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JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

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781 Easy
781 of 995
Analyzing +4 | -1
Let m and M be respectively the minimum and maximum values of ∣cos⁡2x1+sin⁡2xsin⁡2x1+cos⁡2xsin⁡2xsin⁡2xcos⁡2xsin⁡2x1+sin⁡2x∣.\left| {\begin{array}{cccccccccccccccccccc} {{{\cos }^2}x}&{1 + {{\sin }^2}x}&{\sin 2x}\\ {1 + {{\cos }^2}x}&{{{\sin }^2}x}&{\sin 2x}\\ {{{\cos }^2}x}&{{{\sin }^2}x}&{1 + \sin 2x} \end{array}} \right|. Then, the ordered pair (m,M)(m,M) is equal to
View answer and explanation
Correct option: B
(−3,−1)( - 3, - 1)
Option B ((-3, -1)) is correct because the minimum and maximum values of the determinant are found by simplifying the determinant and analyzing the range of the sine function.
782 Easy
782 of 995
Understanding +4 | -1
Consider the following statements

I. To every rectangular matrix A=[aij]A=\left[a_{i j}\right] of order nn, we can associate a number (real or complex) called determinant of A.A.

II. Determinant is a function which associates each square matrix with a unique number (real or complex).
View answer and explanation
Correct option: B
Only II is true
Option B is correct because only statement II is true; determinants are defined only for square matrices.
783 Easy
783 of 995
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.
784 Easy
784 of 995
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}.
785 Easy
785 of 995
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.
786 Easy
786 of 995
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.
787 Medium
787 of 995
Analyzing +4 | -1
If A=(3−342−340−11)A=\left(\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right) then value of trace of the matrix adj⁡(adj⁡A)\operatorname{adj}(\operatorname{adj} A) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the adj(adj A) simplifies to A, and the trace of A is the sum of its diagonal elements.
788 Easy
788 of 995
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.
789 Easy
789 of 995
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.
790 Easy
790 of 995
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].
791 Easy
791 of 995
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.
792 Easy
792 of 995
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.
793 Medium
793 of 995
Analyzing +4 | -1
In Δ  ABC,if∣1ab1ca1bc∣=0\Delta \;ABC,\mathrm{if}\left| {\begin{array}{cccccccccccccccccccc} 1&a&b\\ 1&c&a\\ 1&b&c \end{array}} \right| = 0, then

sin⁡2A+sin⁡2B+sin⁡2C{\sin ^2}A + {\sin ^2}B + {\sin ^2}C is equal to
View answer and explanation
Correct option: B
94\frac{9}{4}
Option B (9/4) is correct because the determinant being zero implies the triangle is equilateral, and substituting 60 degrees into the expression yields 9/4.
794 Medium
794 of 995
Analyzing +4 | -1
Let pp be an odd prime number and TpT_p be the following set of 2×22 \times 2 matrices Tp={A=[abca]:a,b,c∈{0,1,2,…,(p−1)}}T_p=\left\{A=\left[\begin{array}{ll}a & b \\ c & a\end{array}\right]: a, b, c \in\{0,1,2, \ldots,(p-1)\}\right\}
The number of AA in TpT_p such that AA is either symmetric or skew-symmetric or both, and det (A)(A) divisible by pp is
View answer and explanation
Correct option: B
2(p−1)2(p-1)
The correct answer is R−{2,−3}R-\{2,-3\} because the system has only the trivial solution when the determinant of the coefficient matrix is non-zero, which means k≠2k \neq 2 and k≠−3k \neq -3.
795 Easy
795 of 995
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.
796 Easy
796 of 995
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.
797 Medium
797 of 995
Understanding +4 | -1
Consider the following statements

I. ∣A∣|{A}| is also called modulus of square matrix A.A.

II. Every matrix has determinant.
View answer and explanation
Correct option: D
Neither I nor II is true
Option D (Neither I nor II is true) is correct because |A| represents the determinant, not the modulus, and only square matrices have determinants.
798 Easy
798 of 995
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.
799 Easy
799 of 995
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.
800 Easy
800 of 995
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.
801 Medium
801 of 995
Analyzing +4 | -1
The value of the determinant ∣cos⁡α−sin⁡α1sin⁡αcos⁡α1cos⁡(α+β)−sin⁡(α+β)1∣\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ \cos (\alpha+\beta) & -\sin (\alpha+\beta) & 1\end{array}\right| is
View answer and explanation
Correct option: A
Independent of α\alpha
Option A (Δ∈[1−2,1+2]\Delta \in[1-\sqrt{2}, 1+\sqrt{2}]) is correct because the range of the expression 1+sin(beta)-cos(beta) falls within this interval.
802 Easy
802 of 995
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.
803 Medium
803 of 995
Application +4 | -1
The solutions of the equation ∣x2−125x−12x∣=0,\left| {\begin{array}{cccccccccccccccccccc} x&2&{ - 1}\\ 2&5&x\\ { - 1}&2&x \end{array}} \right| = 0, are
View answer and explanation
Correct option: A
3,−13, - 1
Option A (3, -1) is correct because these are the roots of the quadratic equation obtained by expanding the determinant.
804 Easy
804 of 995
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.
805 Medium
805 of 995
Application +4 | -1
If A=[−1111−1111−1]A=\left[\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right], then the value of Det⁡(AB+2I)\operatorname{Det}(A B+2I) where B=adj⁡AB=\operatorname{adj} A and II is identify matrix of order 3
View answer and explanation
Correct option: C
216
Option C (216) is correct because by using the properties of adjoint and determinant, the determinant of the given expression can be simplified and calculated.
806 Easy
806 of 995
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}}.
807 Medium
807 of 995
Understanding +4 | -1
If ∣a1b1c1a2b2c2a3b3c3∣≠0\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right| \neq 0, then the number of solutions of the system of equations a1x+b1y+c1z=0,a2x+b2y+c2z=0a_1 x+b_1 y+c_1 z=0, a_2 x+b_2 y+c_2 z=0 and a3x+b3y+c3z=0a_3 x+b_3 y+c_3 z=0 is
View answer and explanation
Correct option: B
Only one unique solution
Option B (Only one unique solution) is correct because the determinant of the coefficient matrix is non-zero, implying a unique (trivial) solution for the homogeneous system.
808 Medium
808 of 995
Analyzing +4 | -1
If maximum and minimum values of the determinant ∣1+Sin⁡2xCos⁡2xSin⁡2xSin⁡2x1+Cos⁡2xSin⁡2xSin⁡2xCos⁡2x1+Sin⁡2x∣\left|\begin{array}{lll}1+\operatorname{Sin}^2 x & \operatorname{Cos}^2 x & \operatorname{Sin} 2 x \\ \operatorname{Sin}^2 x & 1+\operatorname{Cos}^2 x & \operatorname{Sin} 2 x \\ \operatorname{Sin}^2 x & \operatorname{Cos}^2 x & 1+\operatorname{Sin} 2 x\end{array}\right| are α\alpha and β\beta respectively, then which of the statements are FALSE ?
View answer and explanation
Correct option: D
A triangle can be constructed having it's sides as α,β\alpha, \beta and α−β\alpha-\beta
Option D (A triangle can be constructed having it's sides as α,β\alpha, \beta and α−β\alpha-\beta) is false because the triangle inequality is not satisfied with the given values of α\alpha and β\beta.
809 Easy
809 of 995
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.
810 Easy
810 of 995
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].
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