The value of the determinant $\Delta = \left| {\begin{array}{*{20}{c}} {2{a_1}{b_1}}&{{a_1}{b_2} + {a_2}{b_1}}&{{a_1}{b_3} + {a_3}{b_1}}\\ {{a_1}{b_2} + {a_2}{b_1}}&{2{a_2}{b_2}}&{{a_2}{b_3} + {a_3}{b_2}}\\ {{a_1}{b_3} + {a_3}{b_1}}&{{a_3}{b_2} + {a_2}{b_3}}&{2{a_3}{b_3}} \end{array}} \right|$ is

$2{a_1}{a_2}{a_3}{b_1}{b_2}{b_3}$

${a_1}{a_2}{a_3}{b_1}{b_2}{b_3}$

If $A=\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right],$ then $\mathop {\lim }\limits_{n \to \infty } \frac{{{A^n}}}{n}$ is (where $\theta \in R)$

$\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right]$

$\left[\begin{array}{rr}0 & 1 \\ 0 & -1\end{array}\right]$

If $\left[ {\begin{array}{*{20}{c}} {2\;\,\,\;1}\\ {3\;\;\,2} \end{array}} \right]A\left[ {\begin{array}{*{20}{c}} { - 3\;\;\;\;\;\;\;2}\\ {5\;\;\;\;\, - 3} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {1\;\;\,\,0}\\ {0\;\;\,\,1} \end{array}} \right],\;$ then $A$ is equal to

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

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

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

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

The inverse of the matrix $\left[ {\begin{array}{*{20}{c}} 7&{ - 3}&{ - 3}\\ { - 1}&1&0\\ { - 1}&0&1 \end{array}} \right]$ is

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

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

$\left[ {\begin{array}{*{20}{c}} 1&3&1\\ 4&3&8\\ 3&4&1 \end{array}} \right]$

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

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JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

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931 Medium

931 of 995

Understanding +4 | -1

Let

A

be a

3 \times 3

matrix such that

A^2-5 A+7 I=O

.

Statement

-I: A^{-1}=\frac{1}{7}(5 I-A)

.

Statement - II : The polynomial

A^3-2 A^2-3 A+I

can be reduced to

5(A-4 I)

. Then

View answer and explanation

Correct option: C

Both the statements are true.

Option A (Reflexive but not symmetric) is correct because the relation contains all (a,a) pairs, but lacks symmetry as (1,2) is present but (2,1) is not.

932 Easy

932 of 995

Application +4 | -1

\omega

is a complex cube root of unity, then

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

is equal to

View answer and explanation

Correct option: C

Option C (0) is correct because applying the property

1 + \omega + {\omega ^2} = 0

simplifies the determinant to zero.

933 Easy

933 of 995

Application +4 | -1

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

is a matrix satisfying the equation

x^3-3 x+1=0

, then

View answer and explanation

Correct option: C

A

is non-singular matrix

Option C (

A

is non-singular matrix) is correct because the given equation implies that

A

has an inverse, making it non-singular.

934 Medium

934 of 995

Application +4 | -1

f(\theta)=\left|\begin{array}{ccc}1 & \cos \theta & 1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1\end{array}\right|

and

A

and

B

are respectively the maximum and the minimum values of

f(\theta)

, then

(A, B)

is equal to:

View answer and explanation

Correct option: C

(2+\sqrt{2}, 2-\sqrt{2})

Option C (

(2+\sqrt{2}, 2-\sqrt{2})

) is correct because it represents the maximum and minimum values of the function f(θ) after simplification and applying trigonometric identities.

935 Easy

935 of 995

Analyzing +4 | -1

A d j \cdot(A B)-(A d j \cdot B)(A d j \cdot A)=

View answer and explanation

Correct option: C

O

Option C (O) is correct because

Adj(AB) = Adj(B)Adj(A)

, so the expression simplifies to a zero matrix.

936 Easy

936 of 995

Analyzing +4 | -1

The value of the determinant

\Delta = \left| {\begin{array}{cccccccccccccccccccc} {2{a_1}{b_1}}&{{a_1}{b_2} + {a_2}{b_1}}&{{a_1}{b_3} + {a_3}{b_1}}\\ {{a_1}{b_2} + {a_2}{b_1}}&{2{a_2}{b_2}}&{{a_2}{b_3} + {a_3}{b_2}}\\ {{a_1}{b_3} + {a_3}{b_1}}&{{a_3}{b_2} + {a_2}{b_3}}&{2{a_3}{b_3}} \end{array}} \right|

View answer and explanation

Correct option: C

Option C (0) is correct because the determinant can be factored into a product of determinants that results in zero.

937 Easy

937 of 995

Understanding +4 | -1

A

and

B

are two square matrices of the same size, then

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

can hold if and only if

View answer and explanation

Correct option: A

A B=B A

Option A (

A B=B A

) is correct because

(A+B)^2 = A^2 + AB + BA + B^2

, which equals

A^2 + 2AB + B^2

only if

AB = BA

938 Easy

938 of 995

Application +4 | -1

Let

A = \left( {\begin{array}{cccccccccccccccccccc} {[x + 1]}&{[x + 2]}&{[x + 3]}\\ {[x]}&{[x + 3]}&{[x + 3]}\\ {[x]}&{[x + 2]}&{[x + 4]} \end{array}} \right)

where

[t]

denotes the greatest integer less than or equal to

t

. If

\det (A) = 192

then the set of values of

x

is interval

View answer and explanation

Correct option: B

[62, 63)

The correct answer is [62, 63) because the determinant of the given matrix is equal to 192, which leads to [x] = 62, meaning x lies in the interval [62, 63).

939 Easy

939 of 995

Application +4 | -1

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

then

\lim\limits_{n \to \infty } \frac{{{A^n}}}{n}

is (where

\theta \in R)

View answer and explanation

Correct option: A

A zero matrix

Option A (A zero matrix) is correct because as n approaches infinity, the values of cos(nθ) and sin(nθ) oscillate between -1 and 1, making the limit of the fraction approach zero.

940 Easy

940 of 995

Understanding +4 | -1

The sum of the products of the elements of any row of a determinant

A

with the cofactors of the corresponding elements is equal to

View answer and explanation

Correct option: C

\left| A \right|

Option C is correct because the sum of the products of elements of a row with their corresponding cofactors equals the determinant's value.

941 Easy

941 of 995

Understanding +4 | -1

A^3=O

, then

I+A+A^2

equals

View answer and explanation

Correct option: B

(I-A)^{-1}

Option B (

(I-A)^{-1}

) is correct because

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

, which means

I+A+A^2

is the inverse of

(I-A)

942 Medium

942 of 995

Application +4 | -1

The number of distinct real roots of the equaiton,

\left|\begin{array}{ccc}\cos x & \sin x & \sin x \\ \sin x & \cos x & \sin x \\ \sin x & \sin x & \cos x\end{array}\right|=0

in the interval

\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]

is :

View answer and explanation

Correct option: C

Option C (2) is correct. The determinant equation simplifies to

\sin x = 0

\sin x = \cos x

, leading to two distinct real roots within the given interval.

943 Easy

943 of 995

Application +4 | -1

f\left( \theta \right) = \left[ {\begin{array}{cccccccccccccccccccc} {\cos \mathrm{\theta }}&{ - \sin \mathrm{\theta }}&0\\ {\sin \mathrm{\theta }}&{\cos \mathrm{\theta }}&0\\ 0&0&1 \end{array}} \right],\;

then

\left\{ {f{{\left( \theta \right)}^{ - 1}}} \right\}

is equal to

View answer and explanation

Correct option: A

f\left( { - \theta } \right)

The correct answer is A. The inverse of f(

\theta

) is found by replacing

\theta

with -

\theta

in the original matrix.

944 Easy

944 of 995

Application +4 | -1

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

then

{A^5} =

View answer and explanation

Correct option: C

16A

Option C (16A) is correct because when A is a scalar multiple of the identity matrix, raising it to a power involves raising the scalar to that power.

945 Easy

945 of 995

Remembering +4 | -1

A

is an orthogonal matrix, then

{A^{ - 1}}

equals

View answer and explanation

Correct option: B

{A^T}

Option B (

{A^T}

) is correct because for an orthogonal matrix A, its inverse is equal to its transpose, i.e.,

{A^{ - 1}} = {A^T}

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946 Easy

946 of 995

Application +4 | -1

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

is an invertible matrix, then

x

cannot take value

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the matrix is invertible for all the given values of x. The determinant of A is not zero for x = -1, 2, or 3.

947 Easy

947 of 995

Application +4 | -1

The rank of the matrix

A=\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & 4 & 5\end{array}\right|

, is

View answer and explanation

Correct option: B

Option B (2) is correct because the determinant of the matrix is zero, but a 2x2 submatrix has a non-zero determinant, indicating a rank of 2.

948 Easy

948 of 995

Application +4 | -1

\left[ {\begin{array}{cccccccccccccccccccc} {2\;\,\,\;1}\\ {3\;\;\,2} \end{array}} \right]A\left[ {\begin{array}{cccccccccccccccccccc} { - 3\;\;\;\;\;\;\;2}\\ {5\;\;\;\;\, - 3} \end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;\,\,0}\\ {0\;\;\,\,1} \end{array}} \right],\;

then

A

is equal to

View answer and explanation

Correct option: C

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

The correct answer is C. By pre-multiplying and post-multiplying by the inverses of the given matrices, we can isolate A and find its value.

949 Easy

949 of 995

Understanding +4 | -1

If the points

({x_1},{y_1}),\left( {{x_2},{y_2}} \right)

and

\;\left( {{x_3},{y_3}} \right)

are collinear, then the rank of the matrix

\left[ {\begin{array}{cccccccccccccccccccc} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right]

will always be less than

View answer and explanation

Correct option: B

Option B (3) is correct because if the points are collinear, the determinant of the matrix formed by their coordinates is zero, implying the rank is less than 3.

950 Easy

950 of 995

Application +4 | -1

Suppose

A

is any

3 \times 3

non-singular matrix and

(A-3 I)(A-5 I)=O

, where

I=I_3

and

O=O_3

. If

\alpha A+\beta A^{-1}=4 l

, then

\alpha+\beta

is equal to

View answer and explanation

Correct option: A

The correct answer is 8. By manipulating the given equation

(A-3I)(A-5I)=O

and multiplying by

A^{-1}

, we can express

A + 15A^{-1}

in terms of

I

. Then, by dividing by 2, we can match the given form

\alpha A + \beta A^{-1} = 4I

and find that

\alpha + \beta = 8

951 Easy

951 of 995

Application +4 | -1

\mathrm{A}=\left[\begin{array}{cc}4 & x+2 \\ 2 x-3 & x+1\end{array}\right]

is symmetric, then

x=

View answer and explanation

Correct option: B

Option B (5) is correct because for a symmetric matrix, the elements

a_{12}

and

a_{21}

must be equal, which leads to the equation

x+2 = 2x-3

, solving for x gives x=5.

952 Easy

952 of 995

Application +4 | -1

The inverse of the matrix

\left[ {\begin{array}{cccccccccccccccccccc} 7&{ - 3}&{ - 3}\\ { - 1}&1&0\\ { - 1}&0&1 \end{array}} \right]

View answer and explanation

Correct option: D

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

Option D is correct. The inverse of the given matrix is calculated by finding the adjoint of the matrix and dividing it by the determinant, which is 1 in this case.

953 Easy

953 of 995

Application +4 | -1

let

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

where

a, b, c, d \in\{0,1\}

the number of such

\mathrm{A}

so that trace

(\mathrm{A})>\operatorname{det}(\mathrm{A})

View answer and explanation

Correct option: A

Option A (13) is correct because by listing all possible matrices with elements from {0,1} and checking the condition trace(A) > det(A), we find there are 13 such matrices.

954 Easy

954 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: D

At least 7

Option D (At least 7) is correct because there are at least 6 non-singular matrices with the 1s on the diagonal, and at least one more with a different arrangement.

955 Medium

955 of 995

Application +4 | -1

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

, then the determinant of the matrix

\left(A^{2016}-2 A^{2015}-A^{2014}\right)

is :

View answer and explanation

Correct option: D

-25

Option D (-25) is correct. The determinant of the matrix (A²⁰¹⁶-2A²⁰¹⁵-A²⁰¹⁴) is found by factoring out A²⁰¹⁴ and calculating the determinant of the resulting expression.

956 Medium

956 of 995

Analyzing +4 | -1

If the system of equation

x+y+z=16 ; 2 x+5 y+\alpha z=\beta ; x+2 y+3 z=14

has infinitely many solution, then

\alpha+\beta=

View answer and explanation

Correct option: D

Option D (34) is correct because for infinitely many solutions, the determinant of the coefficient matrix must be zero, and the equations must be consistent.

957 Easy

957 of 995

Understanding +4 | -1

A

and

B

are two matrices such that

A B=A

ant

B A=B

, then

B^2

is equal to

View answer and explanation

Correct option: A

B

Option A (

B

) is correct because

B^2 = BB = (BA)B = B(AB) = BA = B

958 Easy

958 of 995

Understanding +4 | -1

A, B

are square matrices of order 3,

A

is non-singular and

A B=O

, then

B

is a

View answer and explanation

Correct option: A

null matrix

Option A (null matrix) is correct because since A is non-singular, it has an inverse. Multiplying AB = O by A^-1 on the left gives B = O.

959 Easy

959 of 995

Application +4 | -1

\mathrm{A}=\left[\begin{array}{cc}a b & b^2 \\ -a^2 & -a b\end{array}\right]

and

\mathrm{A}^n=\mathrm{O}

, then the minimum value of

n

View answer and explanation

Correct option: A

Option A (2) is correct because A² results in a zero matrix, meaning A^n = O for all n >= 2.

960 Easy

960 of 995

Remembering +4 | -1

{I_n}

is the identity matrix of order

n

, then rank of

{I_n}

View answer and explanation

Correct option: B

n

Option B (

n

) is correct because the rank of an identity matrix of order

n

is always

n

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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