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JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

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481 Easy
481 of 995
Application +4 | -1
If a square matrix AA is such that AAT=I=ATAA{A^T} = I = {A^T}A, then ∣A∣\left| A \right| is equal to
View answer and explanation
Correct option: B
±1\pm 1
Option B (±1\pm 1) is correct because the determinant of the given matrix satisfies the condition ∣A∣2=1\left| A \right|^2 = 1, which implies ∣A∣=±1\left| A \right| = \pm 1.
482 Easy
482 of 995
Application +4 | -1
If A=[3    3    33    3    33    3    3],  A4  A = \left[ {\begin{array}{cccccccccccccccccccc} {3\;\;3\;\;3}\\ {3\;\;3\;\;3}\\ {3\;\;3\;\;3} \end{array}} \right],\;{A^4}\; is equal to
View answer and explanation
Correct option: D
729 A
Option D (729 A) is correct because it is the result of repeatedly multiplying the matrix A by itself.
483 Easy
483 of 995
Analyzing +4 | -1
If (1+ax+bx2)4=a0+a1x+a2x2+…+a8x8\left(1+a x+b x^2\right)^4=a_0+a_1 x+a_2 x^2+\ldots+a_8 x^8, where a,b,a0,a1,…,a8∈Ra, b, a_0, a_1, \ldots, a_8 \in R such that a0+a1+a2≠0a_0+a_1+a_2 \neq 0 and ∣a0a1a2a1a2a0a2a0a1∣=0\left|\begin{array}{lll}a_0 & a_1 & a_2 \\ a_1 & a_2 & a_0 \\ a_2 & a_0 & a_1\end{array}\right|=0, then
View answer and explanation
Correct option: B
a=14,b=532a=\frac{1}{4}, b=\frac{5}{32}
The correct answer is B (a=14,b=532a=\frac{{1}}{{4}}, b=\frac{{5}}{{32}}) because substituting the given conditions into the determinant equation leads to these values for a and b.
484 Easy
484 of 995
Understanding +4 | -1
Let AA and BB are two square matrices such that AB=A  AB = A\; and   BA=B,  \;BA = B,\; then A2{A^2} equals to
View answer and explanation
Correct option: B
AA
Option B (A) is correct because given AB = A and BA = B, substituting BA for B in AB = A leads to ABA = A, which simplifies to AA = A, thus A2 = A.
485 Easy
485 of 995
Understanding +4 | -1
If a matrix has 8 elements, then which of the following will not be a possible order of the matrix?
View answer and explanation
Correct option: D
4×44 \times 4
Option D (4×44 \times 4) is correct because a 4×44 \times 4 matrix has 16 elements, not 8.
486 Easy
486 of 995
Analyzing +4 | -1
Let a,  b,  ca,\;b,\;c are positive real numbers. The following system of equations x2a2+y2b2−z2c2=1,x2a2−y2b2+z2c2=1,−x2a2+y2b2+z2c2=1,\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1, - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1, in x,y  x,y\; and zz has
View answer and explanation
Correct option: B
Unique solution
Option B (Unique solution) is correct because the determinant of the coefficient matrix is non-zero, indicating a unique solution exists.
487 Easy
487 of 995
Analyzing +4 | -1
If θ,ϕ ε R\theta, \phi \,\varepsilon \,{R}, then the determinant Δ=∣cos⁡θ−sin⁡θ1sin⁡θcos⁡θ1cos⁡(θ+ϕ)−sin⁡(θ+ϕ)0∣\Delta=\left|\begin{array}{ccc} \cos \theta & -\sin \theta & 1 \\ \sin \theta & \cos \theta & 1 \\ \cos (\theta+\phi) & -\sin (\theta+\phi) & 0 \end{array}\right| lies in the interval
View answer and explanation
Correct option: A
[−2,2][-\sqrt{2}, \sqrt{2}]
Option A ([−2,2][-\sqrt{2}, \sqrt{2}]) is correct because the determinant simplifies to 2sin⁡(ϕ−π/4)\sqrt{2} \sin (\phi-\pi / 4), and the sine function's range is [-1, 1].
488 Easy
488 of 995
Analyzing +4 | -1
Let A=[1    23    4]A = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;2}\\ {3\;\;4} \end{array}} \right] and   B=[  a    00    b]\;B = \left[ {\begin{array}{cccccccccccccccccccc} {\;a\;\;0}\\ {0\;\;b} \end{array}} \right], a,b  N.a,b\;N. then
View answer and explanation
Correct option: C
There exists infinitely many B′sB's such that AB=BAAB = BA
Option C (There exists infinitely many B′sB's such that AB=BAAB = BA) is correct because if a=ba = b, then AB=BAAB = BA, and there are infinitely many possible values for aa and bb in the set of natural numbers.
489 Easy
489 of 995
Understanding +4 | -1
If AA and BB are square matrices of size n×nn \times n such that A2−B2=(A−B)(A+B),{A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right), then which of the following will be always true ?
View answer and explanation
Correct option: A
AB=BAAB = BA
Option A (AB=BAAB = BA) is correct because the given equation expands to show that AB must equal BA.
490 Easy
490 of 995
Application +4 | -1
If A=[0−ii0],B=[100−1]A=\left[\begin{array}{cc}0 & -i \\ i & 0\end{array}\right], B=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right], then AB+BAA B+B A is
View answer and explanation
Correct option: A
Null matrix
Option A (Null matrix) is correct because when AB and BA are calculated and added, the result is a zero matrix.
491 Easy
491 of 995
Application +4 | -1
If AA is non-singular and (A−2I)(A−4I)=O(A-2 I)(A-4 I)=O, then 16A+43A−1=\frac{1}{6} A+\frac{4}{3} A^{-1}=
View answer and explanation
Correct option: A
II
Option A (II) is correct because simplifying the given equation (A−2I)(A−4I)=O(A-2I)(A-4I)=O and multiplying by A−1A^{-1} leads to 16A+43A−1=I\frac{1}{6}A + \frac{4}{3}A^{-1} = I.
492 Medium
492 of 995
Application +4 | -1
The value of ∣xpqpxqpqx∣\left| {\begin{array}{cccccccccccccccccccc} x&p&q\\ p&x&q\\ p&q&x \end{array}} \right| is
View answer and explanation
Correct option: B
(x−p)(x−q)(x+p+q)\left( {x - p} \right)\left( {x - q} \right)\left( {x + p + q} \right)
Option B ((x - p)(x - q)(x + p + q)) is correct because by applying row operations and factoring, the determinant simplifies to this expression.
493 Medium
493 of 995
Application +4 | -1
The integer represented by the determinant [215342511678364954]\left[ {\begin{array}{cccccccccccccccccccc} {215}&{342}&{511}\\ 6&7&8\\ {36}&{49}&{54} \end{array}} \right] is exactly divisible by
View answer and explanation
Correct option: C
20
Option C (20) is correct because after calculating the determinant, the resulting integer (6140) is exactly divisible by 20.
494 Easy
494 of 995
Application +4 | -1
If A=[aij]4×4A=\left[a_{i j}\right]_{4 \times 4}, such that aij={2, when i=j0, when i≠ja_{i j}=\left\{\begin{array}{l}2, \text { when } i=j \\ 0, \text { when } i \neq j\end{array}\right., then {det⁡(adj⁡(adj⁡A))7}\left\{\frac{\operatorname{det}(\operatorname{adj}(\operatorname{adj} A))}{7}\right\} is [when {⋅}\{\cdot\} represents fractional part function]
View answer and explanation
Correct option: A
17\frac{1}{7}
Option A (17\frac{{1}}{{7}}) is correct because it is the fractional part of the given expression after calculating the determinant.
495 Easy
495 of 995
Analyzing +4 | -1
If A=[2454810−6−12−15],  A = \left[ {\begin{array}{cccccccccccccccccccc} 2&4&5\\ 4&8&{10}\\ { - 6}&{ - 12}&{ - 15} \end{array}} \right],\; the rank of AA is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because after row operations, the matrix has only one non-zero row, indicating a rank of 1.
496 Easy
496 of 995
Application +4 | -1
If A5=OA^5=O such that An≠IA^n \neq I for 1≤n≤41 \leq n \leq 4, then (I−A)−1(I-A)^{-1} is equal to
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because none of the given options satisfy the condition (I-A)-1 when A5 = O and A^n != I for 1 <= n <= 4.
497 Medium
497 of 995
Application +4 | -1
If A=[1201]A=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\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] and P=BTABP=B^T A B then BP2016BTB P^{2016} B^T is equal to
View answer and explanation
Correct option: B
[1403201]\left[\begin{array}{cc}1 & 4032 \\ 0 & 1\end{array}\right]
Option B (\left[ {\begin{{array}}{{cc}} 1 & 4032 \\ 0 & 1 \end{{array}}} \right]\left[ {\begin{{array}}{{cc}} 1 & 4032 \\ 0 & 1 \end{{array}}} \right]) is correct because after simplifying the matrix expression, the result is a matrix with 1, 4032, 0, and 1 as its elements.
498 Easy
498 of 995
Application +4 | -1
Let B=[264101−11−1]B=\left[\begin{array}{rrr}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]. If the product BAB A has inverse C=[−101113202]C=\left[\begin{array}{rrr}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right], then A−1A^{-1} equals
View answer and explanation
Correct option: C
[−3−5−50022146]\left[\begin{array}{rrr}-3 & -5 & -5 \\ 0 & 0 & 2 \\ 2 & 14 & 6\end{array}\right]
Option A (\left[\begin{{array}}{{rrr}}-3 & -5 & -5 \\ 0 & 0 & 2 \\ 2 & 14 & 6\end{{array}}\right]\left[\begin{{array}}{{rrr}}-3 & -5 & -5 \\ 0 & 0 & 2 \\ 2 & 14 & 6\end{{array}}\right]) is correct because it is the result of multiplying matrix C by matrix B.
499 Easy
499 of 995
Application +4 | -1
Suppose AA is square matrix such that A3=IA^3=I, then (A+I)3+(A−I)3−6A(A+I)^3+(A-I)^3-6 A equals
View answer and explanation
Correct option: B
2I2 I
Option B (2I2 I) is correct because simplifying the expression (A+I)3+(A−I)3−6A(A+I)^3+(A-I)^3-6 A using A3=IA^3=I results in 2I2I.
500 Easy
500 of 995
Application +4 | -1
There are two numbers xx making the value of the determinant ∣1−252x−1042x∣\left|\begin{array}{ccc}1 & -2 & 5 \\ 2 & \mathrm{x} & -1 \\ 0 & 4 & 2 \mathrm{x}\end{array}\right| equal to 86 . The sum of these two numbers, is
View answer and explanation
Correct option: A
-4
Option A, -4, is the correct answer because it represents the sum of the roots of the quadratic equation derived from the determinant condition.
501 Easy
501 of 995
Application +4 | -1
Let f(α)=[cos⁡α−sin⁡α0sin⁡αcos⁡α0001],f\left( \alpha \right) = \left[ {\begin{array}{cccccccccccccccccccc} {\cos \alpha }&{ - \sin \alpha }&0\\ {\sin \alpha }&{\cos \alpha }&0\\ 0&0&1 \end{array}} \right], where α∈R.\alpha \in R. Then, (F(α))−1{\left( {F\left( \alpha \right)} \right)^{ - 1}} is equal to
View answer and explanation
Correct option: A
F(−α)F\left( { - \alpha } \right)
Option A (F(−α)F(-α)) is correct because the inverse of F(α) is found by replacing α with -α, which results in the identity matrix when multiplied.
502 Easy
502 of 995
Application +4 | -1
The number of the solutions of the equation log⁡(x2−\log \left(x^2-\right. 6x+7)=log⁡(x−3)6 x+7)=\log (x-3) is
View answer and explanation
Correct option: B
5
Option A (5) is incorrect. The equation log⁡(x2−6x+7)=log⁡(x−3)\log(x^2-6x+7) = \log(x-3) has only one solution, x=5, after considering the domain restrictions.
503 Easy
503 of 995
Understanding +4 | -1
Let A{A} be a 2×22 \times 2 matrix

Assertion: adj⁡(adj⁡A)=A\operatorname{adj}(\operatorname{adj} A)=A

Reason: ∣adj⁡A∣=∣A∣|\operatorname{adj} {A}|=|{A}|
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Option B (Assertion is correct, reason is correct; reason is not a correct explanation for assertion) is correct because both statements are true, but the reason doesn't explain the assertion.
504 Easy
504 of 995
Analyzing +4 | -1
Let N{N} denote the number that turns up when a fair die is rolled. If the probability that the system of equations

x+y+z=12x+Ny+2z=23x+3y+Nz=3\begin{array}{l} x+y+z=1 \\ 2 x+N y+2 z=2 \\ 3 x+3 y+N z=3 \end{array}

has unique solution is k6\frac{k}{6}, then the sum of value of kk and all possible values of NN is
View answer and explanation
Correct option: B
20
Option B (20) is correct because it calculates the sum of k and all possible values of N based on the determinant condition for a unique solution.
505 Easy
505 of 995
Understanding +4 | -1
If A\mathrm{A} and B\mathrm{B} are two matrices such that A+B\mathrm{A}+\mathrm{B} and AB\mathrm{AB} are both defined, then
View answer and explanation
Correct option: B
A\mathrm{A} and B\mathrm{B} are square matrices of same order
Option B (A and B are square matrices of same order) is correct because for A+B and AB to be defined, both matrices must be square and have the same order.
506 Easy
506 of 995
Application +4 | -1
Matrix AA such that A2=2A−I{A^2} = 2A - I, where II is the identity matrix. Then, for n≥2,  Ann \ge 2,\;{A^n} is equal to
View answer and explanation
Correct option: A
nA−(n−1)InA - \left( {n - 1} \right)I
Option A (nA−(n−1)InA - \left( {n - 1} \right)I) is correct because it represents the general form of An{A^n} derived through inductive reasoning, starting from the given relation A2=2A−I{A^2} = 2A - I.
507 Easy
507 of 995
Analyzing +4 | -1
The number of values of kk for which the linear equations 4x+ky+2z=0,   kx+4y+z=0,  2x+2y+z=04x + ky + 2z = 0,\,\,\,kx + 4y + z = 0,\,\,2x + 2y + z = 0 Possess a non - zero solutions is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the determinant of the coefficient matrix must be zero for non-zero solutions, leading to a quadratic equation with two solutions for kk.
508 Medium
508 of 995
Analyzing +4 | -1
If Δr=∣rxn(n+1)22r−1yn23r−2zn(3n−1)2∣,\Delta_r=\left|\begin{array}{ccc}r & x & \frac{n(n+1)}{2} \\ 2 r-1 & y & n^2 \\ 3 r-2 & z & \frac{n(3 n-1)}{2}\end{array}\right|, then ∑r=1nΔr\sum\limits_{r = 1}^n {{\Delta _r}} is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because the sum of the determinants results in a determinant with two identical columns, making its value zero.
509 Easy
509 of 995
Analyzing +4 | -1
Let A=[1432].A=\left[\begin{array}{ll}1 & 4 \\ 3 & 2\end{array}\right]. If θ\theta is the angle between the two non-zero column vectors XX such that AX=λXA X=\lambda X for some scalar λ,\lambda, then tan⁡θ\tan \theta is equal to
View answer and explanation
Correct option: C
7
Option C (7) is correct because after finding the eigenvalues and corresponding eigenvectors, the cosine of the angle between the eigenvectors is used to find the tangent.
510 Easy
510 of 995
Understanding +4 | -1
AA is a square matrix of order 4 and II is a unit matrix, then it is true that
View answer and explanation
Correct option: B
det (2A)=16\left( {2A} \right) = 16 det (A)\left( A \right)
Option B is correct because when a constant multiplies a matrix inside a determinant, the constant is raised to the power of the matrix's order.
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