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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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91 Easy
91 of 922
Analyzing +4 | -1
If the system of linear equations
2x + 2ay + az = 0
2x + 3by + bz = 0
2x + 4cy + cz = 0 where a,b,c∈Ra,b,c \in R are non-zero and distinct; has a non-zero solution, then
View answer and explanation
Correct option: B
1a,1b,1c\frac{1}{a},\frac{1}{b},\frac{1}{c} are in A.P
The correct answer is B (1a,1b,1c\frac{1}{a},\frac{1}{b},\frac{1}{c} are in A.P) because the determinant of the coefficient matrix must be zero for a non-zero solution, leading to the given relationship between a, b, and c.
92 Easy
92 of 922
Analyzing +4 | -1
Let a,  b,  ca,\;b,\;c be any real numbers. Suppose that there are real numbers x,  y,  zx,\;y,\;z not all zero such that x=cy+bz,  y=az+cx,  x = cy + bz,\;y = az + cx,\; and   z=bx+ay  \;z = bx + ay\; have non-zero solution. Then, a2+b2+c2+2  abc  {a^2} + {b^2} + {c^2} + 2\;abc\; is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the determinant of the coefficient matrix must be zero for non-zero solutions, leading to the given equation equaling 1.
93 Easy
93 of 922
Understanding +4 | -1
If A=(23−i−i3+iπ7+ii7−ie)A=\left(\begin{array}{ccc}2 & 3-i & -i \\ 3+i & \pi & 7+i \\ i & 7-i & e\end{array}\right), then AA is
View answer and explanation
Correct option: B
Hermitian
Option B (Hermitian) is correct because A∗=AA^* = A, meaning the conjugate transpose of A is equal to A.
94 Easy
94 of 922
Application +4 | -1
If for aa matrix A,∣A∣=6A,|A|=6 and adj⁡A=[1−24411−1k0]\operatorname{adj} A=\left[\begin{array}{rrr}1 & -2 & 4 \\ 4 & 1 & 1 \\ -1 & k & 0\end{array}\right], then kk is equal to
View answer and explanation
Correct option: B
0
Option D (2) is correct because by calculating the determinant of adj A and using the property |adj A| = |A|^(n-1), we can solve for k.
95 Easy
95 of 922
Application +4 | -1
The number of symmetric matrices of order 3, with all the entries from the set {0,1,2,3,4,5,6,7,8,9}\{0,1,2,3,4,5,6,7,8,9\} is
View answer and explanation
Correct option: B
10610^{6}
Option B (10610^6) is correct because a symmetric matrix of order 3 has 6 independent entries, each of which can be any of the 10 digits.
96 Easy
96 of 922
Application +4 | -1
If the adjoint of a 3×33 \times 3 matrix P\mathrm{P} is [144217113]\left[\begin{array}{ccc}1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3\end{array}\right], then the possible values of the determinant of P\mathrm{P} are
View answer and explanation
Correct option: A
±2\pm 2
Option A (±2\pm 2) is correct because the determinant of the adjoint of a matrix P is equal to the square of the determinant of P, which leads to the possible values of ±2\pm 2.
97 Easy
97 of 922
Analyzing +4 | -1
Let Δ=∣1+x1y11+x1y21+x1y31+x2y11+x2y21+x2y31+x3y11+x3y21+x3y3∣\Delta = \left| {\begin{array}{cccccccccccccccccccc} {1 + {x_1}{y_1}}&{1 + {x_1}{y_2}}&{1 + {x_1}{y_3}}\\ {1 + {x_2}{y_1}}&{1 + {x_2}{y_2}}&{1 + {x_2}{y_3}}\\ {1 + {x_3}{y_1}}&{1 + {x_3}{y_2}}&{1 + {x_3}{y_3}} \end{array}} \right|, then value of Δ\Delta is
View answer and explanation
Correct option: D
0
The correct answer is D (0) because the determinant can be simplified using column operations, resulting in two proportional or identical columns.
98 Easy
98 of 922
Analyzing +4 | -1
The values of λ\lambda and μ\mu such that the system of equations x+y+z=6,  3x+5y+5z=26,  x+2y+λz=μx + y + z = 6,\,\,3x + 5y + 5z = 26,\,\,x + 2y + \lambda z = \mu has no solution, are
View answer and explanation
Correct option: D
λ=2, μ≠10\lambda = 2,\,\mu \ne 10
Option D (λ=2, μ≠10\lambda = 2,\,\mu \ne 10) is correct because for the system to have no solution, the determinant of the coefficients must be zero, and the equations must be inconsistent.
99 Easy
99 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} equals
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because none of the provided options satisfy the condition for being the inverse of (I-A) given A5=OA^5 = O and An≠IA^n \neq I for 1≤n≤41 \leq n \leq 4.
100 Easy
100 of 922
Application +4 | -1
If P=[i0−i0−ii−ii0]P = \left[ {\begin{array}{cccccccccccccccccccc} i&0&{ - i}\\ 0&{ - i}&i\\ { - i}&i&0 \end{array}} \right] and Q=[−ii00i−i]Q = \left[ {\begin{array}{cccccccccccccccccccc} { - i}&i\\ 0&0\\ i&{ - i} \end{array}} \right], then PQPQ is equal to
View answer and explanation
Correct option: B
[2−2−11−11]\left[ {\begin{array}{cccccccccccccccccccc} 2&{ - 2}\\ { - 1}&1\\ { - 1}&1 \end{array}} \right]
Option B (\left[ {\begin{{array}}{{{*{20}{c}}}} 2&{ - 2}\\ { - 1}&1\\ { - 1}&1 \end{{array}}} \right]\left[ {\begin{{array}}{{{*{20}{c}}}} 2&{ - 2}\\ { - 1}&1\\ { - 1}&1 \end{{array}}} \right]) is correct because it is the result of the matrix multiplication of P and Q.
101 Easy
101 of 922
Analyzing +4 | -1
Δ=∣a+xbcbx+cacax+b∣.\Delta=\left|\begin{array}{ccc}a+x & b & c \\ b & x+c & a \\ c & a & x+b\end{array}\right|. Which of the following is a factor for the above determinant?
View answer and explanation
Correct option: B
x+(a+b+c)x+(a+b+c)
Option B (x+(a+b+c)x+(a+b+c)) is correct because applying a column operation allows factoring out this expression, indicating it's a factor of the determinant.
102 Easy
102 of 922
Application +4 | -1
Let AA be a matrix of order 3 and let Δ\Delta denotes the value of determinant AA. Then determinant (−2A)=(-2 A)=
View answer and explanation
Correct option: A
−8Δ-8 \Delta
Option A (-8 Δ\Delta) is correct because when a matrix A of order 3 is multiplied by -2, its determinant is multiplied by (−2)3=−8(-2)^3 = -8.
103 Easy
103 of 922
Application +4 | -1
Let A=[0−tan⁡(α/2)tan⁡(α/2)0]α≠(2n+1)π,n∈I,(I+A)(I−A)−1A=\left[\begin{array}{cc}0 & -\tan (\alpha / 2) \\ \tan (\alpha / 2) & 0\end{array}\right]\alpha \neq(2 n+1) \pi, n \in {I},(I+A)(I-A)^{-1} equals
View answer and explanation
Correct option: A
[cos⁡α−sin⁡αsin⁡αcos⁡α]\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]
Option A (\begin{{bmatrix}}\cos \alpha & -\sin \alpha \\\\ \sin \alpha & \cos \alpha \end{{bmatrix}}\begin{{bmatrix}}\cos \alpha & -\sin \alpha \\\\ \sin \alpha & \cos \alpha \end{{bmatrix}}) is correct because it results from simplifying the matrix expression (I+A)(I−A)−1(I+A)(I-A)^{-1} using trigonometric identities.
104 Easy
104 of 922
Application +4 | -1
If AA and BB are square matrices of same order such that A+B=ABA+B=A B, then
View answer and explanation
Correct option: C
AB=BAA B=B A
Option C (AB=BAA B=B A) is correct because from A+B=ABA+B=A B, it can be shown that (I−A)(I−B)=I(I-A)(I-B) = I, implying AB=BAA B = B A.
105 Easy
105 of 922
Analyzing +4 | -1
Let A=(10004−1012−3)A=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right). Then the sum of the diagonal elements of the matrix (A+I)11(A+I)^{11} is equal to
View answer and explanation
Correct option: D
4097
Option D (4097) is correct because the sum of the diagonal elements (trace) of the matrix (A+I)11(A+I)^{11} is calculated using the property A2=AA^2 = A.

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
106 Easy
106 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 α\alpha is equal to
View answer and explanation
Correct option: A
±3\pm 3
Option A (±3\pm 3) is correct because it satisfies the determinant equation ∣A3∣=125|A^3| = 125 given the matrix A.
107 Easy
107 of 922
Application +4 | -1
If matrix A=[ay]2×2A=\left[a_y\right]_{2 \times 2}, where ay=1a_y=1, if i≠ji \neq j and 0 if i=ji=j, then A2A^2 is equal to
View answer and explanation
Correct option: A
ll
Option A (ll) is correct because the given matrix A, when squared, results in the identity matrix, denoted as l.
108 Easy
108 of 922
Understanding +4 | -1
Let AA be a 3×33 \times 3 matrix and S={(xyz)x,y,z∈R}S=\left\{\left(\begin{array}{l} x \\ y \\ z \end{array}\right) x, y, z \in {R}\right\}

Define f:S→Sf: S \rightarrow S by f(xyz)=A(xyz)f\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=A\left(\begin{array}{l} x \\ y \\ z \end{array}\right)

Suppose f(xyz)=(000)⇒x=y=z=0f\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right) \Rightarrow x=y=z=0. Then
View answer and explanation
Correct option: A
ff is one-to-one
Option A (ff is one-to-one) is correct because the condition f(xyz)=(000)⇒x=y=z=0f(\begin{array}{l}x \\ y \\ z\end{array})=\left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right) \Rightarrow x=y=z=0 implies that the null space of A contains only the zero vector, meaning that the linear transformation is injective (one-to-one).
109 Easy
109 of 922
Analyzing +4 | -1
If D=∣11111+x1111+y∣D = \left| {\begin{array}{cccccccccccccccccccc} 1&1&1\\ 1&{1 + x}&1\\ 1&1&{1 + y} \end{array}} \right| for x≠0,y≠0,x \ne 0,y \ne 0, then DD is
View answer and explanation
Correct option: B
Divisible by both xx and yy
Option B (Divisible by both xx and yy) is correct because the determinant simplifies to xyxy, which is clearly divisible by both xx and yy.
110 Easy
110 of 922
Application +4 | -1
The roots of the equation ∣14201−2512x5x2∣=0\left| {\begin{array}{cccccccccccccccccccc} 1&4&{20}\\ 1&{ - 2}&5\\ 1&{2x}&{5{x^2}} \end{array}} \right| = 0
View answer and explanation
Correct option: B
−1,2- 1,2
Option B (−1,2- 1,2) is correct because these are the roots of the quadratic equation obtained after simplifying the determinant.
111 Easy
111 of 922
Application +4 | -1
The number of distinct values of a 2×22 \times 2 determinant whose entries are from the set {−1,0,1}\{-1,0,1\}, is
View answer and explanation
Correct option: C
5
Option C (5) is correct because there are 5 distinct possible values for the 2x2 determinant.
112 Easy
112 of 922
Application +4 | -1
If A=[1          2−4−1],  then  A−1A = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;\;\;\;2}\\ { - 4 - 1} \end{array}} \right],\;\mathrm{then}\;{A^{ - 1}} is
View answer and explanation
Correct option: A
17[−1    −24            1]\frac{1}{7}\left[ {\begin{array}{cccccccccccccccccccc} { - 1\,\,\,\, - 2}\\ {4\;\;\;\;\;\;1} \end{array}} \right]
Option A (\frac{1}{7}\left[ {\begin{{array}}{*{20}{c}} { - 1\,\,\,\, - 2}\\ {4\,\,\,\,\,\,1} \end{{array}} \right]\frac{1}{7}\left[ {\begin{{array}}{*{20}{c}} { - 1\,\,\,\, - 2}\\ {4\,\,\,\,\,\,1} \end{{array}} \right]) is correct because it is the inverse matrix calculated using the formula A-1 = (1/det(A)) * adj(A).
113 Easy
113 of 922
Application +4 | -1
Let P(x)=∣x+1231x+2312x+3∣,P(x)=\left|\begin{array}{ccc} x+1 & 2 & 3 \\ 1 & x+2 & 3 \\ 1 & 2 & x+3 \end{array}\right|, the product of zeros of P(x)P(x) is
View answer and explanation
Correct option: A
0
The correct answer is A (0) because the product of the zeros of the polynomial P(x) is equal to -P(0), which is 0 in this case.
114 Easy
114 of 922
Application +4 | -1
If A=[685423971]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 and skew-symmetric matrix, CC, then BB is
View answer and explanation
Correct option: A
[667625751]\left[ {\begin{array}{cccccccccccccccccccc} 6&6&7\\ 6&2&5\\ 7&5&1 \end{array}} \right]
Option A is correct. The symmetric matrix B is calculated as (A + A')/2, where A' is the transpose of A.
115 Easy
115 of 922
Application +4 | -1
If A=[x      −23              7]and  A−1=[734      117−334      217],  A = \left[ {\begin{array}{cccccccccccccccccccc} {x\;\;\; - 2}\\ {3\;\;\;\;\;\;\;7} \end{array}} \right]\mathrm{and}\;{A^{ - 1}} = \left[ {\begin{array}{cccccccccccccccccccc} {\frac{7}{{34}}\;\;\;\frac{1}{{17}}}\\ {\frac{{ - 3}}{{34}}\;\;\;\frac{2}{{17}}} \end{array}} \right],\; then the value of xx is
View answer and explanation
Correct option: D
4
Option D (4) is correct because by finding the inverse of matrix A and comparing it to the given inverse, we can solve for x.
116 Easy
116 of 922
Analyzing +4 | -1
The number of q∈(0,4π)q \in(0,4 \pi) for which the system of linear equations

3(sin⁡3θ)x−y+z=23(\sin 3 \theta) x-y+z=2

3(cos⁡2θ)x+4y+3z=33(\cos 2 \theta) x+4 y+3 z=3

6x+7y+7z=96 x+7 y+7 z=9

has no solution, is
View answer and explanation
Correct option: B
7
Option B (7) is correct because there are 7 values of θ\theta in the interval (0,4π)(0, 4\pi) for which the system of linear equations has no solution.
117 Easy
117 of 922
Application +4 | -1
If A=[−1−2−221−22−21]\mathrm{A}=\left[\begin{array}{ccc}-1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1\end{array}\right], then adj⁡A=\operatorname{adj} \mathrm{A}=
View answer and explanation
Correct option: D
3AT3 \mathrm{A}^{\mathrm{T}}
Option B is correct. By successively reducing the powers of A using the given equation, we can express A5 in the form aA + bI and then calculate 2a - 3b.
118 Easy
118 of 922
Application +4 | -1
If AA and BB are two non-singular matrices of the same order such that Br=IB^r=I for some positive integer r>1.r>1. Then A−1Br−1A−A−1B−1AA^{-1} B^{r-1} A-A^{-1} B^{-1} A is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because given Br=IB^r = I, it implies Br−1=B−1B^{r-1} = B^{-1}, hence A−1Br−1A−A−1B−1A=0A^{-1}B^{r-1}A - A^{-1}B^{-1}A = 0.
119 Easy
119 of 922
Analyzing +4 | -1
If AA is 3×33 \times 3 matrix and u\mathbf{u} is a vector. If AuA \mathbf{u} and u\mathbf{u} are orthogonal for all real uu, then matrix AA is aa
View answer and explanation
Correct option: A
singular
The correct answer is A (singular) because if AuA \mathbf{u} and u\mathbf{u} are orthogonal for all real u\mathbf{u}, then the determinant of A must be zero, implying A is singular.
120 Easy
120 of 922
Analyzing +4 | -1
If A1,A2,A3,…,A2n−1A_1, A_2, A_3, \ldots, A_{2 n-1} are n skew-symmetric matrices of same order, then B=∑r=1n(2r−1)(A2r−1)2r−1B = \sum\limits_{r = 1}^n {\left( {2r - 1} \right){{\left( {{A_{2r - 1}}} \right)}^{2r - 1}}} will be
View answer and explanation
Correct option: B
Skew-symmetric
Option B (Skew-symmetric) is correct because when taking the transpose of the given expression, it results in the negative of the original expression, indicating a skew-symmetric matrix.
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