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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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181 Easy
181 of 922
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.
182 Easy
182 of 922
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.
183 Easy
183 of 922
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.
184 Easy
184 of 922
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.
185 Easy
185 of 922
Analyzing +4 | -1
The determinant Δ=∣cos⁡(α+β)−sin⁡(α+β)cos⁡2βsin⁡αcos⁡αsin⁡β−sin⁡αsin⁡αcos⁡β∣\Delta = \left| {\begin{array}{cccccccccccccccccccc} {\cos \left( {\alpha + \beta } \right)}&{ - \sin \left( {\alpha + \beta } \right)}&{\cos 2\beta }\\ {\sin \alpha }&{\cos \alpha }&{\sin \beta }\\ { - \sin \alpha }&{\sin \alpha }&{\cos \beta } \end{array}} \right| is independent of
View answer and explanation
Correct option: A
α\alpha
Option A (α\alpha) is correct because after applying row operations, the determinant simplifies to an expression independent of α\alpha.
186 Easy
186 of 922
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.
187 Easy
187 of 922
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.
188 Easy
188 of 922
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.
189 Easy
189 of 922
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.
190 Easy
190 of 922
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.
191 Easy
191 of 922
Application +4 | -1
In a symmetric matrix of order ' 12 ' maximum number of different elements are -
View answer and explanation
Correct option: A
78
Option A (78) is correct because the maximum number of different elements in a symmetric matrix of order 'n' is given by the formula (n2 + n) / 2.
192 Easy
192 of 922
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.
193 Easy
193 of 922
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.
194 Easy
194 of 922
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.
195 Easy
195 of 922
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.

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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 922
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.
197 Easy
197 of 922
Application +4 | -1
If P=[cos⁡(π/6)sin⁡(π/6)−sin⁡(π/6)cos⁡(π/6)],A=[1101]P=\left[\begin{array}{cc} \cos (\pi / 6) & \sin (\pi / 6) \\ -\sin (\pi / 6) & \cos (\pi / 6) \end{array}\right], \quad A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] and Q=PAP′Q=P A P^{\prime} then P′Q2019PP^{\prime} Q^{2019} P is equal to
View answer and explanation
Correct option: B
[1201901]\left[\begin{array}{cc}1 & 2019 \\ 0 & 1\end{array}\right]
Option B is correct. The expression simplifies to A2019A^{2019}, which, using the binomial theorem, results in the matrix [1201901]\left[\begin{array}{cc}1 & 2019 \\ 0 & 1\end{array}\right].
198 Easy
198 of 922
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.
199 Easy
199 of 922
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.
200 Easy
200 of 922
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.
201 Easy
201 of 922
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.
202 Easy
202 of 922
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.
203 Easy
203 of 922
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).
204 Easy
204 of 922
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.
205 Easy
205 of 922
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.
206 Easy
206 of 922
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.
207 Easy
207 of 922
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.
208 Easy
208 of 922
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.
209 Easy
209 of 922
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.
210 Easy
210 of 922
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.
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