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

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

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

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

The inverse matrix of $A = \left[ {\begin{array}{*{20}{c}} {0\;\;1\;\;2}\\ {1\;\;\,2\;\;3}\\ {3\;\;1\;\;1} \end{array}} \right]$ is

$\left[ {\begin{array}{*{20}{c}} {\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\ { - 4}&3&{ - 1}\\ {\frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}} \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} {\frac{1}{2}}&{ - 4}&{\frac{5}{2}}\\ 1&{ - 6}&3\\ 1&2&{ - 1} \end{array}} \right]$

$\frac{1}{2}\left[ {\begin{array}{*{20}{c}} 1&2&3\\ 3&2&1\\ 4&2&3 \end{array}} \right]$

$\frac{1}{2}\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&{ - 1}\\ { - 8}&6&{ - 2}\\ 5&{ - 3}&1 \end{array}} \right]$

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JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

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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.

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Showing 961-990 of 995 questions

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961 Easy

961 of 995

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

962 Easy

962 of 995

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.

963 Medium

963 of 995

Analyzing +4 | -1

Let

a, b, c

be positive and not all equal, the value of the determinant

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

View answer and explanation

Correct option: B

-ive

Option B (-ive) is correct because the determinant can be expressed as a negative value involving squares, given the conditions

a, b, c

are positive and not all equal.

964 Easy

964 of 995

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.

965 Easy

965 of 995

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).

966 Easy

966 of 995

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.

967 Easy

967 of 995

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.

968 Medium

968 of 995

Analyzing +4 | -1

A, B

and

C

denote the angles of a triangle, then

\Delta=\left|\begin{array}{ccc}-1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -2\end{array}\right|

is independent of

View answer and explanation

Correct option: B

B

Option B (

B

) is correct because after simplification, the determinant's value is expressed in terms of

A

and

C

only, making it independent of

B

969 Easy

969 of 995

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.

970 Easy

970 of 995

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.

971 Easy

971 of 995

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.

972 Easy

972 of 995

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.

973 Easy

973 of 995

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

974 Easy

974 of 995

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}

975 Easy

975 of 995

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.

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

976 Easy

976 of 995

Analyzing +4 | -1

A B=A

and

B A=B

, where

A

and

B

are square matrices, then

View answer and explanation

Correct option: A

B^2=B

and

A^2=A

The correct answer is

B^2=B

and

A^2=A

because given

AB=A

and

BA=B

, it can be shown that

A^2 = A

and

B^2 = B

977 Easy

977 of 995

Analyzing +4 | -1

The number of values of

k

for which the system of equations

(k+1) x+8 y=4 k ; k x+(k+3) y=3 k-1

has no solution, is

View answer and explanation

Correct option: A

Option A (1) is correct because the system has no solution when the determinant of the coefficients is zero and at least one of the other determinants is non-zero, which occurs for one value of

k

978 Easy

978 of 995

Understanding +4 | -1

A

and

B

are two matrices such that both

A + B

and

AB

are defined, then

View answer and explanation

Correct option: C

Both

A

and

B

are of same order

n \times n

Option C (Both

A

and

B

are of same order

n \times n

) is correct because both

A + B

and

AB

being defined requires

A

and

B

to be square matrices of the same order.

979 Easy

979 of 995

Application +4 | -1

Let

A=\left[a_{i y}\right]_{3 \times 3}

be such that

a_{i j}=\left[\begin{array}{ll}3 \text {; when } & \hat{i}=\hat{j} \\ 0 ; & \hat{i} \neq \hat{j}\end{array}\right.

, then

\left\{\frac{\operatorname{det}(\operatorname{adj}(\operatorname{adj} A))}{5}\right\}

equals: (where

\{\cdot\}

denotes fractional part function)

View answer and explanation

Correct option: A

0.2

Option B (1) is correct because the expression simplifies to sin(π/2) + cos(π/2) which equals 1.

980 Easy

980 of 995

Application +4 | -1

A = \left[ {\begin{array}{cccccccccccccccccccc} 1&1\\ 1&1 \end{array}} \right]

and

n \in N,

then

{A^n}

is equal to

View answer and explanation

Correct option: B

{2^{n - 1}}\;A

Option B is correct. By observing the pattern of A raised to successive powers, it can be generalized that A^n = 2^(n-1) * A.

981 Easy

981 of 995

Application +4 | -1

Let

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

and suppose that

\operatorname{det}(A)=2

, then

\operatorname{det}(B)

equals, where

B=\left[\begin{array}{ccc}4 x & 2 a & -p \\ 4 y & 2 b & -q \\ 4 z & 2 c & -r\end{array}\right]

View answer and explanation

Correct option: C

-16

Option C (-16) is correct because applying determinant properties such as row swapping and scalar multiplication leads to det(B) = -16.

982 Easy

982 of 995

Analyzing +4 | -1

If the system of linear equations

x + y + z = 5,\,\,\,x + 2y + 2z = 6,\,\,\,x + 3y + \lambda z = \mu ,\,\,\,\,\left( {\lambda ,\mu \in R} \right),

has infinitely many solutions, then the value of

\lambda + \mu

View answer and explanation

Correct option: C

Option C (10) is correct because for the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the equations must be consistent, leading to

\lambda = 3

and

\mu = 7

, so

\lambda + \mu = 10

983 Easy

983 of 995

Application +4 | -1

The inverse matrix of

A = \left[ {\begin{array}{cccccccccccccccccccc} {0\;\;1\;\;2}\\ {1\;\;\,2\;\;3}\\ {3\;\;1\;\;1} \end{array}} \right]

View answer and explanation

Correct option: A

\left[ {\begin{array}{cccccccccccccccccccc} {\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\ { - 4}&3&{ - 1}\\ {\frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}} \end{array}} \right]

The correct answer is A. The inverse of a matrix is found by dividing the adjoint by the determinant. After calculating the determinant and adjoint, the inverse is found to be option A.

984 Medium

984 of 995

Application +4 | -1

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

and

A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]

and

Q=

P A P^T

and

x=P^T Q^{2005} P

then

x

is equal to

View answer and explanation

Correct option: A

\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]

Option A (

\left[\begin{array}{cc}1 & 2005 \\0 & 1\end{array}\right]

) is correct because

X = A^{2005}

and

A^{2005}

is calculated by observing the pattern of powers of A.

985 Easy

985 of 995

Understanding +4 | -1

Let

A

be a skew-symmetric matrix of odd order, then

\left| A \right|

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the determinant of any skew-symmetric matrix of odd order is always zero.

986 Easy

986 of 995

Analyzing +4 | -1

a \ne b \ne c

, the value of

x

which satisfies the equation

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

View answer and explanation

Correct option: A

x = 0

Option A (

x = 0

) is correct because substituting

x = 0

into the determinant makes it equal to zero.

987 Easy

987 of 995

Analyzing +4 | -1

Let three matrices

A=\left[\begin{array}{ll}2 & 1 \\ 4 & 1\end{array}\right], B=\left[\begin{array}{ll}3 & 4 \\ 2 & 3\end{array}\right]

and

C=\left[\begin{array}{rr}3 & -4 \\ -2 & 3\end{array}\right],

then

\operatorname{tr}(A)+\operatorname{tr}\left(\frac{A B C}{2}\right)+\operatorname{tr}\left(\frac{A(B C)^2}{4}\right)

+\operatorname{tr}\left(\frac{A(B C)^3}{8}\right)+\ldots+\infty

equals to

View answer and explanation

Correct option: D

Option D (6) is correct because after simplifying the expression using the property BC=I, the trace of the resulting geometric series converges to 6.

988 Easy

988 of 995

Understanding +4 | -1

A

and

B

two are

3 \times 3

matrices, then which one of the following is not true:

View answer and explanation

Correct option: B

(A B)^{\prime}=A^{\prime} B^{\prime}

Option B (

(A B)^{\prime}=A^{\prime} B^{\prime}

) is not true because the transpose of a product of matrices is the product of their transposes in reverse order, i.e.,

(A B)^{\prime}=B^{\prime} A^{\prime}

989 Easy

989 of 995

Application +4 | -1

\mathrm{A}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]

, then

\mathrm{A}+\mathrm{A}^{\prime}=I

, if the value of

\alpha

View answer and explanation

Correct option: B

\frac{\pi}{3}

Option B (

\frac{{\pi}}{{3}}

) is correct because when A + A' = I, it leads to

2\cos{\alpha} = 1

, which gives

\alpha = \frac{{\pi}}{{3}}

990 Easy

990 of 995

Application +4 | -1

\log _3 2+\log _3\left(2^x-7 / 2\right)=2 \log _3\left(2^x-5\right)

, then the value of

x

View answer and explanation

Correct option: A

Option A (3) is correct because it is the only value of x that satisfies the given logarithmic equation and the domain restrictions.

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