The set of all $2 \times 2$ matrices which commute with the $\left[ {\begin{array}{*{20}{c}} 1&1 \\ 1&0 \end{array}} \right]$ with respect to matrix multplication is

$\left\{\left[\begin{array}{ll}p & q \\ r & r\end{array}\right]: p, q, r \in R\right\}$

$\left\{\left[\begin{array}{ll}p & q \\ 4 & r\end{array}\right]: p, q, r \in R\right\}$

$\left\{\left[\begin{array}{cc}p-q & p \\ q & r\end{array}\right]: p, q, r \in R\right\}$

One factor of $\left| {\begin{array}{*{20}{c}} {{a^2} + x}&{ab}&{ac}\\ {ab}&{{b^2} + x}&{cb}\\ {ca}&{cb}&{{c^2} + x} \end{array}} \right|$ is

$\left( {{a^2} + x} \right)\left( {{b^2} + x} \right)\left( {{c^2} + x} \right)$

By elementary transformation method, the inverse of $\left[ {\begin{array}{*{20}{c}} 1&2&3\\ 2&3&4\\ 3&4&6 \end{array}} \right]$ is

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

The value of the determinant $\left| {\begin{array}{*{20}{c}} 1&{\cos \left( {{\rm{\alpha }} - {\rm{\beta }}} \right)}&{{\rm{cos\alpha }}}\\ {\cos \left( {{\rm{\alpha }} - {\rm{\beta }}} \right)}&1&{{\rm{cos\beta }}}\\ {\cos {\rm{\alpha }}}&{{\rm{cos\beta }}}&1 \end{array}} \right|$ is

${{\rm{\alpha }}^2} - {{\rm{\beta }}^2}$

${{\rm{\alpha }}^2} + {{\rm{\beta }}^2}$

Back to MATRICES

Your Progress

0 / 26 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • MATRICES

Matrices practice · Hard

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 →

1 Hard

1 of 26

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.

2 Hard

2 of 26

Application +4 | -1

For positive numbers

x,y

and

z

, the numerical value of the determinant

\left| {\begin{array}{cccccccccccccccccccc} 1&{{{\log }_x}y}&{{{\log }_x}z}\\ {{{\log }_y}x}&1&{{{\log }_y}z}\\ {{{\log }_z}x}&{{{\log }_z}y}&1 \end{array}} \right|

View answer and explanation

Correct option: A

Option A (0) is correct because the determinant simplifies to zero using logarithm properties.

3 Hard

3 of 26

Application +4 | -1

S=\left\{x \in[0,2 \pi]:\left|\begin{array}{ccc}0 & \cos x & -\sin x \\ \sin x & 0 & \cos x \\ \cos x & \sin x & 0\end{array}\right|=0\right\}

then

\sum_{x \in S} \tan \left(\frac{\pi}{3}+x\right)

is equal to

View answer and explanation

Correct option: C

-4-2 \sqrt{3}

Option C (

-4-2 \sqrt{3}

) is correct because it is the sum of the tangents of the angles obtained from the given condition.

4 Hard

4 of 26

Analyzing +4 | -1

If the three linear equations

\begin{array}{l} x + 4ay + az = 0\\ x + 3by + bz = 0\\ x + 2cy + cz = 0 \end{array}

Have a non-trivial solution, where

a \ne 0,\;b \ne 0,\;c \ne 0,\;

then

ab + bc

is equal to

View answer and explanation

Correct option: A

2ac

Option A (

2ac

) is correct because for a non-trivial solution, the determinant of the coefficient matrix must be zero, which simplifies to ab + bc = 2ac.

5 Hard

5 of 26

Application +4 | -1

\left|\begin{array}{ccc}\sum\limits_{{k=0}}^{n-2} 1 & n(n-1) & n^2 \\ \sum\limits_{{k=1}}^n 1 & (n+1)(n-1) & n(n+1) \\ \sum\limits_{{k=1}}^{n-1} 1 & n^2+1 & n^2\end{array}\right|=72

, then

n

is equal to

View answer and explanation

Correct option: C

Option C (8) is correct because solving the determinant equation leads to n(n+1) = 72, which gives n = 8.

6 Hard

6 of 26

Application +4 | -1

The set of all

2 \times 2

matrices which commute with the

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

with respect to matrix multplication is

View answer and explanation

Correct option: D

\left\{\left[\begin{array}{cc}p & q \\ q & p-q\end{array}\right]: p, q \in R\right\}

Option D is the correct answer because it represents the set of all 2x2 matrices that commute with the given matrix, derived by equating AB and BA and solving for the matrix elements.

7 Hard

7 of 26

Application +4 | -1

A,B,C

be the angles of a triangle, then

\left| {\begin{array}{cccccccccccccccccccc} { - 1}&{\cos C}&{\cos B}\\ {\cos C}&{ - 1}&{\cos A}\\ {\cos B}&{\cos A}&{ - 1} \end{array}} \right|

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because the determinant simplifies to zero using trigonometric identities and the property A + B + C = π.

8 Hard

8 of 26

Analyzing +4 | -1

a, b

and

c

are sides of

\triangle A B C

such that

\left|\begin{array}{ccc}c & b \cos B+c \beta & a \cos A+b \alpha+c \gamma \\ a & c \cos B+a \beta & b \cos A+c \alpha+a \gamma \\ b & a \cos B+b \beta & c \cos A+a \alpha+b \gamma\end{array}\right|=0

(where

\alpha, \beta, \gamma \in R^{+}

and

\angle A, \angle B, \angle C \neq \frac{\pi}{2}

\triangle A B C

View answer and explanation

Correct option: B

an equilateral

Option B (an equilateral) is correct because after applying determinant properties and trigonometric identities, the condition leads to

a=b=c

, implying an equilateral triangle.

9 Hard

9 of 26

Analyzing +4 | -1

The value of the determinant

\left|\begin{array}{ccc}1 & \sin (\alpha-\beta) \theta & \cos (\alpha-\beta) \theta \\ a & \sin \alpha \theta & \cos \alpha \theta \\ a^2 & \sin (\alpha-\beta) \theta & \cos (\alpha-\beta) \theta\end{array}\right|

is independent of

View answer and explanation

Correct option: A

\alpha

Option A (

\alpha

) is correct because the determinant simplifies to an expression independent of

\alpha

10 Hard

10 of 26

Analyzing +4 | -1

Let

m

be a positive integer and

\Delta_r=\left|\begin{array}{ccc}2 r-1 & { }^m C_r & 1 \\ m^2-1 & 2^m & m+1 \\ \sin ^2\left(m^2\right) & \sin ^2(m) & \sin \left(m^2\right)\end{array}\right|

then the value of

\sum\limits_{{r=0}}^m \Delta r

is given by

View answer and explanation

Correct option: A

The correct answer is B (Arithmetic Progression) because if

f'(x) = 0

, then the determinant must be zero, which implies a relationship between

a^2, b^2,

and

c^2

indicating they are in Arithmetic Progression.

11 Hard

11 of 26

Application +4 | -1

If the matrix

M_r

is given by

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

. Then the value of

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

...

+\operatorname{det}\left(\mathrm{M}_{2009}\right)

View answer and explanation

Correct option: C

(2009)^2

Option C (

(2009)^2

) is correct because the determinant of

M_r

2r-1

, and the sum of the determinants from

r=1

r=2009

is the sum of the first 2009 odd numbers, which equals

2009^2

12 Hard

12 of 26

Application +4 | -1

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

and

\phi(x)=(1+x)(1-x)^{-1}

and

\phi(A)=-\lambda A

, then find the value of

\lambda.

View answer and explanation

Correct option: A

Option A (1) is correct because after calculating

\phi(A) = (I+A)(I-A)^{-1}

, it simplifies to

-A

, which means

\lambda = 1

in the equation

\phi(A) = -\lambda A

13 Hard

13 of 26

Analyzing +4 | -1

\left|\begin{array}{ccc} \mathrm{a}^2 & b^2 & c^2 \\ (a+\lambda)^2 & (b+\lambda)^2 & (c+\lambda)^2 \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2 \end{array}\right|=k \lambda\left|\begin{array}{ccc} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{array}\right|, \lambda \neq 0

then

\mathrm{k}

is equal to:

View answer and explanation

Correct option: C

4 \lambda^2

Option C (

4 \lambda^2

) is correct because after applying row operations and simplifying, the determinant equals

4\lambda^2

times the original determinant, thus k =

4 \lambda^2

14 Hard

14 of 26

Application +4 | -1

f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} 1&{2\left( {x - 1} \right)}&{3\left( {x - 1} \right)\left( {x - 2} \right)}\\ {x - 1}&{\left( {x - 1} \right)\left( {x - 2} \right)}&{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}\\ x&{x\left( {x - 1} \right)}&{x\left( {x - 1} \right)\left( {x - 2} \right)} \end{array}} \right|

Then, the value of

f\left( {49} \right)

View answer and explanation

Correct option: C

Option C (0) is correct because after applying column operations, the determinant becomes zero, making f(x) = 0 for all x.

15 Hard

15 of 26

Analyzing +4 | -1

One factor of

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

View answer and explanation

Correct option: A

{x^2}

Option A (

{x^2}

) is correct because after applying row and column operations, the determinant simplifies to an expression that contains

{x^2}

as a factor.

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 Hard

16 of 26

Analyzing +4 | -1

The number of positive integral solution of the equation

\left|\begin{array}{ccc}x^3+1 & x^2 y & x^2 z \\ x y^2 & y^3+1 & y^2 z \\ x z^2 & y z^2 & z^3+1\end{array}\right|=11

View answer and explanation

Correct option: B

The correct answer is B (3) because solving the determinant equation leads to x³ + y³ + z³ = 10, which has three positive integral solutions.

17 Hard

17 of 26

Application +4 | -1

Let

\alpha \,\,\,\mathrm{and}\,\,\,\,\,\beta

be the roots of equation

{x^2} - 6x - 2 = 0.

{a_n} = {\alpha ^n} - {\beta ^n},

for

n \ge 1

then the value of

\frac{{{a_{10}} - 2{a_8}}}{{2{a_9}}}

is equal to

View answer and explanation

Correct option: C

Option C (3) is correct because, after substituting the roots into the given expression and simplifying using the quadratic equation, the value reduces to 3.

18 Hard

18 of 26

Application +4 | -1

The unique solution of

x+3 y+4 z=8,2 x+y+2 z=5,5 x+y+z=7

, is

View answer and explanation

Correct option: A

x=1, y=1, z=1

Option A is correct. Substituting x=1, y=1, and z=1 into the given system of equations satisfies all three equations.

19 Hard

19 of 26

Analyzing +4 | -1

Let

a, b, c \in R

such that no two of them are equal and satisfy

\left|\begin{array}{ccc}2 a & b & c \\ b & c & 2 a \\ c & 2 a & b\end{array}\right|=0

, the equation

24 a x^2+4 b x+c=0

has

View answer and explanation

Correct option: A

atleast one root in

\left[0, \frac{1}{2}\right]

Option A (atleast one root in

\left[0, \frac{1}{2}\right]

) is correct because by applying Rolle's theorem, we can show that there is at least one root in the given interval.

20 Hard

20 of 26

Analyzing +4 | -1

a, b, c

are real, then

f(x)=\left|\begin{array}{ccc}x+a^2 & a b & a c \\ a b & x+b^2 & b c \\ a c & b c & x+c^2\end{array}\right|

is decreasing in:

View answer and explanation

Correct option: A

\left(-\frac{2}{3}\left(a^2+b^2+c^2\right), 0\right)

Option A is correct because the function's derivative is negative in the interval

\left(-\frac{{2}}{{3}}\left(a^2+b^2+c^2\right), 0\right)

, indicating a decreasing function.

21 Hard

21 of 26

Application +4 | -1

By elementary transformation method, the inverse of

\left[ {\begin{array}{cccccccccccccccccccc} 1&2&3\\ 2&3&4\\ 3&4&6 \end{array}} \right]

View answer and explanation

Correct option: A

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

Option A

\left[ {\begin{{array}}{{{20}{c}}} { - 2}&0&1\\\\0&3&{ - 2}\\\1&{ - 2}&1 \end{{array}}} \right]

is correct because it is the inverse of the given matrix, obtained through elementary row transformations.

22 Hard

22 of 26

Analyzing +4 | -1

A, B

and

C

are angles of a triangle, the value of

\left|\begin{array}{lll}e^{2 i A} & e^{-i C} & e^{-i B} \\ e^{-i C} & e^{2 i B} & e^{-i A} \\ e^{-i B} & e^{-i B A} & e^{2 i C}\end{array}\right|

is (where

\left.i=\sqrt{-1}\right)

View answer and explanation

Correct option: D

-4

Option D (-4) is correct because the determinant simplifies to -4 after applying properties of determinants and using the fact that A+B+C =

\pi

23 Hard

23 of 26

Analyzing +4 | -1

\left| {\begin{array}{cccccccccccccccccccc} \alpha &x&x&x\\ x&\beta &x&x\\ x&x&\gamma &x\\ x&x&x&\delta \end{array}} \right| = f\left( x \right) - xf'\left( x \right)

then

f\left( x \right)

is equal to

View answer and explanation

Correct option: A

\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)\left( {x - \delta } \right)

Option A (

\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)\left( {x - \delta } \right)

) is correct because after applying row operations and comparing with the given expression, it matches the form of

f(x)

24 Hard

24 of 26

Analyzing +4 | -1

Given that

b^2-a c<0, a>0.

The value of

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

View answer and explanation

Correct option: C

Negative

Option C (Negative) is correct because the determinant simplifies to an expression that is negative given the conditions

b^2 - ac < 0

and

a > 0

25 Hard

25 of 26

Analyzing +4 | -1

Let

f(x)=\left|\begin{array}{ccc}1+\sin ^{2} x & \cos ^{2} x & \sin 2 x \\ \sin ^{2} x & 1+\cos ^{2} x & \sin 2 x \\ \sin ^{2} x & \cos ^{2} x & 1+\sin 2 x\end{array}\right|

x \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right]

. If

\alpha

and

\beta

respectively are the maximum and the minimum values of

f

, then

View answer and explanation

Correct option: D

\beta^{2}-2 \sqrt{\alpha}=\frac{19}{4}

Option D (

\beta^{{2}}-2 \sqrt{{\alpha}}=\frac{{19}}{{4}}

) is correct because after simplifying the determinant, finding the max and min values, and substituting, the equation holds true.

26 Hard

26 of 26

Analyzing +4 | -1

The value of the determinant

\left| {\begin{array}{cccccccccccccccccccc} 1&{\cos \left( {\mathrm{\alpha } - \mathrm{\beta }} \right)}&{\mathrm{cos\alpha }}\\ {\cos \left( {\mathrm{\alpha } - \mathrm{\beta }} \right)}&1&{\mathrm{cos\beta }}\\ {\cos \mathrm{\alpha }}&{\mathrm{cos\beta }}&1 \end{array}} \right|

View answer and explanation

Correct option: A

Option A (0) is correct because the given determinant can be simplified to 0 using trigonometric identities.

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 26 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.

Continue Your Practice

📚 Switch Set (HARD)

Chapter Sets

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

0 /26

0✓ 0✗

0% Accuracy