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JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 181-210 of 995 questions

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181 Medium
181 of 995
Analyzing +4 | -1
Let S1S_{1} and S2S_{2} be respectively the sets of a∈R−{0}a \in \mathbb{R}-\{0\} for which the system of linear equations

ax+2ay−3az=1(2a+1)x+(2a+3)y+(a+1)z=2(3a+5)x+(a+5)y+(a+2)z=3\begin{array}{l} a x+2 a y-3 a z=1 \\ (2 a+1) x+(2 a+3) y+(a+1) z=2 \\ (3 a+5) x+(a+5) y+(a+2) z=3 \end{array}

has unique solution and infinitely many solutions. Then
View answer and explanation
Correct option: B
S1=R−{0}S_{1}=\mathbb{R}-\{0\} and S2=ΦS_{2}=\Phi
Option B is correct because the system has a unique solution for all a∈R−{0}a \in \mathbb{{R}} - \{0\}, and no values of aa result in infinitely many solutions.
182 Medium
182 of 995
Analyzing +4 | -1
The system of equations
x+y+z=5x+y+z=5

x+2y+3z=9x+2 y+3 z=9

x+3y+λz=μx+3 y+\lambda z=\mu
has a unique solution if
View answer and explanation
Correct option: B
λ≠5\lambda \neq 5
The correct answer is λ≠5\lambda \neq 5 because the system has a unique solution if the determinant of the coefficient matrix is non-zero, which leads to λ≠5\lambda \neq 5.
183 Easy
183 of 995
Understanding +4 | -1
If A=(aij)3×3A=\left(a_{i j}\right)_{3 \times 3} where aij=cos⁡(i+j)a_{i j}=\cos (i+j), then
View answer and explanation
Correct option: A
AA is symmetric
Option A (AA is symmetric) is correct because aij=cos⁡(i+j)=cos⁡(j+i)=ajia_{ij} = \cos(i+j) = \cos(j+i) = a_{ji}, which satisfies the condition for a symmetric matrix.
184 Easy
184 of 995
Application +4 | -1
If α,β,γ\alpha ,\beta ,\gamma are the cube roots of 8, then ∣αβγβγαγαβ∣=\left| {\begin{array}{cccccccccccccccccccc} \alpha &\beta &\gamma \\ \beta &\gamma &\alpha \\ \gamma &\alpha &\beta \end{array}} \right| =
View answer and explanation
Correct option: A
0
Option A (0) is correct because the determinant is zero due to the sum of the cube roots of 8 being zero.
185 Easy
185 of 995
Application +4 | -1
Let AA be a 3×33 \times 3 matrix such that

A[123023011]=[001100010]\mathrm{A}\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right]

Then A−1\mathrm{A}^{-1} is:
View answer and explanation
Correct option: A
[312302101]\left[\begin{array}{lll}3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1\end{array}\right]
Option A (\left[\begin{{lll}}3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1\end{{lll}}\right]\left[\begin{{lll}}3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1\end{{lll}}\right]) is correct. By performing column operations on the matrix equation to transform the matrix multiplied by A into the identity matrix, we can directly find A−1A^{-1}.
186 Medium
186 of 995
Application +4 | -1
If A=[cos⁡θsin  θ−sin⁡θcos⁡θ]A = \left[ {\begin{array}{cccccccccccccccccccc} {\cos \mathrm{\theta }}&{\mathrm{sin\;\theta }}\\ { - \sin \mathrm{\theta }}&{\cos \mathrm{\theta }} \end{array}} \right], then for n∈N,  Ann \in N,\;{A^n} is equal to
View answer and explanation
Correct option: B
[cos  nθsin⁡  nθ−sin  nθcos⁡  nθ]\left[ {\begin{array}{cccccccccccccccccccc} {\mathrm{cos\;}n\mathrm{\theta }}&{\sin \;n\mathrm{\theta }}\\ { - \mathrm{sin\;}n\mathrm{\theta }}&{\cos \;n\mathrm{\theta }} \end{array}} \right]
Option B is correct because by induction, An{A^n} results in a matrix with cos(nθ) and sin(nθ) terms.
187 Easy
187 of 995
Application +4 | -1
If A=[2−2−4−1341−2x]A = \left[ {\begin{array}{cccccccccccccccccccc} 2&{ - 2}&{ - 4}\\ { - 1}&3&4\\ 1&{ - 2}&x \end{array}} \right] is an idempotent matrix, then xx is equal to
View answer and explanation
Correct option: C
−3- 3
Option C (-3) is correct. For a matrix to be idempotent, A2 = A. Solving the matrix multiplication and comparing elements leads to x = -3.
188 Medium
188 of 995
Analyzing +4 | -1
If Dr=∣r1n(n+1)22r−14n22r−152n−1∣,{D_r} = \left| {\begin{array}{cccccccccccccccccccc} r&1&{\frac{{n\left( {n + 1} \right)}}{2}}\\ {2r - 1}&4&{{n^2}}\\ {{2^{r - 1}}}&5&{{2^n} - 1} \end{array}} \right|, then the value of ∑r=0nDr\sum\nolimits_{r = 0}^n {{D_r}} is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of the determinants results in a determinant with two identical columns.
189 Easy
189 of 995
Understanding +4 | -1
If AA is a non-singular matrix of order 3 , then ∣adj⁡A∣=∣A∣n|\operatorname{adj} A|=|A|^{\mathrm{n}}. Here the value of nn is
View answer and explanation
Correct option: A
2
Option A (2) is correct because for a non-singular matrix A of order 3, |adj A| = |A|^(3-1) = |A|^2, thus n = 2.
190 Easy
190 of 995
Application +4 | -1
ω\omega is an imaginary cube root of unity and ∣x+ω2ω1ωω21+x1x+ωω2∣=0\left| {\begin{array}{cccccccccccccccccccc} {x + {\omega ^2}}&\omega &1\\ \omega &{{\omega ^2}}&{1 + x}\\ 1&{x + \omega }&{{\omega ^2}} \end{array}} \right| = 0, then one of the value of xx is
View answer and explanation
Correct option: B
0
Option B (0) is correct because substituting x = 0 into the determinant makes the first column zero, thus the determinant equals zero.
191 Easy
191 of 995
Application +4 | -1
If A=[2x      0x            x]A = \left[ {\begin{array}{cccccccccccccccccccc} {2x\;\;\;0}\\ {x\;\;\;\;\;\;x} \end{array}} \right] and   A−1=[1              0−1          2]\;{A^{ - 1}} = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;\;\;\;\;\;0}\\ { - 1\;\;\;\;\;2} \end{array}} \right], then xx equals
View answer and explanation
Correct option: D
12\frac{1}{2}
Option D (12\frac{1}{2}) is correct because solving for x using the inverse matrix properties yields x = 1/2.
192 Easy
192 of 995
Application +4 | -1
Let MM be a 3×33 \times 3 non-singular matrix with det⁡(M)=α\operatorname{det}(M)=\alpha.If M−1M^{-1} adj(adj M)=kI\left.M\right)=k I, then the value of kk is
View answer and explanation
Correct option: B
α\alpha
Option B (α\alpha) is correct because the value of k is equal to the determinant of M, which is α\alpha, after simplifying the expression using properties of adjoint and determinant.
193 Medium
193 of 995
Application +4 | -1
If A=(α−100),B=(α+100)A=\left(\begin{array}{c}\alpha-1 \\ 0 \\ 0\end{array}\right), B=\left(\begin{array}{c}\alpha+1 \\ 0 \\ 0\end{array}\right) be two matrices, then ABTA B^{T} is a non-zero matrix for ∣α∣|\alpha| not equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because AB^T is a non-zero matrix when the absolute value of alpha is not equal to 1.
194 Easy
194 of 995
Application +4 | -1
If A=[1a01]A=\left[\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right], then AnA^n (where n∈Nn \in N ) equals
View answer and explanation
Correct option: A
[1na01]\left[\begin{array}{ll}1 & n a \\ 0 & 1\end{array}\right]
Option A [1na01]\left[\begin{array}{ll}1 & n a \\ 0 & 1\end{array}\right] is correct because the pattern of AnA^n follows this form, which can be proven by induction.
195 Easy
195 of 995
Application +4 | -1
If A=[00−10−10−100]A = \left[ {\begin{array}{cccccccccccccccccccc} 0&0&{ - 1}\\ 0&{ - 1}&0\\ { - 1}&0&0 \end{array}} \right]. The only correct statement about the matrix AA is
View answer and explanation
Correct option: D
A2=I{A^2} = I
Option D (A2=I{A^2} = I) is correct because when matrix A is multiplied by itself, the result is the identity matrix I.

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
196 Easy
196 of 995
Application +4 | -1
If P=[1α3133244]P=\left[\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right] is the adjoint of a 3×33 \times 3 matrix AA and ∣A∣=4|A|=4, then α\alpha is equal to
View answer and explanation
Correct option: B
11
Option B (11) is correct because by calculating the determinant of the adjoint matrix P and using the property |adj A| = |A|^(n-1), we can solve for alpha.
197 Easy
197 of 995
Application +4 | -1
If A=[0    1    01    0    00    0    1],  then  A−1A = \left[ {\begin{array}{cccccccccccccccccccc} {0\;\;1\;\;0}\\ {1\;\;0\;\;0}\\ {0\;\;0\;\;1} \end{array}} \right],\;\mathrm{then}\;{A^{ - 1}} is equal to
View answer and explanation
Correct option: B
AA
Option B (AA) is correct because the given matrix AA is an involution matrix, meaning A2=IA^2 = I, and therefore A−1=AA^{-1} = A.
198 Medium
198 of 995
Application +4 | -1
Distance of line y=∣x+1xxxx+2xxxx+3∣y=\left|\begin{array}{ccc} x+1 & x & x \\ x & x+2 & x \\ x & x & x+3 \end{array}\right| from the origin is
View answer and explanation
Correct option: C
6122\frac{6}{\sqrt{122}}
The correct answer is C (6122\frac{6}{\sqrt{122}}) because the equation of the line is found by evaluating the determinant, and then the distance from the origin is calculated using the formula.
199 Easy
199 of 995
Analyzing +4 | -1
Let AA be a 2×22 \times 2 matrix with non-zero entries and let A2=I,A^2=I, where II is 2×22 \times 2 identity matrix. Define Tr⁡(A)=\operatorname{Tr}(A)= sum of diagonal elements of AA and ∣A∣=|A|= determinant of matrix A.A.

Statement-1 Tr⁡(A)=0\operatorname{Tr}(A)=0

Statement-2 ∣A∣=1|A|=1
View answer and explanation
Correct option: B
Statement-1 is true, Statement-2 is false.
Option B (Statement-1 is true, Statement-2 is false) is correct because the trace of A is 0, but the determinant of A is -1, not 1.
200 Easy
200 of 995
Analyzing +4 | -1
If matrix A=[aij]3×3,A=\left[a_{i j}\right]_{3 \times 3}, matrix B=[bij]3×3,B=\left[b_{i j}\right]_{3 \times 3}, where aij+aji=0a_{i j}+a_{j i}=0 and bij−bji=0,b_{i j}-b_{j i}=0, then A4⋅B3A^4 \cdot B^3 is
View answer and explanation
Correct option: B
Singular
Option B (Singular) is correct because since A is skew-symmetric, its determinant is 0, making the determinant of A4 * B3 also 0, hence singular.
201 Easy
201 of 995
Application +4 | -1
If A=[5a−b32]A = \left[ {\begin{array}{cccccccccccccccccccc} {5a}&{ - b}\\ 3&2 \end{array}} \right] and A adj A=AATA\,\mathrm{adj}\,A = A{A^T}, then 5a+b5a + b is equal to
View answer and explanation
Correct option: B
5
The correct answer is B (5) because it is found by using the property A adj A=AATA\,\mathrm{adj}\,A = A{A^T} and solving the resulting matrix equation.
202 Easy
202 of 995
Application +4 | -1
If A=[22−32]A=\left[\begin{array}{cc}2 & 2 \\ -3 & 2\end{array}\right] and B=[0−110]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: A
[2−223]\left[\begin{array}{cc}2 & -2 \\ 2 & 3\end{array}\right]
Option A [2−223]\left[\begin{array}{cc}2 & -2 \\2 & 3\end{array}\right] is correct because (B−1A−1)−1=AB\left(B^{-1} A^{-1}\right)^{-1} = AB, and calculating AB gives the result.
203 Medium
203 of 995
Application +4 | -1
Given that matrix A=[x321y422z]A=\left[\begin{array}{lll}x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z\end{array}\right]. If xyz=2013x y z=2013 and 8x+4y+3z=20128 x+4 y+3 z=2012 then A(adj⁡A)A(\operatorname{adj} A) is equal to
View answer and explanation
Correct option: C
[290002900029]\left[\begin{array}{ccc}29 & 0 & 0 \\ 0 & 29 & 0 \\ 0 & 0 & 29\end{array}\right]
Option C (\left[\begin{{array}}{{ccc}}29 & 0 & 0 \\\\ 0 & 29 & 0 \\\\ 0 & 0 & 29\end{{array}}\right]\left[\begin{{array}}{{ccc}}29 & 0 & 0 \\\\ 0 & 29 & 0 \\\\ 0 & 0 & 29\end{{array}}\right]) is correct because A(adj⁡A)=∣A∣IA(\operatorname{{adj}} A) = |A|I, and the determinant of A, |A|, is 29.
204 Easy
204 of 995
Application +4 | -1
If A=[0sin⁡αsin⁡α0]A = \left[ {\begin{array}{cccccccccccccccccccc} 0&{\sin \alpha }\\ {\sin \alpha }&0 \end{array}} \right] and det⁡(A2−12l)=0,\det \left( {{A^2} - \frac{1}{2}l} \right) = 0, then a possible value of α\alpha is
View answer and explanation
Correct option: C
π4\frac{\pi }{4}
Option C (π4\frac{\pi}{{4}}) is correct because it satisfies the determinant equation after substituting the matrix A and identity matrix.
205 Medium
205 of 995
Application +4 | -1
The value of ∣a1x1+b1y1a1x2+b1y2a1x3+b1y3a2x1+b2y1a2x2+b2y2a2x3+b2y3a3x1+b3y1a3x2+b3y2a3x3+b3y3∣,\left|\begin{array}{llllll}a_1x_1+b_1y_1 & a_1 x_2+b_1 y_2 & a_1 x_3+b_1 y_3 \\ a_2 x_1+b_2y_1 & a_2 x_2+b_2 y_2 & a_2 x_3+b_2 y_3 \\ a_3x_1+b_3y_1 & a_3 x_2+b_3y_2 & a_3 x_3+b_3 y_3\end{array}\right|, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the determinant can be expressed as a product of two determinants, each having a column of zeros, resulting in a zero determinant.
206 Medium
206 of 995
Analyzing +4 | -1
Let Δ=∣−bcb2+bcc2+bca2+ac−acc2+aca2+abb2+ab−ab∣\quad \Delta=\left|\begin{array}{ccc}-b c & b^2+b c & c^2+b c \\ a^2+a c & -a c & c^2+a c \\ a^2+a b & b^2+a b & -a b\end{array}\right| and the equation px3+qx2+rx+s=0p x^3+q x^2+r x+s=0 has roots a,b,ca, b, c, where a,b,c∈R+a, b, c \in R^{+}, then if Δ=27\Delta=27 and a2+b2+c2=2a^2+b^2+c^2=2 find 2pq+1\frac{\sqrt{2} p}{q}+1.
View answer and explanation
Correct option: C
01-Feb
The correct answer is 0.5 because after simplifying the determinant and using the given conditions, we can find the value of 2pq+1\frac{\sqrt{2} p}{q}+1.
207 Easy
207 of 995
Application +4 | -1
Let P=[3/21/2−1/23/2],A=[1101]P=\left[\begin{array}{cc}\sqrt{3} / 2 & 1 / 2 \\ -1 / 2 & \sqrt{3} / 2\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] and Q=PAP′Q=P A P^{\prime}, then P′Q2015PP^{\prime} Q^{2015} P is
View answer and explanation
Correct option: C
[1201501]\left[\begin{array}{cc}1 & 2015 \\ 0 & 1\end{array}\right]
Option C (\begin{{bmatrix}}1 & 2015 \\\\ 0 & 1 \end{{bmatrix}}\begin{{bmatrix}}1 & 2015 \\\\ 0 & 1 \end{{bmatrix}}) is correct because it results from simplifying the matrix expression P′Q2015PP'Q^{2015}P using properties of matrix inverses and binomial theorem.
208 Easy
208 of 995
Application +4 | -1
If P=[3212−1232],A=[1101]P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], \quad A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] and Q=PAPTQ=P A P^T, then PTQ2015PP^T Q^{2015} P, is
View answer and explanation
Correct option: D
[1201501]\left[\begin{array}{cc}1 & 2015 \\ 0 & 1\end{array}\right]
Option D ([1201501]\left[\begin{array}{cc}1 & 2015 \\ 0 & 1\end{array}\right]) is correct because after simplification, PTQ2015PP^T Q^{2015} P equals A2015A^{2015}, which results in the given matrix.
209 Easy
209 of 995
Understanding +4 | -1
Which one of the following statements is true?
View answer and explanation
Correct option: C
If A′=A\mathrm{A}^{\prime}=\mathrm{A}, then A\mathrm{A} is a square matrix
Option C (If A' = A, then A is a square matrix) is correct because the transpose of a matrix is only equal to the original matrix if it's a square matrix.
210 Easy
210 of 995
Application +4 | -1
If C=2cos⁡θC = 2\cos \mathrm{\theta }, then the value of the determinant Δ=∣C101C161C∣\Delta = \left| {\begin{array}{cccccccccccccccccccc} C&1&0\\ 1&C&1\\ 6&1&C \end{array}} \right| is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the determinant simplifies to 8cos⁡3θ−4cos⁡,θ+68{\cos ^3}{\theta } - 4\cos \\,\mathrm{\theta } + 6, which doesn't match any of the other options.
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