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JEE • MATHEMATICS • MATRICES

Matrices practice · Medium

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211 Medium
211 of 408
Application +4 | -1
The integer represented by the determinant [215342511678364954]\left[ {\begin{array}{cccccccccccccccccccc} {215}&{342}&{511}\\ 6&7&8\\ {36}&{49}&{54} \end{array}} \right] is exactly divisible by
View answer and explanation
Correct option: C
20
Option C (20) is correct because after calculating the determinant, the resulting integer (6140) is exactly divisible by 20.
212 Medium
212 of 408
Application +4 | -1
If f(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right] Then
S-I : f(x).f(y)=f(x+y)f(x) . f(y)=f(x+y).
S-II: f(−x)=0f(-x)=0 is invertible.
View answer and explanation
Correct option: B
S-I True, S-II True
Option B (S-I True, S-II True) is correct because f(x)⋅f(y)=f(x+y)f(x) \cdot f(y) = f(x+y) and f(−x)f(-x) is invertible since its determinant is non-zero.
213 Medium
213 of 408
Application +4 | -1
If A=[1201]A=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] and B=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]B=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right] and P=BTABP=B^T A B then BP2016BTB P^{2016} B^T is equal to
View answer and explanation
Correct option: B
[1403201]\left[\begin{array}{cc}1 & 4032 \\ 0 & 1\end{array}\right]
Option B (\left[ {\begin{{array}}{{cc}} 1 & 4032 \\ 0 & 1 \end{{array}}} \right]\left[ {\begin{{array}}{{cc}} 1 & 4032 \\ 0 & 1 \end{{array}}} \right]) is correct because after simplifying the matrix expression, the result is a matrix with 1, 4032, 0, and 1 as its elements.
214 Medium
214 of 408
Analyzing +4 | -1
If the number of matrices A=[aij]2×2A=\left[a_{i j}\right]_{2 \times 2} where aij∈{−2,−1,0,1,2}a_{i j} \in\{-2,-1,0,1,2\} such that ∑aij=0\sum a_{i j}=0 is kk, then
View answer and explanation
Correct option: B
k=80k=80 if AA is non skew symmetric matrix
The correct answer is B (k=80k=80 if AA is non skew symmetric matrix) because the total number of matrices with the given condition is 85, and the number of skew-symmetric matrices is 5, so the number of non-skew-symmetric matrices is 80.
215 Medium
215 of 408
Application +4 | -1
Consider the matrix A=[4315]{A}=\left[\begin{array}{ll} 4 & 3 \\ 1 & 5 \end{array}\right]
On applying elementary row operation R2→R2−nR1R_2 \rightarrow R_2-n R_1 it becomes [43−11−4]\left[\begin{array}{cc}4 & 3 \\ -11 & -4\end{array}\right], then the value of n=n=
View answer and explanation
Correct option: A
3
Option A (3) is correct because substituting n=3 into the row operation yields the given transformed matrix.
216 Medium
216 of 408
Analyzing +4 | -1
If ∣b+cc+aa+ba+bb+cc+ac+aa+bb+c∣=k∣abccabdca∣,\left| {\begin{array}{cccccccccccccccccccc} {b + c}&{c + a}&{a + b}\\ {a + b}&{b + c}&{c + a}\\ {c + a}&{a + b}&{b + c} \end{array}} \right| = k\left| {\begin{array}{cccccccccccccccccccc} a&b&c\\ c&a&b\\ d&c&a \end{array}} \right|, then the value of kk, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because by applying column operations to the left-hand side determinant, it simplifies to 2 times the determinant on the right-hand side, thus k=2.
217 Medium
217 of 408
Analyzing +4 | -1
If Δr=∣rxn(n+1)22r−1yn23r−2zn(3n−1)2∣,\Delta_r=\left|\begin{array}{ccc}r & x & \frac{n(n+1)}{2} \\ 2 r-1 & y & n^2 \\ 3 r-2 & z & \frac{n(3 n-1)}{2}\end{array}\right|, then ∑r=1nΔr\sum\limits_{r = 1}^n {{\Delta _r}} is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because the sum of the determinants results in a determinant with two identical columns, making its value zero.
218 Medium
218 of 408
Application +4 | -1
The system of equations 6x+5y+λz=0,3x−y+4z=0,x+2y−3z=06 x+5 y+\lambda z=0,3 x-y+4 z=0, x+2 y-3 z=0 has non-trivial solutions for
View answer and explanation
Correct option: C
λ=−5\lambda=-5
Option C (λ=−5\lambda=-5) is correct because it satisfies the condition for non-trivial solutions of the system of equations.
219 Medium
219 of 408
Application +4 | -1
Let A=[aij],B=[bij]A=\left[a_{i j}\right], B=\left[b_{i j}\right] be two 3×33 \times 3 matrices such that bij=λi+j−2aijb_{i j}=\lambda^{i+j-2} a_{i j} and ∣B∣=81|B|=81, then the value of ∣A∣|A|, if λ=3\lambda=3, is
View answer and explanation
Correct option: A
19\frac{1}{9}
Option A (1/9) is correct because it is the value of |A| calculated using the given relationship between the elements of matrices A and B and their determinants.
220 Medium
220 of 408
Application +4 | -1
If A=[12−2−1]\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -2 & -1\end{array}\right], then (I+A)(I−A)−1=(\mathrm{I}+\mathrm{A})(\mathrm{I}-\mathrm{A})^{-1}=
View answer and explanation
Correct option: B
[01−1−1]\left[\begin{array}{cc}0 & 1 \\ -1 & -1\end{array}\right]
Option B is correct. The expression (I+A)(I-A)(-1) simplifies to the matrix [[0, 1], [-1, -1]] after matrix addition, subtraction, and inversion.
221 Medium
221 of 408
Understanding +4 | -1
If AA is a 3×33 \times 3 skew-symmetric matrix, then trace of AA is equal to
View answer and explanation
Correct option: D
∣A∣|A|
Option D (∣A∣|A|) is correct because the trace of a skew-symmetric matrix is always 0, and the determinant of a 3x3 skew-symmetric matrix is also 0.
222 Medium
222 of 408
Application +4 | -1
Suppose α,β,γ,θ∈R\alpha, \beta, \gamma, \theta \in {R} and A(α,β,γ,θ)=∣cos⁡(α+θ)sin⁡(α+θ)1cos⁡(β+θ)sin⁡(β+θ)1cos⁡(γ+θ)sin⁡(γ+θ)1∣A(\alpha, \beta, \gamma, \theta)=\left|\begin{array}{lll} \cos (\alpha+\theta) & \sin (\alpha+\theta) & 1 \\ \cos (\beta+\theta) & \sin (\beta+\theta) & 1 \\ \cos (\gamma+\theta) & \sin (\gamma+\theta) & 1 \end{array}\right| Numerical value of A(−π2,0,π2,2π13)A\left(-\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{2 \pi}{13}\right) is
View answer and explanation
Correct option: C
2
Option D (None of these) is incorrect. The determinant A is independent of θ\theta and its value is 2.
223 Medium
223 of 408
Analyzing +4 | -1
If a1,b2,c3≠0a_1, b_2, c_3 \neq 0 Δ1=∣a12+b1+c1a1a2+b2+c2a1a3+b3+c3b1b2+c1b22+c2b2b3+c3c3c1c3c2c32∣\Delta_1=\left|\begin{array}{ccc}a_1^2+b_1+c_1 & a_1 a_2+b_2+c_2 & a_1 a_3+b_3+c_3 \\ b_1 b_2+c_1 & b_2^2+c_2 & b_2 b_3+c_3 \\ c_3 c_1 & c_3 c_2 & c_3^2\end{array}\right| and Δ2=∣a1b1c1a2b2c2a3b3c3∣\Delta_2=\left|\begin{array}{ccc}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|, then Δ1Δ2\frac{\Delta_1}{\Delta_2} equal to
View answer and explanation
Correct option: A
a1b2c3a_1 b_2 c_3
Option A (a1b2c3a_1 b_2 c_3) is correct because after determinant operations, the ratio simplifies to this product.
224 Medium
224 of 408
Application +4 | -1
If Ar=(rr−1r−1r)where r is a natural number then ∣A1∣+∣A2∣+∣A3∣+……+∣A2006∣\mathrm{A}_{\mathrm{r}}=\left(\begin{array}{cc}\mathrm{r} & \mathrm{r}-1 \\ \mathrm{r}-1 & \mathrm{r}\end{array}\right) \text {where } \mathrm{r} \text { is a natural number then }\left|\mathrm{A}_{1}\right|+ \left|A_{2}\right|+\left|A_{3}\right|+\ldots \ldots+\left|A_{2006}\right| must be equal to-
View answer and explanation
Correct option: B
(2006)2(2006)^{2}
Option B ((2006)2(2006)^2) is correct because the determinant of ArA_r simplifies to 2r−12r-1, and the sum of 2r−12r-1 from r=1 to 2006 is (2006)2(2006)^2.
225 Medium
225 of 408
Analyzing +4 | -1
Let S={(a11a12a21a22):aij∈{0,1,2},a11=a22}S=\left\{\left(\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right): a_{i j} \in\{0,1,2\}, a_{11}=a_{22}\right\}
Then the number of non-singular matrices in the set S\mathrm{S} is :
View answer and explanation
Correct option: D
20
Option D (20) is correct because by counting the total number of matrices and subtracting the singular ones, we find 20 non-singular matrices.
226 Medium
226 of 408
Analyzing +4 | -1
The system of equations 3x−2y+z=0,λx−14y+15z=0,x+2y−3z=03x - 2y + z = 0,\lambda x - 14y + 15z = 0,x + 2y - 3z = 0 has a solution other than x=y=z=0x = y = z = 0 then λ\lambda is equal to
View answer and explanation
Correct option: D
5
Option D (5) is correct because the system has non-trivial solutions only if the determinant of the coefficient matrix is zero, leading to λ = 5.
227 Medium
227 of 408
Application +4 | -1
If ω\omega is an imaginary cube root of unity, then the value of ∣1+ωω2−ω1+ω2ω−ω2ω2+ωω−ω2∣\left| {\begin{array}{cccccccccccccccccccc} {1 + \omega }&{{\omega ^2}}&{ - \omega } \\ {1 + {\omega ^2}}&\omega &{ - {\omega ^2}} \\ {{\omega ^2} + \omega }&\omega &{ - {\omega ^2}} \end{array}} \right| is equal to
View answer and explanation
Correct option: D
−3ω2-3 \omega^2
Option D (-3 w2) is correct because after applying column operations and using the property 1 + w + w2 = 0, the determinant simplifies to -3 w2.
228 Medium
228 of 408
Application +4 | -1
If A=[2294]A=\left[\begin{array}{ll}2 & 2 \\ 9 & 4\end{array}\right] and I=[1001]I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] then 20A−1=20 A^{-1}=
View answer and explanation
Correct option: A
2A−12I2 A-12 I
Option A (log⁡e2\log_e^2) is correct because the limit of the Riemann sum can be expressed as a definite integral, which evaluates to ln(2).
229 Medium
229 of 408
Analyzing +4 | -1
Let a,b,c∈Ra, b, c \in R such that abc=2a b c=2 if A=(3abcb3caca3b)A=\left(\begin{array}{ccc}3 a & b & c \\ b & 3 c & a \\ c & a & 3 b\end{array}\right) is such that AAT=16IA A^T=16 I and ∣det⁡A∣|\operatorname{det} A| is a positive real number, then the value (a3+b3+c3)2\left(a^3+b^3+c^3\right)^2 equals
View answer and explanation
Correct option: C
4
Option C (4) is correct because by using the given conditions AAT=16IA A^T = 16I and abc=2abc=2, we can find the value of (a3+b3+c3)2(a^3+b^3+c^3)^2.
230 Medium
230 of 408
Application +4 | -1
If A=[02yzxy−zx−yz]A=\left[\begin{array}{ccc}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right] satisfies A′=A−1A^{\prime}=A^{-1}, then
View answer and explanation
Correct option: B
x=±1/2,y=±1/6,z=±1/3x= \pm 1 / \sqrt{2}, y= \pm 1 / \sqrt{6}, z= \pm 1 / \sqrt{3}
Option B (x=±1/2,y=±1/6,z=±1/3x= \pm 1 / \sqrt{2}, y= \pm 1 / \sqrt{6}, z= \pm 1 / \sqrt{3}) is correct because these values satisfy the condition A′=A−1A' = A^{-1}, which implies AA′=IAA' = I.
231 Medium
231 of 408
Application +4 | -1
∣a+b+2cabc2a+b+cbcaa+2b+c∣\left| {\begin{array}{cccccccccccccccccccc} {a + b + 2c}&a&b\\ c&{2a + b + c}&b\\ c&a&{a + 2b + c} \end{array}} \right| is equal to
View answer and explanation
Correct option: D
2(a+b+c)32{\left( {a + b + c} \right)^3}
Option D (2(a+b+c)32(a + b + c)^3) is correct because applying column and row operations simplifies the determinant to 2(a+b+c)32(a + b + c)^3.
232 Medium
232 of 408
Analyzing +4 | -1
If Δ(x)=∣1cos⁡x1−cos⁡x1+sin⁡xcos⁡x1+sin⁡x−cos⁡xsin⁡xsin⁡x1∣\Delta \left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} 1&{\cos x}&{1 - \cos x}\\ {1 + \sin x}&{\cos x}&{1 + \sin x - \cos x}\\ {\sin x}&{\sin x}&1 \end{array}} \right|, then ∫0π/2Δ(x)dx\int \nolimits_0^{\pi /2} \Delta \left( x \right)dx is equal to
View answer and explanation
Correct option: D
−12- \frac{1}{2}
The correct answer is D (−12- \frac{1}{2}). The integral of the determinant, after simplification, evaluates to -1/2.
233 Medium
233 of 408
Application +4 | -1
If P=[3212−1232],A=[1101]\mathrm{P}=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] and Q=PAPT\mathrm{Q}=\mathrm{PAP}^{\mathrm{T}}, then PTQ2015P\mathrm{P}^{\mathrm{T}} \mathrm{Q}^{2015} \mathrm{P} is ;
View answer and explanation
Correct option: C
[1201501]\left[\begin{array}{cc}1 & 2015 \\ 0 & 1\end{array}\right]
Option C (\left[\begin{{array}}{{cc}}1 & 2015 \\ 0 & 1\end{{array}}\right]\left[\begin{{array}}{{cc}}1 & 2015 \\ 0 & 1\end{{array}}\right]) is correct because after simplifying the expression PTQ2015PP^T Q^{2015} P, it reduces to A2015A^{2015}, which can be easily calculated.
234 Medium
234 of 408
Analyzing +4 | -1
If f(x),g(x)f\left( x \right),\mathrm{g}\left( x \right) and h(x)h\left( x \right) are three polynomials of degree 2 and Δ(x)=∣f(x)g(x)h(x)f′(x)g′(x)h′(x)f′′(x)g′′(x)h′′(x)∣\mathrm{\Delta }\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} {f\left( x \right)}&{\mathrm{g}\left( x \right)}&{h\left( x \right)}\\ {f'\left( x \right)}&{\mathrm{g'}\left( x \right)}&{h'\left( x \right)}\\ {f''\left( x \right)}&{\mathrm{g''}\left( x \right)}&{h''\left( x \right)} \end{array}} \right|, then Δ(x)\mathrm{\Delta }\left( x \right) is polynomial of degree
View answer and explanation
Correct option: C
At most 2
Option C (At most 2) is correct because the determinant is a polynomial of degree at most 2 after simplification.
235 Medium
235 of 408
Analyzing +4 | -1
If A=[cos⁡2  αcos⁡  α  sin⁡  αcos⁡  α  sin⁡  α      sin2α]  A = \left[ {\begin{array}{cccccccccccccccccccc} {{{\cos }^2}\;\alpha }&{\cos \;\alpha \;\sin \;\alpha }\\ {\cos \;\alpha \;\sin \;\alpha }&{\mathrm{\;\;\;si}{\mathrm{n}^2}\alpha } \end{array}} \right]\; and A=[cos⁡2  βcos⁡  α  sin⁡  βcos⁡  β  sin⁡  β      sin2β]A = \left[ {\begin{array}{cccccccccccccccccccc} {{{\cos }^2}\;\beta }&{\cos \;\alpha \;\sin \;\beta }\\ {\cos \;\beta \;\sin \;\beta }&{\mathrm{\;\;\;si}{\mathrm{n}^2}\beta } \end{array}} \right] are two matrices such that the product ABAB is null matrix, then α−β\alpha - \beta is
View answer and explanation
Correct option: C
An odd multiple of π/2\pi /2
Option C (An odd multiple of π/2\pi /2) is correct because the product AB being a null matrix implies that cos⁡(α−β)=0\cos(\alpha - \beta) = 0, which means α−β\alpha - \beta is an odd multiple of π/2\pi /2.
236 Medium
236 of 408
Application +4 | -1
If matrix A=[aij]2×2A=\left[a_{i j}\right]_{2 \times 2}, where aij={1 if i≠j0 if i=ja_{i j}=\left\{\begin{array}{ll}1 & \text { if } i \neq j \\ 0 & \text { if } i=j\end{array}\right., then A2A^2 is equal to
View answer and explanation
Correct option: A
I
Option A (I) is correct because when the given matrix A is squared, the result is the identity matrix I.
237 Medium
237 of 408
Analyzing +4 | -1
The value of the determinant ∣10!11!12!11!12!13!12!13!14!∣\left| {\begin{array}{cccccccccccccccccccc} {10!}&{11!}&{12!}\\ {11!}&{12!}&{13!}\\ {12!}&{13!}&{14!} \end{array}} \right| is
View answer and explanation
Correct option: C
2(10!11!12!)2\left( {10!11!12!} \right)
Option C (2(10!11!12!)2(10!11!12!)) is correct because the determinant simplifies to this value after factoring and applying row operations.
238 Medium
238 of 408
Application +4 | -1
If 2x−y=[3−30332]2 x-y=\left[\begin{array}{ccc}3 & -3 & 0 \\ 3 & 3 & 2\end{array}\right] and 2y+x=[415−14−4]2 y+x=\left[\begin{array}{ccc}4 & 1 & 5 \\ -1 & 4 & -4\end{array}\right], then
View answer and explanation
Correct option: B
x=[2−11120]x=\left[\begin{array}{ccc}2 & -1 & 1 \\ 1 & 2 & 0\end{array}\right]
Option B (x=[2−11120]x=\left[\begin{array}{ccc}2 & -1 & 1 \\ 1 & 2 & 0\end{array}\right]) is correct because solving the system of matrix equations yields this result for matrix x.
239 Medium
239 of 408
Analyzing +4 | -1
If a=1+2+4+…a = 1 + 2 + 4 + \ldots to nn terms, b=1+3+9+…b = 1 + 3 + 9 + \ldots to nn terms and c=1+5+25+…c = 1 + 5 + 25 + \ldots to nn terms, then ∣a2b4c2222n3n5n∣\left| {\begin{array}{cccccccccccccccccccc} a&{2b}&{4c}\\ 2&2&2\\ {{2^n}}&{{3^n}}&{{5^n}} \end{array}} \right| equals
View answer and explanation
Correct option: C
0
Option C (0) is correct because after simplification and applying row operations, two rows of the determinant become identical, resulting in a zero determinant.
240 Medium
240 of 408
Application +4 | -1
If A=[3−342−340−11]A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right] then A−1A^{-1} is equal to :
View answer and explanation
Correct option: C
A3A^3
Option C (38\frac{3}{8}) is correct because it is the probability of exactly two missiles hitting the target, considering the probability of being destroyed and the probability of hitting the target if not destroyed.
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