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JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

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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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Showing 31-60 of 995 questions
31 Easy
31 of 995
Understanding +4 | -1
Let AA and BB are the non-singular square matrices, then which of the following is always correct?
View answer and explanation
Correct option: B
(AB)′=B′A′(A B)^{\prime}=B^{\prime} A^{\prime}
Option B ((AB)′=B′A′(A B)^{\prime}=B^{\prime} A^{\prime}) is correct because the transpose of a product of matrices is the product of their transposes in reverse order.
32 Easy
32 of 995
Analyzing +4 | -1
Let A={X=(x,y,z)T:PX=0\mathrm{A}=\left\{\mathrm{X}=(x, y, z)^{\mathrm{T}}: \mathrm{PX}=0\right. and x2+y2+z2=1}\left.x^2+y^2+z^2=1\right\}, where P=[121−23−419−1]P=\left[\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1\end{array}\right], then the set A :
View answer and explanation
Correct option: D
contains exactly two elements
The correct answer is contains exactly two elements because the system has infinitely many solutions, but the condition x²+y²+z²=1 restricts the solution set to two points.
33 Easy
33 of 995
Understanding +4 | -1
Consider the matrix A=[2519−735−25/21231−517]A=\left[\begin{array}{cccc}2 & 5 & 19 & -7 \\ 35 & -2 & 5 / 2 & 12 \\ \sqrt{3} & 1 & -5 & 17\end{array}\right]

Now, consider the following statements

I. The order of the matrix is 4×34 \times 3 and number of elements is 12 .

II. The elements a13,a21,a33{a}_{13}, {a}_{21}, {a}_{33} are respectively 19,35,−519,35,-5. Choose the correct option.
View answer and explanation
Correct option: B
Only II is true
Option B (Only II is true) is correct because the order of the matrix is 3x4, and the specified elements match the given values.
34 Medium
34 of 995
Application +4 | -1
The value of θ\theta lying between θ=0\theta = 0 and π2\frac{\pi }{2} and satisfying the equation ∣1+sin⁡2θcos⁡2θ4sin⁡4θsin⁡2θ1+cos⁡2θ4sin⁡4θsin⁡2θcos⁡2θ1+4sin⁡4θ∣\left| {\begin{array}{cccccccccccccccccccc} {1 + {{\sin }^2}\mathrm{\theta }}&{{{\cos }^2}\mathrm{\theta }}&{4\sin 4\mathrm{\theta }}\\ {{{\sin }^2}\mathrm{\theta }}&{1 + {{\cos }^2}\mathrm{\theta }}&{4\sin 4\mathrm{\theta }}\\ {{{\sin }^2}\mathrm{\theta }}&{{{\cos }^2}\mathrm{\theta }}&{1 + 4\sin 4\mathrm{\theta }} \end{array}} \right| is
View answer and explanation
Correct option: A
7π24\frac{{7\pi }}{{24}}
Option A (7π24\frac{{7\pi}}{{24}}) is correct because it satisfies the given determinant equation within the specified range of θ\theta after simplification and solving for θ\theta.
35 Easy
35 of 995
Analyzing +4 | -1
The system of equations αx+y+z=α−1,   x+αy+z=α−1,   and x+y+αz=α−1 has no solution, if α is \alpha x + y + z = \alpha - 1,\,\,\,x + \alpha y + z = \alpha - 1,\,\,{\text{ and }}x + y + \alpha z = \alpha - 1{\text{ has no solution, if }}\alpha {\text{ is }}
View answer and explanation
Correct option: C
-2
Option C (-2) is correct because when α=−2\alpha = -2, the system of equations has no solution, as the determinant is zero and the equations are inconsistent.
36 Easy
36 of 995
Application +4 | -1
Let A=[21012−10−12]A=\left[\begin{array}{ccc}2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2\end{array}\right]. If ∣adj⁡(adj⁡(adj2A⁡))∣=(16)n|\operatorname{adj}(\operatorname{adj}(\operatorname{adj2A}))|=(16)^{n}, then nn is equal to
View answer and explanation
Correct option: B
10
Option B (10) is correct because it correctly applies the property of adjoints of adjoints of adjoints and determinants.
37 Easy
37 of 995
Application +4 | -1
If A=[0100]A=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right] and AB=OA B=O, then B=B=
View answer and explanation
Correct option: D
[−1000]\left[\begin{array}{cc}-1 & 0 \\ 0 & 0\end{array}\right]
Option D is correct. Given AB = O, and A is a non-zero matrix, B must be a matrix such that when multiplied by A, the result is a zero matrix. Option D satisfies this condition.
38 Easy
38 of 995
Analyzing +4 | -1
If the system of equations 2x+3y−z=0,x+ky−2z=0 and 2x−y+z=02x + 3y - z = 0,x + ky - 2z = 0{\kern 1pt} \,\mathrm{and}\,2x - y + z = 0 has a non-trivial solution (x, y, z), then xy+yz+zx+k\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k is equal to
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because after solving for kk and finding the ratios xy\frac{x}{y}, yz\frac{y}{z}, and zx\frac{z}{x}, their sum plus kk equals 12\frac{1}{2}.
39 Medium
39 of 995
Analyzing +4 | -1
The value of k∈Rk \in R, for which the following system of linear equations 3x−y+4z=3,x+2y−3z=−2,6x+5y+kz=−33 x-y+4 z=3, x+2 y-3 z=-2,6 x+5 y+k z=-3 has infinitely many solutions, is
View answer and explanation
Correct option: A
-5
Option A (-5) is correct because the value of k is found by ensuring the system of linear equations has infinitely many solutions, which requires the equations to be consistent and dependent.
40 Medium
40 of 995
Application +4 | -1
Matrix A=[x321y422z]A=\left[\begin{array}{ccc}x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z\end{array}\right], if xyz=60x y z=60 and 8x+4y+3z8 x+4 y+3 z =20=20, then A(adj⁡A)A(\operatorname{adj} A) is equal to
View answer and explanation
Correct option: C
[680006800068]\left[\begin{array}{ccc}68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68\end{array}\right]
Option C is correct because the determinant can be expanded and simplified to find the value of |A|, and then A(adj A) = |A|I.
41 Easy
41 of 995
Application +4 | -1
If A=[1−253]A=\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right], then A+ATA+A^T equals:
View answer and explanation
Correct option: A
[2336]\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]
Option A (\left[\begin{{array}}{{ll}}2 & 3 \\ 3 & 6\end{{array}}\right]\left[\begin{{array}}{{ll}}2 & 3 \\ 3 & 6\end{{array}}\right]) is correct because it is the result of adding the matrix A to its transpose.
42 Easy
42 of 995
Analyzing +4 | -1
For the equations x+2y+3z=1,  2x+y+3z=2  x + 2y + 3z = 1,\;2x + y + 3z = 2\; and   5x+5y+9z=4\;5x + 5y + 9z = 4
View answer and explanation
Correct option: A
There is only one solution
Option A (There is only one solution) is correct because the determinant of the coefficient matrix is non-zero, indicating a unique solution.
43 Easy
43 of 995
Application +4 | -1
If 3X+2Y=I3 \mathrm{X}+2 \mathrm{Y}=\mathrm{I} and 2X−Y=O2 \mathrm{X}-\mathrm{Y}=\mathrm{O}, where I\mathrm{I} and O\mathrm{O} are unit and null matrices of order 3 respectively, then
View answer and explanation
Correct option: C
X=(17)I,Y=(27)I\mathrm{X}=\left(\frac{1}{7}\right) \mathrm{I}, \mathrm{Y}=\left(\frac{2}{7}\right) \mathrm{I}
Option C (X = (1/7)I, Y = (2/7)I) is correct because solving the system of equations for X and Y yields these values.
44 Easy
44 of 995
Application +4 | -1
Let a,ba, b and cc be three real numbers satisfying
[abc][197827737]=[000]   (1)\left[\begin{array}{lll} a & b & c \end{array}\right]\left[\begin{array}{lll} 1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \end{array}\right] \,\,\,(1)

If the point P(a,b,c)P(a, b, c) with reference to (1), lies on the plane 2x+y+z=12 x+y+z=1, then the value of 7a+b+c7 a+b+c is
View answer and explanation
Correct option: D
6
Option D (6) is correct because solving the system of equations and substituting into the plane equation gives k=1k=1, leading to 7a+b+c=67a+b+c = 6.
45 Easy
45 of 995
Analyzing +4 | -1
Let pp be an odd prime number and TpT_p be the following set of 2×22 \times 2 matrices Tp={A=[abca]:a,b,c∈{0,1,2,…,(p−1)}}T_p=\left\{A=\left[\begin{array}{ll}a & b \\ c & a\end{array}\right]: a, b, c \in\{0,1,2, \ldots,(p-1)\}\right\}
The number of AA in TpT_p such that det⁡(A)\det (A) is not divisible by pp
View answer and explanation
Correct option: D
p3−p2p^3-p^2
Option D (p3−p2p^3-p^2) is correct because it represents the number of matrices in TpT_p whose determinant is not divisible by pp.
46 Easy
46 of 995
Application +4 | -1
If P=[cos⁡π6sin⁡π6−sin⁡π6cos⁡π6],A=[1101]P=\left[\begin{array}{rr}\cos \frac{\pi}{6} & \sin \frac{\pi}{6} \\ -\sin \frac{\pi}{6} & \cos \frac{\pi}{6}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] and Q=PAPT,Q=P A P^T, then PTQ2007PP^T Q^{2007} P is equal to
View answer and explanation
Correct option: B
[1200701]\left[\begin{array}{cc}1 & 2007 \\ 0 & 1\end{array}\right]
Option B ([1 2007
0 1]) is correct because using the properties of matrix transpose and the given matrices, we can simplify the expression to A2007, which can be easily calculated.
47 Easy
47 of 995
Application +4 | -1
If AA is a square matrix such that A  (adj  A)=[400040004],A\;\left( {\mathrm{adj\;}A} \right) = \left[ {\begin{array}{cccccccccccccccccccc} 4&0&0\\ 0&4&0\\ 0&0&4 \end{array}} \right], then ∣adj  A∣=\left| {\mathrm{adj}\;A} \right| =
View answer and explanation
Correct option: B
16
Option B (16) is correct because |adj A| = |A|^(n-1) and A(adj A) = |A|I, so |A| = 4 and |adj A| = 4(3-1) = 16.
48 Easy
48 of 995
Application +4 | -1
If A=[1log⁡balog⁡ab1]A = \left[ {\begin{array}{cccccccccccccccccccc} 1&{{{\log }_b}a}\\ {{{\log }_a}b}&1 \end{array}} \right], then ∣A∣\left| A \right| is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because the determinant of the matrix A is calculated as 1 - (log_a b)(log_b a) which simplifies to 1 - 1 = 0.
49 Easy
49 of 995
Application +4 | -1
Let A2−A+I=OA^2-A+I=O, then inverse of AA is
View answer and explanation
Correct option: A
I−AI-A
Option A (I−AI-A) is correct because multiplying AA by (I−A)(I-A) results in the identity matrix, thus (I−A)(I-A) is the inverse of AA.
50 Easy
50 of 995
Understanding +4 | -1
If AA is a skew symmetric matrix of odd order, then its inverse is
View answer and explanation
Correct option: D
Non existent
Option D (Non existent) is correct because a skew-symmetric matrix of odd order has a determinant of 0, making it non-invertible.
51 Easy
51 of 995
Application +4 | -1
If X=[3−41−1]X=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right], then the value of XnX^n is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because substituting n=2 into options A, B, and C does not yield the correct X2 matrix.
52 Easy
52 of 995
Application +4 | -1
If A=[0320]A=\left[\begin{array}{ll}0 & 3 \\ 2 & 0\end{array}\right] and A−1=λ(A)A^{-1}=\lambda(A), then λ\lambda
View answer and explanation
Correct option: D
16\frac{1}{6}
Option D (16\frac{1}{6}) is correct because by finding the inverse of A and comparing it to λ(A)\lambda(A), we can solve for λ\lambda and find it to be 16\frac{1}{6}.
53 Easy
53 of 995
Application +4 | -1
If ω\omega is a non-real cube root of unity and nn is not a multiple of 3, then
Δ=∣1ωnω2nω2n1ωnωnω2n1∣\Delta=\left|\begin{array}{ccc}1 & \omega^n & \omega^{2 n} \\ \omega^{2 n} & 1 & \omega^n \\ \omega^n & \omega^{2 n} & 1\end{array}\right| is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because when n is not a multiple of 3, 1+ωn+ω2n=01 + \omega^n + \omega^{2n} = 0, making the determinant equal to zero.
54 Easy
54 of 995
Application +4 | -1
The matrix A=[1−3−4−1341−3−4]A=\left[\begin{array}{rrr}1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4\end{array}\right] is nilpotent of index:
View answer and explanation
Correct option: A
2
Option A (2) is correct because the matrix A squared results in a null matrix, indicating a nilpotent index of 2.
55 Easy
55 of 995
Application +4 | -1
If matrix A=[10−1345067]aA=\left[\begin{array}{ccc}1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & 6 & 7\end{array}\right] a
and its inverse is denoted by
A−1=[a11a12a13a21a22a23a31a32a33]\mathrm{A}^{-1}=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right], then the value of a23=a_{23}=
View answer and explanation
Correct option: D
25\frac{2}{5}
Option D (25\frac{2}{5}) is correct because it is calculated by finding the cofactor of the element in the matrix and dividing by the determinant of the matrix.
56 Easy
56 of 995
Application +4 | -1
If A=[2−3−41]A=\left[\begin{array}{cc}2 & -3 \\ -4 & 1\end{array}\right], then adj (3A2+12A)\left(3 A^2+12 A\right) is equal to:
View answer and explanation
Correct option: C
[51638472]\left[\begin{array}{ll}51 & 63 \\ 84 & 72\end{array}\right]
Option D (\left[\begin{{ll}}51 & 84 \\ 63 & 72\end{{ll}}\right]\left[\begin{{ll}}51 & 84 \\ 63 & 72\end{{ll}}\right]) is correct. By calculating 3A2+12A3A^2 + 12A and then finding its adjugate, we arrive at the correct matrix.
57 Easy
57 of 995
Analyzing +4 | -1
Which of the following matrices can NOT be obtained from the matrix [−121−1]\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right] by a single elementary row operation?
View answer and explanation
Correct option: C
[−12−27]\left[\begin{array}{ll}-1 & 2 \\ -2 & 7\end{array}\right]
The matrix in option C [−12−27]\left[\begin{array}{ll}-1 & 2 \\ -2 & 7\end{array}\right] cannot be obtained from the given matrix by a single elementary row operation.
58 Easy
58 of 995
Application +4 | -1
If A=[4     23     4],∣adj  A∣A = \left[ {\begin{array}{cccccccccccccccccccc} {4\;\;\,2}\\ {3\;\;\,4} \end{array}} \right],\left| {\mathrm{adj}\;A} \right| is equal to
View answer and explanation
Correct option: C
10
Option C (10) is correct because the determinant of the adjugate of a 2x2 matrix is equal to the determinant of the original matrix.
59 Easy
59 of 995
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.
60 Easy
60 of 995
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.
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