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JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

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211 Easy
211 of 995
Application +4 | -1
If AA is a non-singular 3×33 \times 3 matrix and BB is its adjoint such that ∣B∣=64|B|=64, then ∣A∣=|A|=
View answer and explanation
Correct option: C
±8\pm 8
Option C (±8\pm 8) is correct because ∣adjA∣=∣A∣n−1|adj A| = |A|^{n-1}, and since ∣B∣=∣adjA∣=64|B| = |adj A| = 64 and A is a 3x3 matrix, ∣A∣2=64|A|^2 = 64, thus ∣A∣=±8|A| = \pm 8.
212 Easy
212 of 995
Application +4 | -1
Let AA be a 3×33 \times 3 matrix such that det⁡(A)=−2\operatorname{det}(A)=-2. Then det (−2A−1)\left(-2 A^{-1}\right) is equal to
View answer and explanation
Correct option: A
4
Option A (4) is correct because det(-2A⁻¹) can be calculated using the properties of determinants, specifically det(kA) = kⁿdet(A) and det(A⁻¹) = 1/det(A).
213 Easy
213 of 995
Application +4 | -1
Matrix MrM_r is defined as Mr=[rr−1r−1r],r∈NM_r=\left[\begin{array}{cc}r & r-1 \\ r-1 & r\end{array}\right], r \in N.
Then det⁡(M1)+det⁡.(M2)+……..+det⁡.(M2009)=\operatorname{det}\left(M_1\right)+\operatorname{det} .\left(M_2\right)+\ldots \ldots . .+\operatorname{det} .\left(M_{2009}\right)= ?
View answer and explanation
Correct option: B
(2009)2(2009)^2
Option B, (2009)2(2009)^2, is correct because the sum of the determinants simplifies to the sum of the first 2009 odd numbers, which is (2009)2(2009)^2.
214 Easy
214 of 995
Analyzing +4 | -1
Let the homogeneous system of linear equations px+y+z=0,  x+qy+z=0,  px + y + z = 0,\;x + qy + z = 0,\; and   x+y+rz=0,  \;x + y + rz = 0,\; where   p,q,  r≠1,\;p,q,\;r \ne 1, have a non-zero solution, then the value of 11−p+11−q+11−r\frac{1}{{1 - p}} + \frac{1}{{1 - q}} + \frac{1}{{1 - r}} is
View answer and explanation
Correct option: D
1
Option B (1) is correct because the condition for a non-zero solution leads to an equation that simplifies to the given expression equaling 1.
215 Easy
215 of 995
Application +4 | -1
The real value of kk for which the system of equations 2k  x−2y+3z=0,  x+ky+2z=0,  2x+kz=0,2k\;x - 2y + 3z = 0,\;x + ky + 2z = 0,\;2x + kz = 0, has non-trivial solution is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the determinant of the coefficient matrix must be zero for a non-trivial solution, leading to k=2.
216 Easy
216 of 995
Application +4 | -1
Let pp is a non-singular matrix such that 1+p+p2+...+pn=O1 + p + {p^2} + ... + {p^n} = O (OO denotes the null matrix), then p−1{p^{ - 1}} is
View answer and explanation
Correct option: A
pn{p^n}
Option A (pn{p^n}) is correct because it is derived by manipulating the given equation and using the property that pp is non-singular.
217 Easy
217 of 995
Analyzing +4 | -1
Let AA and BB to be two invertible matrices of order 3×3.3 \times 3. If det⁡(ABAT)=8\det (AB{A^T}) = 8\mathrm{ } and det⁡(AB−1)=8,\det (A{B^{ - 1}}) = 8, then det⁡(BA−1BT)\det (B{A^{ - 1}}{B^T}) is equal to
View answer and explanation
Correct option: C
116\frac{1}{16}
Option C (116\frac{1}{16}) is correct because it is the result of correctly applying determinant properties and solving the resulting equations.
218 Easy
218 of 995
Analyzing +4 | -1
If a2+b2+c2=−2a^2+b^2+c^2=-2 and f(x)=∣1+a2x(1+b2)x(1+c2)x(1+a2)x1+b2x(1+c2)x(1+a2)x(1+b2)x1+c2x∣,f(x)=\left|\begin{array}{ccc}1+a^2 x & \left(1+b^2\right) x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & 1+b^2 x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & \left(1+b^2\right) x & 1+c^2 x\end{array}\right|, then f(x)f(x) is a polynomial of degree
View answer and explanation
Correct option: A
2
Option A (2) is correct because after simplification, f(x) becomes a quadratic polynomial (1-x)2, thus the degree is 2.
219 Easy
219 of 995
Analyzing +4 | -1
The number of solutions of the system of equations 2x+y−z=7,x−3y+2z=1,x+4y−3z=52x + y - z = 7,x - 3y + 2z = 1,x + 4y - 3z = 5 is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the equations are inconsistent, leading to no solution.
220 Easy
220 of 995
Analyzing +4 | -1
If Δ=∣cos⁡α−sin⁡α1sin⁡αcos⁡α1cos⁡(α+β)−sin⁡(α+β)1∣,\Delta=\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ \cos (\alpha+\beta) & -\sin (\alpha+\beta) & 1\end{array}\right|, then
View answer and explanation
Correct option: A
Δ∈[1−2,1+2]\Delta \in[1-\sqrt{2}, 1+\sqrt{2}]
The correct answer is A (Δ∈[1−2,1+2]\Delta \in[1-\sqrt{2}, 1+\sqrt{2}]) because the determinant simplifies to 1+sin⁡β−cos⁡β1 + \sin \beta - \cos \beta, and the range of sin⁡β−cos⁡β\sin \beta - \cos \beta is [−2,2][-\sqrt{2}, \sqrt{2}].
221 Easy
221 of 995
Application +4 | -1
If the matrix [abcd]\left[ {\begin{array}{cccccccccccccccccccc} a&b\\ c&d \end{array}} \right] is commutative w.r.t multiplication with the matrix [1101],\left[ {\begin{array}{cccccccccccccccccccc} 1&1\\ 0&1 \end{array}} \right], then
View answer and explanation
Correct option: C
c=0,d=ac = 0,d = a
Option C (c=0,d=ac = 0, d = a) is correct because it satisfies the condition for matrix commutativity with the given matrix.
222 Easy
222 of 995
Analyzing +4 | -1
The total number of matrices
A=[02y12xy−12x−y1],(x,y∈R,x≠y)A = \left[ {\begin{array}{cccccccccccccccccccc} 0&{2y}&1\\ {2x}&y&{ - 1}\\ {2x}&{ - y}&1 \end{array}} \right],(x,y \in R,x \ne y) for which ATA=3I3{A^T}A = 3{I_3} is
View answer and explanation
Correct option: B
4
Option B (4) is correct because there are 4 possible matrices that satisfy the given condition ATA=3I3{A^T}A = 3{I_3}.
223 Medium
223 of 995
Application +4 | -1
Let A=[100110111]A=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1\end{array}\right] and B=A20B=A^{20}. Then the sum of the elements of the first column of BB is?
View answer and explanation
Correct option: C
231
Option C (231) is correct because it involves matrix exponentiation and pattern recognition.
224 Easy
224 of 995
Analyzing +4 | -1
The set of equations λx−y+(cos⁡θ)z=0\lambda x-y+(\cos \theta) z=0, 3x+y+2z=0,(cos⁡θ)x+y+2z=0,0≤θ≤2π3 x+y+2 z=0,(\cos \theta) x+y+2 z=0,0 \leq \theta \leq 2 \pi, has nontrivial solution(s)
View answer and explanation
Correct option: A
for no value of λ\lambda and θ\theta
Option A (for no value of λ\lambda and θ\theta) is correct because the condition for a non-trivial solution leads to cos⁡θ=3\cos \theta = 3 or cos⁡θ=−2\cos \theta = -2, which are impossible, so there are no values of λ\lambda and θ\theta that give non-trivial solutions.
225 Easy
225 of 995
Understanding +4 | -1
If A,BA, B are two n×nn \times n non-singular matrices, then
View answer and explanation
Correct option: A
ABA B is non-singular
Option A (ABA B is non-singular) is correct because if AA and BB are non-singular, then ∣A∣≠0|A| \neq 0 and ∣B∣≠0|B| \neq 0, so ∣AB∣=∣A∣∣B∣≠0|AB| = |A||B| \neq 0, which means ABAB is non-singular.

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
226 Easy
226 of 995
Analyzing +4 | -1
The value of the determinant Δ=∣sin⁡2αsin⁡(α+β)sin⁡(α+γ)sin⁡(β+γ)sin⁡2βsin⁡(γ+β)(sin⁡γ+α)sin⁡(γ+β)sin⁡2γ∣,\Delta=\left|\begin{array}{rrr}\sin 2 \alpha & \sin (\alpha+\beta) & \sin (\alpha+\gamma) \\ \sin (\beta+\gamma) & \sin 2 \beta & \sin (\gamma+\beta) \\ (\sin \gamma+\alpha) & \sin (\gamma+\beta) & \sin 2 \gamma\end{array}\right|, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the determinant can be expressed as a product of two determinants, each of which is zero.
227 Easy
227 of 995
Application +4 | -1
The determinant Δ=∣11+ii1+ii1i11+i∣ equals\Delta=\left|\begin{array}{ccc} 1 & 1+i & i \\ 1+i & i & 1 \\ i & 1 & 1+i \end{array}\right| \text { equals}
View answer and explanation
Correct option: D
2(i−1)2(i-1)
Option D (2(i-1)) is correct because after simplifying the determinant using column operations, the result is 2(i-1).
228 Easy
228 of 995
Application +4 | -1
If A=[100010ab−1]A = \left[ {\begin{array}{cccccccccccccccccccc} 1&0&0\\ 0&1&0\\ a&b&{ - 1} \end{array}} \right], then A2{A^2} is equal to
View answer and explanation
Correct option: B
Unit matrix
Option B (Unit matrix) is correct because when matrix A is multiplied by itself, the result is the identity matrix.
229 Easy
229 of 995
Application +4 | -1
If ω\omega is a complex cube root of unity and A=[ω    00      ω]A = \left[ {\begin{array}{cccccccccccccccccccc} {\omega \;\;0}\\ {0\;\;\;\omega } \end{array}} \right], then A50  {A^{50}}\; is
View answer and explanation
Correct option: A
ω2A{\omega ^2}A
The correct answer is A. Since ω\omega is a complex cube root of unity, ω3=1\omega^3 = 1. Using this property, A50A^{50} can be simplified to ω2A\omega^2 A.
230 Easy
230 of 995
Analyzing +4 | -1
Let AA and BB be 3×33 \times 3 matrices of real numbers, where AA is symmetric, B is skew-symmetric, and (A+B)(A−B)=(A−B)(A+B)(A+B)(A-B)=(A-B)(A+B) if (AB)T=(−1)nAB(A B)^T=(-1)^n A B, then
View answer and explanation
Correct option: D
nn is an odd natural number
The correct answer is nn is an odd natural number because the given conditions lead to (−1)n=−1(-1)^n = -1, which is only true for odd n.
231 Easy
231 of 995
Application +4 | -1
If ∣x218x∣=∣62186∣\left|\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right|, then x\mathrm{x} is cqual to
View answer and explanation
Correct option: B
±6\pm 6
Option B (±6\pm 6) is correct because solving the equation derived from the determinant equality yields two possible values for x: 6 and -6.
232 Medium
232 of 995
Application +4 | -1
Matrix AA is such that A2=2A−IA^2=2 A-I when II is unit matrix then AnA^{{n}} is equal to
View answer and explanation
Correct option: A
nA−(n−1)In A-(n-1) I
Option A (nA−(n−1)InA-(n-1)I) is correct because by observing the pattern in the first few powers of A, we can deduce the general formula for A^n.
233 Easy
233 of 995
Application +4 | -1
The value of Δ=∣1ab+c1bc+a1ca+b∣,\Delta=\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|, is
View answer and explanation
Correct option: D
0
Option D (0) is correct because applying the column operation C2 -> C2 + C3 results in two identical columns, making the determinant zero.
234 Easy
234 of 995
Understanding +4 | -1
If ∣1+ax1+bx1+cx1+a1x1+b1x1+c1x1+a2x1+b2x1+c2x∣=A0+A1x+A2x2+A3x3,then  A0\left| {\begin{array}{cccccccccccccccccccc} {1 + ax}&{1 + bx}&{1 + cx}\\ {1 + {a_1}x}&{1 + {b_1}x}&{1 + {c_1}x}\\ {1 + {a_2}x}&{1 + {b_2}x}&{1 + {c_2}x} \end{array}} \right| = {A_0} + {A_1}x + {A_2}{x^2} + {A_3}{x^3},\mathrm{then}\,\,{A_0} is equal to
View answer and explanation
Correct option: B
0
Option B is correct because when x=0, the determinant has two identical rows, making its value zero.
235 Easy
235 of 995
Analyzing +4 | -1
The system of equations 2x+3y+2z=12 x+3 y+2 z=1, 4x+6y+2z=1,−x+y+2z=34 x+6 y+2 z=1,-x+y+2 z=3 has
View answer and explanation
Correct option: A
a unique solution
Option A is correct because the determinant of the coefficient matrix is non-zero, indicating a unique solution.
236 Easy
236 of 995
Analyzing +4 | -1
The matrix [5103−2−46−2−2b]\left[ {\begin{array}{cccccccccccccccccccc} 5&{10}&3\\ { - 2}&{ - 4}&6\\ { - 2}&{ - 2}&b \end{array}} \right] is a singular matrix, if bb is equal to
View answer and explanation
Correct option: D
For any value of bb
The correct answer is D (For any value of b) because the determinant of the matrix is always zero regardless of the value of b, making it a singular matrix.
237 Easy
237 of 995
Analyzing +4 | -1
If Δ1=∣b+ca−bac+ab−cba+bc−ac∣\Delta_1=\left|\begin{array}{lll}b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c\end{array}\right| and Δ2=∣abcbcacab∣, then Δ1−Δ2 equal\Delta_2=\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right| \text {, then } \Delta_1-\Delta_2 \text { equal}
View answer and explanation
Correct option: A
0
Option A (0) is correct because after applying column operations, the determinant Δ1\Delta_1 simplifies to Δ2\Delta_2, making the difference zero.
238 Easy
238 of 995
Application +4 | -1
Let AA be the set of all 3×33 \times 3 symmetric matrices all of whose entries are either 0 or 1 such that five of these entries are 1 and four are 0 . The number of matrices in set AA, is
View answer and explanation
Correct option: A
12
Option A (12) is correct because there are 3 matrices with three 1s on the diagonal and 9 matrices with one 1 on the diagonal, summing to 12.
239 Medium
239 of 995
Application +4 | -1
The value of ∣199019911992199119921993199219931994∣\left| {\begin{array}{cccccccccccccccccccc} {1990}&{1991}&{1992}\\ {1991}&{1992}&{1993}\\ {1992}&{1993}&{1994} \end{array}} \right| is
View answer and explanation
Correct option: D
0
Option D (0) is correct because applying column operations results in two identical columns, making the determinant zero.
240 Easy
240 of 995
Understanding +4 | -1
If AA is an invertible matrix, then det⁡(A−1)\det \left( {{A^{ - 1}}} \right) is equal to
View answer and explanation
Correct option: B
1det⁡(A)\frac{1}{{\det \left( A \right)}}
Option B is correct because the determinant of the inverse of a matrix is the reciprocal of the determinant of the original matrix.
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