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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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721 Easy
721 of 922
Application +4 | -1
If A=[6    8    54    2    39    7    1]A = \left[ {\begin{array}{cccccccccccccccccccc} {6\;\;8\;\;5}\\ {4\;\;2\;\;3}\\ {9\;\;7\;\;1} \end{array}} \right] is the sum of a symmetric matrix BB skew-symmetric matrix CC, then BB is
View answer and explanation
Correct option: A
[6    6    76    2    57    5    1]\left[ {\begin{array}{cccccccccccccccccccc} {6\;\;6\;\;7}\\ {6\;\;2\;\;5}\\ {7\;\;5\;\;1} \end{array}} \right]
Option A is correct because the symmetric part of a matrix is calculated as (A + A')/2, which results in the given matrix.
722 Easy
722 of 922
Remembering +4 | -1
If AA is an orthogonal matrix, then A−1A^{-1}, equals
View answer and explanation
Correct option: B
A′A^{\prime}
Option B (A′A^{\prime}) is correct because for an orthogonal matrix A, the inverse is equal to its transpose, i.e., A−1=A′A^{-1} = A^{\prime}.
723 Easy
723 of 922
Application +4 | -1
If U=[2−3      4],X=[0    2    3],V=[321]and  Y=[224],then  UV+XYU = \left[ {2 - 3\;\;\;4} \right],X = \left[ {0\;\;2\;\;3} \right],V = \left[ {\begin{array}{cccccccccccccccccccc} 3\\ 2\\ 1 \end{array}} \right]\mathrm{and}\;Y = \left[ {\begin{array}{cccccccccccccccccccc} 2\\ 2\\ 4 \end{array}} \right],\mathrm{then}\;UV + XY
View answer and explanation
Correct option: A
[20]\left[ {20} \right]
Option A ([20]\left[ {20} \right]) is correct because the matrix multiplication and addition result in a 1x1 matrix with the value 20.
724 Easy
724 of 922
Understanding +4 | -1
Assertion: Δ=∣a1a2a3b1b2b3ka1ka2ka3∣=0\Delta=\left|\begin{array}{ccc}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ k a_1 & \mathrm{ka}_2 & \mathrm{ka}_3\end{array}\right|=0

Reason: If corresponding elements of any two rows of a determinant are proportional, then its value is zero.
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion.
Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because the determinant is zero when two rows are proportional.
725 Easy
725 of 922
Analyzing +4 | -1
If the system of linear equations
2x+3y−z=−22x + 3y - z = - 2
x+y+z=4x + y + z = 4
x−y+∣λ∣z=4λ−4x - y + \left| \lambda \right|z = 4\lambda - 4
where λ∈R,\lambda \in R, has no solution, then
View answer and explanation
Correct option: B
λ=−7\lambda = - 7
Option B (λ=−7\lambda = - 7) is correct because it results in the system having no solution, as the determinant is zero and the equations are inconsistent.
726 Easy
726 of 922
Understanding +4 | -1
The square matrix A=[aij]A=\left[a_{i j}\right] given by aij=(i−j)3a_{i j}=(i-j)^3, is a
View answer and explanation
Correct option: B
skew-symmetric matrix
Option B (skew-symmetric matrix) is correct because the matrix elements satisfy the condition aij=−ajia_{ij} = -a_{ji} for all ii and jj.
727 Easy
727 of 922
Application +4 | -1
If 2A+3B=[2  −1    43            2    5]and  A+2B=[5    0    31    6    2],  then    B  is2A + 3B = \left[ {\begin{array}{cccccccccccccccccccc} {2\; - 1\;\;4}\\ {3\;\;\;\;\;\;2\;\;5} \end{array}} \right]\mathrm{and\;}A + 2B = \left[ {\begin{array}{cccccccccccccccccccc} {5\;\;0\;\;3}\\ {1\;\;6\;\;2} \end{array}} \right],\;\mathrm{then}\;\;B\;\mathrm{is}
View answer and explanation
Correct option: C
[812−110−1]\left[ {\begin{array}{cccccccccccccccccccc} 8&1&2\\ { - 1}&{10}&{ - 1} \end{array}} \right]
Option A is correct because solving the system of linear equations for matrices A and B results in the matrix B = [812−110−1]\left[ {\begin{array}{cccccccccccccccccccc} 8&1&2\\ { - 1}&{10}&{ - 1} \end{array}} \right].
728 Easy
728 of 922
Analyzing +4 | -1
If PP is an orthogonal matrix and Q=PAPTQ=P A P^T and B=PTQ1000P,B=P^T Q^{1000} P, then B−1B^{-1} is, where AA is involutory matrix
View answer and explanation
Correct option: C
II
Option C (II) is correct because after simplification using the properties of orthogonal and involutory matrices, B−1B^{-1} simplifies to the identity matrix.
729 Easy
729 of 922
Understanding +4 | -1
If AA is an orthogonal matrix, then ∣A∣|A| is
View answer and explanation
Correct option: C
±1\pm 1
Option C (±1\pm 1) is correct because for an orthogonal matrix A, ATA=IA^T A = I, so ∣ATA∣=∣I∣=1|A^T A| = |I| = 1, which means ∣A∣2=1|A|^2 = 1, thus ∣A∣=±1|A| = \pm 1.
730 Easy
730 of 922
Analyzing +4 | -1
The values of λ\lambda and μ\mu for which the system of linear equations

x+y+z=2x+2y+3z=5x+3y+λz=μ\begin{aligned} &x+y+z=2 \\ &x+2 y+3 z=5 \\ &x+3 y+\lambda z=\mu\end{aligned}

has infinitely many solutions are respectively
View answer and explanation
Correct option: A
5 and 8
Option C (6 and 8) is incorrect. Option A (5 and 8) is correct because these values make the determinant and related determinants zero, leading to infinitely many solutions.
731 Easy
731 of 922
Application +4 | -1
Let A=[21012−10−12].A=\left[\begin{array}{ccc}2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2\end{array}\right]. If ∣adj⁡(adj⁡(adj⁡2A))∣=(16)n,|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 A))|=(16)^n, then nn is equal to
View answer and explanation
Correct option: A
10
Option A (10) is correct because by using the properties of adjoint and determinants, we find that n = 10.
732 Easy
732 of 922
Understanding +4 | -1
For any square matrix AA with real number entries, consider the following statements.

Assertion : AA + A′{A}^{\prime} is a symmetric matrix.

Reason: A−A′{A-A}^{\prime} is a skew-symmetric matrix.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Option B is correct because both the assertion and the reason are true statements about matrix properties, but the reason doesn't directly explain the assertion.
733 Easy
733 of 922
Analyzing +4 | -1
Let AA and BB be 3×33 \times 3 real matrices, such that AA is symmetric matrix and BB is skew-symmetric matrix. Then, the system of linear equations (A2B2−B2A2)X=0({A^2}{B^2} - {B^2}{A^2})X = 0, where XX is a 3×13 \times 1 column matrix of unknown variables and OO is a 3×13 \times 1 null matrix, has
View answer and explanation
Correct option: C
Infinitely many solutions
Option C (Infinitely many solutions) is correct because the matrix (A2B2−B2A2)({A^2}{B^2} - {B^2}{A^2}) is skew-symmetric, implying its determinant is zero, leading to infinitely many solutions for the homogeneous system.
734 Easy
734 of 922
Analyzing +4 | -1
Consider the system of equations in x,  y,  zx,\;y,\;z as
xsin⁡3θ−y+z=0xcos⁡2θ+4y+3z=0\begin{array}{l} x\sin 3\theta - y + z = 0\\ x\cos 2\theta + 4y + 3z = 0 \end{array}
and   2x+7y+7z=0\;2x + 7y + 7z = 0
If this system has a non-trivial solution, then for integer nn, values of θ\theta are given by
View answer and explanation
Correct option: C
π(n+(−1)n6)\pi \left( {n + \frac{{{{\left( { - 1} \right)}^n}}}{6}} \right)
Option C (π(n+(−1)n6)\pi \left( {n + \frac{{{{\left( { - 1} \right)}^n}}}{6}} \right)) is correct because the values of θ\theta that satisfy the condition for a non-trivial solution are given by this expression.
735 Easy
735 of 922
Application +4 | -1
If AA is a 3×33 \times 3 non-singular matrix, then adj⁡(adj⁡A)\operatorname{adj}(\operatorname{adj} A) is equal to
View answer and explanation
Correct option: A
∣A∣A|A| A
Option A (∣A∣A|A| A) is the correct answer because it is the result of applying the properties of adjoint and determinants of matrices.
736 Easy
736 of 922
Application +4 | -1
Inverse of the matrix [cos⁡2θ  −−sin⁡2θsin⁡2θ  cos⁡2θ]\left[ {\begin{array}{cccccccccccccccccccc} {\cos 2\mathrm{\theta }\;--\sin 2\mathrm{\theta }}\\ {\sin 2\mathrm{\theta }\;\cos 2\mathrm{\theta }} \end{array}} \right] is
View answer and explanation
Correct option: D
[cos⁡2θ  sin⁡2θ−sin⁡2θ  cos⁡2θ]\left[ {\begin{array}{cccccccccccccccccccc} {\cos 2\mathrm{\theta }\;\sin 2\mathrm{\theta }}\\ { - \sin 2\mathrm{\theta }\;\cos 2\mathrm{\theta }} \end{array}} \right]
Option D is correct because the inverse of a rotation matrix is its transpose, which involves changing the sign of the sine term.
737 Easy
737 of 922
Application +4 | -1
If the matrix Mr{M_r} is given by Mr=[rr−1r−1r]r=1,2,3...,{M_r} = \left[ {\begin{array}{cccccccccccccccccccc} r&{r - 1}\\ {r - 1}&r \end{array}} \right]r = 1,2,3..., then the value of det (M1)+det⁡(M2)+...+det⁡(M2008)\left( {{M_1}} \right) + \det \left( {{M_2}} \right) + ... + \det ({M_{2008}}) is
View answer and explanation
Correct option: C
(2008)2{\left( {2008} \right)^2}
Option C ((2008)2(2008)^2) is correct because the determinant of MrM_r simplifies to 2r−12r - 1, and the sum of the first 2008 odd numbers is (2008)2(2008)^2.
738 Easy
738 of 922
Application +4 | -1
[712921][345]+2[42]\left[\begin{array}{lll}7 & 1 & 2 \\ 9 & 2 & 1\end{array}\right]\left[\begin{array}{l}3 \\ 4 \\ 5\end{array}\right]+2\left[\begin{array}{l}4 \\ 2\end{array}\right] is equal to
View answer and explanation
Correct option: A
[4344]\left[\begin{array}{l}43 \\ 44\end{array}\right]
Option A (\left[\begin{{array}}{{l}}43 \\ 44\end{{array}}\right]\left[\begin{{array}}{{l}}43 \\ 44\end{{array}}\right]) is correct because performing the matrix multiplication and addition yields the result \left[\begin{{array}}{{l}}43 \\ 44\end{{array}}\right]\left[\begin{{array}}{{l}}43 \\ 44\end{{array}}\right].
739 Easy
739 of 922
Analyzing +4 | -1
If O(A)  =2×3,  O(B)  =3×2,  O\left( A \right)\; = 2 \times 3,\;O\left( B \right)\; = 3 \times 2,\; and   O(c)=3×3,  \;O\left( c \right) = 3 \times 3,\; Which one of the following is not defined?
View answer and explanation
Correct option: A
C  (A+B′)C\;\left( {A + B'} \right)
Option A (C(A+B′)C(A + B')) is correct because the dimensions of the matrices do not allow for addition and multiplication in that order.
740 Easy
740 of 922
Analyzing +4 | -1
If ∣y+zxyz+xzxx+yyz∣=k(x+y+z)(x−z)2,\left| {\begin{array}{cccccccccccccccccccc} {y + z}&x&y\\ {z + x}&z&x\\ {x + y}&y&z \end{array}} \right| = k\left( {x + y + z} \right){\left( {x - z} \right)^2}, then kk is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because after applying row operations and simplifying, the value of k is found to be 1.
741 Easy
741 of 922
Application +4 | -1
If A2−A+I=0,  {A^2} - A + I = 0,\; then the inverse of AA is
View answer and explanation
Correct option: A
I−AI - A
Option A (I−AI - A) is the correct inverse of A, derived by manipulating the given equation A2−A+I=0{A^2} - A + I = 0 and multiplying by A−1{A^{-1}}.
742 Easy
742 of 922
Understanding +4 | -1
If AA is a 2×22 \times 2 unitary matrix, then ∣A∣|A| is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because for a unitary matrix AA, ∣A∣|A| is a complex number with magnitude 1, i.e., ∣A∣=eiθ|A| = e^{i\theta} for some real θ\theta.
743 Easy
743 of 922
Application +4 | -1
For all values of θ∈(0,π/2),\theta \in(0, \pi / 2), the determinant of the matrix [−2tan⁡θ+sec⁡2θ3−sin⁡θcos⁡θsin⁡θ−3−43]\left[\begin{array}{ccc}-2 & \tan \theta+\sec ^2 \theta & 3 \\ -\sin \theta & \cos \theta & \sin \theta \\ -3 & -4 & 3\end{array}\right] always lies in the interval
View answer and explanation
Correct option: B
[3,5][3,5]
Option B ([3,5][3,5]) is correct because the determinant simplifies to 3cos⁡θ+4sin⁡θ3 \cos \theta + 4 \sin \theta, which can be expressed as 5sin⁡(θ+α)5 \sin(\theta + \alpha) where α=tan⁡−1(34)\alpha = \tan^{-1}(\frac{3}{4}), and its range for θ∈(0,π2)\theta \in (0, \frac{\pi}{2}) is [3,5][3, 5].
744 Easy
744 of 922
Application +4 | -1
Let A(02qrpq−rp−qr)A\left( {\begin{array}{cccccccccccccccccccc} 0&{2q}&r\\ p&q&{ - r}\\ p&{ - q}&r \end{array}} \right). If AAT=I3,A{A^T} = {I_3}, then ∣p∣\left| p \right| is
View answer and explanation
Correct option: A
15\frac{1}{{\sqrt 5 }}
Option B (12\frac{1}{{\sqrt 2 }}) is correct because it's the absolute value of p derived from the matrix equation AAT=I3A{A^T} = {I_3}.
745 Easy
745 of 922
Application +4 | -1
If A=[0    10    0],IA = \left[ {\begin{array}{cccccccccccccccccccc} {0\;\;1}\\ {0\;\;0} \end{array}} \right],I is the unit matrix of order 2 and a,ba,b are arbitrary constants, then (aI+bA)2{\left( {aI + bA} \right)^2} is equal to
View answer and explanation
Correct option: B
a2I+2abA{a^2}I + 2abA
Option B (a2I+2abA{a^2}I + 2abA) is correct because when expanding the square and using the property that A² = 0, we arrive at this result.
746 Easy
746 of 922
Application +4 | -1
If [1−tan⁡θtan⁡θ1][1tan⁡θ−tan⁡θ1]−1=[a−bba],\left[ {\begin{array}{cccccccccccccccccccc} 1&{ - \tan \theta }\\ {\tan \theta }&1 \end{array}} \right]{\left[ {\begin{array}{cccccccccccccccccccc} 1&{\tan \theta }\\ { - \tan \theta }&1 \end{array}} \right]^{ - 1}} = \left[ {\begin{array}{cccccccccccccccccccc} a&{ - b}\\ b&a \end{array}} \right], then
View answer and explanation
Correct option: B
a=cos⁡2  θ,  b=sin⁡2  θa = \cos 2\;\theta ,\;b = \sin 2\;\theta
Option B (a=cos⁡2;θ,;b=sin⁡2;θa = \cos 2\\;\theta ,\\;b = \sin 2\\;\theta) is correct because after simplifying the matrix expression using trigonometric identities, the resulting matrix elements match these values.
747 Easy
747 of 922
Understanding +4 | -1
Which one of the following is true always for any two non-singular matrices AA and BB of same order?
View answer and explanation
Correct option: D
(AB)−1=B−1A−1{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}
Option D ((AB)−1=B−1A−1{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}) is correct because this is a fundamental property of the inverse of a product of matrices.
748 Easy
748 of 922
Application +4 | -1
If A=[ab],B=[−b−a]\mathrm{A}=\left[\begin{array}{ll}a & b\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}-b & -a\end{array}\right] and C=[a−a]\mathrm{C}=\left[\begin{array}{c}a \\ -a\end{array}\right], then the correct statement is
View answer and explanation
Correct option: C
AC=BC\mathrm{AC}=\mathrm{BC}
Option C (AC = BC) is correct because when the matrices are multiplied, the resulting matrices are equal.
749 Easy
749 of 922
Application +4 | -1
If A=[α22α]A=\left[\begin{array}{ll}\alpha & 2 \\ 2 & \alpha\end{array}\right] and ∣A3∣=125\left|A^3\right|=125 then the value of α\alpha is
View answer and explanation
Correct option: C
±3\pm 3
Option C (±3\pm 3) is correct because by using the property ∣A3∣=∣A∣3|A^3| = |A|^3 and solving the resulting equation, we find α=±3\alpha = \pm 3.
750 Easy
750 of 922
Application +4 | -1
cos⁡θ[cos⁡θ        sin⁡θ−sin⁡θ          cos⁡θ]+sin⁡θ[sin⁡θ−cos⁡θcos⁡θ        sin⁡θ]\cos \theta \left[ {\begin{array}{cccccccccccccccccccc} {\cos \theta \;\;\;\;\sin \theta }\\ { - \sin \theta \;\;\;\;\;\cos \theta } \end{array}} \right] + \sin \theta \left[ {\begin{array}{cccccccccccccccccccc} {\sin \theta - \cos \theta }\\ {\cos \theta \;\;\;\;\sin \theta } \end{array}} \right] is equal to
View answer and explanation
Correct option: D
[1001]\left[ {\begin{array}{cccccccccccccccccccc} 1&0\\ 0&1 \end{array}} \right]
Option D \left[ {\begin{{array}}{{*{20}{c}}}1&0\\0&1\end{{array}}} \right]\left[ {\begin{{array}}{{*{20}{c}}}1&0\\0&1\end{{array}}} \right] is correct because after performing the matrix multiplication and addition, and using the trigonometric identity cos⁡2θ+sin⁡2θ=1{{\cos }^2}\theta + {{\sin }^2}\theta = 1, the result is the identity matrix.
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