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JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

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361 Easy
361 of 995
Understanding +4 | -1
Assertion : If A=[2312]A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] and B=[2−3−12]B=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right], then BB is the inverse of A{A}.

Reason: If AA is a square matrix of order mm and if there exists another square matrix B{B} of the same order m{m}, such that AB=BA=I{AB}={BA}={I}, then B{B} is called the inverse of A{A}.
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion.
Option A is correct because the assertion is true, the reason is true, and the reason correctly explains the assertion about inverse matrices.
362 Easy
362 of 995
Understanding +4 | -1
If A\mathrm{A} is an invertible matrix of order 2 , then det⁡.(A−1)\operatorname{det} .\left(\mathrm{A}^{-1}\right) is equal to :
View answer and explanation
Correct option: B
1det⁡.(A)\frac{1}{\operatorname{det} .(\mathrm{A})}
Option B (1/det.(A)) is correct because the determinant of the inverse of a matrix is the reciprocal of the determinant of the original matrix.
363 Easy
363 of 995
Analyzing +4 | -1
The value of Δ=∣sin⁡αcos⁡αsin⁡(α+δ)sin⁡βcos⁡βsin⁡(β+δ)sin⁡γcos⁡γsin⁡(γ+δ)∣,\Delta=\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{array}\right|, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because after applying a column operation, one column becomes all zeros, making the determinant zero.
364 Easy
364 of 995
Analyzing +4 | -1
Let AA be a 2×22 \times 2 matrix with det⁡(A)=−1\operatorname{det}(A)=-1 and det⁡\operatorname{det} ((A+I)(Adj⁡(A)+I))=4((A+I)(\operatorname{Adj}(A)+I))=4. Then the sum of the diagonal elements of AA can be
View answer and explanation
Correct option: B
22
Option B (22) is correct because it follows from using the property det(A adj(A)) = det(A)I and solving the equation to find the sum of the diagonal elements of A.
365 Easy
365 of 995
Application +4 | -1
Let A(x1,y1),B(x2,y2)A\left(x_1, y_1\right), B\left(x_2, y_2\right) and C(x3,y3)C\left(x_3, y_3\right) be vertices of an equilateral triangle whose side is 4 units. Let Δ=∣x1y11x2y21x3y31∣,\Delta=\left|\begin{array}{lll} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right|, then Δ2\Delta^2 is equal to
View answer and explanation
Correct option: C
192
The correct answer is C (192) because the area of the equilateral triangle can be used to find the value of the determinant, and then squaring it gives 192.
366 Medium
366 of 995
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.
367 Easy
367 of 995
Analyzing +4 | -1
Let A=[aij]A=\left[a_{i j}\right] be a square matrix of order 3 such that aij=2j−ia_{i j}=2^{j-i}, for all i,j=1,2,3i, j=1,2,3. Then, the matrix A2+A3A^2+A^3 +…+A10+\ldots+A^{10} is equal to :
View answer and explanation
Correct option: A
(310−32)A\left(\frac{3^{10}-3}{2}\right) A
Option A ((310−32)A\left(\frac{{3^{10}-3}}{{2}}\right) A) is correct because the matrix A2=3AA^2 = 3A, and therefore A2+A3+...+A10A^2 + A^3 + ... + A^{10} simplifies to a geometric series with the first term 3A3A and common ratio 3, resulting in the given expression.
368 Easy
368 of 995
Application +4 | -1
If a,b,c,a,b,c, are in A.P., then the value of ∣x+1x+2x+ax+2x+3x+bx+3x+4x+c∣,\left| {\begin{array}{cccccccccccccccccccc} {x + 1}&{x + 2}&{x + a}\\ {x + 2}&{x + 3}&{x + b}\\ {x + 3}&{x + 4}&{x + c} \end{array}} \right|, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because when a, b, and c are in A.P., applying row operations results in a row of zeros, making the determinant zero.
369 Easy
369 of 995
Analyzing +4 | -1
For the system of linear equations x−2y=1,    x−y+kz=−2,    ky+4z=6,    k∈R,x - 2y = 1,\,\,\,\,x - y + kz = - 2,\,\,\,\,ky + 4z = 6,\,\,\,\,k \in R,
Consider the following statements
A) The system has unique solution, if k≠2,    k≠−2k \ne 2,\,\,\,\,k \ne - 2
B) The system has unique solution, if k=−2k = - 2.
C) The system has unique solution, if k=2k = 2.
D) The system has no solution, if k=2k = 2.
E) The system has infinite number of solutions, if k≠−2.k \ne - 2.
Which of the following statements are correct?
View answer and explanation
Correct option: D
(A) and (D)
Option D ((A) and (D)) is correct because the system has a unique solution when k≠2k \ne 2 and k≠−2k \ne -2, and it has no solution when k=2k = 2.
370 Easy
370 of 995
Application +4 | -1
If [1101]⋅[1201]⋅[1301]⋅⋅⋅⋅[1n−101]=[17801],\left[ {\begin{array}{cccccccccccccccccccc} 1&1\\ 0&1 \end{array}} \right] \cdot \left[ {\begin{array}{cccccccccccccccccccc} 1&2\\ 0&1 \end{array}} \right] \cdot \left[ {\begin{array}{cccccccccccccccccccc} 1&3\\ 0&1 \end{array}} \right] \cdot \cdot \cdot \cdot \left[ {\begin{array}{cccccccccccccccccccc} 1&{n - 1}\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc} 1&{78}\\ 0&1 \end{array}} \right], then the inverse of [1n01]\left[ {\begin{array}{cccccccccccccccccccc} 1&n\\ 0&1 \end{array}} \right] is
View answer and explanation
Correct option: B
[1−1301]\left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 13}\\ 0&1 \end{array}} \right]
Option B ([1−1301]\left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 13}\\ 0&1 \end{array}} \right]) is correct because, by multiplying the matrices and solving for n, we can find the inverse of the given matrix.
371 Hard
371 of 995
Analyzing +4 | -1
If a,ba, b and cc are sides of △ABC\triangle A B C such that
∣cbcos⁡B+cβacos⁡A+bα+cγaccos⁡B+aβbcos⁡A+cα+aγbacos⁡B+bβccos⁡A+aα+bγ∣=0\left|\begin{array}{ccc}c & b \cos B+c \beta & a \cos A+b \alpha+c \gamma \\ a & c \cos B+a \beta & b \cos A+c \alpha+a \gamma \\ b & a \cos B+b \beta & c \cos A+a \alpha+b \gamma\end{array}\right|=0 (where α,β,γ∈R+\alpha, \beta, \gamma \in R^{+}and ∠A,∠B,∠C≠π2\angle A, \angle B, \angle C \neq \frac{\pi}{2} ), △ABC\triangle A B C is
View answer and explanation
Correct option: B
an equilateral
Option B (an equilateral) is correct because after applying determinant properties and trigonometric identities, the condition leads to a=b=ca=b=c, implying an equilateral triangle.
372 Easy
372 of 995
Application +4 | -1
The value of ∣a2−ab−ac−abb2−bccabc−c2∣,\left|\begin{array}{ccc}a^2 & -a b & -a c \\ -a b & b^2 & -b c \\ c a & b c & -c^2\end{array}\right|, is
View answer and explanation
Correct option: D
4a2b2c24 a^2 b^2 c^2
Option D (4a2b2c24 a^2 b^2 c^2) is correct because after taking out a, b, and c from rows and columns, the determinant simplifies to 4a2b2c24 a^2 b^2 c^2.
373 Easy
373 of 995
Understanding +4 | -1
If two matrices A\mathrm{A} and B\mathrm{B} are of order p×qp \times q and r×sr \times s respectively, can be subtracted only, if
View answer and explanation
Correct option: C
p=r,q=sp=r, q=s
Option C (p=r, q=s) is correct because matrices can only be subtracted if they have the same order.
374 Easy
374 of 995
Analyzing +4 | -1
The system of equation
αx+y+z=α−1x+αy+z=α−1x+y+αz=α−1\begin{array}{l} \alpha x + y + z = \alpha - 1\\ x + \alpha y + z = \alpha - 1\\ x + y + \alpha z = \alpha - 1 \end{array}
Has no solution, if α\alpha is
View answer and explanation
Correct option: D
−2- 2
Option D (-2) is correct because the system has no solution when the determinant of the coefficient matrix is zero, and alpha is not equal to 1.
375 Easy
375 of 995
Analyzing +4 | -1
Given that xyz=−1,x y z=-1, the value of the determinant ∣xx21+x3yy21+y3zz21+z3∣,\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the determinant can be split into two determinants, and given xyz=−1xyz = -1, they cancel each other out.
376 Easy
376 of 995
Understanding +4 | -1
Which one of the following is correct?
View answer and explanation
Correct option: B
Skew-symmetric matrix of odd order is singular
Option B (Skew-symmetric matrix of odd order is singular) is correct because the determinant of a skew-symmetric matrix of odd order is always zero, making it singular.
377 Medium
377 of 995
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.
378 Easy
378 of 995
Application +4 | -1
If log⁡3x(x+2)=1\log _3 x(x+2)=1, then xx is equal to
View answer and explanation
Correct option: D
1 or -3
Option D (1 or -3) is correct because these are the solutions to the quadratic equation derived from the logarithmic equation.
379 Easy
379 of 995
Remembering +4 | -1
If In{I_n} is the identity matrix of order nn, then (In)−1{\left( {{I_n}} \right)^{ - 1}} is equal to
View answer and explanation
Correct option: B
In{I_n}
Option B (In{I_n}) is correct because the inverse of an identity matrix is the identity matrix itself.
380 Easy
380 of 995
Analyzing +4 | -1
If PP is a 3×33 \times 3 matrix such that PT=2P+IP^T=2 P+I, where PTP^T is the transpose of PP and II is 3×33 \times 3 identity, then there exists a column matrix X=[xyx]≠[000]X=\left[\begin{array}{l}x \\ y \\ x\end{array}\right] \neq\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right] such that PX=P X=
View answer and explanation
Correct option: D
−X-X
Option D (−X-X) is correct because, given PT=2P+IP^T = 2P + I, we can derive that P=−IP = -I, which leads to PX=−XPX = -X.
381 Easy
381 of 995
Understanding +4 | -1
Consider the following statements

I. If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes.

II. If any two rows (or columns) of a determinant are interchanged, then the value of the determinant remains same.
View answer and explanation
Correct option: A
Only I is true
Option A (Only I is true) is correct because interchanging rows or columns in a determinant changes its sign.
382 Medium
382 of 995
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.
383 Easy
383 of 995
Understanding +4 | -1
If A\mathrm{A} is a square matrix A+AT\mathrm{A}+\mathrm{A}^{\mathrm{T}} is symmetric matrix, then A−AT=\mathrm{A}-\mathrm{A}^{\mathrm{T}}=
View answer and explanation
Correct option: C
Skew symmetrix matrix
Option C (Skew symmetrix matrix) is correct because if A+Aᵀ is symmetric, then A-Aᵀ is skew-symmetric.
384 Easy
384 of 995
Analyzing +4 | -1
Let a,b,ca, b, c be any real numbers. Suppose that there are real numbers x,y,zx, y, z not all zero such that x=cy+bz,y=az+cxx=c y+b z, y=a z+c x and z=bx+ayz=b x+a y. Then, a2+b2+c2+2abca^2+b^2+c^2+2 a b c is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because the given equations imply that the determinant of the coefficient matrix is zero, leading to a2+b2+c2+2abc=1a^2+b^2+c^2+2abc = 1.
385 Medium
385 of 995
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.
386 Easy
386 of 995
Application +4 | -1
If AA is a 3×33 \times 3 non-singular matrix such that AAT=ATAA{A^T} = {A^T}A and B=A−1AT,B = {A^{ - 1}}{A^T}, then BBTB{B^T} is equal to
View answer and explanation
Correct option: B
II
Option D (π\pi) is correct because by applying trigonometric identities and L'Hôpital's rule, the limit simplifies to π\pi.
387 Easy
387 of 995
Remembering +4 | -1
IfII is unit matrix of order 10, then the determinant of II is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because the determinant of any identity matrix, regardless of its order, is always 1.
388 Medium
388 of 995
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.
389 Medium
389 of 995
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.
390 Easy
390 of 995
Application +4 | -1
If one root of determinant ∣x372x276x∣=0\left| {\begin{array}{cccccccccccccccccccc} x&3&7\\ 2&x&2\\ 7&6&x \end{array}} \right| = 0, is −9,- 9, then the other two roots are
View answer and explanation
Correct option: A
2,72,7
Option A (2,7) is correct because these are the other two roots of the cubic equation obtained from the determinant.
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