The inverse of the matrix $\left[ {\begin{array}{*{20}{c}} 7&{ - 3}&{ - 3}\\ { - 1}&1&0\\ { - 1}&0&1 \end{array}} \right]$ is

$\left[ {\begin{array}{*{20}{c}} 1&3&3\\ 1&4&3\\ 1&3&4 \end{array}} \right]$

$\;\left[ {\begin{array}{*{20}{c}} 1&1&1\\ 3&4&3\\ 3&3&4 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 1&3&1\\ 4&3&8\\ 3&4&1 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 1&1&1\\ 3&3&4\\ 3&4&3 \end{array}} \right]$

$\left| {\begin{array}{*{20}{c}} {2xy}&{{x^2}}&{{y^2}}\\ {{x^2}}&{{y^2}}&{2xy}\\ {{y^2}}&{2xy}&{{x^2}} \end{array}} \right|$ is equal to

${\left( {{x^2} + {y^2}} \right)^3}$

$ - {\left( {{x^2} + {y^2}} \right)^3}$

If $\left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&{ - 2}&{ - 2}\\ 1&3&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 3\\ 4 \end{array}} \right],\;$ then $\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right]$ is equal to

$\left[ {\begin{array}{*{20}{c}} 0\\ 1\\ 1 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 5\\ { - 2}\\ 1 \end{array}} \right]$

If $c = 2\cos {\rm{\theta }},$ then the value of the determinant $\Delta = \left| {\begin{array}{*{20}{c}} c&1&0\\ 1&c&1\\ 6&1&c \end{array}} \right|$ is

$\frac{{\sin 4{\rm{\theta }}}}{{\sin {\rm{\theta }}}}$

$\frac{{2{{\sin }^2}2{\rm{\theta }}}}{{\sin {\rm{\theta }}}}$

$4{\cos ^2}{\rm{\theta }}(2\cos {\rm{\theta }} - 1)$

Back to MATRICES

Your Progress

0 / 922 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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 871-900 of 922 questions

871 Easy

871 of 922

Understanding +4 | -1

If the points

({x_1},{y_1}),\left( {{x_2},{y_2}} \right)

and

\;\left( {{x_3},{y_3}} \right)

are collinear, then the rank of the matrix

\left[ {\begin{array}{cccccccccccccccccccc} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right]

will always be less than

View answer and explanation

Correct option: B

Option B (3) is correct because if the points are collinear, the determinant of the matrix formed by their coordinates is zero, implying the rank is less than 3.

872 Easy

872 of 922

Application +4 | -1

Suppose

A

is any

3 \times 3

non-singular matrix and

(A-3 I)(A-5 I)=O

, where

I=I_3

and

O=O_3

. If

\alpha A+\beta A^{-1}=4 l

, then

\alpha+\beta

is equal to

View answer and explanation

Correct option: A

The correct answer is 8. By manipulating the given equation

(A-3I)(A-5I)=O

and multiplying by

A^{-1}

, we can express

A + 15A^{-1}

in terms of

I

. Then, by dividing by 2, we can match the given form

\alpha A + \beta A^{-1} = 4I

and find that

\alpha + \beta = 8

873 Easy

873 of 922

Analyzing +4 | -1

For the system of equations

\begin{array}{l} x+y+z=6 \\ x+2 y+\alpha z=10 \end{array}

x+3 y+5 z=\beta

,
which one of the following is NOT true?

View answer and explanation

Correct option: D

System has a unique solution for

\alpha=3, \beta \neq 14

Option D (System has a unique solution for

\alpha=3, \beta \neq 14

) is correct because when

\alpha = 3

, the determinant is 0, so there's no unique solution.

874 Easy

874 of 922

Application +4 | -1

Let

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

and

\mathrm{U}_1, \mathrm{U}_2, \mathrm{U}_3

are column matrices such that

\mathrm{AU}_1=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right] ; \mathrm{AU}_2=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right]

;

\mathrm{AU}_3=\left[\begin{array}{l}2 \\ 3 \\ 1\end{array}\right].

\text{U}

3 \times 3

matrix having as columns

\text{U}_1, \text{U}_2, \text{U}_3

in the sequence, then

|\text{U}|=

View answer and explanation

Correct option: D

Option D (3) is correct because by expressing AU = B, where B is a matrix formed by the column matrices, and then using determinants, |U| simplifies to 3.

875 Easy

875 of 922

Application +4 | -1

\mathrm{A}=\left[\begin{array}{cc}4 & x+2 \\ 2 x-3 & x+1\end{array}\right]

is symmetric, then

x=

View answer and explanation

Correct option: B

Option B (5) is correct because for a symmetric matrix, the elements

a_{12}

and

a_{21}

must be equal, which leads to the equation

x+2 = 2x-3

, solving for x gives x=5.

876 Easy

876 of 922

Application +4 | -1

The inverse of the matrix

\left[ {\begin{array}{cccccccccccccccccccc} 7&{ - 3}&{ - 3}\\ { - 1}&1&0\\ { - 1}&0&1 \end{array}} \right]

View answer and explanation

Correct option: D

\left[ {\begin{array}{cccccccccccccccccccc} 1&3&3\\ 1&4&3\\ 1&3&4 \end{array}} \right]

Option D is correct. The inverse of the given matrix is calculated by finding the adjoint of the matrix and dividing it by the determinant, which is 1 in this case.

877 Easy

877 of 922

Application +4 | -1

let

A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]

where

a, b, c, d \in\{0,1\}

the number of such

\mathrm{A}

so that trace

(\mathrm{A})>\operatorname{det}(\mathrm{A})

View answer and explanation

Correct option: A

Option A (13) is correct because by listing all possible matrices with elements from {0,1} and checking the condition trace(A) > det(A), we find there are 13 such matrices.

878 Easy

878 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: D

At least 7

Option D (At least 7) is correct because there are at least 6 non-singular matrices with the 1s on the diagonal, and at least one more with a different arrangement.

879 Easy

879 of 922

Analyzing +4 | -1

The value of

\left| {\begin{array}{cccccccccccccccccccc} 1&1&1\\ {{{\left( {{2^x} + {2^{ - x}}} \right)}^2}}&{{{\left( {{3^x} + {3^{ - x}}} \right)}^2}}&{{{\left( {{5^x} + {5^{ - x}}} \right)}^2}}\\ {{{\left( {{2^x} - {2^{ - x}}} \right)}^2}}&{{{\left( {{3^x} - {3^{ - x}}} \right)}^2}}&{{{\left( {{5^x} - {5^{ - x}}} \right)}^2}} \end{array}} \right|

View answer and explanation

Correct option: A

Option A (0) is correct because after applying row operations, two rows of the determinant become identical, resulting in a zero determinant.

880 Easy

880 of 922

Application +4 | -1

A, B

and

C

are three matrices conformable for multiplication, then

(A C B)^{-1}

is equal to

View answer and explanation

Correct option: B

B^{-1} C^{-1} A^{-1}

The correct answer is

B^{-1} C^{-1} A^{-1}

because the inverse of a product of matrices is the product of their inverses in reverse order.

881 Easy

881 of 922

Analyzing +4 | -1

\left| {\begin{array}{cccccccccccccccccccc} {2xy}&{{x^2}}&{{y^2}}\\ {{x^2}}&{{y^2}}&{2xy}\\ {{y^2}}&{2xy}&{{x^2}} \end{array}} \right|

is equal to

View answer and explanation

Correct option: D

- {\left( {{x^3} + {y^3}} \right)^2}

Option D (

- {\left( {{x^3} + {y^3}} \right)^2}

) is correct because the determinant simplifies to this expression after expansion and simplification.

882 Easy

882 of 922

Understanding +4 | -1

A

and

B

are two matrices such that

A B=A

ant

B A=B

, then

B^2

is equal to

View answer and explanation

Correct option: A

B

Option A (

B

) is correct because

B^2 = BB = (BA)B = B(AB) = BA = B

883 Easy

883 of 922

Understanding +4 | -1

A, B

are square matrices of order 3,

A

is non-singular and

A B=O

, then

B

is a

View answer and explanation

Correct option: A

null matrix

Option A (null matrix) is correct because since A is non-singular, it has an inverse. Multiplying AB = O by A^-1 on the left gives B = O.

884 Easy

884 of 922

Application +4 | -1

\mathrm{A}=\left[\begin{array}{cc}a b & b^2 \\ -a^2 & -a b\end{array}\right]

and

\mathrm{A}^n=\mathrm{O}

, then the minimum value of

n

View answer and explanation

Correct option: A

Option A (2) is correct because A² results in a zero matrix, meaning A^n = O for all n >= 2.

885 Easy

885 of 922

Remembering +4 | -1

{I_n}

is the identity matrix of order

n

, then rank of

{I_n}

View answer and explanation

Correct option: B

n

Option B (

n

) is correct because the rank of an identity matrix of order

n

is always

n

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

886 Easy

886 of 922

Analyzing +4 | -1

\Delta = \left| {\begin{array}{cccccccccccccccccccc} 1&2&3\\ 2&5&7\\ 3&9&{13} \end{array}} \right|

and

\Delta ' = \left| {\begin{array}{cccccccccccccccccccc} 7&{20}&{29}\\ 2&5&7\\ 3&9&{13} \end{array}} \right|

, then

View answer and explanation

Correct option: C

\Delta ' = \Delta

Option C (

\Delta ' = \Delta

) is correct because applying row operations to

\Delta

results in

\Delta'

, indicating they are equal.

887 Easy

887 of 922

Understanding +4 | -1

If a square matrix

A

is involutory, then

A^{2 n+1}

is equal to

View answer and explanation

Correct option: B

A

Option B (

A

) is correct because for an involutory matrix

A

A^2 = I

, and therefore

A^{2n+1} = A

888 Easy

888 of 922

Analyzing +4 | -1

If the system of equations

\left( {^n{C_3}} \right)x + \left( {^n{C_4}} \right)y + 35z = 0,\,\,\,\,\,\left( {^n{C_4}} \right)x + 35y + \left( {^n{C_3}} \right)z = 0

and

35x + \left( {^n{C_3}} \right)y + \left( {^n{C_4}} \right)z = 0

has a non-trivial solution, then the value of

n

is equal to

\left( {\forall \,\,n \in N,\,\,\,n \geqslant 4} \right)

View answer and explanation

Correct option: B

Option B, 7, is correct because the determinant of the coefficient matrix must be zero for a non-trivial solution, which leads to n = 7.

889 Easy

889 of 922

Application +4 | -1

\left[ {\begin{array}{cccccccccccccccccccc} 1&1&1\\ 1&{ - 2}&{ - 2}\\ 1&3&1 \end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc} 0\\ 3\\ 4 \end{array}} \right],\;

then

\left[ {\begin{array}{cccccccccccccccccccc} x\\ y\\ z \end{array}} \right]

is equal to

View answer and explanation

Correct option: B

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

Option B (

\left[ {\begin{{array}}{*{20}{c}} 1\\ 2\\ { - 3} \end{{array}}} \right]

) is correct because it satisfies the given matrix equation.

890 Easy

890 of 922

Application +4 | -1

Let adj

A=\left[\begin{array}{ccc}4 & 0 & 2 \\ 2 & 0 & -3 \\ 3 & 4 & 5\end{array}\right]

and

B=(\operatorname{adj}(\operatorname{adj} A))^T

, if

|A|=\alpha

and

|B|=\beta

, then

\left|\frac{\sqrt{\beta}}{\alpha}\right|

is:

View answer and explanation

Correct option: B

Option B (8) is correct because it is derived from the properties of adjoint matrices and determinants, where |adj(adj A)| = |A|^(n-1)² and |adj A| = |A|^(n-1).

891 Easy

891 of 922

Analyzing +4 | -1

The least value of the product

x y z

for which the determinant

\left|\begin{array}{lll}x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z\end{array}\right|

is non-negative, is :

View answer and explanation

Correct option: D

-8

Option D (-8) is correct because the least value of the product xyz is found to be -8 by analyzing the determinant inequality.

892 Easy

892 of 922

Understanding +4 | -1

The product of two orthogonal matrices is

View answer and explanation

Correct option: A

Orthogonal

Option A (Orthogonal) is correct because the product of two orthogonal matrices is always an orthogonal matrix.

893 Easy

893 of 922

Application +4 | -1

c = 2\cos \mathrm{\theta },

then the value of the determinant

\Delta = \left| {\begin{array}{cccccccccccccccccccc} c&1&0\\ 1&c&1\\ 6&1&c \end{array}} \right|

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the determinant simplifies to

8\cos^3{\theta} - 4\cos{\theta} + 6

, which doesn't match any of the given options.

894 Easy

894 of 922

Understanding +4 | -1

A = \left[ {{a_{ij}}} \right]

is a skew-symmetric matrix of order

n

, then

{a_{ii}} =

View answer and explanation

Correct option: B

0 for all

i = 1,2, \ldots ,n

Option B (0 for all

i = 1,2, \ldots ,n

) is correct because in a skew-symmetric matrix, the diagonal elements are always zero.

895 Easy

895 of 922

Understanding +4 | -1

If in the determinant

\Delta=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|, \mathrm{A}_1, \mathrm{~B}_1, \mathrm{C}_1

etc. be the co-factors of

a_1, b_1, c_1

etc., then which of the following relations is incorrect?

View answer and explanation

Correct option: B

a_2 \mathrm{~A}_2+b_2 \mathrm{R}_2+c_2 C_2=\Lambda

Option D (

a_1 A_2 + b_1 B_2 + c_1 C_2 = \Delta

) is incorrect because the sum of the products of elements of one row with the cofactors of another row is zero, not the determinant.

896 Easy

896 of 922

Understanding +4 | -1

{A}

and

{B}

are square matrices of order

{n} \times {n}

such that

A^2-B^2=(A-B)(A+B)

, then which of the following will always be true?

View answer and explanation

Correct option: C

{AB}={BA}

Option C (

{AB}={BA}

) is correct because the given condition

A^2-B^2=(A-B)(A+B)

implies that the matrices A and B commute, meaning AB = BA.

897 Easy

897 of 922

Application +4 | -1

Let

A = \left( {\begin{array}{cccccccccccccccccccc} {\cos \,\alpha }&{ - \sin \,\alpha }\\ {\sin \,\alpha }&{\cos \,\alpha } \end{array}} \right),\,\,(\alpha \in R)

such that

{A^{32}} = \left( {\begin{array}{cccccccccccccccccccc} 0&{ - 1}\\ 1&0 \end{array}} \right)

. Then, a value of

\alpha

View answer and explanation

Correct option: C

\frac{\pi }{{64}}

Option C is correct because it accurately determines the value of alpha that satisfies the given matrix condition.

898 Easy

898 of 922

Analyzing +4 | -1

Let

A = \left( {\begin{array}{cccccccccccccccccccc} 1&0&0\\ 0&1&1\\ 1&0&0 \end{array}} \right).

Then

{A^{2025}} - {A^{2020}}

is equal to

View answer and explanation

Correct option: A

{A^6} - A

Option A (

{A^6} - A

) is correct because by finding a pattern in the powers of A, we can simplify the expression

{A^{2025}} - {A^{2020}}

and show it equals

{A^6} - A

899 Easy

899 of 922

Analyzing +4 | -1

If in a

\Delta A B C,\left|\begin{array}{lll}1 & a & b \\ 1 & c & a \\ 1 & b & c\end{array}\right|=0,

then the value of

\sin ^2 A+\sin ^2 B+\sin ^2 C,

View answer and explanation

Correct option: A

\frac{9}{4}

Option A (

\frac{9}{4}

) is correct because the given condition implies that the triangle is equilateral, and for an equilateral triangle,

\sin^2 A + \sin^2 B + \sin^2 C = \frac{9}{4}

900 Easy

900 of 922

Application +4 | -1

One root of the equation

\left| {\begin{array}{cccccccccccccccccccc} {3x - 8}&3&3\\ 3&{3x - 8}&3\\ 3&3&{3x - 8} \end{array}} \right| =

View answer and explanation

Correct option: B

\frac{2}{3}

Option B (

\frac{2}{3}

) is correct because it is one of the roots of the equation obtained after simplifying the given determinant.

Revision notes and key takeaways

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

Showing 871-900 of 922 questions

Continue Your Practice

📚 Switch Set (EASY)

Chapter Sets

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

0 /922

0✓ 0✗

0% Accuracy