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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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Showing 31-60 of 922 questions
31 Easy
31 of 922
Application +4 | -1
Let A=(1−1121−3111)A=\left(\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right). and B=(422−50α1−23)B=\left(\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right). If BB is the inverse of matrix AA, then α\alpha is
View answer and explanation
Correct option: A
5
Option A (5) is correct because when B is the inverse of A, their product AB equals the identity matrix, which allows us to solve for alpha.
32 Easy
32 of 922
Understanding +4 | -1
Let AA and BB are the non-singular square matrices, then which of the following is always correct?
View answer and explanation
Correct option: B
(AB)′=B′A′(A B)^{\prime}=B^{\prime} A^{\prime}
Option B ((AB)′=B′A′(A B)^{\prime}=B^{\prime} A^{\prime}) is correct because the transpose of a product of matrices is the product of their transposes in reverse order.
33 Easy
33 of 922
Analyzing +4 | -1
Let A={X=(x,y,z)T:PX=0\mathrm{A}=\left\{\mathrm{X}=(x, y, z)^{\mathrm{T}}: \mathrm{PX}=0\right. and x2+y2+z2=1}\left.x^2+y^2+z^2=1\right\}, where P=[121−23−419−1]P=\left[\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1\end{array}\right], then the set A :
View answer and explanation
Correct option: D
contains exactly two elements
The correct answer is contains exactly two elements because the system has infinitely many solutions, but the condition x²+y²+z²=1 restricts the solution set to two points.
34 Easy
34 of 922
Understanding +4 | -1
Consider the matrix A=[2519−735−25/21231−517]A=\left[\begin{array}{cccc}2 & 5 & 19 & -7 \\ 35 & -2 & 5 / 2 & 12 \\ \sqrt{3} & 1 & -5 & 17\end{array}\right]

Now, consider the following statements

I. The order of the matrix is 4×34 \times 3 and number of elements is 12 .

II. The elements a13,a21,a33{a}_{13}, {a}_{21}, {a}_{33} are respectively 19,35,−519,35,-5. Choose the correct option.
View answer and explanation
Correct option: B
Only II is true
Option B (Only II is true) is correct because the order of the matrix is 3x4, and the specified elements match the given values.
35 Easy
35 of 922
Application +4 | -1
If A=[sin⁡α−cos⁡α0cos⁡αsin⁡α0001]A=\left[\begin{array}{ccc}\sin \alpha & -\cos \alpha & 0 \\ \cos \alpha & \sin \alpha & 0 \\ 0 & 0 & 1\end{array}\right] then A−1{A}^{-1} is equal to:
View answer and explanation
Correct option: C
adj⁡A\operatorname{adj} {A}
Option C (adj A) is correct because for a matrix A with determinant 1, the inverse of A is equal to the adjugate of A.
36 Easy
36 of 922
Analyzing +4 | -1
The system of equations αx+y+z=α−1,   x+αy+z=α−1,   and x+y+αz=α−1 has no solution, if α is \alpha x + y + z = \alpha - 1,\,\,\,x + \alpha y + z = \alpha - 1,\,\,{\text{ and }}x + y + \alpha z = \alpha - 1{\text{ has no solution, if }}\alpha {\text{ is }}
View answer and explanation
Correct option: C
-2
Option C (-2) is correct because when α=−2\alpha = -2, the system of equations has no solution, as the determinant is zero and the equations are inconsistent.
37 Easy
37 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⁡(adj2A⁡))∣=(16)n|\operatorname{adj}(\operatorname{adj}(\operatorname{adj2A}))|=(16)^{n}, then nn is equal to
View answer and explanation
Correct option: B
10
Option B (10) is correct because it correctly applies the property of adjoints of adjoints of adjoints and determinants.
38 Easy
38 of 922
Application +4 | -1
If A=[0100]A=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right] and AB=OA B=O, then B=B=
View answer and explanation
Correct option: D
[−1000]\left[\begin{array}{cc}-1 & 0 \\ 0 & 0\end{array}\right]
Option D is correct. Given AB = O, and A is a non-zero matrix, B must be a matrix such that when multiplied by A, the result is a zero matrix. Option D satisfies this condition.
39 Easy
39 of 922
Analyzing +4 | -1
If the system of equations 2x+3y−z=0,x+ky−2z=0 and 2x−y+z=02x + 3y - z = 0,x + ky - 2z = 0{\kern 1pt} \,\mathrm{and}\,2x - y + z = 0 has a non-trivial solution (x, y, z), then xy+yz+zx+k\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k is equal to
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because after solving for kk and finding the ratios xy\frac{x}{y}, yz\frac{y}{z}, and zx\frac{z}{x}, their sum plus kk equals 12\frac{1}{2}.
40 Easy
40 of 922
Application +4 | -1
Let
A=[abb2−a2−ab]A=\left[\begin{array}{cc} a b & b^2 \\ -a^2 & -a b \end{array}\right]
where ab≠0a b \neq 0. Then
View answer and explanation
Correct option: B
A2=OA^2=O
Option B (A2=OA^2=O) is correct because the square of the given matrix A results in a zero matrix.
41 Easy
41 of 922
Application +4 | -1
If A=[1−253]A=\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right], then A+ATA+A^T equals:
View answer and explanation
Correct option: A
[2336]\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]
Option A (\left[\begin{{array}}{{ll}}2 & 3 \\ 3 & 6\end{{array}}\right]\left[\begin{{array}}{{ll}}2 & 3 \\ 3 & 6\end{{array}}\right]) is correct because it is the result of adding the matrix A to its transpose.
42 Easy
42 of 922
Analyzing +4 | -1
For the equations x+2y+3z=1,  2x+y+3z=2  x + 2y + 3z = 1,\;2x + y + 3z = 2\; and   5x+5y+9z=4\;5x + 5y + 9z = 4
View answer and explanation
Correct option: A
There is only one solution
Option A (There is only one solution) is correct because the determinant of the coefficient matrix is non-zero, indicating a unique solution.
43 Easy
43 of 922
Application +4 | -1
If 3X+2Y=I3 \mathrm{X}+2 \mathrm{Y}=\mathrm{I} and 2X−Y=O2 \mathrm{X}-\mathrm{Y}=\mathrm{O}, where I\mathrm{I} and O\mathrm{O} are unit and null matrices of order 3 respectively, then
View answer and explanation
Correct option: C
X=(17)I,Y=(27)I\mathrm{X}=\left(\frac{1}{7}\right) \mathrm{I}, \mathrm{Y}=\left(\frac{2}{7}\right) \mathrm{I}
Option C (X = (1/7)I, Y = (2/7)I) is correct because solving the system of equations for X and Y yields these values.
44 Easy
44 of 922
Application +4 | -1
If the system of equation
2x+5y+8z=0x+4y+7z=06x+4y−λz=0\begin{array}{l} 2 x+5 y+8 z=0 \\ x+4 y+7 z=0 \\ 6 x+4 y-\lambda z=0 \end{array} has a non trivial solution then λ\lambda is equal to -
View answer and explanation
Correct option: B
-2
Option D (11) is correct because K is equal to the determinant of A, which calculates to 11.
45 Easy
45 of 922
Application +4 | -1
Let a,ba, b and cc be three real numbers satisfying
[abc][197827737]=[000]   (1)\left[\begin{array}{lll} a & b & c \end{array}\right]\left[\begin{array}{lll} 1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \end{array}\right] \,\,\,(1)

If the point P(a,b,c)P(a, b, c) with reference to (1), lies on the plane 2x+y+z=12 x+y+z=1, then the value of 7a+b+c7 a+b+c is
View answer and explanation
Correct option: D
6
Option D (6) is correct because solving the system of equations and substituting into the plane equation gives k=1k=1, leading to 7a+b+c=67a+b+c = 6.

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
46 Easy
46 of 922
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 det⁡(A)\det (A) is not divisible by pp
View answer and explanation
Correct option: D
p3−p2p^3-p^2
Option D (p3−p2p^3-p^2) is correct because it represents the number of matrices in TpT_p whose determinant is not divisible by pp.
47 Easy
47 of 922
Application +4 | -1
If P=[cos⁡π6sin⁡π6−sin⁡π6cos⁡π6],A=[1101]P=\left[\begin{array}{rr}\cos \frac{\pi}{6} & \sin \frac{\pi}{6} \\ -\sin \frac{\pi}{6} & \cos \frac{\pi}{6}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] and Q=PAPT,Q=P A P^T, then PTQ2007PP^T Q^{2007} P is equal to
View answer and explanation
Correct option: B
[1200701]\left[\begin{array}{cc}1 & 2007 \\ 0 & 1\end{array}\right]
Option B ([1 2007
0 1]) is correct because using the properties of matrix transpose and the given matrices, we can simplify the expression to A2007, which can be easily calculated.
48 Easy
48 of 922
Application +4 | -1
If AA is a square matrix such that A  (adj  A)=[400040004],A\;\left( {\mathrm{adj\;}A} \right) = \left[ {\begin{array}{cccccccccccccccccccc} 4&0&0\\ 0&4&0\\ 0&0&4 \end{array}} \right], then ∣adj  A∣=\left| {\mathrm{adj}\;A} \right| =
View answer and explanation
Correct option: B
16
Option B (16) is correct because |adj A| = |A|^(n-1) and A(adj A) = |A|I, so |A| = 4 and |adj A| = 4(3-1) = 16.
49 Easy
49 of 922
Application +4 | -1
If A=[1log⁡balog⁡ab1]A = \left[ {\begin{array}{cccccccccccccccccccc} 1&{{{\log }_b}a}\\ {{{\log }_a}b}&1 \end{array}} \right], then ∣A∣\left| A \right| is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because the determinant of the matrix A is calculated as 1 - (log_a b)(log_b a) which simplifies to 1 - 1 = 0.
50 Easy
50 of 922
Application +4 | -1
Let A2−A+I=OA^2-A+I=O, then inverse of AA is
View answer and explanation
Correct option: A
I−AI-A
Option A (I−AI-A) is correct because multiplying AA by (I−A)(I-A) results in the identity matrix, thus (I−A)(I-A) is the inverse of AA.
51 Easy
51 of 922
Understanding +4 | -1
If AA is a skew symmetric matrix of odd order, then its inverse is
View answer and explanation
Correct option: D
Non existent
Option D (Non existent) is correct because a skew-symmetric matrix of odd order has a determinant of 0, making it non-invertible.
52 Easy
52 of 922
Application +4 | -1
If X=[3−41−1]X=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right], then the value of XnX^n is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because substituting n=2 into options A, B, and C does not yield the correct X2 matrix.
53 Easy
53 of 922
Application +4 | -1
If A=[0320]A=\left[\begin{array}{ll}0 & 3 \\ 2 & 0\end{array}\right] and A−1=λ(A)A^{-1}=\lambda(A), then λ\lambda
View answer and explanation
Correct option: D
16\frac{1}{6}
Option D (16\frac{1}{6}) is correct because by finding the inverse of A and comparing it to λ(A)\lambda(A), we can solve for λ\lambda and find it to be 16\frac{1}{6}.
54 Easy
54 of 922
Application +4 | -1
If ω\omega is a non-real cube root of unity and nn is not a multiple of 3, then
Δ=∣1ωnω2nω2n1ωnωnω2n1∣\Delta=\left|\begin{array}{ccc}1 & \omega^n & \omega^{2 n} \\ \omega^{2 n} & 1 & \omega^n \\ \omega^n & \omega^{2 n} & 1\end{array}\right| is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because when n is not a multiple of 3, 1+ωn+ω2n=01 + \omega^n + \omega^{2n} = 0, making the determinant equal to zero.
55 Easy
55 of 922
Analyzing +4 | -1
If Δ1=∣xbbaxbaax∣{\Delta _1} = \left| {\begin{array}{cccccccccccccccccccc} x&b&b\\ a&x&b\\ a&a&x \end{array}} \right| and Δ2=∣xbax∣{\Delta _2} = \left| {\begin{array}{cccccccccccccccccccc} x&b\\ a&x \end{array}} \right| are the given determinants, then
View answer and explanation
Correct option: B
ddx(Δ1)=3Δ2\frac{d}{{dx}}\left( {{\Delta _1}} \right) = 3{\Delta _2}
Option B (ddx(Δ1)=3Δ2\frac{d}{{dx}}\left( {{\Delta _1}} \right) = 3{\Delta _2}) is correct because the derivative of Δ1\Delta_1 with respect to x is three times Δ2\Delta_2.
56 Easy
56 of 922
Application +4 | -1
The matrix A=[1−3−4−1341−3−4]A=\left[\begin{array}{rrr}1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4\end{array}\right] is nilpotent of index:
View answer and explanation
Correct option: A
2
Option A (2) is correct because the matrix A squared results in a null matrix, indicating a nilpotent index of 2.
57 Easy
57 of 922
Application +4 | -1
If matrix A=[10−1345067]aA=\left[\begin{array}{ccc}1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & 6 & 7\end{array}\right] a
and its inverse is denoted by
A−1=[a11a12a13a21a22a23a31a32a33]\mathrm{A}^{-1}=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right], then the value of a23=a_{23}=
View answer and explanation
Correct option: D
25\frac{2}{5}
Option D (25\frac{2}{5}) is correct because it is calculated by finding the cofactor of the element in the matrix and dividing by the determinant of the matrix.
58 Easy
58 of 922
Application +4 | -1
If A=[2−3−41]A=\left[\begin{array}{cc}2 & -3 \\ -4 & 1\end{array}\right], then adj (3A2+12A)\left(3 A^2+12 A\right) is equal to:
View answer and explanation
Correct option: C
[51638472]\left[\begin{array}{ll}51 & 63 \\ 84 & 72\end{array}\right]
Option D (\left[\begin{{ll}}51 & 84 \\ 63 & 72\end{{ll}}\right]\left[\begin{{ll}}51 & 84 \\ 63 & 72\end{{ll}}\right]) is correct. By calculating 3A2+12A3A^2 + 12A and then finding its adjugate, we arrive at the correct matrix.
59 Easy
59 of 922
Analyzing +4 | -1
Which of the following matrices can NOT be obtained from the matrix [−121−1]\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right] by a single elementary row operation?
View answer and explanation
Correct option: C
[−12−27]\left[\begin{array}{ll}-1 & 2 \\ -2 & 7\end{array}\right]
The matrix in option C [−12−27]\left[\begin{array}{ll}-1 & 2 \\ -2 & 7\end{array}\right] cannot be obtained from the given matrix by a single elementary row operation.
60 Easy
60 of 922
Application +4 | -1
If A=[4     23     4],∣adj  A∣A = \left[ {\begin{array}{cccccccccccccccccccc} {4\;\;\,2}\\ {3\;\;\,4} \end{array}} \right],\left| {\mathrm{adj}\;A} \right| is equal to
View answer and explanation
Correct option: C
10
Option C (10) is correct because the determinant of the adjugate of a 2x2 matrix is equal to the determinant of the original matrix.
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