If $f\left( {\rm{\theta }} \right) = \left| {\begin{array}{*{20}{c}} {{{\cos }^2}{\rm{\theta }}}&{\cos {\rm{\theta }}\sin {\rm{\theta }}}&{ - {\rm{sin\theta }}}\\ {\cos {\rm{\theta }}\sin {\rm{\theta }}}&{{{\sin }^2}{\rm{\theta }}}&{{\rm{cos\theta }}}\\ {\sin {\rm{\theta }}}&{ - \cos {\rm{\theta }}}&0 \end{array}} \right|{\rm{.}}$. Then, for all $\theta $

$f\left( {\rm{\theta }} \right) = 1$

$f\left( {\rm{\theta }} \right) = 0$

$f\left( {\rm{\theta }} \right) = - 1$

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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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Showing 1-30 of 922 questions

1 Easy

1 of 922

Analyzing +4 | -1

\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

and

a, b, c

are real numbers then the line

a x+b y+c=0

passes through the fixed point is

View answer and explanation

Correct option: B

(1,1)

Option B ((1,1)) is correct because the determinant being zero implies a+b+c=0, which means the line ax+by+c=0 passes through (1,1).

2 Easy

2 of 922

Analyzing +4 | -1

The number of solutions of the system of equations

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because the system of equations is inconsistent, meaning there are no solutions that satisfy all equations simultaneously.

3 Easy

3 of 922

Analyzing +4 | -1

The value of the determinant

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

, is

View answer and explanation

Correct option: A

The correct answer is A (0) because simplifying the determinant leads to two identical columns, making the determinant zero.

4 Easy

4 of 922

Application +4 | -1

\omega

is the cube root of unity, then

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because after applying column operations and using the property of cube roots of unity (

1 + \omega + \omega^2 = 0

), the determinant simplifies to zero.

5 Easy

5 of 922

Application +4 | -1

For any matrix

A

of order

2 \times 2

, if

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

, then

|A|

is equal to

View answer and explanation

Correct option: C

Option C (10) is correct because

A(\operatorname{adj} A) = |A|I

, and given

A(\operatorname{adj} A) = 10I

, therefore

|A| = 10

6 Easy

6 of 922

Analyzing +4 | -1

f\left( \mathrm{\theta } \right) = \left| {\begin{array}{cccccccccccccccccccc} {{{\cos }^2}\mathrm{\theta }}&{\cos \mathrm{\theta }\sin \mathrm{\theta }}&{ - \mathrm{sin\theta }}\\ {\cos \mathrm{\theta }\sin \mathrm{\theta }}&{{{\sin }^2}\mathrm{\theta }}&{\mathrm{cos\theta }}\\ {\sin \mathrm{\theta }}&{ - \cos \mathrm{\theta }}&0 \end{array}} \right|\mathrm{.}

. Then, for all

\theta

View answer and explanation

Correct option: B

f\left( \mathrm{\theta } \right) = 1

Option B (

f\left( \mathrm{\theta } \right) = 1

) is correct because the determinant simplifies to 1 for all values of

\theta

7 Easy

7 of 922

Application +4 | -1

The inverse of the matrix

A=\left[\begin{array}{ccc}1 & 0 & 0 \\ a & 1 & 0 \\ b & c & 1\end{array}\right]

View answer and explanation

Correct option: A

\left[\begin{array}{ccc}1 & 0 & 0 \\ -a & 1 & 0 \\ a c-b & -c & 1\end{array}\right]

Option B (

\frac{1}{2}(3 I-A)

) is correct because it is derived by multiplying the given equation by

A^{-1}

and rearranging the terms.

8 Easy

8 of 922

Application +4 | -1

E\left( \theta \right) = \left[ {\begin{array}{cccccccccccccccccccc} {\cos \theta }&{\sin \theta }\\ { - \sin \theta }&{\cos \theta } \end{array}} \right],

then

E\left( \alpha \right)E\left( \beta \right)

is equal to

View answer and explanation

Correct option: C

E\left( {\alpha + \beta } \right)

Option C (E(alpha + beta)) is correct because the multiplication of the matrices results in a matrix of the same form with the angle being the sum of alpha and beta.

9 Easy

9 of 922

Application +4 | -1

\left[\begin{array}{rr} 3 & -2 \\ 3 & 0 \\ 2 & 4 \end{array}\right]\left[\begin{array}{ll} x & x \\ y & y \end{array}\right]=\left[\begin{array}{cc} 3 & 3 \\ 3 x & 3 x \\ 10 & 10 \end{array}\right]

then the integral part of

x+y

View answer and explanation

Correct option: A

Option A is correct. By performing the matrix multiplication and solving the resulting system of equations, we find x = 2 and y = 3/2, so the integral part of x+y is 3.

10 Easy

10 of 922

Analyzing +4 | -1

x,\,y,\,z

are in arithmetic progression with common difference

d,\,\,x\, \ne \,3d

, and the determinant of the matrix

\left[ {\begin{array}{cccccccccccccccccccc} 3&{4\sqrt 2 }&x\\ 4&{5\sqrt 2 }&y\\ 5&k&z \end{array}} \right]

is zero, then the value of

{k^2}

View answer and explanation

Correct option: A

Option A (72) is correct because by using properties of determinants and arithmetic progressions, we can find the value of k².

11 Easy

11 of 922

Analyzing +4 | -1

The number of

3 \times 3

non-singular matrices with four entries as 1 and all other entries as 0, is

View answer and explanation

Correct option: B

At least 7

Option B (At least 7) is correct because there are more than 6 non-singular 3x3 matrices that can be formed with four entries as 1 and the rest as 0.

12 Easy

12 of 922

Application +4 | -1

If the value of a third order determinant is 11, then the value of the square of the determinant formed by the cofactors will be

View answer and explanation

Correct option: D

14641

Option D (14641) is correct because the square of the determinant formed by the cofactors is equal to the square of the original determinant raised to the power of (n-1), where n is the order of the determinant.

13 Easy

13 of 922

Analyzing +4 | -1

A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right], 6 A^{-1}=A^2+c A+d I

, then

({c}, {d})

View answer and explanation

Correct option: A

(-6,11)

The correct answer is (-6, 11) because by substituting the given matrix A into the equation

6A^{-1} = A^2 + cA + dI

and solving for c and d, we find that c = -6 and d = 11.

14 Easy

14 of 922

Application +4 | -1

2 X+\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]=\left[\begin{array}{ll}3 & 8 \\ 7 & 2\end{array}\right]

, then

X=

View answer and explanation

Correct option: D

\left[\begin{array}{cc}1 & 3 \\ 2 & -1\end{array}\right]

Option A is correct. After calculating the inverse of A and squaring it, we can compare the result with the given expression to find the value of f(theta).

15 Easy

15 of 922

Application +4 | -1

\;\left[ {\begin{array}{cccccccccccccccccccc} {x - y - z}\\ { - y + z}\\ z \end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc} 0\\ 5\\ 3 \end{array}} \right],\;

than the values of

x,y

and

z

are respectively

View answer and explanation

Correct option: B

1,\mathrm{ } - 2,\mathrm{ }3

Option B (

1, -2, 3

) is correct because it satisfies the given matrix equation by solving the system of linear equations.

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16 Easy

16 of 922

Application +4 | -1

A = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;2\;\;2}\\ {2\;\;1\;\;2}\\ {2\;\;2\;\;1} \end{array}} \right],\;\mathrm{then}\;{A^2} - 4\;A

is equal to

View answer and explanation

Correct option: D

5{I_3}

Option D (

5{I_3}

) is correct because after calculating

{A^2}

and subtracting

4A

, the result is

5{I_3}

17 Easy

17 of 922

Application +4 | -1

A=\left[\begin{array}{cc} -4 & 1 \\ 3 & 2 \end{array}\right] \text { and } f(x)=x^2-2 x+3

then

f(A)

View answer and explanation

Correct option: C

\left[\begin{array}{rr}30 & -4 \\ -12 & 6\end{array}\right]

Option C is correct. We evaluate f(A) by substituting the matrix A into the polynomial and performing the matrix operations.

18 Easy

18 of 922

Analyzing +4 | -1

f(x)

satisfies the equation

\left|\begin{array}{ccc} f(x-3) & f(x+4) & f\left((x+1)(x-2)-(x-1)^2\right) \\ 5 & 4 & -5 \\ 5 & 6 & 15 \end{array}\right|=0

for all real

x

, then-

View answer and explanation

Correct option: C

f(x)

is periodic and is of period 7

Option C is correct because the given equation implies that

f(x+4) = f(x-3)

, which leads to

f(x+7) = f(x)

, indicating a period of 7.

19 Easy

19 of 922

Understanding +4 | -1

A

and

B

are two square matrices such that

A B=I

, then which of the following is not true?

View answer and explanation

Correct option: D

A^2=B

Option D (

A^2=B

) is not necessarily true because

AB = I

implies that

A

and

B

are inverses of each other, so

BA = I

A^{-1} = B

, and

B^{-1} = A

, but not necessarily

A^2 = B

20 Easy

20 of 922

Analyzing +4 | -1

The value of

\left|\begin{array}{ccc}2 y_1 z_1 & y_1 z_2+y_2 z_1 & y_1 z_3+y_3 z_1 \\ y_1 z_2+y_2 z_1 & 2 y_2 z_2 & y_2 z_3+y_3 z_2 \\ y_1 z_3+y_3 z_1 & y_2 z_3+y_3 z_2 & 2 y_3 z_3\end{array}\right|,

View answer and explanation

Correct option: D

Option D (0) is the correct answer because the determinant can be factored into a product of two determinants, each of which is zero.

21 Easy

21 of 922

Analyzing +4 | -1

Let

A=\left[a_j\right]

be a

3 \times 3

matrix and let

A_1

denote the matrix of the cofactors of elements of matrix

A

and

A_2

be the matrix of cofactors of elements of matrix

A_1

and so on. If

A_{n}

denote the matrix of cofactors of elements of matrix

A_{n-1},

then

\left|A_n\right|

equals

View answer and explanation

Correct option: B

|A|^{2^n}

Option B (

|A|^{{2^n}}

) is correct because the determinant of the matrix of cofactors is raised to the power of 2 in each successive matrix, resulting in an exponent of

2^n

22 Easy

22 of 922

Analyzing +4 | -1

Let p be an odd prime number and

T_p

be the following set of matrices

T_p=\left\{A=\left[\begin{array}{ll}a & b \\ c & a\end{array}\right]: a, b, c \in\{0,1,2, \ldots, p-1\}\right\}

The number of

A

T_p

such that the trace of

A

is not divisible by

p

but

\operatorname{det}(A)

is divisible by

p

, is

View answer and explanation

Correct option: C

(p-1)^2

The correct answer is

-X

because the given condition

P^T = 2P + I

leads to

P = -I

, and therefore

PX = -X

23 Easy

23 of 922

Analyzing +4 | -1

Let

A

and

B

be two

3 \times 3

non-zero real matrices such that

A B

is a zero matrix. Then

View answer and explanation

Correct option: B

the system of linear equations

A X=0

has infinitely many solutions

Option B (the system of linear equations

A X=0

has infinitely many solutions) is correct because since

AB = 0

, the determinant of A must be zero, leading to infinitely many solutions for the homogeneous system.

24 Easy

24 of 922

Analyzing +4 | -1

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

and

\alpha+\beta=-2

, then

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

is equal to :

View answer and explanation

Correct option: C

Option C (14) is correct because after finding λ, α, and β using the given conditions, substituting them into the expression yields 14.

25 Easy

25 of 922

Application +4 | -1

Let

S = \left\{ {\sqrt n :1 \leqslant n \leqslant 50\,\,{\text{and}}\,\,n\,\,{\text{is}}\,\,{\text{odd}}} \right\}.

Let

a \in S\,\,{\text{and}}\,\,A = \left[ {\begin{array}{cccccccccccccccccccc} 1&0&a \\ { - 1}&1&0 \\ { - a}&0&1 \end{array}} \right]

\sum\limits_{a \in S} {\det \left( {adj\,A} \right)} = 100\lambda ,

then

\lambda

is equal to:

View answer and explanation

Correct option: B

221

Option B (221) is correct because the sum of the determinants of the adjoint matrices is calculated and simplified to find the value of lambda.

26 Easy

26 of 922

Analyzing +4 | -1

Let

a - 2b + c = 1.

f(x) = \left| {\begin{array}{cccccccccccccccccccc} {x + a}&{x + 2}&{x + 1}\\ {x + b}&{x + 3}&{x + 2}\\ {x + c}&{x + 4}&{x + 3} \end{array}} \right|,

then

View answer and explanation

Correct option: B

f(50) = 1

Option B (

f(50) = 1

) is correct because applying row operations to the determinant simplifies it to a constant value of 1, independent of

x

27 Easy

27 of 922

Analyzing +4 | -1

If the system of equations

a x+b y+c=0,

b x+c y+a=0, cx+a y+b=0

has infinitely many solutions then the system of equations

\begin{aligned} &(b+c) x+(c+a) y+(a+b) z=0 \\ &(c+a) x+(a+b) y+(b+c) z=0 \\ &(a+b) x+(b+c) y+(c+a) z=0\end{aligned}

has

View answer and explanation

Correct option: C

Infinite number of solutions

The system has infinitely many solutions because the determinant of the coefficient matrix of the second system is zero, given the condition from the first system.

28 Easy

28 of 922

Analyzing +4 | -1

A=\left[\begin{array}{lll}a & b & c \\ x & y & z \\ p & q & r\end{array}\right], B=\left[\begin{array}{rrr}q & -b & y \\ -p & a & -x \\ r & -c & z\end{array}\right]

and if

A

is invertible, then which of the following is not true?

View answer and explanation

Correct option: A

|A|=|B|

Option A (|A|=|B|) is correct because the determinant of B is actually -|A|, not |A|.

29 Easy

29 of 922

Understanding +4 | -1

The value of

\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|,

View answer and explanation

Correct option: A

(a-b)(b-c)(c-a)

Option A (

(a-b)(b-c)(c-a)

) is correct because the determinant is a Vandermonde determinant, which has a known factorization.

30 Easy

30 of 922

Analyzing +4 | -1

\Delta = \left| {\begin{array}{cccccccccccccccccccc} {x - 2}&{2x - 3}&{3x - 4}\\ {2x - 3}&{3x - 4}&{4x - 5}\\ {3x - 5}&{5x - 8}&{10x - 17} \end{array}} \right| = A{x^3} + B{x^2} + Cx + D

, then

B + C

is equal to

View answer and explanation

Correct option: C

-3

Option C -3 is correct because, after expanding the determinant and comparing coefficients, the sum of B and C is found to be -3.

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