If $A = \left[ {\begin{array}{*{20}{c}} 3&2\\ 0&1 \end{array}} \right],\;$ then ${\left( {{A^{ - 1}}} \right)^3}$ is equal to

$\frac{1}{{27}}\left[ {\begin{array}{*{20}{c}} 1&{ - 26}\\ 0&{27} \end{array}} \right]$

$\frac{1}{{27}}\left[ {\begin{array}{*{20}{c}} { - 1}&{26}\\ 0&{27} \end{array}} \right]$

$\frac{1}{{27}}\left[ {\begin{array}{*{20}{c}} 1&{ - 26}\\ 0&{ - 27} \end{array}} \right]$

$\frac{1}{{27}}\left[ {\begin{array}{*{20}{c}} { - 1}&{ - 26}\\ 0&{ - 27} \end{array}} \right]$

If $A = \left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 5 }}}&{\frac{2}{{\sqrt 5 }}}\\ {\frac{{ - 2}}{{\sqrt 5 }}}&{\frac{1}{{\sqrt 5 }}} \end{array}} \right)$, $B = \left( {\begin{array}{*{20}{c}} 1&0\\ i&1 \end{array}} \right)$, $i = \sqrt { - 1} $ and $Q = {A^T}BA$ then the inverse of matrix $A\,{Q^{2021}}{A^T}$ is equal to

$\left( {\begin{array}{*{20}{c}} 1&0\\ { - 2021i}&1 \end{array}} \right)$

$\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 5 }}}&{ - 2021}\\ {2021}&{\frac{1}{{\sqrt 5 }}} \end{array}} \right)$

$\left( {\begin{array}{*{20}{c}} 1&0\\ {2021i}&1 \end{array}} \right)$

$\left( {\begin{array}{*{20}{c}} 1&{ - 2021i}\\ 0&1 \end{array}} \right)$

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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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Showing 301-330 of 922 questions

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301 Easy

301 of 922

Application +4 | -1

Let the numbers 2, b, c be in an AP and

A = \left[ {\begin{array}{cccccccccccccccccccc} 1&1&1\\ 2&b&c\\ 4&{{b^2}}&{{c^2}} \end{array}} \right].\mathrm{ If}\det (A) \in [2,16],\mathrm{then}\,\mathrm{ }c

lies in the interval

View answer and explanation

Correct option: C

\left[ {4,6} \right]

Option C ([4,6]) is correct because the value of c lies in this interval after solving for the common difference of the AP and using the determinant condition.

302 Easy

302 of 922

Application +4 | -1

A root of 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: C

Option C (0) is correct because expanding the determinant leads to an equation where x=0 is a root.

303 Easy

303 of 922

Application +4 | -1

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

then

{\left( {{A^{ - 1}}} \right)^3}

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it is the result of cubing the inverse of the given matrix.

304 Easy

304 of 922

Analyzing +4 | -1

Let

\alpha

be a root of the equation

{x^2} + x + 1 = 0

and the matrix

A = \frac{1}{{\sqrt 3 }}\left[ {\begin{array}{cccccccccccccccccccc} 1&1&1\\ 1&\alpha &{{\alpha ^2}}\\ 1&{{\alpha ^2}}&{{\alpha ^4}} \end{array}} \right]

, then the matrix

{A^{31}}

is equal to

View answer and explanation

Correct option: A

{A^3}

Option A (

{A^3}

) is correct because after calculating

A^2

and using the property

{\alpha^3} = 1

, we find that

{A^4} = I

, so

{A^{31}} = {A^{28}} \cdot {A^3} = {A^3}

305 Easy

305 of 922

Application +4 | -1

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

, then

A^2-5 A-I

equals

View answer and explanation

Correct option: B

I

Option B (

I

) is correct because after calculating

A^2 - 5A

, the result is

2I

. Subtracting

I

from

2I

yields

I

306 Easy

306 of 922

Analyzing +4 | -1

a, b

and

c

are non-zero real numbers and if the equations

(a-1) x=y+z,(b-1) y=z+x

(c-1) z=x+y

has a non-trivial solution, then

a b+b c+c a

equals to

View answer and explanation

Correct option: B

{abc}

Option B (

{abc}

) is correct because, for a non-trivial solution, the determinant of the coefficient matrix must be zero, leading to the equation

ab + bc + ca = abc

307 Easy

307 of 922

Application +4 | -1

Let

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

be a matrix, then (determinant of A)

\times

(adjoint of inverse A) is equal to

View answer and explanation

Correct option: B

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

Option B is correct. The expression simplifies to the original matrix A, as the determinant of A times the adjoint of the inverse of A equals A.

308 Easy

308 of 922

Application +4 | -1

\omega

be a complex cube root of unity, then

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{ - {\omega ^2}/2}\\ 1&1&1\\ 1&{ - 1}&0 \end{array}} \right|

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because simplifying the determinant using the property

1 + \omega + \omega^2 = 0

results in a zero determinant.

309 Easy

309 of 922

Application +4 | -1

Let

\left[ x \right]

represent the greatest integer less than or equal to

x

, then the value of the determinant

\left| {\begin{array}{cccccccccccccccccccc} {\left[ e \right]}&{\left[ \pi \right]}&{\left[ {{\pi ^2} - 6} \right]}\\ {\left[ \pi \right]}&{\left[ {{\pi ^2} - 6} \right]}&{\left[ e \right]}\\ {\left[ {{\pi ^2} - 6} \right]}&{\left[ e \right]}&{\left[ \pi \right]} \end{array}} \right|

View answer and explanation

Correct option: A

- 8

Option A (-8) is correct because substituting the integer values of e, pi, and pi² - 6 into the determinant and evaluating it yields -8.

310 Easy

310 of 922

Application +4 | -1

A

is a matrix such that

{A^2} = A + 1,\;

where

I

is the unit matrix, then

{A^5}\;

is equal to

View answer and explanation

Correct option: C

5A + 3I

Option C (

5A + 3I

) is correct because repeated application of the given relation

{A^2} = A + I

simplifies

{A^5}

5A + 3I

311 Easy

311 of 922

Application +4 | -1

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

and

B=\left[\begin{array}{ll}1 & 0 \\ 5 & 1\end{array}\right]

, then the value of

a

for which

A^2=B

View answer and explanation

Correct option: D

No real values

Option D (No real values) is correct because when

A^2

is calculated, the resulting matrix requires

\alpha^2 = 1

and

\alpha + 1 = 5

, which is impossible for any real value of

\alpha

312 Easy

312 of 922

Analyzing +4 | -1

S

is the set of distinct values of

'b'

for which the following system of linear equations

x+y+z=1

x+a y+z=1

a x+b y+z=0

has no solution, then

S

View answer and explanation

Correct option: D

a singleton set

The correct answer is a singleton set because the system has no solution only when a=b=1, resulting in a single value for 'b'.

313 Easy

313 of 922

Application +4 | -1

a_1, a_2, a_3, \ldots, a_n, \ldots

are in G.P., then the value of the determinant

\left|\begin{array}{ccc} \log a_n & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+1} & \log a_{n+2} & \log a_{n+3} \\ \log a_{n+2} & \log a_{n+3} & \log a_{n+4} \end{array}\right|

View answer and explanation

Correct option: C

Option C (0) is correct because the rows of the determinant are linearly dependent due to the properties of geometric progressions.

314 Easy

314 of 922

Remembering +4 | -1

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

, then the trace of the matrix

A

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because the trace of a matrix is the sum of its diagonal elements, which in this case is 1 + 1 + (-1) = 1.

315 Easy

315 of 922

Application +4 | -1

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

and

A^2=\left[\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right]

, then

View answer and explanation

Correct option: C

\alpha=a^2+b^2, \beta=2 a b

Option C (

\alpha=a^2+b^2, \beta=2 a b

) is correct because when A is multiplied by itself, the resulting matrix elements are

\alpha = a^2 + b^2

and

\beta = 2ab

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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

316 Easy

316 of 922

Application +4 | -1

\mathrm{A}=\left[\begin{array}{cc}1 & m \\ 0 & 1\end{array}\right]

, then

\mathrm{A}^n=

View answer and explanation

Correct option: B

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

Option B is correct. By observing the pattern of A raised to successive powers, we can deduce that A^n results in the matrix where the top right element is mn.

317 Easy

317 of 922

Application +4 | -1

A = \left( {\begin{array}{cccccccccccccccccccc} {\frac{1}{{\sqrt 5 }}}&{\frac{2}{{\sqrt 5 }}}\\ {\frac{{ - 2}}{{\sqrt 5 }}}&{\frac{1}{{\sqrt 5 }}} \end{array}} \right)

B = \left( {\begin{array}{cccccccccccccccccccc} 1&0\\ i&1 \end{array}} \right)

i = \sqrt { - 1}

and

Q = {A^T}BA

then the inverse of matrix

A\,{Q^{2021}}{A^T}

is equal to

View answer and explanation

Correct option: B

\left( {\begin{array}{cccccccccccccccccccc} 1&0\\ { - 2021i}&1 \end{array}} \right)

Option B is correct because the inverse of the matrix

A{Q^{2021}}{A^T}

simplifies to the inverse of

B^{2021}

, which is

\begin{{pmatrix}} 1 & 0 \\ -2021i & 1 \end{{pmatrix}}

318 Easy

318 of 922

Analyzing +4 | -1

The value of

\Delta=\left|\begin{array}{lll}b-c & c-a & a-b \\ c-a & a-b & b-c \\ a-b & b-c & c-a\end{array}\right|

View answer and explanation

Correct option: B

Option B (0) is correct because the determinant's first column becomes all zeros after applying the column operation, resulting in a zero determinant.

319 Easy

319 of 922

Application +4 | -1

If a square matrix

A

is such that

A{A^{T\;}} = I = {A^T}A,\;\mathrm{then}\left| A \right|\mathrm{is\;equal\;to}

View answer and explanation

Correct option: B

\pm 1

Option B (

\pm 1

) is correct because if

A{A^{T}} = I

, then

|A||{A^T}| = |I|

, which simplifies to

|A|^2 = 1

, thus

|A| = \pm 1

320 Easy

320 of 922

Application +4 | -1

Let

A=\left[\begin{array}{ll}2 & 2 \\ 9 & 4\end{array}\right]

and

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

, then

10 A^{-1}

is equal to

View answer and explanation

Correct option: D

A-6 I

Option D (A-6I) is correct because it represents the expression for 10A⁻¹ in terms of A and the identity matrix I, derived by finding the inverse of A.

321 Easy

321 of 922

Application +4 | -1

\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]=O

then value of

x

View answer and explanation

Correct option: A

-2

-14

Option A is correct. By multiplying the matrices and solving the resulting quadratic equation, we find the values of x to be -2 and -14.

322 Easy

322 of 922

Understanding +4 | -1

The inverse of a skew symmetric matrix (if it exists) is

View answer and explanation

Correct option: B

a skew symmetric matrix

Option B (a skew symmetric matrix) is correct because the inverse of a skew-symmetric matrix is also skew-symmetric.

323 Easy

323 of 922

Application +4 | -1

Suppose

a, b, c \in {R}

. Let

\Delta=\left|\begin{array}{ccc} 3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c \end{array}\right|

Then

\Delta

equals

View answer and explanation

Correct option: A

3(a+b+c)(b c+c a+a b)

The correct answer is A (

3(a+b+c)(b c+c a+a b)

) because by applying column and row operations, the determinant can be simplified to this expression.

324 Easy

324 of 922

Application +4 | -1

A = \left[ {\begin{array}{cccccccccccccccccccc} 3&4\\ 2&4 \end{array}} \right],\;B = \left[ {\begin{array}{cccccccccccccccccccc} { - 2}&{ - 2}\\ 0&{ - 2} \end{array}} \right]

, then

{\left( {A + B} \right)^{ - 1}}

is equal to

View answer and explanation

Correct option: D

None of the above

Option D (None of the above) is correct because the inverse of (A+B) exists and is not a skew-symmetric matrix, nor is it equal to the sum of the inverses of A and B.

325 Easy

325 of 922

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], A=\left[\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right], Q=p A p^T

and

p^T Q^{2012} p=\left[\begin{array}{ll}\alpha & \gamma \\ \beta & \alpha\end{array}\right]

, then the value of

\alpha+\beta

View answer and explanation

Correct option: C

Option C (1) is correct because the value of

\alpha + \beta

is calculated using matrix operations and properties of rotation matrices.

326 Easy

326 of 922

Application +4 | -1

Let

A = \left[ {\begin{array}{cccccccccccccccccccc} 1&2\\ { - 1}&4 \end{array}} \right]

. If

{A^{ - 1}} = \alpha I + \beta A,\,\,\alpha ,\,\,\beta \in R

I

is a

2 \times 2

identify matrix, then

4\left( {\alpha - \beta } \right)

is equal to

View answer and explanation

Correct option: D

Option D (4) is correct because by finding the inverse of matrix A and comparing it with the given equation, we can determine the values of alpha and beta, and then calculate 4(alpha - beta).

327 Easy

327 of 922

Analyzing +4 | -1

If the system of linear equations

\begin{aligned} &2 x+y-z=7 \\ &x-3 y+2 z=1 \\ &x+4 y+\delta z=k, \text { where } \delta, k \in \mathbf{R} \end{aligned}

has infinitely many solutions, then

\delta+k

is equal to:

View answer and explanation

Correct option: B

Option B (3) is correct because, for the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero, leading to

\delta = -3

, and then solving for

k

gives

k = 6

, so

\delta + k = 3

328 Easy

328 of 922

Application +4 | -1

1,\omega ,{\omega ^2}

are the cube roots of unity, then

\Delta = \left| {\begin{array}{cccccccccccccccccccc} 1&{{\omega ^n}}&{{\omega ^{2n}}}\\ {{\omega ^n}}&{{\omega ^{2n}}}&1\\ {{\omega ^{2n}}}&1&{{\omega ^n}} \end{array}} \right|

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because expanding the determinant and using the property

\omega^3 = 1

results in the determinant being equal to zero.

329 Easy

329 of 922

Analyzing +4 | -1

If the system of equations

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

has a non-zero solution, then the possible values of

k

are

View answer and explanation

Correct option: D

-1,1

Option D (-1,1) is correct because these values of k make the determinant of the coefficient matrix zero, leading to a non-zero solution.

330 Easy

330 of 922

Application +4 | -1

A

is a square matrix such that

A^2=A

and

B=I-A

, then

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

View answer and explanation

Correct option: A

A

Option A (

A

) is correct because simplifying the expression

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

using the given conditions

A^2=A

and

B=I-A

results in

A

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