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JEE • MATHEMATICS • MATRICES

Matrices practice · Medium

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391 Medium
391 of 408
Analyzing +4 | -1
Let a1,a2,a3,…,a10a_1, a_2, a_3, \ldots, a_{10} be in GP. with ai>0a_i>0 for i=1,2,…,10i=1,2, \ldots, 10 and S\mathrm{S} be the set of pairs (r, k), r,k∈Nr, k \in N (the set of natural numbers) for which

∣log⁡ea1ra2klog⁡ea2ra3klog⁡ea3ra4klog⁡ea4ra5klog⁡ea5ra6klog⁡ea6ra7klog⁡ea7ra8klog⁡ea8ra9klog⁡ea9ra10k∣=0\left|\begin{array}{llll} \log _e \mathrm{a}_1^{\mathrm{r}} \mathrm{a}_2^{\mathrm{k}} & \log _e \mathrm{a}_2^{\mathrm{r}} \mathrm{a}_3^{\mathrm{k}} & \log _e \mathrm{a}_3^{\mathrm{r}} \mathrm{a}_4^{\mathrm{k}} \\ \log _e \mathrm{a}_4^{\mathrm{r}} \mathrm{a}_5^{\mathrm{k}} & \log _e \mathrm{a}_5^{\mathrm{r}} \mathrm{a}_6^{\mathrm{k}} & \log _e \mathrm{a}_6^{\mathrm{r}} \mathrm{a}_7^{\mathrm{k}} \\ \log _e \mathrm{a}_7^{\mathrm{r}} \mathrm{a}_8^{\mathrm{k}} & \log _e \mathrm{a}_8^{\mathrm{r}} \mathrm{a}_9^{\mathrm{k}} & \log _e \mathrm{a}_9^{\mathrm{r}} \mathrm{a}_{10}^{\mathrm{k}} \end{array}\right|=0
Then the number of elements in S, is :
View answer and explanation
Correct option: B
infinitelymany
Option B (infinitelymany) is correct because the determinant becomes zero regardless of the values of r and k, as the first two columns become proportional after simplification.
392 Medium
392 of 408
Analyzing +4 | -1
If AB=0\mathrm{AB}=0, then for the matrices A=[cos⁡2θcos⁡θsin⁡θcos⁡θsin⁡θsin⁡2θ]and  B=[cos⁡2ϕcos⁡ϕsin⁡ϕcos⁡ϕsin⁡ϕsin⁡2ϕ],θ−ϕA = \left[ {\begin{array}{cccccccccccccccccccc} {{{\cos }^2}\theta }&{\cos \theta \sin \theta } \\ {\cos \theta \sin \theta }&{{{\sin }^2}\theta } \end{array}} \right]{\text{and}}\,\,B = \left[ {\begin{array}{cccccccccccccccccccc} {{{\cos }^2}\phi }&{\cos \phi \sin \phi } \\ {\cos \phi \sin \phi }&{{{\sin }^2}\phi } \end{array}} \right],\theta - \phi is
View answer and explanation
Correct option: A
an odd number of π2\frac{\pi}{2}
Option A (an odd number of π2\frac{\pi}{2}) is correct because the condition AB=0 implies that cos⁡(θ−ϕ)=0\cos(\theta - \phi) = 0, which means θ−ϕ\theta - \phi is an odd multiple of π2\frac{\pi}{2}.
393 Medium
393 of 408
Application +4 | -1
If A=[−4−131]A=\left[\begin{array}{cc}-4 & -1 \\ 3 & 1\end{array}\right], then the determinant of the matrix (A2016−2A2015−A2014)\left(A^{2016}-2 A^{2015}-A^{2014}\right) is :
View answer and explanation
Correct option: D
-25
Option D (-25) is correct. The determinant of the matrix (A2016-2A2015-A2014) is found by factoring out A2014 and calculating the determinant of the resulting expression.
394 Medium
394 of 408
Analyzing +4 | -1
If the system of equation x+y+z=16;2x+5y+αz=β;x+2y+3z=14x+y+z=16 ; 2 x+5 y+\alpha z=\beta ; x+2 y+3 z=14 has infinitely many solution, then α+β=\alpha+\beta=
View answer and explanation
Correct option: D
34
Option D (34) is correct because for infinitely many solutions, the determinant of the coefficient matrix must be zero, and the equations must be consistent.
395 Medium
395 of 408
Application +4 | -1
The matrix of the transformation reflection in the line x+y=0x+y=0 is
View answer and explanation
Correct option: D
[0−1−10]\left[\begin{array}{cc}0 & -1 \\ -1 & 0\end{array}\right]
Option D [0−1−10]\left[\begin{array}{cc}0 & -1 \\ -1 & 0\end{array}\right] is correct because it represents the transformation matrix for reflection in the line x+y=0x+y=0.
396 Medium
396 of 408
Analyzing +4 | -1
If f(x)=∣x−32x2−183x3−81x−52x2−504x3−500123∣,f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} {x - 3}&{2{x^2} - 18}&{3{x^3} - 81}\\ {x - 5}&{2{x^2} - 50}&{4{x^3} - 500}\\ 1&2&3 \end{array}} \right|, then f(1).f(3)+f(3).f(5)+f(5).f(1)f\left( 1 \right).f\left( 3 \right) + f\left( 3 \right).f\left( 5 \right) + f\left( 5 \right).f\left( 1 \right) is equal to
View answer and explanation
Correct option: B
f(3)f\left( 3 \right)
The correct answer is B (f(3)f\left( 3 \right)). Since f(3) = 0 and f(5) = 0, the expression simplifies to 0, which is equal to f(3).
397 Medium
397 of 408
Analyzing +4 | -1
In ΔABC\Delta ABC if ∣1ab1ca1bc∣=0,\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 (94\frac{9}{4}) is correct because the condition implies that the triangle is equilateral, and the sum of squares of sines of 60 degrees is 94\frac{9}{4}.
398 Medium
398 of 408
Analyzing +4 | -1
Let a,b,ca, b, c be positive and not all equal, the value of the determinant ∣abcbcacab∣,\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|, is
View answer and explanation
Correct option: B
−ive-ive
Option B (-ive) is correct because the determinant can be expressed as a negative value involving squares, given the conditions a,b,ca, b, c are positive and not all equal.
399 Medium
399 of 408
Analyzing +4 | -1
If a,b,ca,b,c are unequal what is the condition that the value of the determinant, Δ≡∣aa2a3+1bb2b3+1cc2c3+1∣\mathrm{\Delta } \equiv \left| {\begin{array}{cccccccccccccccccccc} a&{{a^2}}&{{a^3} + 1}\\ b&{{b^2}}&{{b^3} + 1}\\ c&{{c^2}}&{{c^3} + 1} \end{array}} \right| is 0?
View answer and explanation
Correct option: A
1+abc=01 + abc = 0
Option A (1+abc=01 + abc = 0) is correct because the condition for the determinant to be zero is 1+abc=01 + abc = 0 when a, b, and c are unequal.
400 Medium
400 of 408
Analyzing +4 | -1
If A,BA, B and CC denote the angles of a triangle, then Δ=∣−1cos⁡Ccos⁡Bcos⁡C−1cos⁡Acos⁡Bcos⁡A−2∣\Delta=\left|\begin{array}{ccc}-1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -2\end{array}\right| is independent of
View answer and explanation
Correct option: B
BB
Option B (BB) is correct because after simplification, the determinant's value is expressed in terms of AA and CC only, making it independent of BB.
401 Medium
401 of 408
Application +4 | -1
If [1/250x1/25]={[50−a5]−1}2\left[\begin{array}{cc}1 / 25 & 0 \\ x & 1 / 25\end{array}\right]=\left\{\left[\begin{array}{cc}5 & 0 \\ -a & 5\end{array}\right]^{-1}\right\}^2, then the value of xx is
View answer and explanation
Correct option: C
2a/1252 a / 125
Option C (2a/1252a/125) is correct because after computing the inverse and squaring it, the resulting matrix matches the given form with x = 2a/125.
402 Medium
402 of 408
Application +4 | -1
If A=[1101]\mathrm{A}=\left[\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right] and B=[3212−1232]\mathrm{B}=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{-1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], then (BB⊤A)5\left(B B^{\top} A\right)^5 is equal to :
View answer and explanation
Correct option: C
[1501]\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]
Option C [1501]\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right] is correct because by simplifying the expression and using the properties of matrix multiplication, we can find the result.
403 Medium
403 of 408
Application +4 | -1
If A=[sin⁡θ−cos⁡θcos⁡θsin⁡θ]\mathrm{A}=\left[\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta\end{array}\right], then A−1=\mathrm{A}^{-1}=
View answer and explanation
Correct option: B
[sin⁡θcos⁡θ−cos⁡θsin⁡θ]\left[\begin{array}{cc}\sin \theta & \cos \theta \\ -\cos \theta & \sin \theta\end{array}\right]
Option B ([sin⁡θcos⁡θ−cos⁡θsin⁡θ]\left[\begin{array}{cc}\sin \theta & \cos \theta \\ -\cos \theta & \sin \theta\end{array}\right]) is correct because it is the inverse matrix calculated using the adjoint and determinant.
404 Medium
404 of 408
Application +4 | -1
If k[−1222−1222−1]\mathrm{k} \left[\begin{array}{ccc}-1 & 2 & 2 \\ 2 & -1 & 2 \\ 2 & 2 & -1\end{array}\right] is an orthogonal matrix then k\mathrm{k} is equal to-
View answer and explanation
Correct option: C
1/31 / 3
Option C (1/3) is correct because for the given matrix to be orthogonal, the value of k must be 1/3.
405 Medium
405 of 408
Analyzing +4 | -1
Statement 1: If the system of equations x+ky+3z=0x+k y+3 z=0, 3x+ky−2z=0,2x+3y−4z=03 x+k y-2 z=0,2 x+3 y-4 z=0 has a non-trivial solution, then the value of kk is 312\frac{31}{2}.

Statement 2: A system of three homogeneous equations in three variables has a non trivial solution if the determinant of the coefficient matrix is zero.
View answer and explanation
Correct option: A
Statement 1 is false, Statement 2 is true.
Option A (Statement 1 is false, Statement 2 is true) is correct because the value of k calculated from the determinant is not 31/2, and Statement 2 is a fundamental property.
406 Medium
406 of 408
Application +4 | -1
If P=[3212−1232]P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right] and A=[1101]A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] and Q=Q= PAPTP A P^T and x=PTQ2005Px=P^T Q^{2005} P then xx is equal to
View answer and explanation
Correct option: A
[1200501]\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]
Option A ([1200501]\left[\begin{array}{cc}1 & 2005 \\0 & 1\end{array}\right]) is correct because X=A2005X = A^{2005} and A2005A^{2005} is calculated by observing the pattern of powers of A.
407 Medium
407 of 408
Analyzing +4 | -1
If α,β,γ\alpha, \beta, \gamma are different from 1 and are the roots of ax3+bx2+cx+d=0a x^3+b x^2+c x+d=0 and (β−γ)(γ−α)(α−β)=25/2(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)=25 / 2, then the determinant Δ=∣α1−αβ1−βγ1−γαβγα2β2γ2∣\Delta=\left|\begin{array}{ccc}\frac{\alpha}{1-\alpha} & \frac{\beta}{1-\beta} & \frac{\gamma}{1-\gamma} \\ \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2\end{array}\right| equals
View answer and explanation
Correct option: D
−25d2(a+b+c+d)\frac{{ - 25d}}{{2(a + b + c + d)}}
Option D (−25d2(a+b+c+d)\frac{{ - 25d}}{{2(a + b + c + d)}}) is correct because after simplification and using Vieta's formulas, the determinant equals this expression.
408 Medium
408 of 408
Application +4 | -1
If lr2+mr2+nr2=1l_r^2+m_r^2+n_r^2=1, where r=1,2,3r=1,2,3 and l1l2+m1m2+n1n2=0…l_1 l_2+m_1 m_2+n_1 n_2=0\ldots etc, then, Δ=∣l1m1n1l2m2n2l3m3n3∣2\Delta=\left|\begin{array}{lll}l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3\end{array}\right|^2 is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because the given conditions imply that the rows (or columns) of the determinant form an orthonormal basis, and the square of the determinant of an orthonormal matrix is 1.
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