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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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271 Easy
271 of 922
Application +4 | -1
If C=2cos⁡θ,\mathrm{C}=2 \cos \theta, then the value of the determinant Δ=∣C101C161C∣,\Delta=\left|\begin{array}{lll}C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C\end{array}\right|, is
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because the determinant evaluates to 8cos⁡3θ−4cos⁡θ+68 \cos^3 \theta - 4 \cos \theta + 6, which doesn't match any of the given options.
272 Easy
272 of 922
Application +4 | -1
If AA and BB are two square matrix such that A2B=BAA^{2} B=B A and if (AB)10=AkB10(A B)^{10}=A^{k} B^{10}, then kk is
View answer and explanation
Correct option: B
1023
Option B (1023) is correct because the pattern (AB)n=A(2n−1)Bn(AB)^n = A^{(2^n - 1)}B^n leads to k=210−1=1023k = 2^{10} - 1 = 1023 for n=10.
273 Easy
273 of 922
Analyzing +4 | -1
Study the following statement and select the correct option:
Statement-I: If a,b,ca, b, c are in A.P the system of equations 3x+4y+5z=a,4x+5y+6z=b,5x+6y+7z=c3 x+4 y+5 z=a, 4 x+5 y+6 z=b, 5 x+6 y+7 z=c is consistent

Statement-II: If ∣A∣≠0|A| \neq 0, the system of equations AX=BA X=B is consistent
View answer and explanation
Correct option: B
Statement-1 is true, statement-2 is true, statement-2 is not correct explanation of statement-1
Option B (Statement-1 is true, statement-2 is true, statement-2 is not correct explanation of statement-1) is correct because both statements are individually true, but statement-2 doesn't explain statement-1.
274 Easy
274 of 922
Application +4 | -1
Let A=(2−102)A=\left(\begin{array}{cc}2 & -1 \\ 0 & 2\end{array}\right). If B=I−5C1(adj⁡A)+5C2(adj⁡A)2−B=I-{ }^5 C_1(\operatorname{adj} A)+{ }^5 C_2(\operatorname{adj} A)^2- .....−5C5(adj⁡A)5-{ }^5 C_5(\operatorname{adj} A)^5, then the sum of all elements of the matrix BB is
View answer and explanation
Correct option: C
−7-7
Option C (-7) is correct because after simplifying the expression and calculating the matrix power, the sum of the elements of the resulting matrix is -7.
275 Easy
275 of 922
Application +4 | -1
Let At=(13225t47−t−6)A_t=\left(\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6\end{array}\right), then the value(s) of tt for which inverse of AtA_t does not exist.
View answer and explanation
Correct option: C
2,−32, -3
Option C (2, -3) is correct because the inverse of At does not exist when the determinant of At is zero, which occurs at t = 2 and t = -3.
276 Easy
276 of 922
Application +4 | -1
The matrix [101210311]\left[\begin{array}{lll}1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 1 & 1\end{array}\right] is
View answer and explanation
Correct option: B
singular
Option B (singular) is correct because the determinant of the matrix is 0, indicating that the matrix is singular.
277 Easy
277 of 922
Application +4 | -1
If aij=(i−2j)23a_{i j}=\frac{(i-2 j)^2}{3}, then A=[aij]2×2=\mathrm{A}=\left[a_{i j}\right]_{2 \times 2}=
View answer and explanation
Correct option: A
[133043]\left[\begin{array}{ll}\frac{1}{3} & 3 \\ 0 & \frac{4}{3}\end{array}\right]
Option A is correct because the matrix A is constructed by substituting the given values of i and j into the formula and calculating each element.
278 Easy
278 of 922
Application +4 | -1
If α\alpha is a complex root of the equation x2+x+1=0x^2+x+1=0 and A=13[1111αα21α2α]A=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha\end{array}\right], then A31A^{31} is equal to
View answer and explanation
Correct option: C
A3A^3
Option C (A3A^3) is correct because, after calculating A², we find that A⁴ = I, which simplifies A³¹ to A³.
279 Easy
279 of 922
Analyzing +4 | -1
If the system of equations 2x−y+z=0,x−2y+z=02 x-y+z=0, x-2 y+z=0, tx−y+2z=0t x-y+2 z=0 has infinitely many solutions and f(x)f(x) be a continuous function, such that f(5+x)+f(x)=2f(5+x)+f(x)=2, then ∫0−2tf(x)dx\int_0^{-2 t} f(x) d x is equal to
View answer and explanation
Correct option: B
−2t-2 t
The correct answer is B (-2t) because solving the system of equations gives t=5, and the integral evaluates to -10, which is -2t.
280 Easy
280 of 922
Analyzing +4 | -1
If a0,a1,a2,a3,a4a_0, a_1, a_2, a_3, a_4 are in A.P. with the common difference dd,
the value of ∣a1a2a1a0a2a3a2a1a3a4a3a2∣\left|\begin{array}{lll}a_1 a_2 & a_1 & a_0 \\ a_2 a_3 & a_2 & a_1 \\ a_3 a_4 & a_3 & a_2\end{array}\right| is
View answer and explanation
Correct option: A
2d42 d^4
Option A is correct because by applying properties of arithmetic progressions and determinant simplification, the value of the determinant is found to be 2d42d^4.
281 Easy
281 of 922
Application +4 | -1
If A=[122230012]A = \left[ {\begin{array}{cccccccccccccccccccc} 1&2&2\\ 2&3&0\\ 0&1&2 \end{array}} \right] and adj  A=[6−2−6−42xy−1−1],\mathrm{adj}\;A = \left[ {\begin{array}{cccccccccccccccccccc} 6&{ - 2}&{ - 6}\\ { - 4}&2&x\\ y&{ - 1}&{ - 1} \end{array}} \right], then x+y=x + y =
View answer and explanation
Correct option: A
6
Option A (6) is correct because adj(A) * A = |A| * I. Calculating |A| and performing the multiplication allows us to solve for x and y, then x + y.
282 Easy
282 of 922
Application +4 | -1
If X=[3−41−1]X = \left[ {\begin{array}{cccccccccccccccccccc} 3&{ - 4}\\ 1&{ - 1} \end{array}} \right], then the value of Xn{X^n} is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because none of the given options satisfy the condition when n=2 after calculating X2.
283 Easy
283 of 922
Understanding +4 | -1
Let AA be a skew-symmetric matrix of order nn then
View answer and explanation
Correct option: B
∣A∣=0|A|=0 if nn is odd
Option B (∣A∣=0|A|=0 if nn is odd) is correct because the determinant of a skew-symmetric matrix of odd order is always zero.
284 Easy
284 of 922
Application +4 | -1
If A=[3201]A=\left[\begin{array}{ll}3 & 2 \\ 0 & 1\end{array}\right], then A−3A^{-3} is
View answer and explanation
Correct option: C
127[1−26027]\frac{1}{27}\left[\begin{array}{cc}1 & -26 \\ 0 & 27\end{array}\right]
Option C (127[1−26027]\frac{1}{27}\left[\begin{array}{cc}1 & -26 \\ 0 & 27\end{array}\right]) is correct because by calculating A-1 and then cubing it, we arrive at the given matrix.
285 Easy
285 of 922
Remembering +4 | -1
If A=[1  −5        70          7        911          8      9]A = \left[ {\begin{array}{cccccccccccccccccccc} {1\; - 5\;\;\;\;7}\\ {0\;\;\;\;\;7\;\;\;\;9}\\ {11\;\;\;\;\;8\;\;\;9} \end{array}} \right], then trace of matrix AA is
View answer and explanation
Correct option: A
17
Option A (17) is correct because the trace of a matrix is the sum of its diagonal elements, which is 1 + 7 + 9 = 17.
286 Easy
286 of 922
Application +4 | -1
If A=[n000n000n]A=\left[\begin{array}{lll}n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n\end{array}\right] and B=[a1a2a3b1b2b3c1c2c3]B=\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right], then ABA B equal to
View answer and explanation
Correct option: B
nBn B
Option B (nBnB) is correct because multiplying a matrix B by a scalar matrix A (nI) results in each element of B being multiplied by n.
287 Easy
287 of 922
Application +4 | -1
If A=[1111]A=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right], then A100=A^{100}=
View answer and explanation
Correct option: A
299A2^{99} {A}
Option A is correct. By observing the pattern of A raised to increasing powers, it becomes clear that A(100) = 2(99) * A.
288 Easy
288 of 922
Application +4 | -1
Let X=[x1x2x3],  A=[1−12201321]X = \left[ {\begin{array}{cccccccccccccccccccc} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right],\;A = \left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 1}&2\\ 2&0&1\\ 3&2&1 \end{array}} \right] and B=[314]B = \left[ {\begin{array}{cccccccccccccccccccc} 3\\ 1\\ 4 \end{array}} \right]. If AX=BAX = B, then XX is equal to
View answer and explanation
Correct option: D
[−123]\left[ {\begin{array}{cccccccccccccccccccc} { - 1}\\ 2\\ 3 \end{array}} \right]
Option D ([-1], [2], [3]) is correct because it is the solution to the matrix equation AX = B, obtained by finding the inverse of A and multiplying it by B.
289 Easy
289 of 922
Understanding +4 | -1
If AA and BB are 2×22 \times 2 matrices, then which of the following is true?
View answer and explanation
Correct option: C
(A−B)(A+B)=A2+AB−BA−B2(\mathrm{A}-\mathrm{B})(\mathrm{A}+\mathrm{B})=\mathrm{A}^2+\mathrm{AB}-\mathrm{BA}-\mathrm{B}^2
Option C, (A−B)(A+B)=A2+AB−BA−B2(\mathrm{A}-\mathrm{B})(\mathrm{A}+\mathrm{B})=\mathrm{A}^2+\mathrm{AB}-\mathrm{BA}-\mathrm{B}^2, is correct because matrix multiplication is not commutative, so AB is not necessarily equal to BA.
290 Easy
290 of 922
Application +4 | -1
Let A=[1000110−24]A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right]

If
A−1=16(A2+xA+yI)A^{-1}=\frac{1}{6}\left(A^2+x A+y I\right) where x,yx, y are scalars and II is 3×33 \times 3 unit matrix, then
x,yx, y are, respectively
View answer and explanation
Correct option: B
−6,11-6,11
Option B (-6, 11) is correct because it is found by using the Cayley-Hamilton theorem to find the inverse of the matrix A.
291 Easy
291 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, and therefore is not invertible.
292 Easy
292 of 922
Understanding +4 | -1
Consider the following statements

I. For multiplication of two matrices AA and BB, the number of columns in AA should be less than the number of rows in B{B}.

II. For getting the elements of the product matrix, we take rows of AA and column of BB, multiply them elementwise and take the sum.

Choose the correct option.
View answer and explanation
Correct option: B
Only II is true
Option B (Only II is true) is correct because for matrix multiplication, the number of columns in the first matrix must equal the number of rows in the second.
293 Easy
293 of 922
Application +4 | -1
If A3=0\mathrm{A}^3=0 then I+A+A2\mathrm{I}+\mathrm{A}+\mathrm{A}^2 equals:-
View answer and explanation
Correct option: B
(I−A)−1({I}-{A})^{-1}
Option B (I−A)−1({I}-{A})^{-1} is correct because it utilizes the property that if A3 = 0, then the inverse of (I-A) is I + A + A2.
294 Easy
294 of 922
Analyzing +4 | -1
If AA and BB are square matrices such that A2006=0A^{2006}=0 and AB=A+B,A B=A+B, then det⁡(B)\operatorname{det}(B) equals to
View answer and explanation
Correct option: B
0
Option B (0) is correct because given A2006 = 0, det(A) = 0, which leads to det(B) = 0.
295 Easy
295 of 922
Application +4 | -1
If A=[2225]{A}=\left[\begin{array}{ll}2 & 2 \\ 2 & 5\end{array}\right] and B=[21−12]{B}=\left[\begin{array}{cc}2 & 1 \\ -1 & 2\end{array}\right], then B−1AB={B}^{-1} {AB}=
View answer and explanation
Correct option: A
[1006]\left[\begin{array}{ll}1 & 0 \\ 0 & 6\end{array}\right]
Option A ([1006]\left[\begin{array}{ll}1 & 0 \\ 0 & 6\end{array}\right]) is correct because calculating B−1ABB^{-1}AB results in this diagonal matrix.
296 Easy
296 of 922
Understanding +4 | -1
Choose the correct answer in the following questions : If A,BA, B are symmetric matrices of same order then AB−BA\mathrm{AB}-\mathrm{BA} is a
View answer and explanation
Correct option: A
Skew-symmetric matrix
Option A (Skew-symmetric matrix) is correct because if A and B are symmetric, then (AB - BA)' = -(AB - BA), which is the definition of a skew-symmetric matrix.
297 Easy
297 of 922
Understanding +4 | -1
If I=(1001),J=(01−10)I=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right), J=\left(\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right), and B=(cos⁡θsin⁡θ−sin⁡θcos⁡θ)B=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right), then BB equals
View answer and explanation
Correct option: A
(cos⁡θ)I+(sin⁡θ)J(\cos \theta) I+(\sin \theta) J
Option A ((cos⁡θ\cos \theta)I + (sinθsin \theta)J) is correct because it accurately represents the given matrix B as a linear combination of the identity matrix I and the matrix J.
298 Easy
298 of 922
Application +4 | -1
The matrix A=[1212−12−12]A = \left[ {\begin{array}{cccccccccccccccccccc} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\ { - \frac{1}{{\sqrt 2 }}}&{ - \frac{1}{{\sqrt 2 }}} \end{array}} \right] is
View answer and explanation
Correct option: C
Nilpotent
Option C (Nilpotent) is correct because A squared equals the zero matrix, which is the definition of a nilpotent matrix.
299 Easy
299 of 922
Application +4 | -1
If [2174]A[−325−3]=[1001]\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right] A\left[\begin{array}{rr}-3 & 2 \\ 5 & -3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right], then matrix AA equals
View answer and explanation
Correct option: A
[75−11−8]\left[\begin{array}{cc}7 & 5 \\ -11 & -8\end{array}\right]
Option A ([75−11−8]\left[\begin{array}{cc}7 & 5 \\ -11 & -8\end{array}\right]) is correct because it is the result of calculating the inverse matrices and performing the matrix multiplication.
300 Easy
300 of 922
Application +4 | -1
If P=[3212−1232],A=[1101]{P}=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], {A}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] and Q=PAP⊤{Q}={PA} {P}^{\top}, then P⊤(Q2005)P{P}^{\top}\left({Q}^{2005}\right) {P} is equal to
View answer and explanation
Correct option: A
[1200501]\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]
Option A ([1200501]\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]) is correct because by using the properties of matrix transpose and repeated multiplication, we can simplify the expression to obtain the given matrix.
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