If $\left| {\begin{array}{*{20}{c}} a&b&0\\ 0&a&b\\ b&0&a \end{array}} \right| = 0,$ then

$a$ is one of the cube roots of unity

$b$ is one of the cube roots of unity

$\left( {\frac{a}{b}} \right)$ is one of the cube roots of unity

If $ab + bc + ca = 0$ and $\left| {\begin{array}{*{20}{c}} {a - x}&c&b\\ c&{b - x}&a\\ b&a&{c - x} \end{array}} \right| = 0,$ then one of the value of $x$ is

${\left[ {\frac{3}{2}\left( {{a^2} + {b^2} + {c^2}} \right)} \right]^{1/2}}$

${\left[ {\frac{1}{2}\left( {{a^2} + {b^2} + {c^2}} \right)} \right]^{1/2}}$

If $F\left( x \right) = \left[ {\begin{array}{*{20}{c}} {\cos x}&{ - \sin x}&0\\ {\sin x}&{\cos x}&0\\ 0&0&1 \end{array}} \right]\;{\rm{and\;}}G\left( y \right) = \left[ {\begin{array}{*{20}{c}} {\cos x}&0&{\sin x}\\ 0&1&0\\ { - \sin x}&0&{\cos x} \end{array}} \right],$ then ${\left[ {F\left( x \right)G\left( y \right)} \right]^{ - 1}}$ is equal to

$F\left( {{x^{ - 1}}} \right)G\left( {{y^{ - 1}}} \right)$

$G\left( {{y^{ - 1}}} \right)F\left( {{x^{ - 1}}} \right)$

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JEE • MATHEMATICS • MATRICES

Matrices practice · Medium

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241 Medium

241 of 408

Application +4 | -1

\left[\begin{array}{cc}\cos \frac{2 \pi}{5} & -\sin \frac{2 \pi}{5} \\ \sin \frac{2 \pi}{5} & \cos \frac{2 \pi}{5}\end{array}\right]^n=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]

, then the least positive integer

n

View answer and explanation

Correct option: D

Option D (5) is correct because the matrix represents a rotation by

\frac{2\pi}{5}

, and it takes 5 such rotations to return to the identity matrix.

242 Medium

242 of 408

Analyzing +4 | -1

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

, then which one of the following statements is not correct?

View answer and explanation

Correct option: A

\mathrm{A}^{2}+\mathrm{I}=\mathrm{A}\left(\mathrm{A}^{2}-\mathrm{I}\right)

Option A (

\mathrm{A}^{2}+\mathrm{I}=\mathrm{A}(\mathrm{A}^{2}-\mathrm{I})

) is not correct because it simplifies to

0 = -2A

, which is false.

243 Medium

243 of 408

Analyzing +4 | -1

The equation

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

where

a,b,c

are different, is satisfied by

View answer and explanation

Correct option: C

x = \frac{1}{3}\left( {a + b + c} \right)

The correct answer is C (

x = \frac{{1}}{{3}}\left( {a + b + c} \right)

) because substituting this value into the determinant makes it equal to zero.

244 Medium

244 of 408

Analyzing +4 | -1

The values of

\alpha

for which the system of equations

\begin{array}{l} x + y + z = 1\\ x + 2y + 4z = \mathrm{\alpha }\\ x + 4y + 10z = {\mathrm{\alpha }^2} \end{array}

Is consistent, are

View answer and explanation

Correct option: C

1,2

Option C (1,2) is correct because these are the values of

\alpha

for which the system of equations is consistent, obtained by setting the determinant of the augmented matrix to zero.

245 Medium

245 of 408

Analyzing +4 | -1

\omega

is a cube root of unity, then for polynomial is

\left| {\begin{array}{cccccccccccccccccccc} {x + 1}&\omega &{{\omega ^2}}\\ \omega &{x + {\omega ^2}}&1\\ {{\omega ^2}}&1&{x + \omega } \end{array}} \right|

View answer and explanation

Correct option: D

Option D (0) is correct because the polynomial evaluates to 0.

246 Medium

246 of 408

Analyzing +4 | -1

If the system of linear equations

x+3 y+7 z=0

-x+4 y+7 z=0

(\sin 3 \theta) x+(\cos 2 \theta) y+2 z=0

has non-trivial solutions, then the number of values of

\theta

lying in the interval

[0, \pi]

, is

View answer and explanation

Correct option: D

more than three

Option D (more than three) is correct because solving the determinant equation leads to four distinct values of θ in the interval [0, π].

247 Medium

247 of 408

Application +4 | -1

A=\left[\begin{array}{cc}\frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \\ \frac{-1}{\sqrt{10}} & \frac{3}{\sqrt{10}}\end{array}\right]

and

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

M=A^{T} B A

, then the matrix

A M^{2023} A^{T}

View answer and explanation

Correct option: A

\left[\begin{array}{cc}1 & -2023 i \\ 0 & 1\end{array}\right]

Option A (

\left[\begin{{array}}{{cc}}1 & -2023 i \\ 0 & 1\end{{array}}\right]

) is correct because after simplifying the expression using the properties of matrices and the given matrices, we find that

AM^{2023}A^T = B^{2023}

, and

B^{2023}

follows a pattern that leads to the correct answer.

248 Medium

248 of 408

Application +4 | -1

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

then

View answer and explanation

Correct option: D

\left( {\frac{a}{b}} \right)

is one of the cube roots of

- 1

Option D (

\left( {\frac{a}{b}} \right)

is one of the cube roots of

- 1

) is correct because the determinant equation simplifies to

a^3 + b^3 = 0

, which means

\left( {\frac{a}{b}} \right)^3 = -1

249 Medium

249 of 408

Analyzing +4 | -1

For the system of linear equations

2 x-y+3 z=5

3 x+2 y-z=7

4 x+5 y+\alpha z=\beta,

which of the following is NOT correct?

View answer and explanation

Correct option: B

The system has infinitely many solutions for

\alpha=-6

and

\beta=9

Option B (The system has infinitely many solutions for

\alpha=-6

and

\beta=9

) is NOT correct because when

\alpha \neq -5

, the system has a unique solution, and

\alpha = -6

satisfies this condition.

250 Medium

250 of 408

Application +4 | -1

\Delta = \left| {\begin{array}{cccccccccccccccccccc} a&{a + b}&{a + b + c}\\ {3a}&{4a + 3b}&{5a + 4b + 3c}\\ {6a}&{9a + 6b}&{11a + 9b + 6c} \end{array}} \right|

, where

a = i,b = \omega ,c = {\omega ^2},

then

\Delta

is equal to

View answer and explanation

Correct option: A

i

Option A (

i

) is correct because after applying row operations and substituting the given values, the determinant simplifies to

i

251 Medium

251 of 408

Application +4 | -1

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

, then

A

View answer and explanation

Correct option: C

A nilpotent matrix

Option C (A nilpotent matrix) is correct because

A^2 = 0

, which satisfies the definition of a nilpotent matrix.

252 Medium

252 of 408

Application +4 | -1

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

, then

A A^T

View answer and explanation

Correct option: B

A singular matrix

Option B (A singular matrix) is correct because the determinant of

AA^T

is zero, indicating that the matrix is singular.

253 Medium

253 of 408

Application +4 | -1

Let

A

and

B

be any two

3 \times 3

matrices. If

A

is symmetric and

\mathrm{B}

is skewsymmetric, then the matrix

\mathrm{AB}-\mathrm{BA}

is:

View answer and explanation

Correct option: B

symmetric

Option B (symmetric) is correct because the transpose of (AB - BA) is equal to (AB - BA), satisfying the definition of a symmetric matrix.

254 Medium

254 of 408

Understanding +4 | -1

A

be a square matrix of order

n

and if

\left| A \right| = D

and

\left| {\mathrm{adj\;A}} \right| = D'

, then

View answer and explanation

Correct option: C

DD' = {D^n}

Option C (

DD' = {D^n}

) is correct because |adj A| = |A|^(n-1) and |A| = D, |adj A| = D', so D' = D^(n-1). Therefore, DD' = D * D^(n-1) = D^n.

255 Medium

255 of 408

Application +4 | -1

\left| {\begin{array}{cccccccccccccccccccc} 1&2&3\\ {{1^3}}&{{2^3}}&{{3^3}}\\ {{1^5}}&{{2^5}}&{{3^5}} \end{array}} \right|

is equal to

View answer and explanation

Correct option: C

6!

Option C (

6!

) is correct because the determinant simplifies to 720, which is equal to 6 factorial.

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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

256 Medium

256 of 408

Application +4 | -1

\left|\begin{array}{rrr} 6 i & -3 i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i \end{array}\right|=x+i y

where

i=\sqrt{-1}

, given

x^2+y^2

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the determinant evaluates to 0 after simplification.

257 Medium

257 of 408

Analyzing +4 | -1

ab + bc + ca = 0

and

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

then one of the value of

x

View answer and explanation

Correct option: A

{\left( {{a^2} + {b^2} + {c^2}} \right)^{1/2}}

Option A (

{\left( {{a^2} + {b^2} + {c^2}} \right)^{1/2}}

) is correct because it is one of the solutions obtained after simplifying the determinant and using the given condition ab + bc + ca = 0.

258 Medium

258 of 408

Application +4 | -1

A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & 3 & 5 \\ 1 & 5 & 12\end{array}\right]

, then

\operatorname{adj}(\operatorname{adj} A)

View answer and explanation

Correct option: C

\left[\begin{array}{ccc}3 & 6 & 9 \\ 3 & 9 & 15 \\ 3 & 15 & 36\end{array}\right]

Option C (

\left[\begin{array}{ccc}3 & 6 & 9 \\ 3 & 9 & 15 \\ 3 & 15 & 36\end{array}\right]

) is correct because adj(adj A) = |A|^(n-2) * A, and since |A| = 3, adj(adj A) = 3A.

259 Medium

259 of 408

Analyzing +4 | -1

D_r=\left|\begin{array}{lll}r & 1 & \frac{n(n+1)}{2} \\ 2 r-1 & 4 & n^2 \\ 2^{r-1} & 5 & 2^n-1\end{array}\right|

, then the value of

\sum\limits_{r = 1}^n {{D_r},}

View answer and explanation

Correct option: A

Option A (0) is correct because the summation of the determinants results in a determinant with two identical columns, making its value zero.

260 Medium

260 of 408

Application +4 | -1

The value of

\left| {\begin{array}{cccccccccccccccccccc} {(a + 1)\left( {a + 2} \right)}&{a + 2}&1\\ {\left( {a + 2} \right)\left( {a + 3} \right)}&{a + 3}&1\\ {\left( {a + 3} \right)\left( {a + 4} \right)}&{a + 4}&1 \end{array}} \right|

View answer and explanation

Correct option: B

- 2

Option B (-2) is correct because after applying row operations and expanding the determinant, the expression simplifies to -2.

261 Medium

261 of 408

Application +4 | -1

The roots of the equation

\left| {\begin{array}{cccccccccccccccccccc} 1&4&{20}\\ 1&{ - 2}&5\\ 1&{2x}&{5{x^2}} \end{array}} \right| = 0

are

View answer and explanation

Correct option: B

- 1,2

Option B (-1, 2) is correct because these are the roots obtained after simplifying the determinant equation and solving for x.

262 Medium

262 of 408

Application +4 | -1

F\left( x \right) = \left[ {\begin{array}{cccccccccccccccccccc} {\cos x}&{ - \sin x}&0\\ {\sin x}&{\cos x}&0\\ 0&0&1 \end{array}} \right]\;\mathrm{and\;}G\left( y \right) = \left[ {\begin{array}{cccccccccccccccccccc} {\cos x}&0&{\sin x}\\ 0&1&0\\ { - \sin x}&0&{\cos x} \end{array}} \right],

then

{\left[ {F\left( x \right)G\left( y \right)} \right]^{ - 1}}

is equal to

View answer and explanation

Correct option: C

G\left( { - y} \right)F\left( { - x} \right)

Option C (

G\left( { - y} \right)F\left( { - x} \right)

) is correct because the inverse of the product of two matrices is the product of their inverses in reverse order.

263 Medium

263 of 408

Understanding +4 | -1

A=\left[a_{i j}\right]

is a matrix of order

4 \times 5

, then the diagonal elements of

\mathrm{A}

are

View answer and explanation

Correct option: D

do not exist

Option D (do not exist) is correct because a diagonal requires a square matrix, and a 4x5 matrix is not square.

264 Medium

264 of 408

Analyzing +4 | -1

Let

A = \left[ {\begin{array}{cccccccccccccccccccc} 1&{\sin \theta }&1\\ { - \sin \theta }&1&{\sin \theta }\\ { - 1}&{ - \sin \theta }&1 \end{array}} \right],

where

0 \le \theta < 2\pi .

Then, which of the following is not correct?

View answer and explanation

Correct option: C

\det \left( A \right) \in \left[ {2,4} \right]

Option C (

\det \left( A \right) \in \left[ {2,4} \right]

) is not correct because the determinant is always within the range [2, 4], so stating it's *not* in that range is false.

265 Medium

265 of 408

Analyzing +4 | -1

The value of

\sum\nolimits_{n = 1}^N {{U_n}}

{U_n} = \left| {\begin{array}{cccccccccccccccccccc} n&1&5\\ {{n^2}}&{2N + 1}&{2N + 1}\\ {{n^3}}&{3{N^2}}&{3N} \end{array}} \right|

, is

View answer and explanation

Correct option: A

The correct answer is A (0) because the determinant simplifies to zero due to identical columns after applying summation and column operations.

266 Medium

266 of 408

Application +4 | -1

The values of

x

for which the given matrix

\left[ {\begin{array}{cccccccccccccccccccc} { - x}&x&2\\ 2&x&{ - x}\\ x&{ - 2}&{ - x} \end{array}} \right]

will be non-singular, are

View answer and explanation

Correct option: B

For all

x

other then 2 and

- 2

Option B (For all x other then 2 and -2) is correct because the matrix is non-singular when its determinant is non-zero, which occurs for all x except 2 and -2.

267 Medium

267 of 408

Analyzing +4 | -1

Consider the system of equations

\alpha x+y+z=p, x+\alpha y+z=q

and

x+y+\alpha z=r

, then the sum of all possible distinct value(s) of

\alpha

for which system does not possess a unique solution is

View answer and explanation

Correct option: C

-1

Option C (-1) is correct because the sum of all possible distinct values of

\alpha

for which the system does not possess a unique solution is -1, derived from the determinant of the coefficient matrix.

268 Medium

268 of 408

Application +4 | -1

Let

A

be a

3 \times 3

matrix such that

|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))|

=12^{4}

. Then

\mid A^{-1} \, \operatorname{adj} \, A \mid

is equal to

View answer and explanation

Correct option: B

2 \sqrt{3}

Option B (

2 \sqrt{3}

) is correct because it is derived from the properties of adjoint and determinant of a matrix.

269 Medium

269 of 408

Application +4 | -1

\mathrm{A}

is a square matrix of order

\mathrm{n}

such that

|\operatorname{adj}(\operatorname{adj} \mathrm{A})|=|\mathrm{A}|^{\mathrm{P} 2}

, (where

\mathrm{P}

is a prime number), then

n

cannot be:

View answer and explanation

Correct option: A

10

Option A (

10

) is correct because if

n=10

, then

n-1 = 9

which is not a prime number, violating the condition

n = P+1

where

P

is prime.

270 Medium

270 of 408

Application +4 | -1

a > 0,b > 0,c > 0

are respectively the

{p^{th}},{q^{th}},{r^{th}}

terms of a GP, then the value of the determinant

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

View answer and explanation

Correct option: B

Option B (0) is correct because expressing log a, log b, and log c in terms of log A and log R and applying column operations results in a zero determinant.

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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