac{\sin x}{x}\end{array}\right|, x \neq 0, \lim \limits_{x \rightarrow 0} f(x)=\lambda+\mu \mathrm{a}+\nu \mathrm{b}.

Then

(\lambda+\mu+v)^2

is equal to :

A 25

B 16

C 9

D 36

Correct Answer

Option B

Solution

\begin{aligned} & \lim _{x \rightarrow 0} f(x)=\left|\begin{array}{ccc} a+1 & 1 & b \\ a & 1+1 & b \\ a & 1 & b+1 \end{array}\right| \\ & =(a+1)(2(b+1)-b)+1(a b-a(b+1))+b a \\ & =(a+1)(b+2)-a+a b \\ & =b+a+2=\lambda+\mu a+v b \\ & \lambda=2, \mu=1, v=1 \Rightarrow(\lambda+\mu+v)^2=16 \end{aligned}

Q185

The values of

m, n

, for which the system of equations

\begin{aligned} & x+y+z=4, \\ & 2 x+5 y+5 z=17, \\ & x+2 y+\mathrm{m} z=\mathrm{n} \end{aligned}

has infinitely many solutions, satisfy the equation :

A

\mathrm{m}^2+\mathrm{n}^2-\mathrm{m}-\mathrm{n}=46

B

\mathrm{m}^2+\mathrm{n}^2+\mathrm{mn}=68

C

\mathrm{m}^2+\mathrm{n}^2-\mathrm{mn}=39

D

\mathrm{m}^2+\mathrm{n}^2+\mathrm{m}+\mathrm{n}=64

Correct Answer

Option C

Solution

The given system of linear equations can be represented as,

\begin{aligned} & \left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 2 & 5 & 5 & 17 \\ 1 & 2 & m & n \end{array}\right) \\ & \sim\left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 0 & 3 & 3 & 9 \\ 0 & 1 & m-1 & n-4 \end{array}\right) \\ & \sim\left(\begin{array}{ccc|c} 1 & 1 & 1 & 4 \\ 0 & 1 & 1 & 3 \\ 0 & 0 & m-2 & n-7 \end{array}\right) \end{aligned}

$\because$ System of equations has infinitely many solutions

\therefore m=2 \& n=7

Which satisfy equation given in option (1).

\text{ (i.e., } 2^2+7^2-14=39 \text{ ) }

Q186

Let

\alpha \beta \neq 0

and

A=\left[\begin{array}{rrr}\beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2 \alpha\end{array}\right]

. If

B=\left[\begin{array}{rrr}3 \alpha & -9 & 3 \alpha \\ -\alpha & 7 & -2 \alpha \\ -2 \alpha & 5 & -2 \beta\end{array}\right]

is the matrix of cofactors of the elements of

A

, then

\operatorname{det}(A B)

is equal to :

A 64

B 343

C 125

D 216

Correct Answer

Option D

Solution

A=\left[\begin{array}{ccc} \beta & \alpha & 3 \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2 \alpha \end{array}\right], B=\left[\begin{array}{ccc} 3 \alpha & -9 & 3 \alpha \\ -\alpha & 7 & -2 \alpha \\ -2 \alpha & 5 & -2 \beta \end{array}\right]

Cofactor of

A

-matrix is

=\left[\begin{array}{ccc} 2 \alpha^2-\alpha \beta & -\left(2 \alpha^2+\beta^2\right) & \alpha^2+\alpha \beta \\ -\left(2 \alpha^2-3 \alpha\right) & (2 \alpha \beta+3 \beta) & -(2 \alpha \beta) \\ \alpha \beta-3 \alpha & -\left(\beta^2-3 \alpha\right) & \beta \alpha-\alpha^2 \end{array}\right]

which is equal to matrix

B

So, by comparing elements of two matrix

\begin{aligned} & \Rightarrow \alpha \beta-3 \alpha=-2 \alpha \\ & \Rightarrow \alpha \beta-\alpha=0 \\ & \Rightarrow \alpha(\beta-1)=0 \\ & \Rightarrow \alpha=0 \text{ or } \beta=1[\because \alpha \text{ cannot be } 0] \\ & \Rightarrow \beta=1 \\ & \text{ and }-\beta^2+3 \alpha=5 \\ & \Rightarrow 3 \alpha=6 \\ & \Rightarrow \alpha=2 \\ & A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4 \end{array}\right] \\ & \operatorname{Det}(A B)=|A||B|=|A|\left|(\operatorname{adj} A)^{\top}\right| \\ & =|A| \cdot|A|^2 \\ & =|A|^3 \\ & =(6-18+18)^3 \\ & =6^3 \\ & =216 \\ \end{aligned}

Q187

If

A

is a square matrix of order 3 such that

\operatorname{det}(A)=3

and

\operatorname{det}\left(\operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 \mathrm{~A})^{-1}\right)\right)\right)\right)\right)=2^{\mathrm{m}} 3^{\mathrm{n}}

, then

\mathrm{m}+2 \mathrm{n}

is equal to :

A 2

B 4

C 3

D 6

Correct Answer

Option B

Solution

\begin{aligned} & |A|=3 \\ & \left|\operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}(2 A)^{-1}\right)\right)\right)\right| \\ & =\left|-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 A)^{-1}\right)\right)\right)\right|^2 \end{aligned}

\begin{aligned} & =4^6\left|-3 \operatorname{adj}\left(\operatorname{aadj}\left((2 A)^{-1}\right)\right)\right|^4 \\ & =4^6 \cdot 3^{12}\left|\operatorname{aadj}\left((2 A)^{-1}\right)\right|^8 \\ & =4^6 \cdot 3^{12} \cdot 3^{24}\left|(2 A)^{-1}\right|^{16} \\ & =4^6 \cdot 3^{36} 2^{-48}\left|A^{-1}\right|^{16} \\ & =\frac{2^{-36} 3^{36}}{3^{16}}=2^{-36} 3^{20} \\ & m=-36 \\ & n=20 \\ & m+2 n=4 \end{aligned}

Q188

For

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

and a natural number

n

, let

A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|

. Then

2 A_{10}-A_8

is

A

4 \alpha+2 \beta

B 0

C

2 n

D

2 \alpha+4 \beta

Correct Answer

Option A

Solution

We are given a determinant $A_r$ defined as:

A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|

We need to find the value of $2 A_{10} - A_8$ .

Let's calculate the determinant $A_r$ .

We can use properties of determinants or expand it directly.

A helpful trick is to use row operations to simplify the determinant before expanding.

Let's perform the following row operations: Replace Row 2 with (Row 2) - 2 $\times$ (Row 1) Replace Row 3 with (Row 3) - 3 $\times$ (Row 1) The determinant becomes:

A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r - 2r & 2 - 2(1) & (n^2-\beta) - 2(\frac{n^2}{2}+\alpha) \\ (3 r-2) - 3r & 3 - 3(1) & \frac{n(3 n-1)}{2} - 3(\frac{n^2}{2}+\alpha)\end{array}\right|

Let's simplify the entries in the new rows: Row 2: First element: $2r - 2r = 0$ Second element: $2 - 2 = 0$ Third element: $(n^2-\beta) - 2(\dfrac{n^2}{2}+\alpha) = n^2 - \beta - n^2 - 2\alpha = -2\alpha - \beta$ Row 3: First element: $(3r-2) - 3r = -2$ Second element: $3 - 3 = 0$ Third element: $\dfrac{n(3 n-1)}{2} - 3(\dfrac{n^2}{2}+\alpha) = \dfrac{3n^2 - n}{2} - \dfrac{3n^2}{2} - 3\alpha = \dfrac{3n^2 - n - 3n^2}{2} - 3\alpha = -\dfrac{n}{2} - 3\alpha$ So the determinant simplifies to:

A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 0 & 0 & -2\alpha - \beta \\ -2 & 0 & -\frac{n}{2} - 3\alpha\end{array}\right|

Now, we can easily calculate this determinant by expanding along the second column, since it has two zeros.

The expansion along the second column is:

A_r = -1 \cdot \left|\begin{array}{cc} 0 & -2\alpha - \beta \\ -2 & -\frac{n}{2} - 3\alpha \end{array}\right|

A_r = -1 \cdot \left( (0) \times (-\frac{n}{2} - 3\alpha) - (-2) \times (-2\alpha - \beta) \right)

A_r = -1 \cdot \left( 0 - (4\alpha + 2\beta) \right)

A_r = -1 \cdot (-4\alpha - 2\beta)

A_r = 4\alpha + 2\beta

Notice that the value of the determinant $A_r$ does not depend on $r$ at all!

It's a constant value determined by $\alpha$ and $\beta$ .

Now we need to find the value of $2 A_{10} - A_8$ .

Since $A_r = 4\alpha + 2\beta$ for any value of $r$ , we have: $A_{10} = 4\alpha + 2\beta$ $A_8 = 4\alpha + 2\beta$ So, $2 A_{10} - A_8 = 2 (4\alpha + 2\beta) - (4\alpha + 2\beta)$ $2 A_{10} - A_8 = 8\alpha + 4\beta - 4\alpha - 2\beta$ $2 A_{10} - A_8 = (8\alpha - 4\alpha) + (4\beta - 2\beta)$ $2 A_{10} - A_8 = 4\alpha + 2\beta$ The value of the expression $2 A_{10} - A_8$ is $4\alpha + 2\beta$ .

Q189

Let

A = \begin{bmatrix} a_{ij} \end{bmatrix} = \begin{bmatrix} \log_5 128 & \log_4 5 \\ \log_5 8 & \log_4 25 \end{bmatrix}

. If

A_{ij}

is the cofactor of

a_{ij}

,

C_{ij} = \sum\limits_{k=1}^{2} a_{ik} A_{jk} , 1 \leq i, j \leq 2

, and

C=[C_{ij}]

, then

8|C|

is equal to :

A 288

B 262

C 222

D 242

Correct Answer

Option D

Solution

To solve the problem, we need to determine the determinant of matrix $A$ : $A = \begin{bmatrix} \log_5 128 & \log_4 5 \\ \log_5 8 & \log_4 25 \end{bmatrix}$ The determinant of $A$ , denoted as $|A|$ , is calculated as: $|A| = (\log_5 128)(\log_4 25) - (\log_4 5)(\log_5 8)$ Evaluating each component: $\log_5 128$ can be simplified using change of base formula: $\log_5 128 = \dfrac{\log_{10}128}{\log_{10}5}$ $\log_4 5$ using change of base: $\log_4 5 = \dfrac{\log_{10}5}{\log_{10}4}$ $\log_5 8$ : $\log_5 8 = \dfrac{\log_{10}8}{\log_{10}5}$ $\log_4 25$ : $\log_4 25 = \dfrac{\log_{10}25}{\log_{10}4}$ Now, substitute these values into $|A|$ : $|A| = \left(\dfrac{\log_{10}128}{\log_{10}5}\right) \left(\dfrac{\log_{10}25}{\log_{10}4}\right) - \left(\dfrac{\log_{10}5}{\log_{10}4}\right) \left(\dfrac{\log_{10}8}{\log_{10}5}\right)$ Next, we find the cofactors of matrix $A$ : $A_{11} = \log_4 25$ $A_{12} = -\log_5 8$ $A_{21} = -\log_4 5$ $A_{22} = \log_5 128$ Then, calculate matrix $C$ whose elements are given by $C_{ij} = \sum\limits_{k=1}^{2} a_{ik} A_{jk}$ : $C_{11} = a_{11}A_{11} + a_{12}A_{12} = |A| = \dfrac{11}{2}$ $C_{12} = a_{11}A_{21} + a_{12}A_{22} = 0$ $C_{21} = a_{21}A_{11} + a_{22}A_{12} = 0$ $C_{22} = a_{21}A_{21} + a_{22}A_{22} = |A| = \dfrac{11}{2}$ Thus, matrix $C$ is: $C = \begin{bmatrix} \dfrac{11}{2} & 0 \\ 0 & \dfrac{11}{2} \end{bmatrix}$ To find $|C|$ , we compute: $|C| = \left(\dfrac{11}{2}\right)\left(\dfrac{11}{2}\right) = \dfrac{121}{4}$ Finally, calculate $8|C|$ : $8|C| = 8 \times \dfrac{121}{4} = 242$

Q190

Let

A = [a_{ij}]

be a

2 \times 2

matrix such that

a_{ij} \in \{0, 1\}

for all

i

and

j

. Let the random variable

X

denote the possible values of the determinant of the matrix

A

. Then, the variance of

X

is:

A

\dfrac{5}{8}

B

\dfrac{1}{4}

C

\dfrac{3}{4}

D

\dfrac{3}{8}

Correct Answer

Option D

Solution

\begin{aligned} & |A|=\left|\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right| \\ & =a_{11} a_{22}-a_{21} a_{12} \\ & =\{-1,0,1\} \end{aligned}

\begin{array}{c|c|c|c} \mathrm{x} & \mathrm{P}_{\mathrm{i}} & \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}} & \mathrm{P}_1 \mathrm{X}_{\mathrm{i}}{ }^2 \\ -1 & \frac{3}{16} & -\frac{3}{16} & \frac{3}{16} \\ 0 & \frac{10}{16} & 0 & 0 \\ 1 & \frac{3}{16} & \frac{3}{16} & \frac{3}{16} \\ \\ & & \sum \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}=0 & \sum \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}{ }^2=\frac{3}{8} \end{array}

\begin{aligned} & \therefore \operatorname{var}(\mathrm{x})=\sum \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}^2-\left(\sum \mathrm{P}_{\mathrm{i}} X_{\mathrm{i}}\right)^2 \\ & =\frac{3}{8}-0=\frac{3}{8} \end{aligned}