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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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Showing 841-870 of 922 questions

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841 Easy
841 of 922
Understanding +4 | -1
Consider the following statements

I. Matrix cannot be reduced to a number.

II. Determinant can be reduced to a number.
View answer and explanation
Correct option: C
Both I and II are true
Option C (Both I and II are true) is correct because a matrix is an arrangement of numbers, while a determinant can be reduced to a single number.
842 Easy
842 of 922
Analyzing +4 | -1
The sum of the real roots of the equation ∣x−6−12−3xx−3−32xx+2∣=0,\left| {\begin{array}{cccccccccccccccccccc} x&{ - 6}&{ - 1}\\ 2&{ - 3x}&{x - 3}\\ { - 3}&{2x}&{x + 2} \end{array}} \right| = 0, is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of the real roots of the given cubic equation, obtained by expanding the determinant, is 0.
843 Easy
843 of 922
Application +4 | -1
If A=[12−1−1122−11]A = \left[ {\begin{array}{cccccccccccccccccccc} 1&2&{ - 1}\\ { - 1}&1&2\\ 2&{ - 1}&1 \end{array}} \right] , then det [adj(adj  A)]\left[ {\mathrm{adj}\left( {\mathrm{adj\;}A} \right)} \right] is equal to
View answer and explanation
Correct option: C
144{14^4}
Option C (144{14^4}) is the correct answer because it follows from the properties of adjoints and determinants of matrices.
844 Easy
844 of 922
Understanding +4 | -1
Let A,B\mathrm{A}, \mathrm{B} and C\mathrm{C} are three matrices of same order. Now, consider the following statements
I. If A=B\mathrm{A}=\mathrm{B}, then AC=BC\mathrm{AC}=\mathrm{BC}
II. If AC=BC\mathrm{AC}=\mathrm{BC}, then A=B\mathrm{A}=\mathrm{B}
Choose the correct option
View answer and explanation
Correct option: A
Only I is true
Option A (Only I is true) is correct because if A=B, then multiplying both sides by C will result in AC=BC. However, the converse is not always true.
845 Easy
845 of 922
Understanding +4 | -1
Consider the following statements

I. Scalar matrix A=[aij]={k;i=j0;i≠jA=\left[a_{i j}\right]=\left\{\begin{array}{ll}k ; & i=j \\ 0 ; & i \neq j\end{array}\right. where kk is a scalar, in an identity matrix when k=1{k}=1.

II. Every identity matrix is not a scalar matrix.

Choose the correct option.
View answer and explanation
Correct option: A
Only I is true
Option A (Only I is true) is correct because a scalar matrix with k=1 is an identity matrix, but not every identity matrix is *not* a scalar matrix.
846 Easy
846 of 922
Understanding +4 | -1
Assertion : [300040007]\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 7\end{array}\right] is a diagonal matrix.

Reason: A=[aij]A=\left[a_{i j}\right] is a square matrix such that aij=0,∀i≠ja_{i j}=0, \forall i \neq j, then A{A} is called diagonal 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 about diagonal matrices.
847 Easy
847 of 922
Analyzing +4 | -1
If AA and BB are two non-zero n×nn \times n matrics such that A2+B=A2BA^{2}+B=A^{2} B, then
View answer and explanation
Correct option: A
A2B=BA2A^{2} B=B A^{2}
Option A (A2B=BA2A^{2} B=B A^{2}) is correct because it is derived from the given equation by algebraic manipulation and using the properties of matrices.
848 Easy
848 of 922
Application +4 | -1
If Aα=[cos⁡αsin⁡α−sin⁡αcos⁡α]A_\alpha=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right], then AαAβA_\alpha A_\beta is equal to
View answer and explanation
Correct option: A
Aα+βA_{\alpha+\beta}
Option A (Aα+βA_{\alpha+\beta}) is correct because the product of the two rotation matrices results in another rotation matrix with an angle equal to the sum of the individual angles.
849 Easy
849 of 922
Application +4 | -1
For non-singular square matrices A,BA,B and CC of the same order, (AB−1C)−1  {\left( {A{B^{ - 1}}C} \right)^{ - 1}}\; is equal to
View answer and explanation
Correct option: D
C−1BA−1{C^{ - 1}}B{A^{ - 1}}
Option D (C−1BA−1{C^{ - 1}}B{A^{ - 1}}) is correct because it follows the rule of inverting a product of matrices: (ABC)-1 = C-1 B-1 A-1.
850 Easy
850 of 922
Analyzing +4 | -1
Let d∈Rd \in R, and A=[−24+d(sin⁡θ)−21(sin⁡θ)+2d5(2sin⁡θ)−d(−sin⁡θ)+2+2d],θ∈[θ,2π].A = \left[ {\begin{array}{cccccccccccccccccccc} { - 2}&{4 + d}&{(\sin \theta ) - 2}\\ 1&{(\sin \theta ) + 2}&d\\ 5&{(2\sin \theta ) - d}&{( - \sin \theta ) + 2 + 2d} \end{array}} \right],\theta \in [\theta ,2\pi ]. If the minimum value of det (A) is 8, then a value of d is
View answer and explanation
Correct option: A
-5
Option A (-5) is correct because the minimum value of det(A) is found by maximizing sin²(θ), which leads to d = -5.
851 Easy
851 of 922
Understanding +4 | -1
If II is a unit matrix, then 3 I will be
View answer and explanation
Correct option: C
A scalar matrix
Option C (A scalar matrix) is correct because multiplying a unit matrix by a scalar results in a scalar matrix.
852 Easy
852 of 922
Application +4 | -1
If f  (x)=x2+4x−5  and  A=[1         24    −3]f\;\left( x \right) = {x^2} + 4x - 5\;\mathrm{and\;}A = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;\,\,\,\;2}\\ {4\;\,\, - 3} \end{array}} \right], then f  (A)f\;\left( A \right) is equal to
View answer and explanation
Correct option: D
[8480]\left[ {\begin{array}{cccccccccccccccccccc} 8&4\\ 8&0 \end{array}} \right]
Option D (\left[ {\begin{{array}}{{{*{20}{c}}}} 8&4\\ 8&0 \end{{array}}} \right]\left[ {\begin{{array}}{{{*{20}{c}}}} 8&4\\ 8&0 \end{{array}}} \right]) is correct because it is the result of substituting matrix A into the given polynomial function and performing the matrix operations.
853 Easy
853 of 922
Application +4 | -1
If ω\omega is a complex cube root of unity, then ∣1ωω2ωω21ω21ω∣\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right| is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because applying the property 1+ω+ω2=01 + \omega + {\omega ^2} = 0 simplifies the determinant to zero.
854 Easy
854 of 922
Application +4 | -1
If A=(aij)3×3A=\left(a_{i j}\right)_{3 \times 3} is a matrix satisfying the equation x3−3x+1=0x^3-3 x+1=0, then
View answer and explanation
Correct option: C
AA is non-singular matrix
Option C (AA is non-singular matrix) is correct because the given equation implies that AA has an inverse, making it non-singular.
855 Easy
855 of 922
Application +4 | -1
Let w≠1w \neq 1 be a cube root of unity and
A=[w00w]A=\left[\begin{array}{cc} w & 0 \\ 0 & w \end{array}\right]

Then A2010A^{2010} is equal to
View answer and explanation
Correct option: C
A3A^3
Option C (A3A^3) is correct because ww is a cube root of unity, so w3=1w^3 = 1, and A2010=(A3)670=I=A3A^{2010} = (A^3)^{670} = I = A^3.

Explore Other Chapters in MATHEMATICS

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs
856 Easy
856 of 922
Analyzing +4 | -1
Adj⋅(AB)−(Adj⋅B)(Adj⋅A)=A d j \cdot(A B)-(A d j \cdot B)(A d j \cdot A)= ?
View answer and explanation
Correct option: C
OO
Option C (O) is correct because Adj(AB)=Adj(B)Adj(A)Adj(AB) = Adj(B)Adj(A), so the expression simplifies to a zero matrix.
857 Easy
857 of 922
Analyzing +4 | -1
The value of the determinant Δ=∣2a1b1a1b2+a2b1a1b3+a3b1a1b2+a2b12a2b2a2b3+a3b2a1b3+a3b1a3b2+a2b32a3b3∣\Delta = \left| {\begin{array}{cccccccccccccccccccc} {2{a_1}{b_1}}&{{a_1}{b_2} + {a_2}{b_1}}&{{a_1}{b_3} + {a_3}{b_1}}\\ {{a_1}{b_2} + {a_2}{b_1}}&{2{a_2}{b_2}}&{{a_2}{b_3} + {a_3}{b_2}}\\ {{a_1}{b_3} + {a_3}{b_1}}&{{a_3}{b_2} + {a_2}{b_3}}&{2{a_3}{b_3}} \end{array}} \right| is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the determinant can be factored into a product of determinants that results in zero.
858 Easy
858 of 922
Understanding +4 | -1
If AA and BB are two square matrices of the same size, then (A+B)2=A2+2AB+B2(A+B)^2=A^2+2 A B+B^2 can hold if and only if
View answer and explanation
Correct option: A
AB=BAA B=B A
Option A (AB=BAA B=B A) is correct because (A+B)2=A2+AB+BA+B2(A+B)^2 = A^2 + AB + BA + B^2, which equals A2+2AB+B2A^2 + 2AB + B^2 only if AB=BAAB = BA.
859 Easy
859 of 922
Application +4 | -1
Let A=([x+1][x+2][x+3][x][x+3][x+3][x][x+2][x+4])A = \left( {\begin{array}{cccccccccccccccccccc} {[x + 1]}&{[x + 2]}&{[x + 3]}\\ {[x]}&{[x + 3]}&{[x + 3]}\\ {[x]}&{[x + 2]}&{[x + 4]} \end{array}} \right) where [t][t] denotes the greatest integer less than or equal to tt. If det⁡(A)=192\det (A) = 192 then the set of values of xx is interval
View answer and explanation
Correct option: B
[62, 63)
The correct answer is [62, 63) because the determinant of the given matrix is equal to 192, which leads to [x] = 62, meaning x lies in the interval [62, 63).
860 Easy
860 of 922
Application +4 | -1
If a=2i+k,b=i+j+k\mathbf{a}=2 \mathbf{i}+\mathbf{k}, \mathbf{b}=\mathbf{i}+\mathbf{j}+\mathbf{k} and c=4i−3j+7k\mathbf{c}=4 \mathbf{i}-3 \mathbf{j}+7 \mathbf{k}. If d×b=c×b\mathbf{d} \times \mathbf{b}=\mathbf{c} \times \mathbf{b} and d.a=0\mathbf{d} . \mathbf{a}=0, then d\mathrm{d} will be
View answer and explanation
Correct option: D
−i−8j+2k-\mathbf{i}-8 \mathbf{j}+2 \mathbf{k}
Option D (-i - 8j + 2k) is correct because by using the given conditions d x b = c x b and d . a = 0, we can solve for the components of vector d.
861 Easy
861 of 922
Application +4 | -1
If A=[cos⁡θsin⁡θ−sin⁡θcos⁡θ],A=\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right], then lim⁡n→∞Ann\lim\limits_{n \to \infty } \frac{{{A^n}}}{n} is (where θ∈R)\theta \in R)
View answer and explanation
Correct option: A
A zero matrix
Option A (A zero matrix) is correct because as n approaches infinity, the values of cos(nθ) and sin(nθ) oscillate between -1 and 1, making the limit of the fraction approach zero.
862 Easy
862 of 922
Understanding +4 | -1
The sum of the products of the elements of any row of a determinant AA with the cofactors of the corresponding elements is equal to
View answer and explanation
Correct option: C
∣A∣\left| A \right|
Option C is correct because the sum of the products of elements of a row with their corresponding cofactors equals the determinant's value.
863 Easy
863 of 922
Understanding +4 | -1
If A3=OA^3=O, then I+A+A2I+A+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 (I−A)(I+A+A2)=I(I-A)(I+A+A^2) = I, which means I+A+A2I+A+A^2 is the inverse of (I−A)(I-A).
864 Easy
864 of 922
Application +4 | -1
If f(θ)=[cos⁡θ−sin⁡θ0sin⁡θcos⁡θ0001],  f\left( \theta \right) = \left[ {\begin{array}{cccccccccccccccccccc} {\cos \mathrm{\theta }}&{ - \sin \mathrm{\theta }}&0\\ {\sin \mathrm{\theta }}&{\cos \mathrm{\theta }}&0\\ 0&0&1 \end{array}} \right],\; then {f(θ)−1}\left\{ {f{{\left( \theta \right)}^{ - 1}}} \right\} is equal to
View answer and explanation
Correct option: A
f(−θ)f\left( { - \theta } \right)
The correct answer is A. The inverse of f(θ\theta) is found by replacing θ\theta with -θ\theta in the original matrix.
865 Easy
865 of 922
Application +4 | -1
If A(α,β)=[cos⁡αsin⁡α0−sin⁡αcos⁡α000eβ]A(\alpha, \beta)=\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^\beta\end{array}\right], then A(α,β)−1A(\alpha, \beta)^{-1} is equal to
View answer and explanation
Correct option: B
A(−α,−β)A(-\alpha,-\beta)
Option B (A(−α,−β)A(-\alpha,-\beta)) is correct because it represents the inverse matrix obtained by changing the signs of α\alpha and β\beta.
866 Easy
866 of 922
Application +4 | -1
If A=[200020002],A = \left[ {\begin{array}{cccccccccccccccccccc} 2&0&0\\ 0&2&0\\ 0&0&2 \end{array}} \right], then A5={A^5} =
View answer and explanation
Correct option: C
16A
Option C (16A) is correct because when A is a scalar multiple of the identity matrix, raising it to a power involves raising the scalar to that power.
867 Easy
867 of 922
Remembering +4 | -1
If AA is an orthogonal matrix, then A−1{A^{ - 1}} equals
View answer and explanation
Correct option: B
AT{A^T}
Option B (AT{A^T}) is correct because for an orthogonal matrix A, its inverse is equal to its transpose, i.e., A−1=AT{A^{ - 1}} = {A^T}.
868 Easy
868 of 922
Application +4 | -1
If A=[4x+22x−3x+1]A=\left[\begin{array}{cc}4 & x+2 \\ 2 x-3 & x+1\end{array}\right] is an invertible matrix, then xx cannot take value
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the matrix is invertible for all the given values of x. The determinant of A is not zero for x = -1, 2, or 3.
869 Easy
869 of 922
Application +4 | -1
The rank of the matrix A=∣123456345∣A=\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & 4 & 5\end{array}\right|, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the determinant of the matrix is zero, but a 2x2 submatrix has a non-zero determinant, indicating a rank of 2.
870 Easy
870 of 922
Application +4 | -1
If [2      13     2]A[−3              25         −3]=[1      00      1],  \left[ {\begin{array}{cccccccccccccccccccc} {2\;\,\,\;1}\\ {3\;\;\,2} \end{array}} \right]A\left[ {\begin{array}{cccccccccccccccccccc} { - 3\;\;\;\;\;\;\;2}\\ {5\;\;\;\;\, - 3} \end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;\,\,0}\\ {0\;\;\,\,1} \end{array}} \right],\; then AA is equal to
View answer and explanation
Correct option: C
[1       01        1]\left[ {\begin{array}{cccccccccccccccccccc} {1\;\;\,\,\,0}\\ {1\;\,\,\,\,\;1} \end{array}} \right]
The correct answer is C. By pre-multiplying and post-multiplying by the inverses of the given matrices, we can isolate A and find its value.
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