If $A^{\prime}$ is the transpose of a square matrix $A$, then

$|A| \neq\left|A^{\prime}\right|$

$|A|+\left|A^{\prime}\right|=0$

$|A|=\left|A^{\prime}\right|$ only when $A$ is symmetric

If $A\left( \theta \right) = \left[ {\begin{array}{*{20}{c}} {1\;\;\;\;\;\;\;\;\;\;\tan {\rm{\theta }}}\\ { - \tan {\rm{\theta }}\;\;\;\;\;\;\;1} \end{array}} \right]$ and $\;AB = I,\;$ then $\left( {{{\sec }^2}\theta } \right)B$ is equal to

$A\left( {\frac{\theta }{2}} \right)$

$A\left( { - \frac{\theta }{2}} \right)$

If $X = \left[ {\begin{array}{*{20}{c}} 3&{ - 4}\\ 1&{ - 1} \end{array}} \right]$, then the value of ${X^n}$ is

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

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

$\left[ {\begin{array}{*{20}{c}} {{3^n}}&{{{\left( { - 4} \right)}^n}}\\ {{1^n}}&{{{\left( { - 1} \right)}^n}} \end{array}} \right]$

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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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271 Easy

271 of 995

Application +4 | -1

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

and

B=\left[\begin{array}{ccc}1 & 0 & 0 \\ -2 & 1 & 0 \\ 7 & -2 & 1\end{array}\right]

then

A B

equals

View answer and explanation

Correct option: A

I

Option A (

I

) is correct because when matrix A and B are multiplied, the result is the identity matrix I.

272 Medium

272 of 995

Application +4 | -1

Let

S=\left\{\left(\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right): a_{i j} \in\{0,1,2\}, a_{11}=a_{22}\right\}

Then the number of non-singular matrices in the set

\mathrm{S}

is :

View answer and explanation

Correct option: A

Option A is correct. By expanding the magnitudes and using the properties of complex numbers, the expression simplifies to

32(x_1^2 + y_1^2)

273 Easy

273 of 995

Understanding +4 | -1

The matrix

A=\left[\begin{array}{cc}i & 1-2 i \\ -1-2 i & 0\end{array}\right]

is which of the following?

View answer and explanation

Correct option: D

skew-hermitian

Option D (skew-hermitian) is correct because the conjugate transpose of the matrix A is equal to -A.

274 Easy

274 of 995

Application +4 | -1

f\left( \alpha \right) = \left| {\begin{array}{cccccccccccccccccccc} 1&\alpha &{{\alpha ^2}}\\ \alpha &{{\alpha ^2}}&1\\ {{\alpha ^2}}&1&a \end{array}} \right|,

then

f\left( {\sqrt[3]{3}} \right)

is equal to

View answer and explanation

Correct option: B

- 4

Option B (-4) is correct because substituting

\alpha = \sqrt[3]{3}

into the determinant expression and simplifying yields -4.

275 Easy

275 of 995

Analyzing +4 | -1

The set of all values of

\lambda

for which the system of linear equations

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

has a non-trivial solution

View answer and explanation

Correct option: A

contains two elements

Option A (contains two elements) is correct because the characteristic equation has two distinct roots (1 and -3), so the set of values of

\lambda

contains two elements.

276 Easy

276 of 995

Analyzing +4 | -1

The values of

a

for which the system of equations

x + y + z = 0,\;x + ay + az = 0,\;x - ay + z = 0

, possesses non-zero solutions, are given by

View answer and explanation

Correct option: B

1, - 1

Option B (

1, -1

) is the correct answer because these are the values of

a

that make the determinant of the coefficient matrix zero, ensuring non-zero solutions.

277 Medium

277 of 995

Application +4 | -1

{B}, {C}

are square matrices of order

{n}

and if

{A}={B}+{C}, {BC}={CB}, {C}^2=0

, then for any positive integer

{N}, {A}^{{N}+1}={B}^{{K}}[{B}+({N}+1) {C}]

, then

{K} /{N}

View answer and explanation

Correct option: A

Option A (1) is correct because by expanding

(B+C)^{N+1}

and using the given conditions, we find that

K=N

, thus

K/N = 1

278 Easy

278 of 995

Remembering +4 | -1

A^{\prime}

is the transpose of a square matrix

A

, then

View answer and explanation

Correct option: B

|A|=\left|A^{\prime}\right|

Option B (

|A|=\left|A^{\prime}\right|

) is correct because the determinant of a matrix is equal to the determinant of its transpose.

279 Easy

279 of 995

Application +4 | -1

A

is an

3 \times 3

non-singular matrix such that

A A^T=A^T A

and

B=A^{-1} A^T

, then

B B^T

equals

View answer and explanation

Correct option: D

I

Option D (I) is correct because by using the given conditions and properties of matrix transpose and inverse,

B B^T

simplifies to the identity matrix I.

280 Easy

280 of 995

Analyzing +4 | -1

The number of values of

k

for which the system of equations

\begin{aligned} &(k+1) x+8 y=4 k \\ &k x+(k+3) y=3 k-1 \end{aligned}

has infinitely many solutions, is

View answer and explanation

Correct option: B

Option B (1) is correct because there is only one value of k (k=1) for which the system has infinitely many solutions, which is found by setting the determinant to zero.

281 Medium

281 of 995

Application +4 | -1

A^2=A

, the

(I+A)^4

equals

View answer and explanation

Correct option: A

I+15 A

Option A (

I+15 A

) is correct because expanding

(I+A)^4

using the binomial theorem and the property

A^2=A

simplifies to

I+15A

282 Easy

282 of 995

Application +4 | -1

A\left( \theta \right) = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;\;\;\;\;\;\;\;\;\tan \mathrm{\theta }}\\ { - \tan \mathrm{\theta }\;\;\;\;\;\;\;1} \end{array}} \right]

and

\;AB = I,\;

then

\left( {{{\sec }^2}\theta } \right)B

is equal to

View answer and explanation

Correct option: C

A\left( { - \theta } \right)

Option C,

A(- \theta)

, is correct because by finding the inverse of A and then multiplying by sec²θ, we obtain A(-θ).

283 Medium

283 of 995

Application +4 | -1

The value of

\Delta = \left| {\begin{array}{cccccccccccccccccccc} a&{a + b}&{a + 2b}\\ {a + 2b}&a&{a + b}\\ {a + b}&{a + 2b}&a \end{array}} \right|

is equal to

View answer and explanation

Correct option: B

9{b^2}\left( {a + b} \right)

Option B (

9{b^2}(a + b)

) is correct because after applying column operations and simplifying the determinant, the result is

9{b^2}(a + b)

284 Easy

284 of 995

Application +4 | -1

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

, then

x

equals

View answer and explanation

Correct option: D

0,-(a+b+c)

Option D (

0,-(a+b+c)

) is correct because these are the values of x that make the determinant equal to zero.

285 Easy

285 of 995

Application +4 | -1

The values of

a

for which the matrix

A=\left(\begin{array}{ccc}a & a^2-1 & -3 \\ a+1 & 2 & a^2+4 \\ -3 & 4 a & -1\end{array}\right)

is symmetric are

View answer and explanation

Correct option: D

Option D (2) is correct because setting A = A' leads to the equations a+1 = a^2-1 and 4a = a²⁺4, which are both satisfied when a = 2.

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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

286 Medium

286 of 995

Application +4 | -1

Let

a,b,c,

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

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

View answer and explanation

Correct option: B

Negative

Option B (Negative) is correct because the determinant simplifies to a negative expression given the conditions on a, b, and c.

287 Easy

287 of 995

Application +4 | -1

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

then the value of the determinant

\Delta=\left|\begin{array}{lll}C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C\end{array}\right|,

View answer and explanation

Correct option: D

None of these

The correct answer is None of these because the determinant evaluates to

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

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

288 Easy

288 of 995

Application +4 | -1

A

and

B

are two square matrix such that

A^{2} B=B A

and if

(A B)^{10}=A^{k} B^{10}

, then

k

View answer and explanation

Correct option: B

1023

Option B (1023) is correct because the pattern

(AB)^n = A^{(2^n - 1)}B^n

leads to

k = 2^{10} - 1 = 1023

for n=10.

289 Easy

289 of 995

Analyzing +4 | -1

Study the following statement and select the correct option:
Statement-I: If

a, b, c

are in A.P the system of equations

3 x+4 y+5 z=a, 4 x+5 y+6 z=b, 5 x+6 y+7 z=c

is consistent

Statement-II: If

|A| \neq 0

, the system of equations

A X=B

is consistent

View answer and explanation

Correct option: B

Statement-1 is true, statement-2 is true, statement-2 is not correct explanation of statement-1

Option B (Statement-1 is true, statement-2 is true, statement-2 is not correct explanation of statement-1) is correct because both statements are individually true, but statement-2 doesn't explain statement-1.

290 Easy

290 of 995

Application +4 | -1

Let

A=\left(\begin{array}{cc}2 & -1 \\ 0 & 2\end{array}\right)

. If

B=I-{ }^5 C_1(\operatorname{adj} A)+{ }^5 C_2(\operatorname{adj} A)^2-

.....

-{ }^5 C_5(\operatorname{adj} A)^5

, then the sum of all elements of the matrix

B

View answer and explanation

Correct option: C

-7

Option C (-7) is correct because after simplifying the expression and calculating the matrix power, the sum of the elements of the resulting matrix is -7.

291 Easy

291 of 995

Application +4 | -1

Let

A_t=\left(\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6\end{array}\right)

, then the value(s) of

t

for which inverse of

A_t

does not exist.

View answer and explanation

Correct option: C

2, -3

Option C (2, -3) is correct because the inverse of At does not exist when the determinant of At is zero, which occurs at t = 2 and t = -3.

292 Easy

292 of 995

Application +4 | -1

The matrix

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

View answer and explanation

Correct option: B

singular

Option B (singular) is correct because the determinant of the matrix is 0, indicating that the matrix is singular.

293 Medium

293 of 995

Understanding +4 | -1

l_1, m_1, n_1 ; l_2, m_2, n_2

and

l_3, m_3, n_3

are direction cosines of three mutually perpendicular lines then, the value of

\left|\begin{array}{lll}l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3\end{array}\right|,

View answer and explanation

Correct option: B

\pm 1

Option B (

\pm 1

) is correct because the determinant of the direction cosines of three mutually perpendicular lines is either 1 or -1.

294 Easy

294 of 995

Application +4 | -1

\alpha

is a complex root of the equation

x^2+x+1=0

and

A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha\end{array}\right]

, then

A^{31}

is equal to

View answer and explanation

Correct option: C

A^3

Option C (

A^3

) is correct because, after calculating A², we find that A⁴ = I, which simplifies A³¹ to A³.

295 Easy

295 of 995

Analyzing +4 | -1

If the system of equations

2 x-y+z=0, x-2 y+z=0

t x-y+2 z=0

has infinitely many solutions and

f(x)

be a continuous function, such that

f(5+x)+f(x)=2

, then

\int_0^{-2 t} f(x) d x

is equal to

View answer and explanation

Correct option: B

-2 t

The correct answer is B (-2t) because solving the system of equations gives t=5, and the integral evaluates to -10, which is -2t.

296 Easy

296 of 995

Application +4 | -1

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

and

\mathrm{adj}\;A = \left[ {\begin{array}{cccccccccccccccccccc} 6&{ - 2}&{ - 6}\\ { - 4}&2&x\\ y&{ - 1}&{ - 1} \end{array}} \right],

then

x + y =

View answer and explanation

Correct option: A

Option A (6) is correct because adj(A) * A = |A| * I. Calculating |A| and performing the multiplication allows us to solve for x and y, then x + y.

297 Easy

297 of 995

Application +4 | -1

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

, then the value of

{X^n}

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because none of the given options satisfy the condition when n=2 after calculating X².

298 Easy

298 of 995

Understanding +4 | -1

Let

A

be a skew-symmetric matrix of order

n

then

View answer and explanation

Correct option: B

|A|=0

n

is odd

Option B (

|A|=0

n

is odd) is correct because the determinant of a skew-symmetric matrix of odd order is always zero.

299 Easy

299 of 995

Application +4 | -1

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

, then

A^{-3}

View answer and explanation

Correct option: C

\frac{1}{27}\left[\begin{array}{cc}1 & -26 \\ 0 & 27\end{array}\right]

Option C (

\frac{1}{27}\left[\begin{array}{cc}1 & -26 \\ 0 & 27\end{array}\right]

) is correct because by calculating A^-1 and then cubing it, we arrive at the given matrix.

300 Easy

300 of 995

Remembering +4 | -1

A = \left[ {\begin{array}{cccccccccccccccccccc} {1\; - 5\;\;\;\;7}\\ {0\;\;\;\;\;7\;\;\;\;9}\\ {11\;\;\;\;\;8\;\;\;9} \end{array}} \right]

, then trace of matrix

A

View answer and explanation

Correct option: A

Option A (17) is correct because the trace of a matrix is the sum of its diagonal elements, which is 1 + 7 + 9 = 17.

Revision notes and key takeaways

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

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