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JEE • MATHEMATICS • MATRICES

Matrices practice · Medium

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151 Medium
151 of 408
Analyzing +4 | -1
If A,B,CA,B,C are the angles of a triangle, then the determinant Δ=∣sin⁡2Asin⁡Csin⁡Bsin⁡Csin⁡2Bsin⁡Asin⁡Bsin⁡Asin⁡2C∣\Delta = \left| {\begin{array}{cccccccccccccccccccc} {\sin 2A}&{\sin C}&{\sin B}\\ {\sin C}&{\sin 2B}&{\sin A}\\ {\sin B}&{\sin A}&{\sin 2C} \end{array}} \right| is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the determinant simplifies to 0 using the sine rule and properties of determinants.
152 Medium
152 of 408
Application +4 | -1
Let A=[1201]A=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] and B=I+adj⁡(A)+(adj⁡A)2+…+(adj⁡A)10B=I+\operatorname{adj}(A)+(\operatorname{adj} A)^2+\ldots+(\operatorname{adj} {A})^{10}. Then, the sum of all the elements of the matrix BB is:
View answer and explanation
Correct option: C
-88
Option C (-88) is correct because it is the sum of all elements of matrix B, which is calculated by summing a series of adjoint matrices.
153 Medium
153 of 408
Application +4 | -1
The matrix AA satisfying A[1501]=[3−160]A\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}3 & -1 \\ 6 & 0\end{array}\right] is
View answer and explanation
Correct option: B
[3−166−30]\left[\begin{array}{rr}3 & -16 \\ 6 & -30\end{array}\right]
Option B ([3−166−30]\left[\begin{array}{rr}3 & -16 \\ 6 & -30\end{array}\right]) is correct because multiplying this matrix by the given matrix results in the specified matrix.
154 Medium
154 of 408
Analyzing +4 | -1
If f(x)=a+bx+cx2f(x)=a+b x+c x^2 and α,β,γ\alpha, \beta, \gamma are roots of x3−1=0x^3-1=0, then ∣abcbcacab∣\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right| is equal to
View answer and explanation
Correct option: C
−f(α)f(β)f(γ)-f(\alpha) f(\beta) f(\gamma)
Option C (-f(α)f(β)f(γ)) is correct because the determinant can be factored into -(a+bω+cω^2)(a+bω^2+cω)(a+b+c) which equals -f(α)f(β)f(γ).
155 Medium
155 of 408
Analyzing +4 | -1
The function f(x)f\left( x \right) given by f(x)=∣x−1  x+1  2x+1x+1  x+3  2x+32x+1  2x−1  4x+1∣f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} {x - 1\;x + 1\;2x + 1}\\ {x + 1\;x + 3\;2x + 3}\\ {2x + 1\;2x - 1\;4x + 1} \end{array}} \right| has
View answer and explanation
Correct option: D
None of above
The correct answer is D (None of above) because the function simplifies to a linear function, and thus has no maximum or minimum points.
156 Medium
156 of 408
Application +4 | -1
The value of ∣b+caabc+abcca+b∣,\left| {\begin{array}{cccccccccccccccccccc} {b + c}&a&a\\ b&{c + a}&b\\ c&c&{a + b} \end{array}} \right|, is
View answer and explanation
Correct option: C
4 abc4\,abc
Option C (4abc4abc) is correct because applying row operations and expanding the determinant results in the expression 4abc.
157 Medium
157 of 408
Analyzing +4 | -1
If a1,a2,a3,……,an,…a_1, a_2, a_3, \ldots \ldots, a_n, \ldots. are in G.P., then the value of the determinant
∣log⁡anlog⁡an+1log⁡an+2log⁡an+3log⁡an+4log⁡an+5log⁡an+6log⁡an+7log⁡an+8∣\left|\begin{array}{ccc}\log a_n & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8}\end{array}\right|, is
View answer and explanation
Correct option: D
0
The correct answer is 0 because the rows of the determinant become linearly dependent when the terms are in a G.P.
158 Medium
158 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=\left[\begin{{array}}{{ccc}}2 & -1 & 1 \\ 1 & 2 & 0\end{{array}}\right]x=\left[\begin{{array}}{{ccc}}2 & -1 & 1 \\ 1 & 2 & 0\end{{array}}\right]) is correct because solving the system of matrix equations gives the stated matrix for x.
159 Medium
159 of 408
Application +4 | -1
If a1,a2.…..,an,…..,{a_1},{a_2}_. \ldots ..,{a_n}, \ldots .., are in GP, then the determinant Δ=∣log⁡anlog⁡an+1log⁡an+2log⁡an+3log⁡an+4log⁡an+5log⁡an+6log⁡an+7log⁡an+8∣\Delta = \left| {\begin{array}{cccccccccccccccccccc} {\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\ {\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\ {\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}} \end{array}} \right| is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because the rows (or columns) of the determinant become linearly dependent when the elements are in arithmetic progression.
160 Medium
160 of 408
Application +4 | -1
If adj B=A,∣P∣=∣Q∣=1\mathrm{B}=\mathrm{A},|\mathrm{P}|=|\mathrm{Q}|=1, then adj (Q−1BP−1)\left(\mathrm{Q}^{-1} \mathrm{BP}^{-1}\right) is
View answer and explanation
Correct option: C
PAQ
Option C (PAQ) is correct because by using the properties of adjoints and inverses, we can simplify the expression to PAQ.
161 Medium
161 of 408
Analyzing +4 | -1
If Δ1=∣xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x∣\Delta_{1}=\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right| and Δ2=∣xsin⁡2θcos⁡2θ−sin⁡2θ−x1cos⁡2θ1x∣,x≠0\Delta_{2}=\left|\begin{array}{ccc}x & \sin 2 \theta & \cos 2 \theta \\ -\sin 2 \theta & -x & 1 \\ \cos 2 \theta & 1 & x\end{array}\right|, x \neq 0; then for all θ∈(0,π2)\theta \in\left(0, \frac{\pi}{2}\right)
View answer and explanation
Correct option: B
Δ1+Δ2=−2x3\Delta_{1}+\Delta_{2}=-2 x^{3}
Option B (Δ1+Δ2=−2x3\Delta_{{1}}+\Delta_{{2}}=-2 x^{{3}}) is correct because after expanding both determinants and summing them, the trigonometric terms cancel out, leaving only −2x3-2x^3.
162 Medium
162 of 408
Understanding +4 | -1
Consider the following relation R\mathrm{R} on the set of real square matrices of order 3 .
R={(A,B)∣A=P−1BPR=\left\{(A, B) \mid A=P^{-1} B P\right. for some invertible matrix P}\left.P\right\}
Statement-1: RR is equivalence relation.

Statement-2 : For any two invertible 3×33 \times 3 matrices MM and
N,(MN)−1=N−1M−1. N,(M N)^{-1}=N^{-1} M^{-1} \text {. }
View answer and explanation
Correct option: B
Statement-1 is true, statement- 2 is true; statement- 2 is not a correct explanation for statement-1
Option B is correct. R is an equivalence relation because it's reflexive, symmetric, and transitive. The second statement is also true, but it doesn't explain why R is an equivalence relation.
163 Medium
163 of 408
Analyzing +4 | -1
If the system equations (λ−1)x+(3λ+1)y+2λz=0(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0, (λ−1)x+(4λ−2)y+(λ+3)z=0(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0, and 2x+(3λ+1)y+3(λ−1)z=02 x+(3 \lambda+1) y+3(\lambda-1) z=0 has non-trivial solutions for some values of λ\lambda, then the ratio x:y:zx: y: z when λ\lambda has the smallest of these values, is
View answer and explanation
Correct option: D
1:1:11: 1: 1
Option D (1: 1: 1) is correct because when λ = 0, the system of equations simplifies, leading to the ratio x:y:z = 1:1:1.
164 Medium
164 of 408
Understanding +4 | -1
If A and B are non singular Matrices of same order then Adj. (AB)(\mathrm{AB}) is
View answer and explanation
Correct option: B
(Adj. B) (Adj.A)
Option B ((Adj. B) (Adj.A)) is correct because Adj(AB) = Adj(B)Adj(A) is a standard property of adjoint matrices.
165 Medium
165 of 408
Application +4 | -1
In the interval [−π4,π4],\left[ { - \frac{\pi }{4},\frac{\pi }{4}} \right], the number of real solutions of the equations
∣sin⁡xcos⁡xcos⁡xcos⁡xsin⁡xcos⁡xcos⁡xcos⁡xsin⁡x∣=0\left| {\begin{array}{cccccccccccccccccccc} {\sin x}&{\cos x}&{\cos x}\\ {\cos x}&{\sin x}&{\cos x}\\ {\cos x}&{\cos x}&{\sin x} \end{array}} \right| = 0 is
View answer and explanation
Correct option: C
1
Option C (1) is the correct answer because the determinant equation simplifies to (sin⁡x−cos⁡x)2(sin⁡x+2cos⁡x)=0(\sin x - \cos x)^2(\sin x + 2\cos x) = 0, and within the given interval, only tan⁡x=1\tan x = 1 is a valid solution, giving one real solution.
166 Medium
166 of 408
Application +4 | -1
The value of ∣441442443445446447449450451∣\left| {\begin{array}{cccccccccccccccccccc} {441}&{442}&{443}\\ {445}&{446}&{447}\\ {449}&{450}&{451} \end{array}} \right| is
View answer and explanation
Correct option: B
0
Option B (0) is correct because the determinant is zero when two columns are identical after applying column operations.
167 Medium
167 of 408
Application +4 | -1
If A=[α011]A = \left[ {\begin{array}{cccccccccccccccccccc} \mathrm{\alpha }&0\\ 1&1 \end{array}} \right] and B=[1051]B = \left[ {\begin{array}{cccccccccccccccccccc} 1&0\\ 5&1 \end{array}} \right], then value of α for which A2=B{A^2} = B is
View answer and explanation
Correct option: D
No real values
Option D (No real values) is correct because there is no real value of alpha that satisfies the condition A2 = B.
168 Medium
168 of 408
Application +4 | -1
Determinant ∣1/aa2bc1/bb2ca1/cc2ab∣\left| {\begin{array}{cccccccccccccccccccc} {1/a}&{{a^2}}&{bc}\\ {1/b}&{{b^2}}&{ca}\\ {1/c}&{{c^2}}&{ab} \end{array}} \right| is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because after multiplying each row by a, b, and c respectively, and factoring out abc, two columns become identical, making the determinant zero.
169 Medium
169 of 408
Analyzing +4 | -1
Let AA and BB be real matrices of the form [α00β]\left[\begin{array}{ll}\alpha & 0 \\ 0 & \beta\end{array}\right] and
[0γδ0], respectively.\left[\begin{array}{ll} 0 & \gamma \\ \delta & 0 \end{array}\right] \text {, respectively.}

Statement 1: AB−BAA B-B A is always an invertible matrix.

Statement 2: AB−BAA B-B A is never an identity matrix
View answer and explanation
Correct option: A
Statement 1 is true, Statement 2 is false.
Statement 1 is true, Statement 2 is false. AB-BA is always invertible, but it can be an identity matrix under certain conditions.
170 Medium
170 of 408
Analyzing +4 | -1
If Δ1=∣111abca2b2c2∣,Δ2=∣1bca1cab1abc∣,\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{array}\right|, \Delta_2=\left|\begin{array}{lll}1 & b c & a \\ 1 & c a & b \\ 1 & a b & c\end{array}\right|, then
View answer and explanation
Correct option: A
Δ1+Δ2=0\Delta_1+\Delta_2=0
The correct answer is A (Δ1+Δ2=0\Delta_1+\Delta_2=0) because Δ2\Delta_2 can be transformed into −Δ1-\Delta_1 by applying row operations and factoring.
171 Medium
171 of 408
Analyzing +4 | -1
If the system of linear equations
x+ay+z=3x+a y+z=3
x+2y+2z=6x+2 y+2 z=6
x+5y+3z=bx+5 y+3 z=b
has no solution, then
View answer and explanation
Correct option: D
a=−1,b≠9a=-1, b \neq 9
Option D (a=−1,b≠9a=-1, b \neq 9) is correct because for no solution, the determinant of the coefficient matrix must be zero, and at least one of the determinants formed by replacing a column with the constants must be non-zero.
172 Medium
172 of 408
Application +4 | -1
If A=[21−12],B=[1011],C=ABATA=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right], C=A B A^T and X=ATC2AX=A^T C^2 A, then det XX is equal to
View answer and explanation
Correct option: C
729
Option C (729) is correct because it is the determinant of matrix X, calculated using the properties of determinants and the given matrices.
173 Medium
173 of 408
Application +4 | -1
Let A=[x2683y2945z2],  B=[2x3522y6142z−3]A = \left[ {\begin{array}{cccccccccccccccccccc} {{x^2}}&6&8 \\ 3&{{y^2}}&9 \\ 4&5&{{z^2}} \end{array}} \right],\,\,B = \left[ {\begin{array}{cccccccccccccccccccc} {2x}&3&5 \\ 2&{2y}&6 \\ 1&4&{2z - 3} \end{array}} \right] . If trace A=trace B{\text{trace}}\,A = {\text{trace}}\,B, then 2x+y+x2x + y + x is equal to
View answer and explanation
Correct option: A
4
Option A (4) is correct because by equating the trace of A and B, we can solve for x, y, and z, and then calculate 2x + y + z.
174 Medium
174 of 408
Analyzing +4 | -1
The repeated factor of the determinant ∣y+zxyz+xzxx+yyz∣,\left| {\begin{array}{cccccccccccccccccccc} {y + z}&x&y\\ {z + x}&z&x\\ {x + y}&y&z \end{array}} \right|, is
View answer and explanation
Correct option: A
z−xz - x
Option A (z−xz - x) is correct because it is a repeated factor obtained after applying row and column operations to simplify the determinant.
175 Medium
175 of 408
Application +4 | -1
If A=[α22α]A = \left[ {\begin{array}{cccccccccccccccccccc} \alpha &2\\ 2&\alpha \end{array}} \right] and ∣A3∣=125,\left| {{A^3}} \right| = 125, then the value of α\alpha is
View answer and explanation
Correct option: C
±3\pm 3
Option C (±3\pm 3) is correct because these are the values of alpha that satisfy the given condition |A3| = 125.
176 Medium
176 of 408
Analyzing +4 | -1
Which of the following is correct?
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because all the other options contain incorrect statements about matrix properties.
177 Medium
177 of 408
Analyzing +4 | -1
If f(x)=∣cos⁡(x+α)cos⁡(x+β)cos⁡(x+γ)sin⁡(x+α)sin⁡(x+β)sin⁡(x+γ)sin⁡(β−γ)sin⁡(γ−α)sin⁡(α−β)∣f(x)=\left|\begin{array}{lll}\cos (x+\alpha) & \cos (x+\beta) & \cos (x+\gamma) \\ \sin (x+\alpha) & \sin (x+\beta) & \sin (x+\gamma) \\ \sin (\beta-\gamma) & \sin (\gamma-\alpha) & \sin (\alpha-\beta)\end{array}\right|, then f(θ)−2f(ϕ)+f(ψ)f(\theta ) - 2f(\phi ) + f(\psi ) is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the function f(x)f(x) is constant, so f(θ)−2f(ϕ)+f(ψ)=0f(\theta) - 2f(\phi) + f(\psi) = 0.
178 Medium
178 of 408
Application +4 | -1
If f(θ)=∣1tan⁡θ1−tan⁡θ1tan⁡θ−1−tan⁡θ1∣f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|, then the set {f(θ):0≤θ≤π2}\left\{f(\theta): 0 \leq \theta \leq \frac{\pi}{2}\right\} is
View answer and explanation
Correct option: B
[2,∞)[2, \infty)
Option B ([2,∞)[2, \infty)) is correct because the function f(θ)f(\theta) simplifies to 2sec⁡2θ2 \sec^2 \theta, and since 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2}, the range of f(θ)f(\theta) is [2,∞)[2, \infty).
179 Medium
179 of 408
Application +4 | -1
If A,B,CA, B, C are the angles of a non right angled triangle ABCA B C, ∣tan⁡A111tan⁡B111tan⁡C∣\left|\begin{array}{ccc}\tan \mathrm{A} & 1 & 1 \\ 1 & \tan \mathrm{B} & 1 \\ 1 & 1 & \tan \mathrm{C}\end{array}\right| is equal to - Then the value of
View answer and explanation
Correct option: B
2
Option C ((a−b)(b−c)(c−a)(a-b)(b-c)(c-a)) is correct because it is the standard factorization of the given Vandermonde determinant.
180 Medium
180 of 408
Application +4 | -1
If A=[1−1102−3210]A = \left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 1}&1\\ 0&2&{ - 3}\\ 2&1&0 \end{array}} \right] and   B=(adj  A),  \;B = \left( {\mathrm{adj}\;A} \right),\; and C=5A,  C = 5A,\; then   ∣adj  B∣∣C∣  \;\frac{{\left| {\mathrm{adj}\;B} \right|}}{{\left| C \right|}}\; is equal to
View answer and explanation
Correct option: D
1
Option D (1) is correct because the ratio simplifies to 1 after applying properties of adjoint and determinants.
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