If $A$ is square matrix, then

${A^T} + A\;$ is skew-symmetric

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

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

If $p = \left[ {\begin{array}{*{20}{c}} 1&0\\ {1/2}&1 \end{array}} \right]$ then ${p^{50}}$ is

$\left[ {\begin{array}{*{20}{c}} 1&0\\ {25}&1 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 1&{50}\\ 0&1 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 1&{25}\\ 0&1 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 1&0\\ {50}&1 \end{array}} \right]$

If $n$ is a natural number. Then ${\left[ {\begin{array}{*{20}{c}} 2&{ - 1}\\ 3&{ - 2} \end{array}} \right]^n},$ is

$\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]$ is $n$ is even

$\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]$ if $n$ is odd

$\left[ {\begin{array}{*{20}{c}} { - 1}&0\\ 0&1 \end{array}} \right]$ if $n$ is a natural number

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

$\left[ {\begin{array}{*{20}{l}} { - 6}&0&0 \\ 0&{ - 6}&0 \\ 0&0&{ - 6} \end{array}} \right]$

$\left[\begin{array}{rrr}0 & 1 / 6 & -1 / 6 \\ 2 / 6 & 1 / 6 & 3 / 6 \\ 3 / 6 & 2 / 6 & 1 / 6\end{array}\right]$

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JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

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661 Easy

661 of 995

Application +4 | -1

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

and

\;B = \left[ {\begin{array}{cccccccccccccccccccc} x\\ y\\ z \end{array}} \right],\;

then

\;B'\left( {AB} \right)

View answer and explanation

Correct option: A

Null matrix

Option A (Null matrix) is correct because the resulting matrix is a 1x1 matrix with a value of 0, making it a null matrix.

662 Easy

662 of 995

Understanding +4 | -1

A

is a square matrix such that

A^3=I

then

A^{-1}

is equal to

View answer and explanation

Correct option: C

A^2

Option C (

A^2

) is correct because if

A^3 = I

, then multiplying both sides by

A^{-1}

gives

A^2 = A^{-1}

663 Easy

663 of 995

Application +4 | -1

A=\left[\begin{array}{cc}\frac{-1+i \sqrt{3}}{2 i} & \frac{-1-i \sqrt{3}}{2 i} \\ \frac{1+i \sqrt{3}}{2 i} & \frac{1-i \sqrt{3}}{2 i}\end{array}\right], i=\sqrt{-1}

and

f(x)=x^2+2,

then

f(A)

equals to

View answer and explanation

Correct option: D

(2+i \sqrt{3})\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]

Option D ((2+i√3)[1 0
0 1]) is correct because substituting the values and simplifying leads to f(A) = (2+i√3)I.

664 Easy

664 of 995

Application +4 | -1

The simultaneous equations

Kx + 2y - z = 1,\left( {K - 1} \right)y - 2z = 2\;

and

\;\left( {K + 2} \right)z = 3\;

have only one solution when

View answer and explanation

Correct option: B

K = - 1

Option B (K = -1) is correct because for the system to have a unique solution, the determinant of the coefficient matrix must be non-zero, which holds true when K = -1.

665 Easy

665 of 995

Understanding +4 | -1

A

is square matrix, then

View answer and explanation

Correct option: A

A + {A^T}

is symmetric

Option A is correct because the transpose of the sum of a matrix and its transpose is equal to the sum itself, indicating symmetry.

666 Hard

666 of 995

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.

667 Medium

667 of 995

Application +4 | -1

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

and

i

is the unit matrix of order 2, then

{A^2}\;

equals

View answer and explanation

Correct option: A

4A - 3I

Option A (4A - 3I) is correct because when A is squared, it results in 4A - 3I, which can be verified through matrix multiplication and simplification.

668 Easy

668 of 995

Understanding +4 | -1

Which of the following is correct statement ?

View answer and explanation

Correct option: B

Identity matrix is a particular case of scalar matrix

Option B is the correct statement because an identity matrix is a special case of a scalar matrix where all diagonal elements are equal to 1.

669 Easy

669 of 995

Application +4 | -1

A

is a square matrix or order

n \times n,

then

\mathrm{adj}\left( {\mathrm{adj\;}A} \right)

is equal to

View answer and explanation

Correct option: C

{\left| A \right|^{n - 2}}A

Option C (

{\left| A \right|^{n - 2}}A

) is correct because it is a standard result for the adjoint of the adjoint of a matrix.

670 Hard

670 of 995

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.

671 Easy

671 of 995

Application +4 | -1

A

is a square matrix, then

\mathrm{adj\;}{A^T} - {\left( {\mathrm{adj\;}A} \right)^T}

is equal to

View answer and explanation

Correct option: C

Null matrix

Option C (Null matrix) is correct because the transpose of the adjoint of a matrix is equal to the adjoint of the transpose of the matrix.

672 Easy

672 of 995

Analyzing +4 | -1

The number of

3 \times 3

non-singular matrices, with four entries as 1 and all other entries as 0, is

View answer and explanation

Correct option: C

Atleast 7

Option C (Atleast 7) is correct because there are at least 6 ways to place the four 1s such that each row and column has at least one 1, and other configurations also exist.

673 Medium

673 of 995

Analyzing +4 | -1

If the system of equations

\left. {\begin{array}{cccccccccccccccccccc} {x - \lambda y - z = 0} \\ {\lambda x - y - z = 0} \\ {x + y - z = 0} \end{array}} \right\}

has a unique solution, then the range of

\lambda

R-\{a, b\}

. Then the value of

\left(a^2+b^2\right)

is:

View answer and explanation

Correct option: B

The correct answer is B (2) because the determinant of the coefficient matrix must be non-zero for a unique solution, leading to

\lambda = \pm 1

, and

a^2 + b^2 = 1 + 1 = 2

674 Easy

674 of 995

Understanding +4 | -1

Let

A=\left(a_{i j}\right)_{3 \times 2}

be a

3 \times 2

matrix with real entries and

B=A A^{\prime}

. Then

View answer and explanation

Correct option: C

B^{-1}

does not exist

Option C (

B^{-1}

does not exist) is correct because the determinant of B is always zero, making B a singular matrix and thus non-invertible.

675 Easy

675 of 995

Application +4 | -1

P=\left[\begin{array}{lll}1 & k & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right]

is the adjoint of a

3 \times 3

matrix

Q

and

\operatorname{det}(Q)=6

, then

k

is equal to

View answer and explanation

Correct option: C

Option C (36) is correct. The determinant of the adjoint of a matrix Q is equal to the determinant of Q squared.

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676 Hard

676 of 995

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.

677 Easy

677 of 995

Analyzing +4 | -1

If the system of equations

x+y+z=6

\,\,\,\,\,2 x+5 y+\alpha z=\beta,\,\,\,

x+2 y+3 z=14

has infinitely many solutions, then

\alpha + \beta

is equal to

View answer and explanation

Correct option: C

Option C (44) is correct because it is the sum of

\alpha

and

\beta

when the system of equations has infinitely many solutions, found by setting the determinant to zero and solving for the parameters.

678 Easy

678 of 995

Analyzing +4 | -1

Let

P=\left[\begin{array}{ccc}3 & -1 & -2 \\ 2 & 0 & a \\ 3 & -5 & 0\end{array}\right]

where

a \in R

, suppose

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

Matrix such that

P Q=\mathrm{KI}_3

for some non-zero

K \in R

.If

q_{23}=\frac{-k}{8}

and

|Q|=\frac{k^2}{2}

, then

a^2+k^2

View answer and explanation

Correct option: B

Prime number

Option B (Prime number) is correct because after solving the equations derived from the given conditions, we find a = -1 and k = 4, so a² + k² = 17, which is a prime number.

679 Medium

679 of 995

Application +4 | -1

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

then

{p^{50}}

View answer and explanation

Correct option: A

\left[ {\begin{array}{cccccccccccccccccccc} 1&0\\ {25}&1 \end{array}} \right]

Option A (

\left[ {\begin{array}{cccccccccccccccccccc}1&0\\{25}&1\end{array}} \right]

) is correct because the pattern of the matrix power leads to the correct element in the lower left corner after 50 iterations.

680 Easy

680 of 995

Application +4 | -1

n

is a natural number. Then

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

View answer and explanation

Correct option: A

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

n

is even

Option A (

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

n

is even) is correct because when

n

is even, the matrix raised to the power of

n

results in the identity matrix.

681 Easy

681 of 995

Application +4 | -1

Let

D=\left|\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right|; 0 \leq \theta<2 \pi,

then

View answer and explanation

Correct option: C

D \in[2,4]

Option C (D ∈ [2,4]) is correct because the determinant simplifies to 2(1 + sin²(theta)), and since 0 <= sin²(theta) <= 1, then 2 <= D <= 4.

682 Easy

682 of 995

Application +4 | -1

A

is a non-singular matrix of order 3, then

adj(adj\,A)

is equal to

View answer and explanation

Correct option: D

\left| A \right|A

Option D (

\left| A \right|A

) is correct because for a non-singular matrix

A

of order 3,

adj(adj A) = |A|A

683 Easy

683 of 995

Application +4 | -1

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

, then adj

\left(3 A^2+12 A\right)

is equal to

View answer and explanation

Correct option: B

\left[\begin{array}{ll}51 & 63 \\ 84 & 72\end{array}\right]

Option B (

\left[\begin{array}{ll}51 & 63 \\ 84 & 72\end{array}\right]

) is correct because after calculating 3A² + 12A, the adjugate of the resulting matrix is the given option.

684 Easy

684 of 995

Understanding +4 | -1

\mathrm{A}

and

\mathrm{B}

are matrices of same order, then

\left(\mathrm{AB}^{\prime}-\mathrm{BA}^{\prime}\right)

is a

View answer and explanation

Correct option: A

skew symmetric matrix

Option A (skew symmetric matrix) is correct because (AB' - BA')' = -(AB' - BA'), which satisfies the definition of a skew-symmetric matrix.

685 Easy

685 of 995

Application +4 | -1

A

is a

3 \times 3

matrix and

|A|=2

, then

\left|3 \operatorname{adj}\left(|3 A| A^2\right)\right|

is equal to

View answer and explanation

Correct option: D

3^{11} \cdot 6^{10}

Option D (

3^{11} \cdot 6^{10}

) is correct because it involves using properties of determinants and adjoints to simplify the expression.

686 Easy

686 of 995

Analyzing +4 | -1

Let

M

and

N

be two

3 \times 3

non-singular skew-symmetric matrices such that

M N=N M

. If

P^T

denotes the transpose of

P

, then

M^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T

is equal to

View answer and explanation

Correct option: C

-M^2

Option C (

-M^2

) is correct because, given the properties of skew-symmetric matrices and the commutative property

MN=NM

, the expression simplifies to

-M^2

687 Easy

687 of 995

Analyzing +4 | -1

If the system of linear equations

\begin{aligned} &x+2 a y+a z=0 \\ &x+3 b y+b z=0 \\ &x+4 c y+c z=0 \end{aligned}

has a non-zero solution, then

a, b, c

View answer and explanation

Correct option: D

Are in H.P.

Option D (Are in H.P.) is correct because the condition for a non-zero solution leads to the relationship

\frac{2}{b} = \frac{1}{a} + \frac{1}{c}

, which defines a harmonic progression.

688 Easy

688 of 995

Application +4 | -1

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

then

(A(adj\,A){A^{ - 1}})A =

View answer and explanation

Correct option: A

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

Option A

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

is correct because

(A(adj\,A){A^{ - 1}})A = |A|I

, and

|A| = 6

, so the result is

6I

which equals the given option.

689 Medium

689 of 995

Analyzing +4 | -1

A=\left|\begin{array}{ccc}\sin (\theta+\alpha) & \cos (\theta+\alpha) & 1 \\ \sin (\theta+\beta) & \cos (\theta+\beta) & 1 \\ \sin (\theta+\gamma) & \cos (\theta+\gamma) & 1\end{array}\right|,

then

View answer and explanation

Correct option: D

A

is independent of

\theta

Option D (

A

is independent of

\theta

) is correct because after applying row operations and simplifying, the determinant's value does not depend on

\theta

690 Easy

690 of 995

Understanding +4 | -1

B

is a non-singular matrix and

A

is a square matrix, then

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

is equal to

View answer and explanation

Correct option: C

\operatorname{det}(A)

Option C (

\operatorname{det}(A)

) is correct because

\operatorname{det}(B^{-1} A B) = \operatorname{det}(B^{-1}) \operatorname{det}(A) \operatorname{det}(B) = \frac{1}{\operatorname{det}(B)} \operatorname{det}(A) \operatorname{det}(B) = \operatorname{det}(A)

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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