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JEE • MATHEMATICS • MATRICES

Matrices practice · Medium

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Showing 61-90 of 408 questions

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61 Medium
61 of 408
Analyzing +4 | -1
The value of determinant ∣(ax+a−x)2(ax−a−x)21(bx+b−x)2(bx−b−x)21(cx+c−x)2(cx−c−x)21∣\left| {\begin{array}{cccccccccccccccccccc} {{{\left( {{a^x} + {a^{ - x}}} \right)}^2}}&{{{\left( {{a^x} - {a^{ - x}}} \right)}^2}}&1\\ {{{\left( {{b^x} + {b^{ - x}}} \right)}^2}}&{{{\left( {{b^x} - {b^{ - x}}} \right)}^2}}&1\\ {{{\left( {{c^x} + {c^{ - x}}} \right)}^2}}&{{{\left( {{c^x} - {c^{ - x}}} \right)}^2}}&1 \end{array}} \right| is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the determinant simplifies to 0 due to identical columns after applying column operations.
62 Medium
62 of 408
Analyzing +4 | -1
Let matrix A=[xy−z123112]A=\left[\begin{array}{ccc}x & y & -z \\ 1 & 2 & 3 \\ 1 & 1 & 2\end{array}\right] where x,y,z∈Nx, y, z \in N. If det⁡(adj⁡(adj⁡A))=28⋅34\operatorname{det}(\operatorname{adj}(\operatorname{adj} A))=2^8 \cdot 3^4 then the number of such matrices AA is

[Note: adj⁡A\operatorname{adj} A denotes adjoint of square matrix AA]
View answer and explanation
Correct option: C
55
Option C (55) is correct because it represents the number of possible matrices A, where the determinant condition leads to a stars and bars problem, resulting in 11C2=55{}^{11}C_2 = 55 solutions.
63 Medium
63 of 408
Analyzing +4 | -1
Let A=(mnpq),d=∣A∣≠0A=\left(\begin{array}{ll}m & n \\ p & q\end{array}\right), d=|A| \neq 0 and ∣A−d(Adj⁡A)∣=0|A-d(\operatorname{Adj} A)|=0. Then
View answer and explanation
Correct option: B
(1+d)2=(m+q)2(1+d)^2=(m+q)^2
Option B ((1+d)2=(m+q)2(1+d)^2=(m+q)^2) is correct because it is derived from the given determinant condition and the properties of the adjugate matrix.
64 Medium
64 of 408
Application +4 | -1
If ω≠1\omega \neq 1 is the complex cube root of unity and matrix H=[ω00ω]H=\left[\begin{array}{ll}\omega & 0 \\ 0 & \omega\end{array}\right], then H70\mathrm{H}^{70} is equal to
View answer and explanation
Correct option: D
H\mathrm{H}
Option D (HH) is correct because H70H^{70} simplifies to H using the property of complex cube roots of unity, where ω3n=1\omega^{3n} = 1.
65 Medium
65 of 408
Application +4 | -1
If ∣x3636x6x3∣=∣2x7x7272x∣=[45x5x4x45]=0,\left| {\begin{array}{cccccccccccccccccccc} x&3&6\\ 3&6&x\\ 6&x&3 \end{array}} \right| = \left| {\begin{array}{cccccccccccccccccccc} 2&x&7\\ x&7&2\\ 7&2&x \end{array}} \right| = \left[ {\begin{array}{cccccccccccccccccccc} 4&5&x\\ 5&x&4\\ x&4&5 \end{array}} \right] = 0, then xx is equal to
View answer and explanation
Correct option: B
−9- 9
Option B (−9- 9) is correct because it is the value of x that makes all three determinants equal to zero.
66 Medium
66 of 408
Understanding +4 | -1
If AA and BB are symmetric matrices of order n(A≠B)n(A \neq B), then
View answer and explanation
Correct option: B
A+BA+B is symmetric
Option B (A+BA+B is symmetric) is correct because the sum of two symmetric matrices is always a symmetric matrix.
67 Medium
67 of 408
Application +4 | -1
The value of determinant ∣a+ba+2ba+3ba+2ba+3ba+4ba+4ba+5ba+6b∣\left| {\begin{array}{cccccccccccccccccccc} {a + b}&{a + 2b}&{a + 3b}\\ {a + 2b}&{a + 3b}&{a + 4b}\\ {a + 4b}&{a + 5b}&{a + 6b} \end{array}} \right| is
View answer and explanation
Correct option: D
0
Option D (0) is correct because row operations result in two proportional rows, making the determinant zero.
68 Medium
68 of 408
Analyzing +4 | -1
If f(x)=∣x+a2x4+13x+b22x4+23x+c23x4+73∣,(f(x)=\left|\begin{array}{ccc}x+a^2 & x^4+1 & 3 \\ x+b^2 & 2 x^4+2 & 3 \\ x+c^2 & 3 x^4+7 & 3\end{array}\right|,( where x≠0)x \neq 0) and f′(x)=0f^{\prime}(x)=0, then a2,b2,c2a^2, b^2, c^2 are in
View answer and explanation
Correct option: B
Arithmetic Progression
The correct answer is Arithmetic Progression because when f′(x)=0f'(x) = 0, the determinant condition implies that a2,b2,c2a^2, b^2, c^2 form an arithmetic progression.
69 Medium
69 of 408
Analyzing +4 | -1
if ∣x2+xx+1x−22x2+3x−13x3x−3x2+2x+32x−12x−1∣=ax−12\left|\begin{array}{ccc}x^2+x & x+1 & x-2 \\ 2 x^2+3 x-1 & 3 x & 3 x-3 \\ x^2+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|=a x-12, then ' aa ' is equal to :
View answer and explanation
Correct option: A
24
Option A (24) is correct because substituting x = -1 into the determinant equation and solving for 'a' yields a = 24.
70 Medium
70 of 408
Application +4 | -1
Solving x−2y−4=0x-2 y-4=0 and 3x−5y+7=03 x-5 y+7=0, we get x=x=_____ and y=y= _____
View answer and explanation
Correct option: D
−34,−19-34,-19
Option D (x=-34, y=-19) is correct because solving the system of linear equations using matrix inversion or other methods yields these values.
71 Medium
71 of 408
Application +4 | -1
The value of ∣log⁡5729log⁡35log⁡527log⁡925∣∣log⁡35log⁡275log⁡59log⁡59∣\left| {\begin{array}{cccccccccccccccccccc} {{{\log }_5}729}&{{{\log }_3}5}\\ {{{\log }_5}27}&{{{\log }_9}25} \end{array}} \right|\left| {\begin{array}{cccccccccccccccccccc} {{{\log }_3}5}&{{{\log }_{27}}5}\\ {{{\log }_5}9}&{{{\log }_5}9} \end{array}} \right| is equal to
View answer and explanation
Correct option: D
log⁡35.log⁡581{\log _3}5.{\log _5}81
Option D (log⁡35.log⁡581{\log _3}5.{\log _5}81) is correct because after simplifying both determinants and multiplying them, the result is equal to log⁡35.log⁡581=4{\log _3}5.{\log _5}81 = 4.
72 Medium
72 of 408
Analyzing +4 | -1
If the system of equations λx1+x2+x3=1,x1+λx2+x3=1,x1+x2+λx3=1\lambda x_1+x_2+x_3=1, x_1+\lambda x_2+x_3=1, x_1+x_2+\lambda x_3=1 is inconsistent, then λ\lambda equals
View answer and explanation
Correct option: D
−2-2
The correct answer is -2 because when λ=−2\lambda = -2, the system of equations becomes inconsistent, leading to no solution.
73 Medium
73 of 408
Analyzing +4 | -1
The value of ∣a2b2c2(a+1)2(b+1)2(c+1)2(a−1)2(b−1)2(c−1)2∣\left| {\begin{array}{cccccccccccccccccccc} {{a^2}}&{{b^2}}&{{c^2}}\\ {{{\left( {a + 1} \right)}^2}}&{{{\left( {b + 1} \right)}^2}}&{{{\left( {c + 1} \right)}^2}}\\ {{{\left( {a - 1} \right)}^2}}&{{{\left( {b - 1} \right)}^2}}&{{{\left( {c - 1} \right)}^2}} \end{array}} \right| is
View answer and explanation
Correct option: A
4∣a2b2c2abc111∣4\left| {\begin{array}{cccccccccccccccccccc} {{a^2}}&{{b^2}}&{{c^2}}\\ a&b&c\\ 1&1&1 \end{array}} \right|
The correct answer is A (4\left| {\begin{{array}}{*{20}{c}} {{{a^2}}}&{{{b^2}}}&{{{c^2}}}\\ a&b&c\\ 1&1&1 \end{{array}}} \right|4\left| {\begin{{array}}{*{20}{c}} {{{a^2}}}&{{{b^2}}}&{{{c^2}}}\\ a&b&c\\ 1&1&1 \end{{array}}} \right|). Row operations are used to simplify the determinant to the given form.
74 Medium
74 of 408
Application +4 | -1
If [x+y+zx+yy+z]=[957],  \left[ {\begin{array}{cccccccccccccccccccc} {x + y + z}\\ {x + y}\\ {y + z} \end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc} 9\\ 5\\ 7 \end{array}} \right],\; then the value of (x,y,z)  (x,y,z)\; is
View answer and explanation
Correct option: C
(2,3,4)\left( {2,3,4} \right)
Option C ((2,3,4)) is correct because these values satisfy the given system of equations when the matrices are equated.
75 Medium
75 of 408
Application +4 | -1
If [5−7−23]X=[−16−672]\left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right]X=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right], then 2×22 \times 2 matrix X=X=
View answer and explanation
Correct option: A
[1−43−2]\left[\begin{array}{ll}1 & -4 \\ 3 & -2\end{array}\right]
Option A ([1−43−2]\left[\begin{array}{ll}1 & -4 \\ 3 & -2\end{array}\right]) is correct because solving the matrix equation for X yields this result.

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
76 Medium
76 of 408
Analyzing +4 | -1
If

a=1+2+4+……a=1+2+4+\ldots \ldots to nn terms

b=1+3+9+……b=1+3+9+\ldots \ldots to n terms

c=1+5+25+……c=1+5+25+\ldots \ldots to nn terms

then ∣a2b4c2222n3n5n∣=\left|\begin{array}{ccc}a & 2 b & 4 c \\ 2 & 2 & 2 \\ 2^n & 3^n & 5^n\end{array}\right|=
View answer and explanation
Correct option: C
0
Option C (0) is correct because the determinant simplifies to zero after applying properties of determinants and recognizing that two rows become identical.
77 Medium
77 of 408
Analyzing +4 | -1
If A=15!6!7![5!6!7!6!7!8!7!8!9!]A=\frac{1}{5 ! 6 ! 7 !}\left[\begin{array}{lll}5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 !\end{array}\right], then ∣adj⁡(adj⁡(2A))∣|\operatorname{adj}(\operatorname{adj}(2 A))| is equal to
View answer and explanation
Correct option: D
2162^{16}
Option D (2162^{16}) is correct because the determinant of adj(adj(2A)) is calculated using the properties of adjoint and determinant, resulting in the value 2162^{16}.
78 Medium
78 of 408
Application +4 | -1
If α,β\alpha, \beta and γ\gamma are the roots of the equation x3+px+q=0x^3+p x+q=0 (with p≠0p \neq 0 and q≠0)q \neq 0), the value of the determinant ∣αβγβγαγαβ∣,\left|\begin{array}{lll}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta\end{array}\right|, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the determinant simplifies to 0 since alpha + beta + gamma = 0.
79 Medium
79 of 408
Analyzing +4 | -1
If ∣xnxn+2x2n1xaaxn+5xa+6x2n+5∣=0,∀x∈R\left|\begin{array}{ccc}x^n & x^{n+2} & x^{2 n} \\ 1 & x^a & a \\ x^{n+5} & x^{a+6} & x^{2 n+5}\end{array}\right|=0, \forall x \in R, where n∈Nn \in N, the value of aa is
View answer and explanation
Correct option: C
n+1n+1
Option C (n+1n+1) is correct because by simplifying the determinant and comparing terms, we find that a=n+1a = n+1.
80 Medium
80 of 408
Analyzing +4 | -1
Let S1S_{1} and S2S_{2} be respectively the sets of a∈R−{0}a \in \mathbb{R}-\{0\} for which the system of linear equations

ax+2ay−3az=1(2a+1)x+(2a+3)y+(a+1)z=2(3a+5)x+(a+5)y+(a+2)z=3\begin{array}{l} a x+2 a y-3 a z=1 \\ (2 a+1) x+(2 a+3) y+(a+1) z=2 \\ (3 a+5) x+(a+5) y+(a+2) z=3 \end{array}

has unique solution and infinitely many solutions. Then
View answer and explanation
Correct option: B
S1=R−{0}S_{1}=\mathbb{R}-\{0\} and S2=ΦS_{2}=\Phi
Option B is correct because the system has a unique solution for all a∈R−{0}a \in \mathbb{{R}} - \{0\}, and no values of aa result in infinitely many solutions.
81 Medium
81 of 408
Analyzing +4 | -1
The system of equations
x+y+z=5x+y+z=5

x+2y+3z=9x+2 y+3 z=9

x+3y+λz=μx+3 y+\lambda z=\mu
has a unique solution if
View answer and explanation
Correct option: B
λ≠5\lambda \neq 5
The correct answer is λ≠5\lambda \neq 5 because the system has a unique solution if the determinant of the coefficient matrix is non-zero, which leads to λ≠5\lambda \neq 5.
82 Medium
82 of 408
Application +4 | -1
If A=[cos⁡θsin  θ−sin⁡θcos⁡θ]A = \left[ {\begin{array}{cccccccccccccccccccc} {\cos \mathrm{\theta }}&{\mathrm{sin\;\theta }}\\ { - \sin \mathrm{\theta }}&{\cos \mathrm{\theta }} \end{array}} \right], then for n∈N,  Ann \in N,\;{A^n} is equal to
View answer and explanation
Correct option: B
[cos  nθsin⁡  nθ−sin  nθcos⁡  nθ]\left[ {\begin{array}{cccccccccccccccccccc} {\mathrm{cos\;}n\mathrm{\theta }}&{\sin \;n\mathrm{\theta }}\\ { - \mathrm{sin\;}n\mathrm{\theta }}&{\cos \;n\mathrm{\theta }} \end{array}} \right]
Option B is correct because by induction, An{A^n} results in a matrix with cos(nθ) and sin(nθ) terms.
83 Medium
83 of 408
Application +4 | -1
Let
A=[111111111]A=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]
and nn be a positive integer. Then
An=[λn−1λn−1λn−1λn−1λn−1λn−1λn−1λn−1λn−1]A^n=\left[\begin{array}{lll} \lambda^{n-1} & \lambda^{n-1} & \lambda^{n-1} \\ \lambda^{n-1} & \lambda^{n-1} & \lambda^{n-1} \\ \lambda^{n-1} & \lambda^{n-1} & \lambda^{n-1} \end{array}\right]
where λ\lambda equals
View answer and explanation
Correct option: B
3
Option B is correct. By calculating A2 and A3, we can observe a pattern that A^n = 3(n-1) * A, thus lambda equals 3.
84 Medium
84 of 408
Analyzing +4 | -1
If Dr=∣r1n(n+1)22r−14n22r−152n−1∣,{D_r} = \left| {\begin{array}{cccccccccccccccccccc} r&1&{\frac{{n\left( {n + 1} \right)}}{2}}\\ {2r - 1}&4&{{n^2}}\\ {{2^{r - 1}}}&5&{{2^n} - 1} \end{array}} \right|, then the value of ∑r=0nDr\sum\nolimits_{r = 0}^n {{D_r}} is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of the determinants results in a determinant with two identical columns.
85 Medium
85 of 408
Application +4 | -1
If A=(α−100),B=(α+100)A=\left(\begin{array}{c}\alpha-1 \\ 0 \\ 0\end{array}\right), B=\left(\begin{array}{c}\alpha+1 \\ 0 \\ 0\end{array}\right) be two matrices, then ABTA B^{T} is a non-zero matrix for ∣α∣|\alpha| not equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because AB^T is a non-zero matrix when the absolute value of alpha is not equal to 1.
86 Medium
86 of 408
Application +4 | -1
A=[12−51]\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right] and A−1=mA+nl\mathrm{A}^{-1}=m \mathrm{A}+n \mathrm{l}, then m=……m=\ldots \ldots and n=……n=\ldots \ldots
View answer and explanation
Correct option: A
−111,211\frac{-1}{11}, \frac{2}{11}
Option A (m = -1/11, n = 2/11) is correct because by comparing the elements of the inverse matrix with the given expression, we can solve for m and n.
87 Medium
87 of 408
Analyzing +4 | -1
If Δr=∣r2r−13r−2n2n−1a12n(n−1)(n−1)212(n−1)(3n−4)∣\Delta_{\mathrm{r}}=\left|\begin{array}{ccc}\mathrm{r} & 2 \mathrm{r}-1 & 3 \mathrm{r}-2 \\ \frac{\mathrm{n}}{2} & \mathrm{n}-1 & \mathrm{a} \\ \frac{1}{2} \mathrm{n}(\mathrm{n}-1) & (\mathrm{n}-1)^2 & \frac{1}{2}(\mathrm{n}-1)(3 \mathrm{n}-4)\end{array}\right|
then the value of ∑r=1n−1Δr\sum_{\mathrm{r}=1}^{\mathrm{n}-1} \Delta_{\mathrm{r}}
View answer and explanation
Correct option: D
is independent of both a and n
Option D (is independent of both a and n) is correct because the summation of the determinants results in a determinant with two identical rows, making its value zero, thus independent of a and n.
88 Medium
88 of 408
Application +4 | -1
Distance of line y=∣x+1xxxx+2xxxx+3∣y=\left|\begin{array}{ccc} x+1 & x & x \\ x & x+2 & x \\ x & x & x+3 \end{array}\right| from the origin is
View answer and explanation
Correct option: C
6122\frac{6}{\sqrt{122}}
The correct answer is C (6122\frac{6}{\sqrt{122}}) because the equation of the line is found by evaluating the determinant, and then the distance from the origin is calculated using the formula.
89 Medium
89 of 408
Application +4 | -1
If matrix A=[32412−1011] and A−1=1Kadj⁡(A), then K is A=\left[\begin{array}{rrr} 3 & 2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{array}\right] \text { and } A^{-1}=\frac{1}{K} \operatorname{adj}(A) \text {, then } \mathrm{K} \text { is }
View answer and explanation
Correct option: D
11
Option D (12) is correct because it represents the number of ways to arrange five 1s and four 0s in a 3x3 symmetric matrix, considering the symmetry constraint.
90 Medium
90 of 408
Application +4 | -1
Given that matrix A=[x321y422z]A=\left[\begin{array}{lll}x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z\end{array}\right]. If xyz=2013x y z=2013 and 8x+4y+3z=20128 x+4 y+3 z=2012 then A(adj⁡A)A(\operatorname{adj} A) is equal to
View answer and explanation
Correct option: C
[290002900029]\left[\begin{array}{ccc}29 & 0 & 0 \\ 0 & 29 & 0 \\ 0 & 0 & 29\end{array}\right]
Option C (\left[\begin{{array}}{{ccc}}29 & 0 & 0 \\\\ 0 & 29 & 0 \\\\ 0 & 0 & 29\end{{array}}\right]\left[\begin{{array}}{{ccc}}29 & 0 & 0 \\\\ 0 & 29 & 0 \\\\ 0 & 0 & 29\end{{array}}\right]) is correct because A(adj⁡A)=∣A∣IA(\operatorname{{adj}} A) = |A|I, and the determinant of A, |A|, is 29.
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