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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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451 Easy
451 of 922
Analyzing +4 | -1
If ∣xx21+x3yy21+y3zz21+z3∣=0\left| {\begin{array}{cccccccccccccccccccc} x&{{x^2}}&{1 + {x^3}}\\ y&{{y^2}}&{1 + {y^3}}\\ z&{{z^2}}&{1 + {z^3}} \end{array}} \right| = 0 and x,y,zx,y,z are all distinct, then xyzxyz is equal to
View answer and explanation
Correct option: A
−1- 1
The correct answer is A (-1). By splitting the determinant and using properties of determinants, we find that xyz = -1.
452 Easy
452 of 922
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.
453 Easy
453 of 922
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.
454 Easy
454 of 922
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.
455 Easy
455 of 922
Application +4 | -1
If
(1−tan⁡θtan⁡θ1)(1tan⁡θ−tan⁡θ1)−1=[a−bba],\left(\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right)\left(\begin{array}{cc} 1 & \tan \theta \\ -\tan \theta & 1 \end{array}\right)^{-1}=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right], then
View answer and explanation
Correct option: B
a=cos⁡2θ,b=sin⁡2θa=\cos 2 \theta, b=\sin 2 \theta
Option B is correct because the given expression simplifies to a rotation matrix, where a=cos⁡2θa = \cos 2\theta and b=sin⁡2θb = \sin 2\theta.
456 Easy
456 of 922
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.
457 Easy
457 of 922
Analyzing +4 | -1
The system of equations
kx+y+z=1x+ky+z=kx+y+kz=k2\begin{array}{l} kx + y + z = 1\\ x + ky + z = k\\ x + y + kz = {k^2} \end{array}
have no solution, if kk equals
View answer and explanation
Correct option: D
−2- 2
Option D (-2) is correct because the system has no solution when k = -2, which makes the determinant of the coefficients matrix zero and at least one of the other determinants non-zero.
458 Easy
458 of 922
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.
459 Easy
459 of 922
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.
460 Easy
460 of 922
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].
461 Easy
461 of 922
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.
462 Easy
462 of 922
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.
463 Easy
463 of 922
Application +4 | -1
If A=[i−i−ii]A=\left[\begin{array}{cc}i & -i \\ -i & i\end{array}\right] and B=[1−1−11]B=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right], then A8A^8 equals
View answer and explanation
Correct option: A
128 B128 \mathrm{~B}
Option A (128B) is correct because after repeated squaring of matrix A and simplification, A8A^8 equals 128B.
464 Easy
464 of 922
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.
465 Easy
465 of 922
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.
466 Easy
466 of 922
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.
467 Easy
467 of 922
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.
468 Easy
468 of 922
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.
469 Easy
469 of 922
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.
470 Easy
470 of 922
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.
471 Easy
471 of 922
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.
472 Easy
472 of 922
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.
473 Easy
473 of 922
Analyzing +4 | -1
Let for A=[123α31112],∣A∣=2A=\left[\begin{array}{lll}1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2\end{array}\right],|A|=2. If ∣2adj⁡(2adj⁡(2A))∣|2 \operatorname{adj}(2 \operatorname{adj}(2 A))| =32n=32^{n}, then 3n+α3 n+\alpha is equal to
View answer and explanation
Correct option: B
11
The correct answer is B (4) because it is derived by using the properties of adjoint and determinant of a matrix.
474 Easy
474 of 922
Application +4 | -1
Let Ak=[cos⁡2αkcos⁡αksin⁡αkcos⁡αksin⁡αksin⁡2αk](k=1,2)A_k=\left[\begin{array}{cc} \cos ^2 \alpha_k & \cos \alpha_k \sin \alpha_k \\ \cos \alpha_k \sin \alpha_k & \sin ^2 \alpha_k \end{array}\right] \quad(k=1,2)
If the difference between α1\alpha_1 and α2\alpha_2 is an odd multiple of π/2\pi / 2, then A1A2A_1 A_2 is a matrix whose sum of all its elements is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the product A1A2A_1 A_2 results in a zero matrix when the difference between α1\alpha_1 and α2\alpha_2 is an odd multiple of π/2\pi / 2.
475 Easy
475 of 922
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.
476 Easy
476 of 922
Analyzing +4 | -1
If ω≠1\omega \neq 1 is a cube root of unity, find the value of ∣1ωω2ω3ωω2ω31ω2ω31ωω31ωω2∣\left|\begin{array}{cccc}1 & \omega & \omega^2 & \omega^3 \\ \omega & \omega^2 & \omega^3 & 1 \\ \omega^2 & \omega^3 & 1 & \omega \\ \omega^3 & 1 & \omega & \omega^2\end{array}\right|
View answer and explanation
Correct option: A
±i33\pm i3 \sqrt{3}
Option A (±i33\pm i3 \sqrt{3}) is correct because after simplifying the determinant using properties of cube roots of unity, the result is an imaginary number.
477 Easy
477 of 922
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.
478 Easy
478 of 922
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.
479 Easy
479 of 922
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.
480 Easy
480 of 922
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.
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