If $P = \left[ {\begin{array}{*{20}{c}} {\sqrt 3 /2\;\;\;\;\;\;1/2}\\ { - 1/2\;\;\,\,\,\sqrt 3 /2} \end{array}} \right],\;A = \left[ {\begin{array}{*{20}{c}} {1\;\;1}\\ {0\;\;1} \end{array}} \right]\;{\rm{and}}\;Q = PA{P^T},$ then ${P^T}{Q^{2005}}$ $P$ is

$\left[ {\begin{array}{*{20}{c}} {1\;\;\;\;\;\;\;\;\;\;2005}\\ {2005\;\;\;\;\;\;\;\;1} \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} {1\;\;\;\;\;\;\;\;\;0}\\ {2005\;\;\;\;\;1} \end{array}} \right]$

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

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JEE • MATHEMATICS • MATRICES

Matrices practice · Medium

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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 1-30 of 408 questions

1 Medium

1 of 408

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

2 Medium

2 of 408

Application +4 | -1

The sum of two non-integral roots of

\left|\begin{array}{lll}x & 2 & 5 \\ 3 & x & 3 \\ 5 & 4 & x\end{array}\right|=0

View answer and explanation

Correct option: B

-5

Option B (-5) is correct because the sum of the non-integral roots of the determinant equation is -5.

3 Medium

3 of 408

Application +4 | -1

Matrix

M_r

is defiend as

M_r=\left[\begin{array}{cc}r & r-1 \\ r-1 & r\end{array}\right], r \in N

value of

\operatorname{det}\left(M_1\right)+\operatorname{det}\left(M_2\right)+\operatorname{det}\left(M_3\right)+\ldots+

\operatorname{det}\left(M_{2007}\right)

View answer and explanation

Correct option: D

2007^2

Option D (

2007^2

) is correct because the sum of the determinants forms an arithmetic series that simplifies to

2007^2

4 Medium

4 of 408

Analyzing +4 | -1

\left|\begin{array}{lll}a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x\end{array}\right|=0

then

x

View answer and explanation

Correct option: C

0,3 \mathrm{a}

Option C (0, 3a) is correct because these are the values of x that make the determinant equal to zero, found by applying row/column operations and solving the resulting equation.

5 Medium

5 of 408

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.

6 Medium

6 of 408

Analyzing +4 | -1

The arbitrary constant on which the value of the Determinant

\left| {\begin{array}{cccccccccccccccccccc} 1&a&{{a^2}}\\ {\cos \left( {p - d} \right)a}&{\cos pa}&{\cos \left( {p - d} \right)a}\\ {\sin \left( {p - d} \right)a}&{\sin pa}&{\sin \left( {p - d} \right)a} \end{array}} \right|

Does not depend, is

View answer and explanation

Correct option: B

p

The correct answer is B (

p

). After simplification, the determinant's value does not depend on

p

7 Medium

7 of 408

Analyzing +4 | -1

a \ne p,b \ne q,c \ne r

and

\left| {\begin{array}{cccccccccccccccccccc} p&b&c\\ {p + a}&{q + b}&{2c}\\ a&b&r \end{array}} \right| = 0,

then

\frac{p}{{p - a}} + \frac{q}{{q - b}} + \frac{r}{{r - c}}

is equal to

View answer and explanation

Correct option: C

The correct answer is C (2). The value of the expression is found by simplifying the given determinant equation and solving for the required sum.

8 Medium

8 of 408

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.

9 Medium

9 of 408

Application +4 | -1

If matrices

{A}_r=\left[\begin{array}{cc}r & r-1 \\ r-1 & r\end{array}\right], r=1,2,3 \ldots

, then

\left|{A}_1\right|+\left|{A}_2\right|+\ldots . .+\left|{A}_{101}\right|=

View answer and explanation

Correct option: B

(101)^2

Option B (

(101)^2

) is correct because the sum of the determinants forms an arithmetic progression, and the sum of the first 101 terms is

(101)^2

10 Medium

10 of 408

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|

11 Medium

11 of 408

Application +4 | -1

Let

f(x)=\left[\begin{array}{lll}1 & x & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]

, then

\left(f(x)^n(n \in N)\right.

is equal to

View answer and explanation

Correct option: C

f(n x)

Option B (

\frac{-3}{5}

) is correct because by solving the trigonometric equation for tan(x) and using the given range for x, we can determine the value of sin(2x).

12 Medium

12 of 408

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.

13 Medium

13 of 408

Application +4 | -1

if A is skew symmetric matrix then I-A is

View answer and explanation

Correct option: D

non singular

Option D (non singular) is correct because if A is a skew-symmetric matrix, then (I-A) is non-singular, meaning its determinant is non-zero and it has an inverse.

14 Medium

14 of 408

Application +4 | -1

The value of

\theta

lying between

\theta = 0

and

\frac{\pi }{2}

and satisfying the equation

\left| {\begin{array}{cccccccccccccccccccc} {1 + {{\sin }^2}\mathrm{\theta }}&{{{\cos }^2}\mathrm{\theta }}&{4\sin 4\mathrm{\theta }}\\ {{{\sin }^2}\mathrm{\theta }}&{1 + {{\cos }^2}\mathrm{\theta }}&{4\sin 4\mathrm{\theta }}\\ {{{\sin }^2}\mathrm{\theta }}&{{{\cos }^2}\mathrm{\theta }}&{1 + 4\sin 4\mathrm{\theta }} \end{array}} \right|

View answer and explanation

Correct option: A

\frac{{7\pi }}{{24}}

Option A (

\frac{{7\pi}}{{24}}

) is correct because it satisfies the given determinant equation within the specified range of

\theta

after simplification and solving for

\theta

15 Medium

15 of 408

Application +4 | -1

M

is a square matrix of order 2 then

-\operatorname{tr}\left(M^2\right)+(\operatorname{tr} M)^2

(\operatorname{tr}(P)

denotes trace of matrix

P)

View answer and explanation

Correct option: B

2|\mathrm{M}|

Option B (2|M|) is correct because the expression simplifies to 2 times the determinant of the matrix M.

Explore Other Chapters in MATHEMATICS

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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

16 Medium

16 of 408

Application +4 | -1

3 \times 3

matrix A satisfies the equation

\mathrm{A}^2-5 \mathrm{A}+7 \mathrm{I}=0

. If

\mathrm{A}^5=a \mathrm{A}+b \mathrm{I}

, then

2 a-3 b=

View answer and explanation

Correct option: B

1453

Option D is correct. By evaluating the greatest integer functions and then calculating the determinant, we find the value to be -8.

17 Medium

17 of 408

Analyzing +4 | -1

The value of

k \in R

, for which the following system of linear equations

3 x-y+4 z=3, x+2 y-3 z=-2,6 x+5 y+k z=-3

has infinitely many solutions, is

View answer and explanation

Correct option: A

-5

Option A (-5) is correct because the value of k is found by ensuring the system of linear equations has infinitely many solutions, which requires the equations to be consistent and dependent.

18 Medium

18 of 408

Application +4 | -1

Matrix

A=\left[\begin{array}{ccc}x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z\end{array}\right]

, if

x y z=60

and

8 x+4 y+3 z

=20

, then

A(\operatorname{adj} A)

is equal to

View answer and explanation

Correct option: C

\left[\begin{array}{ccc}68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68\end{array}\right]

Option C is correct because the determinant can be expanded and simplified to find the value of |A|, and then A(adj A) = |A|I.

19 Medium

19 of 408

Application +4 | -1

A unit matrix for the set of matrices

=\left\{\left[\begin{array}{ccc}x & x & x \\ x & x & x \\ x & x & x\end{array}\right], x \in R, x \neq 0\right\}

under multiplication is

View answer and explanation

Correct option: A

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

Option A is correct because when multiplied by any matrix of the given form, it results in the same matrix, thus acting as a unit matrix.

20 Medium

20 of 408

Analyzing +4 | -1

If the system of equations

x+y=1,(c+2) x+(c+4) y=6

and

(c+2)^2 x+(c+4)^2 y=36

are consistent and

{c}_1, {c}_2\left({c}_1>{c}_2\right)

are two values of

{c}

, then

{c}_1 {c}_2

is divisible by

View answer and explanation

Correct option: D

Option D (8) is correct because the product of the two values of c, which make the system consistent, is divisible by 8.

21 Medium

21 of 408

Analyzing +4 | -1

Let

a,b,c

be such that

\left( {b + c} \right) \ne 0

and

\left| {\begin{array}{cccccccccccccccccccc} a&{a + 1}&{a - 1}\\ { - b}&{b + 1}&{b - 1}\\ c&{c - 1}&{c + 1} \end{array}} \right|

+ {\left| {\begin{array}{cccccccccccccccccccc} {a + 1}&{b + 1}&{c - 1}\\ {a - 1}&{b - 1}&{c + 1}\\ {{{\left( { - 1} \right)}^{n + 2}}a}&{{{\left( { - 1} \right)}^{n + 1}}b}&{{{\left( { - 1} \right)}^n}c} \end{array}} \right|^1} = 0

Then the value of

n

View answer and explanation

Correct option: C

Any odd integer

The correct answer is C (Any odd integer). The given condition holds true only when n is an odd integer, making the second determinant the negative of the first.

22 Medium

22 of 408

Analyzing +4 | -1

The coefficient of

x

f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} x&{1 + \sin x}&{\cos x}\\ 1&{\log \left( {1 + x} \right)}&2\\ {{x^2}}&{1 + {x^2}}&0 \end{array}} \right|, - 1 < x \le 1,

View answer and explanation

Correct option: B

- 2

Option B (-2) is correct because after expanding the determinant and finding the coefficient of x, it results in -2.

23 Medium

23 of 408

Analyzing +4 | -1

{\Delta _r} = \left| {\begin{array}{cccccccccccccccccccc} 1&n&n\\ {2r}&{{n^2} + n + 1}&{{n^2} + n}\\ {2r - 1}&{{n^2}}&{{n^2} + n + 1} \end{array}} \right|

and

\,\sum\nolimits_{r = 1}^n {{\Delta _r} = 56}

, then

n

equals

View answer and explanation

Correct option: C

Option C (7) is correct because by simplifying the summation of determinants and solving the resulting equation, we find that n = 7.

24 Medium

24 of 408

Analyzing +4 | -1

If the system of linear equations:

x_1+2 x_2+3 x_3=6

x_1+3 x_2+5 x_3=9

2 x_1+5 x_2+a x_3=b

is consistent and has infinite number of solutions, then :

View answer and explanation

Correct option: D

a=8, b=15

Option D (

a=8, b=15

) is correct because these values ensure the system is consistent and has infinite solutions.

25 Medium

25 of 408

Application +4 | -1

\left| {\begin{array}{cccccccccccccccccccc} {{b^2} + {c^2}}&{ab}&{ac}\\ {ab}&{{c^2} + {a^2}}&{bc}\\ {ca}&{cb}&{{a^2} + {b^2}} \end{array}} \right| = k{a^2}{b^2}{c^2},

then

k

is equal to

View answer and explanation

Correct option: C

Option C (4) is correct because after applying row and column operations and simplifying, the determinant equals

4a^2b^2c^2

, thus k = 4.

26 Medium

26 of 408

Analyzing +4 | -1

Consider the system of equations:

x+a y=0, y+a z=0

and

z+a x=0

. Then the set of all real values of '

a

' for which the system has a unique solution is:

View answer and explanation

Correct option: B

\mathrm{R}-\{-1\}

Option B (R-{-1}) is correct because the system has a unique solution when the determinant of the coefficient matrix is non-zero, which occurs when a is not equal to -1.

27 Medium

27 of 408

Application +4 | -1

P = \left[ {\begin{array}{cccccccccccccccccccc} {\sqrt 3 /2\;\;\;\;\;\;1/2}\\ { - 1/2\;\;\,\,\,\sqrt 3 /2} \end{array}} \right],\;A = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;1}\\ {0\;\;1} \end{array}} \right]\;\mathrm{and}\;Q = PA{P^T},

then

{P^T}{Q^{2005}}

P

View answer and explanation

Correct option: A

\left[ {\begin{array}{cccccccccccccccccccc} {1\;\;\;\;2005}\\ {0\;\;\;\;\;\;\;\;1} \end{array}} \right]

Option A (

\left[ {\begin{{array}}{*{20}{c}} {1\,\,\,\,2005}\\ {0\,\,\,\,\,\,\,\,1} \end{{array}}} \right]

) is correct because it follows from simplifying the expression and using the properties of matrix powers.

28 Medium

28 of 408

Application +4 | -1

Let

P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 9 & 1 & 0 \\ 27 & 9 & 1\end{array}\right]

and

Q=\left[q_{i j}\right]_{3 \times 3}

be such that

P^5-Q=I

, then

\frac{q_{21}+q_{31}}{q_{32}}

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the matrix

P^5 - Q = I

implies that

Q = P^5 - I

, and the calculation of

Q

involves nilpotent matrices, resulting in the given ratio being 22.

29 Medium

29 of 408

Application +4 | -1

A^3=\left[\begin{array}{cc}8 & -57 \\ 0 & 27\end{array}\right]

, where

A

is an upper triangular matrix with real entries, then

\operatorname{det}\left(5 A-A^2\right)

is equal to

View answer and explanation

Correct option: D

Option D (36) is correct because after finding the matrix A, we can calculate det(5A - A²) which equals 36.

30 Medium

30 of 408

Application +4 | -1

a,b,c

are respectively the

p

th,

q

th,

r

th terms of an AP, then

\left| {\begin{array}{cccccccccccccccccccc} a&p&1\\ b&q&1\\ c&r&1 \end{array}} \right|

is equal to

View answer and explanation

Correct option: C

Option C (0) is correct because after applying column operations, two columns of the determinant become identical, resulting in a zero determinant.

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Key Takeaways

✓ This set contains 408 single-correct-answer MCQs on MATRICES.
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