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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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331 Easy
331 of 922
Application +4 | -1
If [i03−i]+X=[i234+i]−X\left[\begin{array}{cc}i & 0 \\ 3 & -i\end{array}\right]+X=\left[\begin{array}{cc}i & 2 \\ 3 & 4+i\end{array}\right]-X, then XX is equal to
View answer and explanation
Correct option: B
[0102+i]\left[\begin{array}{cc}0 & 1 \\ 0 & 2+i\end{array}\right]
Option B ([0102+i]\left[\begin{array}{cc}0 & 1 \\ 0 & 2+i\end{array}\right]) is correct because it is the result of solving the given matrix equation for XX.
332 Easy
332 of 922
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: D
Δ1+Δ2=−2x3\Delta_1+\Delta_2=-2 x^3
Option D (Δ1+Δ2=−2x3\Delta_1+\Delta_2=-2 x^3) is correct because both Δ1\Delta_1 and Δ2\Delta_2 simplify to −x3-x^3.
333 Easy
333 of 922
Analyzing +4 | -1
For α,β∈R\alpha, \beta \in \mathbb{R}, suppose the system of linear equations

x−y+z=5x-y+z=5

2x+2y+αz=82 x+2 y+\alpha z=8

3x−y+4z=β3 x-y+4 z=\beta

Has infinitely many solutions. Then α\alpha and β\beta are the roots of
View answer and explanation
Correct option: B
x2−18x+56=0x^2-18 x+56=0
Option B (x2−18x+56=0x^2-18 x+56=0) is correct because the values of α\alpha and β\beta that result in infinitely many solutions are the roots of this quadratic equation.
334 Easy
334 of 922
Analyzing +4 | -1
If the trivial solution is the only solution of the system of equations
x−ky+z=0x-k y+z=0

kx+3y−kz=0k x+3 y-k z=0
3x+y−z=03 x+y-z=0, then the set of values of kk, is
View answer and explanation
Correct option: D
R−{2,−3}R-\{2,-3\}
Option D (R−{2,−3}R-\left\{2,-3\right\}) is correct because the system has only the trivial solution when the determinant of the coefficient matrix is non-zero, which occurs when kk is not equal to 2 or -3.
335 Easy
335 of 922
Application +4 | -1
∣15πlog⁡ee55log⁡10105e∣\left| {\begin{array}{cccccccccccccccccccc} 1&5&\mathrm{\pi }\\ {{{\log }_e}e}&5&{\sqrt 5 }\\ {{{\log }_{10}}10}&5&e \end{array}} \right| is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because two columns become identical after simplification, making the determinant zero.
336 Easy
336 of 922
Analyzing +4 | -1
Let a,b,ca, b, c be such that b(a+c)≠0.b(a+c) \neq 0. If ∣aa+1a−1−bb+1b−1cc−1c+1∣+∣a+1b+1c−1a−1b−1c+1(−1)n+2a(−1)n+1b(−1)nc∣=0\left|\begin{array}{rrr}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|+\left|\begin{array}{ccc}a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^n c\end{array}\right|=0 then the value of n,n, is
View answer and explanation
Correct option: C
Any odd integer
Option C (Any odd integer) is correct because the given condition simplifies to 1 + (-1)(n+2) = 0, which is true only when n is an odd integer.
337 Easy
337 of 922
Application +4 | -1
If m[−3      4]+n[4    −3]=[10      −11],  then   3m+7n  m\left[ { - 3\;\;\;4} \right] + n\left[ {4\;\; - 3} \right] = \left[ {10\;\;\; - 11} \right],\;\mathrm{then}\;\,3m + 7n\; is equal to
View answer and explanation
Correct option: D
1
Option D (1) is correct because solving the system of equations for mm and nn and then substituting into 3m+7n3m + 7n yields the value 1.
338 Easy
338 of 922
Analyzing +4 | -1
For which of the following ordered pairs (μ, δ),\left( {\mu ,\,\delta } \right), the system of liner equations x+2y+3z=1,    3x+4y+5z=μ,   4x+4y+4z=δx + 2y + 3z = 1,\,\,\,\,3x + 4y + 5z = \mu ,\,\,\,4x + 4y + 4z = \delta is inconsistent?
View answer and explanation
Correct option: B
(4,3)\left( {4,3} \right)
Option B ((4,3)\left( {4,3} \right)) is correct because for this ordered pair, the determinant of the coefficient matrix is zero, and at least one of the determinants formed by replacing a column with the constants is non-zero, indicating an inconsistent system.
339 Easy
339 of 922
Remembering +4 | -1
The matrix P=[004040400]P=\left[\begin{array}{ccc}0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0\end{array}\right] is
View answer and explanation
Correct option: A
square matrix
Option A (square matrix) is correct because the matrix has the same number of rows and columns (3x3).
340 Easy
340 of 922
Application +4 | -1
If AA is a non-singular matrix of order 3×33 \times 3, then adj⁡(adj⁡A)\operatorname{adj}(\operatorname{adj} A) is equal to
View answer and explanation
Correct option: A
∣A∣A|A| A
Option A (∣A∣A|A| A) is correct because the adjoint of the adjoint of a matrix AA is given by the formula adj⁡(adj⁡A)=∣A∣n−2A\operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A, where nn is the order of the matrix. In this case, n=3n=3.
341 Easy
341 of 922
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.
342 Easy
342 of 922
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.
343 Easy
343 of 922
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.
344 Easy
344 of 922
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.
345 Easy
345 of 922
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.
346 Easy
346 of 922
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.
347 Easy
347 of 922
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.
348 Easy
348 of 922
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.
349 Easy
349 of 922
Application +4 | -1
If A=[44−34];B=[0−110]{A}=\left[\begin{array}{cc}4 & 4 \\ -3 & 4\end{array}\right] ; {B}=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right], then (B−1A−1)−1=\left({B}^{-1} {A}^{-1}\right)^{-1}=
View answer and explanation
Correct option: C
[4−443]\left[\begin{array}{rr}4 & -4 \\ 4 & 3\end{array}\right]
Option C is correct. Using the property (B-1A-1)-1 = AB, we multiply matrix A by matrix B to get the result.
350 Easy
350 of 922
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.
351 Easy
351 of 922
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.
352 Easy
352 of 922
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.
353 Easy
353 of 922
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.
354 Easy
354 of 922
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.
355 Easy
355 of 922
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.
356 Easy
356 of 922
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.
357 Easy
357 of 922
Application +4 | -1
Let A=[1000αβ0βα]A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha\end{array}\right] and ∣2A∣3=221|2 A|^3=2^{21} where α,β∈Z\alpha, \beta \in {Z}, then a value of α\alpha is
View answer and explanation
Correct option: A
5
Option A (5) is correct because solving the determinant equation |2A|^3 = 221 leads to α^2 - β^2 = 16, and α=5 is a possible integer solution.
358 Easy
358 of 922
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.
359 Easy
359 of 922
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.
360 Easy
360 of 922
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.
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