Skip to main content
Back to MATRICES

Your Progress

0 / 995 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 721-750 of 995 questions

Preparing practice and saved progress…

721 Medium
721 of 995
Application +4 | -1
If a1,a2,a3,a5a_1, a_2, a_3, a_5 are consecutive terms of arithmetic progression with common difference 3, then the value of ∣a32a2a1a42a3a2a52a4a3∣\left|\begin{array}{lll}a_3^2 & a_2 & a_1 \\ a_4^2 & a_3 & a_2 \\ a_5^2 & a_4 & a_3\end{array}\right| is
View answer and explanation
Correct option: D
162
Option A (0) is incorrect. Option B (27) is incorrect. Option C (81) is incorrect. Option D (162) is correct because it uses properties of determinants and arithmetic progressions to simplify the expression and find the value.
722 Easy
722 of 995
Analyzing +4 | -1
The number of values of kk for which the system of equations: (k+1)x+8y=4k,kx+(k+3)y=3k−1(k + 1)x + 8y = 4k,kx + (k + 3)y = 3k - 1 has solution is
View answer and explanation
Correct option: B
1
Option B is correct because there is only one value of k (k=3) for which the system of equations has a solution.
723 Easy
723 of 995
Analyzing +4 | -1
Consider the system of equations in x,y,zx, y, z as
xsin⁡3θ−y+z=0x \sin 3 \theta-y+z=0

xcos⁡2θ+4y+3z=0x \cos 2 \theta+4 y+3 z=0

2x+7y+7z=02 x+7 y+7 z=0
If this system has a non-trivial solution, then for any integer nn, values of θ\theta are given
View answer and explanation
Correct option: C
(n+(−1)n6)π\left(n+\frac{(-1)^n}{6}\right) \pi
The correct answer is (n+(−1)n6)π\left(n+\frac{{(-1)^n}}{{6}}\right) \pi because solving the determinant equation leads to θ=nπ\theta = n\pi or θ=nπ+(−1)nπ6\theta = n\pi + (-1)^n \frac{{\pi}}{{6}}.
724 Easy
724 of 995
Analyzing +4 | -1
There are two possible values of AA in the solution of the matrix equation [2A+1−5−4A]−1[A−5B2A−2C]=[14DEF],\left[\begin{array}{rr}2 A+1 & -5 \\ -4 & A\end{array}\right]^{-1}\left[\begin{array}{rr}A-5 & B \\ 2 A-2 & C\end{array}\right]=\left[\begin{array}{rr}14 & D \\ E & F\end{array}\right], where A,B,C,D,E,FA, B, C, D, E, F are real numbers. The absolute value of the difference of these two solutions, is
View answer and explanation
Correct option: D
193\frac{19}{3}
Option D (193\frac{{19}}{{3}}) is correct because solving the matrix equation leads to a quadratic in A, and the absolute difference of the roots is 193\frac{{19}}{{3}}.
725 Easy
725 of 995
Understanding +4 | -1
Consider the following statements

I. If AB{AB} and BA{BA} are both defined, then they must be equal i.e., AB=BAA B=B A.

II. If AB{AB} and BA{BA} are both defined, it is not necessary that AB=BA{AB}={BA}.

Choose the correct option.
View answer and explanation
Correct option: B
Only II is true
Option B (Only II is true) is correct because matrix multiplication is not always commutative, meaning AB is not necessarily equal to BA.
726 Easy
726 of 995
Application +4 | -1
If AA is a square matrix of order 3 such that A2=A^2= 2A2 A, then ∣A∣2|A|^2 is equal to
View answer and explanation
Correct option: B
8∣A∣8|A|
Option B (8∣A∣8|A|) is correct because from the given condition, we can deduce that ∣A∣2=8∣A∣|A|^2 = 8|A|.
727 Easy
727 of 995
Application +4 | -1
If A=[1              xx2      4y]A = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;\;\;\;\;\;x}\\ {{x^2}\;\;\;4y} \end{array}} \right] and   B=[−3          11            0],  adj  A+B=[1        00          1]\;B = \left[ {\begin{array}{cccccccccccccccccccc} { - 3\;\;\;\;\;1}\\ {1\;\;\;\;\;\;0} \end{array}} \right],\;\mathrm{adj}\;A + B = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;\;\;0}\\ {0\;\;\;\;\;1} \end{array}} \right], Then values of xx and yy are
View answer and explanation
Correct option: A
1,11,1
Option A (1,11,1) is correct because substituting x=1x = 1 and y=1y = 1 into the equation adjA+B=Iadj A + B = I satisfies the matrix equality.
728 Easy
728 of 995
Application +4 | -1
Let A=[12−51]and  A−1=xA+yIA = \left[ {\begin{array}{cccccccccccccccccccc} 1&2\\ { - 5}&1 \end{array}} \right]\mathrm{and}\;{A^{ - 1}} = xA + yI, then the value of xx and yy are
View answer and explanation
Correct option: A
x=−111,  y=211x = \frac{{ - 1}}{{11}},\;y = \frac{2}{{11}}
Option A (x=−111,  y=211x = \frac{{ - 1}}{{11}},\;y = \frac{2}{{11}}) is correct because these values satisfy the equation A−1=xA+yI{A^{ - 1}} = xA + yI.
729 Easy
729 of 995
Understanding +4 | -1
Let for any matrix M,  M−1  M,\;{M^{ - 1}}\; exist which of the following is not true?
View answer and explanation
Correct option: B
(M2)−1=(  M−1)2{\left( {{M^2}} \right)^{ - 1}} = {\left( {\;{M^{ - 1}}} \right)^2}
The correct answer is B. The inverse of M squared is the inverse of M raised to the power of -2, not 2.
730 Easy
730 of 995
Application +4 | -1
If xayb=em,xcyd=en,Δ1=∣mbnd∣,x^a y^b=e^m, x^c y^d=e^n, \Delta_1=\left|\begin{array}{ll}m & b \\ n & d\end{array}\right|, Δ2=∣amcn∣\Delta_2=\left|\begin{array}{ll}a & m \\ c & n\end{array}\right| and Δ3=∣abcd∣,\Delta_3=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|, then the values of xx and yy are
View answer and explanation
Correct option: D
eΔ1/Δ3e^{\Delta_1 / \Delta_3} and eΔ2/Δ3e^{\Delta_2 / \Delta_3}
Option D (eΔ1/Δ3e^{\Delta_1 / \Delta_3} and eΔ2/Δ3e^{\Delta_2 / \Delta_3}) is correct because it follows from applying Cramer's rule to solve the system of equations after taking logarithms.
731 Easy
731 of 995
Analyzing +4 | -1
The system of equations kx + y+ z = 1, x+ ky + z = k and x +y +zk = k2{k^2} has no solution, If k is equal to
View answer and explanation
Correct option: D
-2
Option D (-2) is correct because when k = -2, the system of equations has no solution, as the determinant is zero and the equations are inconsistent.
732 Medium
732 of 995
Understanding +4 | -1
If A=[3457]A = \left[ {\begin{array}{cccccccccccccccccccc} 3&4\\ 5&7 \end{array}} \right] then A.(adj  A)A.\left( {\mathrm{adj}\;A} \right) is equal to
View answer and explanation
Correct option: C
∣A∣I\left| A \right|I
Option C (∣A∣I\left| A \right|I) is correct because A.(adj;A)A.\left( {\mathrm{adj\\;}A} \right) is equal to the determinant of A multiplied by the identity matrix.
733 Medium
733 of 995
Application +4 | -1
If a,b,ca, b, c are positive integers such that a>b>ca > b > c and ∣111abca2b2c2∣=−2\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array}\right|=-2 then 3a+7b−10c3 a+7 b-10 c equals
View answer and explanation
Correct option: D
13
Option D (13) is correct because by analyzing the given determinant condition, we can determine the values of a, b, and c, and then calculate 3a + 7b - 10c.
734 Easy
734 of 995
Understanding +4 | -1
The inverse of a skew-symmetric matrix (if it exists) is
View answer and explanation
Correct option: B
A skew-symmetric matrix
Option B (A skew-symmetric matrix) is correct because the inverse of a skew-symmetric matrix, if it exists, is also skew-symmetric.
735 Easy
735 of 995
Application +4 | -1
Let A=(aij)3×3A=\left(a_{i j}\right)_{3 \times 3} be a matrix with aij∈Ca_{i j} \in {C}. Let BB be a matrix obtained by interchanging two columns of A.A. Then det⁡(A+B)\operatorname{det}(A+B) is equal to
View answer and explanation
Correct option: B
0
Option C (A is non-singular matrix) is correct because the given equation can be rearranged to show that A has an inverse, implying it is non-singular.
736 Easy
736 of 995
Understanding +4 | -1
If AA is a square matrix then which one of the following is not a symmetric matrix
View answer and explanation
Correct option: D
A−A′A-A^{\prime}
Option D (A−A′A-A^{\prime}) is correct because A−A′A-A^{\prime} is a skew-symmetric matrix, not a symmetric matrix.
737 Easy
737 of 995
Understanding +4 | -1
If AA is an invertible matrix of order nn, then the determinant of adj(A)adj\left( A \right) is equal to
View answer and explanation
Correct option: C
∣A  ∣n−1  |A\;{\left. \right|^{n - 1}}\;
Option C (∣A∣n−1|A|^{n-1}) is correct because the determinant of the adjugate of a matrix A is equal to the determinant of A raised to the power of (n-1), where n is the order of the matrix.
738 Easy
738 of 995
Analyzing +4 | -1
If the trivial solution is the only solution of the system of equations x−ky+z=0,  kx+3y−kz=0,   3x+y−z=0x - ky + z = 0,\,\,kx + 3y - kz = 0,\,\,\,3x + y - z = 0 Then, the set of values of kk is
View answer and explanation
Correct option: B
R−{2,−3}R-\{2,-3\}
Option B (R−{2,−3}R-\left\{2,-3\right\}) is correct because the trivial solution is the only solution when the determinant of the coefficient matrix is non-zero, which means kk cannot be 2 or -3.
739 Medium
739 of 995
Application +4 | -1
Let A[12−51]A\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right] and A−1=xA+yIA^{-1}=x A+y I, then the value of (2x+3y)(2 x+3 y) is______
View answer and explanation
Correct option: A
0.36
Option A (0.36) is correct because by calculating the inverse of matrix A and comparing it with the given equation, we can find the values of x and y, and then calculate 2x + 3y.
740 Easy
740 of 995
Understanding +4 | -1
If AA is a singular matrix, then adj AA is
View answer and explanation
Correct option: B
singular
Option B (singular) is correct because if A is singular, |A| = 0, and |adj A| = |A|^(n-1) = 0, which means adj A is also singular.
741 Medium
741 of 995
Application +4 | -1
If [12x4−1724−6]\left[ {\begin{array}{cccccccccccccccccccc} 1&2&x\\ 4&{ - 1}&7\\ 2&4&{ - 6} \end{array}} \right] is a singular matrix, then xx is equal to
View answer and explanation
Correct option: C
−3- 3
Option C (-3) is correct because when the determinant of the given matrix is set to zero (since it's a singular matrix), solving for x yields x = -3.
742 Easy
742 of 995
Application +4 | -1
If ∣xyz−xyzx−yz∣=kxyz,\left| {\begin{array}{cccccccccccccccccccc} x&y&z\\ { - x}&y&z\\ x&{ - y}&z \end{array}} \right| = kxyz, then kk is equal to
View answer and explanation
Correct option: C
4
Option C (4) is correct because the determinant simplifies to 4xyz4xyz after applying row operations, thus k=4k=4.
743 Easy
743 of 995
Application +4 | -1
If ω\omega is a complex cube root of unity, then the value of ∣x+1ωω2ωx+ω21ω21x+ω∣\left| {\begin{array}{cccccccccccccccccccc} {x + 1}&\omega &{{\omega ^2}}\\ \omega &{x + {\omega ^2}}&1\\ {{\omega ^2}}&1&{x + \omega } \end{array}} \right| is
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the determinant simplifies to x2, which is not among the provided options.
744 Easy
744 of 995
Application +4 | -1
If A=[0110]A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right], then A2A^2 is equal to
View answer and explanation
Correct option: D
[1001]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]
Option D (\left[\begin{{array}}{{ll}} 1 & 0 \\ 0 & 1 \end{{array}}\right]\left[\begin{{array}}{{ll}} 1 & 0 \\ 0 & 1 \end{{array}}\right]) is correct because the matrix multiplication of A by itself results in the identity matrix.
745 Easy
745 of 995
Analyzing +4 | -1
The value of Δ=∣(ax+a−x)2(ax−a−x)21(ay+a−y)2(ay−a−y)21(az+a−z)2(az−a−z)21∣,\Delta=\left|\begin{array}{lll}\left(a^x+a^{-x}\right)^2 & \left(a^x-a^{-x}\right)^2 & 1 \\ \left(a^y+a^{-y}\right)^2 & \left(a^y-a^{-y}\right)^2 & 1 \\ \left(a^z+a^{-z}\right)^2 & \left(a^z-a^{-z}\right)^2 & 1\end{array}\right|, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because after applying a column operation, two columns become identical, resulting in a zero determinant.
746 Easy
746 of 995
Analyzing +4 | -1
Let SS be the set of all λ∈R\lambda \in R for which the system of linear equations

2x−y+2z=2x−2y+λz=−4x+λy+z=4\begin{aligned} &2 x-y+2 z=2 \\ &x-2 y+\lambda z=-4 \\ &x+\lambda y+z=4 \end{aligned}

has no solution, than the set SS
View answer and explanation
Correct option: B
Contains exactly two elements
Option B (Contains exactly two elements) is correct because the set S contains two values of lambda for which the system has no solution.
747 Easy
747 of 995
Application +4 | -1
In a square matrix AA of order 3 the elements aiia_{i i} 's are the sum of the roots of the equation x2−(a+b)x+ab=0;x^2-(a+b) x+a b=0; ai, i+1a_{i, \,i+1} 's are the product of the roots, ai, i−1a_{i, \,i-1} 's are all unity and the rest of the elements are all zero.

The value of the det⁡(A)\operatorname{det}(A) is equal to
View answer and explanation
Correct option: D
(a2+b2)(a+b)\left(a^2+b^2\right)(a+b)
The correct answer is D, (a2+b2)(a+b)\left(a^2+b^2\right)(a+b), which is the determinant of the given matrix A after expanding and simplifying the expression.
748 Easy
748 of 995
Application +4 | -1
Suppose a∈Ra \in {R}. Let f(x)=∣x+axxxx+axxxx+a∣f(x)=\left|\begin{array}{ccc} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{array}\right| Then f(2x)−f(x)f(2 x)-f(x) is equal to
View answer and explanation
Correct option: A
3xa23 x a^2
The correct answer is A (3xa23 x a^2) because after evaluating f(x) and f(2x), the difference f(2x) - f(x) simplifies to 3xa23 x a^2.
749 Medium
749 of 995
Application +4 | -1
If A=[010001pqr]A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r\end{array}\right] then A3−rA2−qA=A^3-r A^2-q A=
View answer and explanation
Correct option: A
pIp I
Option A (pIpI) is correct because by calculating A2 and A3, and substituting them into the expression A3 - rA2 - qA, we find that it simplifies to pI.
750 Easy
750 of 995
Application +4 | -1
If AA and BB are two square matrices such that B=−A−1BA,B=-A^{-1} B A, then (A+B)2(A+B)^2 is equal to
View answer and explanation
Correct option: B
A2+B2A^2+B^2
Option B (A2+B2) is correct because given B = -A-1BA, it implies AB = -BA, which simplifies the expansion of (A+B)2.
Revision notes and key takeaways

Review MATRICES in your textbook or class notes, then use this set to check your understanding.

Showing 721-750 of 995 questions

Your practice

0 / 995 answered · 995 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.