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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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541 Easy
541 of 922
Understanding +4 | -1
If AA is a symmetric matrix and n  ∈  Nn\; \in \;N, then An  {A^n}\; is
View answer and explanation
Correct option: A
Symmetric matrix
Option A (Symmetric matrix) is correct because if A is symmetric, then A transpose equals A, and (A^n) transpose also equals A^n.
542 Easy
542 of 922
Understanding +4 | -1
If AA is a skew-symmetric matrix of odd order, then ∣adj  A∣\left| {\mathrm{adj}\;A} \right| is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the determinant of a skew-symmetric matrix of odd order is always 0, and therefore the determinant of its adjugate is also 0.
543 Easy
543 of 922
Application +4 | -1
If A=[1     −24       5]  and  f(t)=t2−3t+7,  then  f(A)+[3                6−12      −9]  A = \left[ {\begin{array}{cccccccccccccccccccc} {1\,\,\,\; - 2}\\ {4\,\,\,\;\;5} \end{array}} \right]\mathrm{\;and\;}f\left( t \right) = {t^2} - 3t + 7,\mathrm{\;then\;}f\left( A \right) + \left[ {\begin{array}{cccccccccccccccccccc} {3\;\;\;\;\;\;\;\;6}\\ { - 12\,\,\,\,\; - 9} \end{array}} \right]\mathrm{\;} is equal to
View answer and explanation
Correct option: B
[0000]\left[ {\begin{array}{cccccccccccccccccccc} 0&0\\ 0&0 \end{array}} \right]
The correct answer is B. By calculating f(A) and adding the given matrix, the result is a zero matrix.
544 Easy
544 of 922
Application +4 | -1
If A=[abb2−a2−ab]A=\left[\begin{array}{cc}a b & b^2 \\ -a^2 & -a b\end{array}\right], then A2A^2 is equal
View answer and explanation
Correct option: A
OO
Option A (O) is correct because when A is squared, all elements become zero, resulting in a null matrix.
545 Easy
545 of 922
Analyzing +4 | -1
Let the system of linear equation x+2y+z=2,αx+3y−z=α,−αx+y+2z=−αx + 2y + z = 2,\alpha x + 3y - z = \alpha , - \alpha x + y + 2z = - \alpha be inconsistent. Then α\alpha is equal to:
View answer and explanation
Correct option: D
−72-\frac{7}{2}
Option D (−72\frac{{ - 7}}{2}) is correct because when α=−72\alpha = -\frac{7}{2}, the determinant of the coefficient matrix is zero, and the system becomes inconsistent.
546 Easy
546 of 922
Application +4 | -1
If A=[42−11]A=\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right] and II is the identity matrix of order 2 , then (A−2I)(A−3I)=(\mathrm{A}-2 \mathrm{I})(\mathrm{A}-3 \mathrm{I})=
View answer and explanation
Correct option: B
0
Option 0 (O) is correct because multiplying (A-2I) and (A-3I) results in a zero matrix.
547 Easy
547 of 922
Application +4 | -1
∣1+sin⁡2θsin⁡2θsin⁡2θcos⁡2θ1+cos⁡2θcos⁡2θ4sin⁡4θ4sin⁡4θ1+4sin⁡4θ∣=0,\left| {\begin{array}{cccccccccccccccccccc} {1 + {{\sin }^2}\mathrm{\theta }}&{{{\sin }^2}\mathrm{\theta }}&{{{\sin }^2}\mathrm{\theta }}\\ {{{\cos }^2}\mathrm{\theta }}&{1 + {{\cos }^2}\mathrm{\theta }}&{{{\cos }^2}\mathrm{\theta }}\\ {4\sin 4\mathrm{\theta }}&{4\sin 4\mathrm{\theta }}&{1 + 4\sin 4\mathrm{\theta }} \end{array}} \right| = 0, then sin⁡4θ\sin 4\theta equals to
View answer and explanation
Correct option: C
−12\frac{{ - 1}}{2}
Option C (−12\frac{{ - 1}}{2}) is correct because after applying row and column operations, the equation simplifies to sin⁡4θ=−12\sin 4\theta = - \frac{1}{2}.
548 Easy
548 of 922
Application +4 | -1
The value of xx for which the matrix A=[2/x−121x2x211/x2]A=\left[\begin{array}{ccc}2 / x & -1 & 2 \\ 1 & x & 2 x^2 \\ 1 & 1 / x & 2\end{array}\right] is singular is
View answer and explanation
Correct option: A
±1\pm 1
Option A (±1\pm 1) is correct because these are the values of x that make the determinant of the matrix A equal to zero, thus making it singular.
549 Easy
549 of 922
Application +4 | -1
If [x+y2x+zx−y2z+w]=[47010]\left[\begin{array}{cc}x+y & 2 x+z \\ x-y & 2 z+w\end{array}\right]=\left[\begin{array}{cc}4 & 7 \\ 0 & 10\end{array}\right], then values of x,y,zx, y, z, ware
View answer and explanation
Correct option: A
2,2,3,42,2,3,4
Option A (2,2,3,42,2,3,4) is correct because by comparing the corresponding elements of the matrices, we can solve the system of equations to find x=2, y=2, z=3, and w=4.
550 Easy
550 of 922
Application +4 | -1
If
A=[2x0xx] and A−1=[10−12]A=\left[\begin{array}{cc} 2 x & 0 \\ x & x \end{array}\right] \text { and } A^{-1}=\left[\begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array}\right]
then xx is equal to
View answer and explanation
Correct option: B
1/21 / 2
Option B (1/21/2) is the correct answer because when AA−1=IA A^{-1}=I, solving the matrix multiplication results in x=1/2x = 1/2.
551 Easy
551 of 922
Understanding +4 | -1
Assertion : The matrix A=[912374]A=\left[\begin{array}{lll}9 & 1 & 2 \\ 3 & 7 & 4\end{array}\right] does not possesses any inverse.

Reason : A{A} is not a square matrix.
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion.
Option A is correct because the assertion is true, the reason is true, and the reason correctly explains the assertion. Only square matrices can have inverses.
552 Easy
552 of 922
Application +4 | -1
If A=[0121233x1]A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1\end{array}\right] and A−1=[1/2−1/21/2−43y5/2−3/21/2], thenA^{-1}=\left[\begin{array}{ccc} 1 / 2 & -1 / 2 & 1 / 2 \\ -4 & 3 & y \\ 5 / 2 & -3 / 2 & 1 / 2 \end{array}\right], \text { then}
View answer and explanation
Correct option: A
x=1,y=−1x=1, y=-1
Option A (x=1,y=−1x=1, y=-1) is correct because these values satisfy the matrix multiplication AA−1=IAA^{-1} = I.
553 Easy
553 of 922
Analyzing +4 | -1
x+ky−z=0,  3x−ky−z=0  x + ky - z = 0,\;3x - ky - z = 0\; and x−3y+z=0  x - 3y + z = 0\; has non-zero solution for kk is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because the determinant of the coefficient matrix must be zero for a non-zero solution, which leads to k = 1.
554 Easy
554 of 922
Application +4 | -1
If A−2B=[1537]A-2 B=\left[\begin{array}{ll}1 & 5 \\ 3 & 7\end{array}\right] and 2A−3B=[−2507]2 A-3 B=\left[\begin{array}{rr}-2 & 5 \\ 0 & 7\end{array}\right], then matrix BB is equal to
View answer and explanation
Correct option: A
[−4−5−6−7]\left[\begin{array}{ll}-4 & -5 \\ -6 & -7\end{array}\right]
Option A ([−4−5−6−7]\left[\begin{array}{ll}-4 & -5 \\ -6 & -7\end{array}\right]) is correct because solving the system of matrix equations for B yields this result.
555 Easy
555 of 922
Application +4 | -1
If P=(200020002)\mathrm{P}=\left(\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right) then, det⁡(Adj⁡P)=\operatorname{det}(\operatorname{Adj} P)=
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because det⁡(Adj⁡P)=(det⁡P)n−1\operatorname{det}(\operatorname{Adj} P) = (\operatorname{det} P)^{n-1}, and in this case, it equals (det⁡P)2=64(\operatorname{det} P)^2 = 64, which is not equal to P27P^{27} or P17P^{17} or P6P^6.
556 Easy
556 of 922
Application +4 | -1
If A=[aij]4×4A=\left[a_{i j}\right]_{4 \times 4} such that aij={2,i=j0,i≠ja_{i j}=\left\{\begin{array}{ll}2, & i=j \\ 0, & i \neq j\end{array}\right., then {17det⁡(adj⁡(adj⁡A)}\left\{\frac{1}{7} \operatorname{det}(\operatorname{adj}(\operatorname{adj} A)\}\right. is, where {.} denotes the fractional part function.
View answer and explanation
Correct option: A
17\frac{1}{7}
Option A (1/7) is correct because it represents the fractional part of the expression after simplifying the determinant and using the binomial theorem.
557 Easy
557 of 922
Application +4 | -1
If A=[a000a000a]A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right], then ∣A∣∣adj⁡A∣|A||\operatorname{adj} A| is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the value of ∣A∣∣adjA∣|A||adj A| is a9a^9, which is not among the given options.
558 Easy
558 of 922
Application +4 | -1
If A=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]A=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right], then AT+A=I2A^T+A=I_2, if
View answer and explanation
Correct option: C
θ=2nπ+π3,n∈Z\theta=2 n \pi+\frac{\pi}{3}, n \in Z
Option C (θ=2nπ+π3,n∈Z\theta=2 n \pi+\frac{\pi}{3}, n \in Z) is correct because it satisfies the equation AT+A=I2A^T + A = I_2 by resulting in cos⁡θ=12\cos \theta = \frac{1}{2}.
559 Easy
559 of 922
Application +4 | -1
If AA and BB are two matrices such that AB=BA B=B and BA=AB A=A, then A2+B2A^2+B^2 is equal to
View answer and explanation
Correct option: C
A+BA+B
Option C (A+BA+B) is correct because given AB=B and BA=A, we can show that A2 = A and B2 = B, therefore A2 + B2 = A + B.
560 Easy
560 of 922
Application +4 | -1
If A=[cos⁡θsin⁡θsin⁡θ−cos⁡θ],B=[10−11],C=ABATA=\left[\begin{array}{rr}\cos \theta & \sin \theta \\ \sin \theta & -\cos \theta\end{array}\right], B=\left[\begin{array}{rr}1 & 0 \\ -1 & 1\end{array}\right], C=A B A^T, then ATCnAA^T C^n A equals to (n∈I+)\left(n \in I^{+}\right)
View answer and explanation
Correct option: D
[10−n1]\left[\begin{array}{cc}1 & 0 \\ -n & 1\end{array}\right]
Option D (\left[\begin{{array}}{{cc}}1 & 0 \\ -n & 1\end{{array}}\right]\left[\begin{{array}}{{cc}}1 & 0 \\ -n & 1\end{{array}}\right]) is correct because by simplifying the expression A^T C^n A using the given matrices and properties, it reduces to B^n, which is equal to the given matrix.
561 Easy
561 of 922
Application +4 | -1
If AA is a 2×22 \times 2 matrix and II is an identity matrix of order 2 & ∣A−λI∣=02 \,\&\,|A-\lambda I|=0 gives value of λ\lambda as −1 & 3-1 \,\&\, 3 then trace of A2A^2 is equal to ____
View answer and explanation
Correct option: A
10
Option A (10) is correct because the trace of A2A^2 is the sum of the squares of the eigenvalues, which are 1 and 9, so the trace is 10.
562 Easy
562 of 922
Application +4 | -1
If P=[123234345][−1−2−200−4][−4−5−6001]P = \left[ {\begin{array}{cccccccccccccccccccc} 1&2&3\\ 2&3&4\\ 3&4&5 \end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc} { - 1}&{ - 2}\\ { - 2}&0\\ 0&{ - 4} \end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc} { - 4}&{ - 5}&{ - 6}\\ 0&0&1 \end{array}} \right], then P22{P_{22}} is equal to
View answer and explanation
Correct option: A
40
Option A is correct. The element P_{22} is found by multiplying the corresponding rows and columns of the matrices.
563 Easy
563 of 922
Application +4 | -1
If A=[1221]A=\left[\begin{array}{cc}1 & 2 \\ 2 & 1\end{array}\right] and f(x)=1+x1−xf(x)=\frac{1+x}{1-x}, then f(A)f(A) is
View answer and explanation
Correct option: C
[−1−1−1−1]\left[\begin{array}{ll}-1 & -1 \\ -1 & -1\end{array}\right]
Option C ([−1−1−1−1]\left[\begin{array}{ll}-1 & -1 \\ -1 & -1\end{array}\right]) is correct because by substituting the matrix A into the function f(x) and simplifying, we arrive at this result.
564 Easy
564 of 922
Analyzing +4 | -1
If Δ1=∣xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x∣{\Delta _1} = \left| {\begin{array}{cccccccccccccccccccc} x&{\sin \theta }&{\cos \theta }\\ { - \sin \theta }&{ - x}&1\\ {\cos \theta }&1&x \end{array}} \right| and Δ2=∣xsin⁡2θcos⁡2θ−sin⁡2θ−x1cos⁡2θ1x∣,  x≠0,{\Delta _2} = \left| {\begin{array}{cccccccccccccccccccc} x&{\sin 2\theta }&{\cos 2\theta }\\ { - \sin 2\theta }&{ - x}&1\\ {\cos 2\theta }&1&x \end{array}} \right|,\,\,x \ne 0, then for all θ∈(0,π2)\theta \in \left( {0,\frac{\pi }{2}} \right)
View answer and explanation
Correct option: C
Δ1+Δ2=−2x3{\Delta _1} + {\Delta _2} = - 2{x^3}
Option C (Δ1+Δ2=−2x3\Delta _1 + \Delta _2 = - 2{x^3}) is correct because the sum of the determinants simplifies to -2x3 after expansion.
565 Easy
565 of 922
Understanding +4 | -1
If AA is a non-singular matrix of order 3 , then ∣adj⁡A∣=∣A∣n|\operatorname{adj} A|=|A|^{\mathrm{n}}. Here the value of nn is
View answer and explanation
Correct option: A
2
Option A (2) is correct because for a non-singular matrix A of order 3, |adj A| = |A|^(3-1) = |A|^2.
566 Easy
566 of 922
Understanding +4 | -1
If AA is a singular matrix, then AA adj (A) is a
View answer and explanation
Correct option: B
Zero matrix
The correct answer is B (Zero matrix) because for a singular matrix A, its determinant is 0, and A adj(A) equals |A|I, which results in a zero matrix.
567 Easy
567 of 922
Application +4 | -1
If AA is square matrix of order n,a=n, a= maximum number of distinct entries. If AA is triangular matrix, b=b= maximum number of distinct entries. If AA is a diagonal matrix. c=c= minimum number of zeros. If AA is a triangular matrix if a+5=c+2ba+5=c+2 b
View answer and explanation
Correct option: A
4
Option A (4) is correct because solving the equation a+5=c+2ba+5=c+2b with the given definitions for a, b, and c results in n=4.
568 Easy
568 of 922
Analyzing +4 | -1
If a+x=b+y=c+z+1, a + x = b + y = c + z + 1,\, where a, b, c, x, y, za,\,b,\,c,\,x,\,y,\,z are non-zero distinct real numbers, then ∣xa+yx+ayb+yy+bzc+yz+c∣\left| {\begin{array}{cccccccccccccccccccc} x&{a + y}&{x + a}\\ y&{b + y}&{y + b}\\ z&{c + y}&{z + c} \end{array}} \right| is equal to
View answer and explanation
Correct option: B
y(a−b)y(a - b)
Option B (y(a−b)y(a - b)) is correct because the determinant can be simplified using properties of determinants and the given conditions to arrive at this expression.
569 Easy
569 of 922
Application +4 | -1
If A=[000100010]A=\left[\begin{array}{lll}0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right], then
View answer and explanation
Correct option: D
A3=0A^3=0
Option D (A3=0A^3=0) is correct because when the given matrix A is multiplied by itself three times, the result is a zero matrix.
570 Easy
570 of 922
Application +4 | -1
If a square matrix AA is such that AAT=I=ATA   then    ∣A∣A{A^T} = I = {A^T}A\,\,\,\mathrm{then}\,\,\,\,\left| A \right| then is equal to
View answer and explanation
Correct option: B
± 1\pm \,1
Option B (±1\pm 1) is correct because the determinant of a matrix multiplied by its transpose equals 1, implying the determinant of the matrix is either 1 or -1.
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