Back to MATRICES

Your Progress

0 / 995 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →

Showing 1-30 of 995 questions

1 Easy

1 of 995

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 995

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 Medium

3 of 995

Application +4 | -1

If the equations

2x + 3y + 1 = 0,3x + y - 2 = 0

and

ax + 2y - b = 0

are consistent, then

View answer and explanation

Correct option: A

a - b = 2

Option A (

a - b = 2

) is correct because the consistency condition for the three equations leads to the determinant of the coefficients being zero, which simplifies to

a - b = 2

4 Easy

4 of 995

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.

5 Easy

5 of 995

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.

6 Easy

6 of 995

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.

7 Easy

7 of 995

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.

8 Medium

8 of 995

Application +4 | -1

If matrix

A=[a_{i j}]_{5 \times 5}

such that

a_{i j}=\left\{\begin{array}{ll}2 ; & \text { where } i=j \\ 0 ; & \text { when } i \neq j\end{array}\right.

, then value of

\operatorname{det}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A)))

is equal to

View answer and explanation

Correct option: B

(2)^{320}

Option B (

(2)^{320}

) is correct because the determinant of the adjoint of the adjoint of the adjoint of A can be calculated using the properties of determinants and adjoints.

9 Easy

9 of 995

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.

10 Easy

10 of 995

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.

11 Easy

11 of 995

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.

12 Easy

12 of 995

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.

13 Medium

13 of 995

Application +4 | -1

P = \left[ {\begin{array}{cccccccccccccccccccc} 1&\alpha &3\\ 1&3&3\\ 2&4&4 \end{array}} \right]

is the adjoint of a

3 \times 3

matrix

A

and

\left| A \right| = 4,

then

\alpha

is equal to

View answer and explanation

Correct option: B

Option B (11) is correct because using the properties of adjoint matrices and determinants, the value of

\alpha

is found to be 11.

14 Medium

14 of 995

Application +4 | -1

Let

A=\left(\begin{array}{ccc}0 & 2 \mathrm{q} & r \\ p & q & -r \\ p & -q & r\end{array}\right)

. If

\mathrm{AA}^{\mathrm{T}}=\mathrm{I}_{3}

, then

|\mathrm{p}|

is :

View answer and explanation

Correct option: C

\frac{1}{\sqrt{2}}

Option C (

\frac{{1}}{{\sqrt{{2}}}}

) is correct because after calculating

AA^T

and equating it to

I_3

, we can solve for

|p|

15 Hard

15 of 995

Analyzing +4 | -1

f(x), g(x)

and

h(x)

are polynomials of degree 4 and

\left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ a & b & c \\ p & q & r\end{array}\right|=m x^4+n x^3+r x^2+s x+t

be an identity in

x

, then

\left| {\begin{array}{cccccccccccccccccccc} {f'''(0) - f''(0)}&{g'''(0) - g''(0)}&{h'''(0) - h''(0)} \\ a&b&c \\ p&q&r \end{array}} \right|

is equal to

View answer and explanation

Correct option: D

2(3 n-r)

Option D (

2(3 n-r)

) is correct because by differentiating the given identity and substituting

x=0

, we can evaluate the required determinant.

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs

16 Easy

16 of 995

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.

17 Easy

17 of 995

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}

18 Medium

18 of 995

Understanding +4 | -1

B

is a non-singular matrix and

A

is a square matrix such that

{B^{ - 1}}AB

exists, then det

\left( {{B^{ - 1}}AB} \right)

is equal to

View answer and explanation

Correct option: D

\det \left( A \right)

Option D is correct because the determinant of a product of matrices can be separated, and the inverse cancels out.

19 Easy

19 of 995

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 995

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 995

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 995

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 995

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 995

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.

25 Easy

25 of 995

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

26 Easy

26 of 995

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.

27 Easy

27 of 995

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|.

28 Easy

28 of 995

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.

29 Easy

29 of 995

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.

30 Easy

30 of 995

Application +4 | -1

Let

A=\left(\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right)

. and

B=\left(\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right)

. If

B

is the inverse of matrix

A

, then

\alpha

View answer and explanation

Correct option: A

Option A (5) is correct because when B is the inverse of A, their product AB equals the identity matrix, which allows us to solve for alpha.

Revision notes and key takeaways

Review MATRICES in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 995 single-correct-answer MCQs on MATRICES.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

Showing 1-30 of 995 questions

Continue Your Practice

📚 Switch Set (ALL)

Chapter Sets

ALL 995 MCQs EASY 922 MCQs HARD 26 MCQs MEDIUM 408 MCQs

0 /995

0✓ 0✗

0% Accuracy