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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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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.

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61 Easy
61 of 922
Application +4 | -1
If α,β\alpha, \beta, γ\gamma are the roots of x3+ax2+b=0,x^3+a x^2+b=0, then the value of ∣αβγβγαγαβ∣,\left|\begin{array}{lll}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta\end{array}\right|, is
View answer and explanation
Correct option: C
a3a^3
Option C (a3a^3) is correct because the determinant simplifies to a3a^3 using the relationships between roots and coefficients of the cubic equation.
62 Easy
62 of 922
Understanding +4 | -1
Which of the following

(i) Adjoint of a symmetric matrix is symmetric

(ii) Adjoint of a unit matrix is a unit matrix

(iii) A(adj⁡A)=(adj⁡A)A=∣A∣I\mathrm{A}(\operatorname{adj} \mathrm{A})=(\operatorname{adj} \mathrm{A}) \mathrm{A}=|\mathrm{A}| \mathrm{I} and

(iv) Adjoint of a diagonal matrix is a diagonal matrix, is/are incorrect
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because all the statements (i), (ii), (iii), and (iv) are true.
63 Easy
63 of 922
Application +4 | -1
If A=[1234],A = \left[ {\begin{array}{cccccccccccccccccccc} 1&2\\ 3&4 \end{array}} \right], then I+A+A2+A3+…∞I + A + {A^2} + {A^3} + \ldots \infty equals to
View answer and explanation
Correct option: C
[1/2−1/3−1/20]\left[ {\begin{array}{cccccccccccccccccccc} {1/2}&{ - 1/3}\\ { - 1/2}&0 \end{array}} \right]
The correct answer is C. The sum of the infinite matrix series can be found by using the formula for the sum of an infinite geometric series and then inverting the resulting matrix.
64 Easy
64 of 922
Application +4 | -1
AA and BB are two square matrices such that A2B=A^2 B= BAB A and if (AB)10=AkB10(A B)^{10}=A^k B^{10}, then kk is
View answer and explanation
Correct option: A
1023
Option A (1023) is correct because by repeatedly applying the given condition A2B=BAA^2B = BA, we can derive the formula (AB)n=A2n−1Bn(AB)^n = A^{2^n-1}B^n, and thus find k=210−1=1023k = 2^{10} - 1 = 1023.
65 Easy
65 of 922
Analyzing +4 | -1
If x,yx, y and zz are the integers in AP lying between 1 and 9 and x 51,y 41x\, 51, y\, 41 and z 31z\, 31 are three digits number, the value of ∣543x 51y 41z 31xyz∣\left|\begin{array}{ccc}5 & 4 & 3 \\ x\, 51 & y\, 41 & z\, 31 \\ x & y & z\end{array}\right| is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the determinant simplifies to zero due to the properties of arithmetic progressions and determinant operations.
66 Easy
66 of 922
Analyzing +4 | -1
The value of ∣525354535455545556∣,\left|\begin{array}{lll}5^2 & 5^3 & 5^4 \\ 5^3 & 5^4 & 5^5 \\ 5^4 & 5^5 & 5^6\end{array}\right|, is
View answer and explanation
Correct option: B
0
Option B (0) is correct because the determinant is zero due to two rows (or columns after taking common factors) being identical.
67 Easy
67 of 922
Application +4 | -1
If A=[235−2]A=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right], then A−1A^{-1} is equal to
View answer and explanation
Correct option: D
119A\frac{1}{19} A
Option D (119A\frac{1}{19} A) is correct because the inverse of A is equal to A divided by the determinant of A.
68 Easy
68 of 922
Application +4 | -1
The adjoint matrix of [3−342−340−11]\left[ {\begin{array}{cccccccccccccccccccc} 3&{ - 3}&4\\ 2&{ - 3}&4\\ 0&{ - 1}&1 \end{array}} \right] is
View answer and explanation
Correct option: B
[1−10−23−4−23−3]\left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 1}&0\\ { - 2}&3&{ - 4}\\ { - 2}&3&{ - 3} \end{array}} \right]
Option B is correct. The adjoint of the given matrix is found by calculating the cofactors, forming the cofactor matrix, and then taking the transpose.
69 Easy
69 of 922
Application +4 | -1
If A=[cos⁡αsin⁡α−sin⁡αcos⁡α]A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right], then A2A^2 is equal to
View answer and explanation
Correct option: C
[cos⁡2αsin⁡2α−sin⁡2αcos⁡2α]\left[\begin{array}{cc}\cos 2 \alpha & \sin 2 \alpha \\ -\sin 2 \alpha & \cos 2 \alpha\end{array}\right]
Option C is correct. A2 is found by multiplying matrix A by itself, and using trigonometric identities simplifies the result to the given matrix.
70 Easy
70 of 922
Analyzing +4 | -1
If A=[−4−131]A=\left[\begin{array}{rr}-4 & -1 \\ 3 & 1\end{array}\right], then the determinant of the matrix⁡(A2016−2A2015−A2014)\operatorname{matrix}\left(A^{2016}-2 A^{2015}-A^{2014}\right), is
View answer and explanation
Correct option: D
-25
Option D (-25) is correct. By calculating A2 - 2A - I and then finding its determinant, and using the property |A^n| = |A|^n, the determinant of the given matrix is -25.
71 Easy
71 of 922
Application +4 | -1
If A=[1334]{A}=\left[\begin{array}{ll}1 & 3 \\ 3 & 4\end{array}\right] and A2−5A=kI{A}^2-5 {A}={kI}, then k=k=
View answer and explanation
Correct option: C
5
Option A is correct. By performing the matrix multiplications AB and XY, and then adding the results, we get [20].
72 Easy
72 of 922
Application +4 | -1
If AA is a square matrix of order 3 such that ∣A∣=2,|A|=2, then ∣(adj⁡A−1)−1∣\left|\left(\operatorname{adj} A^{-1}\right)^{-1}\right| is
View answer and explanation
Correct option: C
4
Option C (4) is correct because ∣(adj⁡A−1)−1∣\left|\left(\operatorname{adj} A^{-1}\right)^{-1}\right| simplifies to ∣A∣2|A|^2, and since ∣A∣=2|A|=2, the result is 4.
73 Easy
73 of 922
Understanding +4 | -1
Let A=[4],B=[−1002]A=[4], B=\left[\begin{array}{cc}-1 & 0 \\ 0 & 2\end{array}\right] and C=[−1.100020003]C=\left[\begin{array}{ccc}-1.1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{array}\right] are the matrices. Consider the following statements

I. The matrices A,B{A}, {B} and C{C} are diagonal matrices.

II. The matrices AA, BB and CC are of order 1, 3 and 2, respectively.

Choose the correct option.
View answer and explanation
Correct option: A
I is true and II is false
Option A (I is true and II is false) is correct because all three matrices are diagonal, but their orders are 1, 2, and 3, respectively.
74 Easy
74 of 922
Understanding +4 | -1
The inverse of a skew-symmetric matrix of odd order is
View answer and explanation
Correct option: D
Does not exist
Option D (Does not exist) is correct because a skew-symmetric matrix of odd order has a determinant of 0, making it non-invertible.
75 Easy
75 of 922
Application +4 | -1
If A=[−i00i]A=\left[\begin{array}{cc}-i & 0 \\ 0 & i\end{array}\right], then A′AA^{\prime} A is equal to
View answer and explanation
Correct option: C
−I-I
Option C (-I) is correct because when A' is multiplied by A, the result is a negative identity matrix.
76 Easy
76 of 922
Application +4 | -1
If A=[4x+22x−3x+1]A = \left[ {\begin{array}{cccccccccccccccccccc} 4&{x + 2}\\ {2x - 3}&{x + 1} \end{array}} \right] is symmetric, then x=x =
View answer and explanation
Correct option: B
5
Option B (5) is correct because for a symmetric matrix, the elements Aij and Aji must be equal, leading to the equation x + 2 = 2x - 3, which solves to x = 5.
77 Easy
77 of 922
Application +4 | -1
If A=[2012131−10]A=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right], then A2−5A+6IA^2-5 A+6 I is equal to
View answer and explanation
Correct option: A
[1−1−3−1−1−10−544]\left[\begin{array}{ccc}1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4\end{array}\right]
Option B (0) is correct because when two columns of A are interchanged to form B, det(B) = -det(A), and therefore det(A+B) = 0.
78 Easy
78 of 922
Application +4 | -1
If ω\omega is a cube root of unity and A=[ω00ω]A=\left[\begin{array}{ll} \omega & 0 \\ 0 & \omega\end{array}\right], then A37=……(ω≠1)A^{37}=\ldots \ldots(\omega \neq 1)
View answer and explanation
Correct option: C
AA
Option C (AA) is correct because A37A^{37} simplifies to AA using the property that ω3=1\omega^3 = 1 where ω\omega is a cube root of unity.
79 Easy
79 of 922
Application +4 | -1
If f(x)=∣1x(x+1)2xx(x−1)x(x+1)3x(x−1)x(x−1)(x−2)x(x−1)(x+1)∣f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} 1&x&{\left( {x + 1} \right)}\\ {2x}&{x\left( {x - 1} \right)}&{x\left( {x + 1} \right)}\\ {3x\left( {x - 1} \right)}&{x\left( {x - 1} \right)\left( {x - 2} \right)}&{x\left( {x - 1} \right)\left( {x + 1} \right)} \end{array}} \right| then f(11)f\left( {11} \right) equals
View answer and explanation
Correct option: A
0
Option A (0) is correct because after taking common factors and applying row operations, the determinant simplifies to zero.
80 Easy
80 of 922
Application +4 | -1
If A=[123−5],B=[1002]\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ 3 & -5\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right] and X\mathrm{X} is a matrix such that A−BX\mathrm{A}-\mathrm{BX}, then X\mathrm{X} equals-
View answer and explanation
Correct option: B
12[243−5]\frac{1}{2}\left[\begin{array}{cc}2 & 4 \\ 3 & -5\end{array}\right]
Option B (12[243−5]\frac{1}{2}\left[\begin{array}{cc}2 & 4 \\ 3 & -5 \end{array}\right]) is correct because it involves solving the matrix equation A = BX for X by pre-multiplying by the inverse of B.
81 Easy
81 of 922
Understanding +4 | -1
If [1420]=[xy2z0],y<0\left[\begin{array}{ll}1 & 4 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}x & y^2 \\ z & 0\end{array}\right], y < 0 then x−y+zx-y+z is equal to
View answer and explanation
Correct option: A
5
Option A (5) is correct because by equating the corresponding elements of the matrices, we find x=1x=1, y=−2y=-2 (since y<0y<0), and z=2z=2. Therefore, x−y+z=1−(−2)+2=5x-y+z = 1 - (-2) + 2 = 5.
82 Easy
82 of 922
Understanding +4 | -1
Let AA be a non-singular square matrix. Then, ∣adj  A∣\left| {\mathrm{adj\;}A} \right| is equal to
View answer and explanation
Correct option: B
∣A∣n−1{\left| A \right|^{n - 1}}
Option B (∣A∣n−1\left| A \right|^{n - 1}) is correct because the determinant of the adjoint of a matrix A is equal to the determinant of A raised to the power of (n-1), where n is the order of the matrix.
83 Easy
83 of 922
Application +4 | -1
If 1,ω,ω21, \omega, \omega^2 are the cube roots of unity then Δ=∣1ωnω2nωnω2n1ω2n1ωn∣ equals\Delta=\left|\begin{array}{ccc} 1 & \omega^n & \omega^{2 n} \\ \omega^n & \omega^{2 n} & 1 \\ \omega^{2 n} & 1 & \omega^n \end{array}\right| \text { equals}
View answer and explanation
Correct option: D
0
Option D (0) is correct because the determinant is zero due to the properties of cube roots of unity.
84 Easy
84 of 922
Application +4 | -1
What must be the matrix XX, if 2X+[1234]=[3872]?2X + \left[ {\begin{array}{cccccccccccccccccccc} 1&2\\ 3&4 \end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc} 3&8\\ 7&2 \end{array}} \right]?
View answer and explanation
Correct option: A
[132−1]\left[ {\begin{array}{cccccccccccccccccccc} 1&3\\ 2&{ - 1} \end{array}} \right]
Option A \left[ {\begin{{array}}{{{20}{c}}} 1&3\\\\2&{ - 1} \end{{array}}} \right]\left[ {\begin{{array}}{{{20}{c}}} 1&3\\\\2&{ - 1} \end{{array}}} \right] is correct because it is the matrix X that satisfies the given matrix equation.
85 Easy
85 of 922
Application +4 | -1
Let ∣a−b−c2a2a2bb−a−c2b2c2cc−a−b∣=ks3(\left|\begin{array}{lll}a-b-c & 2 a & 2 a \\ 2 b & b-a-c & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=k s^3( where 2s=a+b+c)2 s=a+b+c). Then value of kk is
View answer and explanation
Correct option: A
8
Option A (8) is correct because the determinant simplifies to 8s38s^3 where 2s=a+b+c2s = a+b+c, thus k=8k=8.
86 Easy
86 of 922
Analyzing +4 | -1
The system of equations

αx+y+z=α−1x+αy+z=α−1x+y+αz=α−1\begin{aligned} &\alpha x+y+z=\alpha-1 \\ &x+\alpha y+z=\alpha-1 \\ &x+y+\alpha z=\alpha-1\end{aligned}

has no solution, if α\alpha is
View answer and explanation
Correct option: D
−2-2
Option D (-2) is correct because when alpha is -2, the determinant of the coefficient matrix is zero, but at least one of the determinants formed by replacing a column with the constants is non-zero, indicating no solution.
87 Easy
87 of 922
Application +4 | -1
If A=[ab0−ba0001],  where  a2+b2=1,A = \left[ {\begin{array}{cccccccccccccccccccc} a&b&0\\ { - b}&a&0\\ 0&0&1 \end{array}} \right],\;\mathrm{where}\;{a^2} + {b^2} = 1, then adj(A)adj\left( A \right) is equal to
(Here, AT  is  the  transpose  of  A){A^T}\mathrm{\;is\;the\;transpose\;of}\;A)
View answer and explanation
Correct option: A
A−1{A^{ - 1}}
Option A (A−1{A^{ - 1}}) is correct because the determinant of A is 1, making the adjugate of A equal to its inverse.
88 Easy
88 of 922
Analyzing +4 | -1
The number of 3×33 \times 3 matrices AA whose are either 0 or 1 and for which the system⁡A[xyz]=[100]\operatorname{system} A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right] has exactly two distinct solutions, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because three planes cannot intersect at exactly two distinct points; the intersection would either be a line or no intersection.
89 Easy
89 of 922
Application +4 | -1
If n=3kn = 3k and 1ω,ω21\omega ,{\omega ^2} are the cube roots of unity, then Δ=∣1ωnω2nω2n1ωnωnω2n1∣\Delta = \left| {\begin{array}{cccccccccccccccccccc} 1&{{\omega ^n}}&{{\omega ^{2n}}}\\ {{\omega ^{2n}}}&1&{{\omega ^n}}\\ {{\omega ^n}}&{{\omega ^{2n}}}&1 \end{array}} \right| has the value
View answer and explanation
Correct option: A
0
Option A (0) is correct because when n = 3k, the determinant simplifies to zero due to the properties of cube roots of unity.
90 Easy
90 of 922
Application +4 | -1
If A=[12221−2a2b]A = \left[ {\begin{array}{cccccccccccccccccccc} 1&2&2\\ 2&1&{ - 2}\\ a&2&b \end{array}} \right] is a matrix satisfying the equation AAT=9I,A{A^T} = 9I, where II is 3×33 \times 3 identity matrix, then the ordered pair (a, b) is equal to
View answer and explanation
Correct option: D
(-2, -1)
Option D (-2, -1) is correct because, after substituting the matrix A into the equation AA^T = 9I and solving the resulting system of equations, the ordered pair (a, b) is found to be (-2, -1).
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