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JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

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Showing 811-840 of 995 questions

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811 Easy
811 of 995
Application +4 | -1
Let A(02qrpq−rp−qr)A\left( {\begin{array}{cccccccccccccccccccc} 0&{2q}&r\\ p&q&{ - r}\\ p&{ - q}&r \end{array}} \right). If AAT=I3,A{A^T} = {I_3}, then ∣p∣\left| p \right| is
View answer and explanation
Correct option: A
15\frac{1}{{\sqrt 5 }}
Option B (12\frac{1}{{\sqrt 2 }}) is correct because it's the absolute value of p derived from the matrix equation AAT=I3A{A^T} = {I_3}.
812 Easy
812 of 995
Application +4 | -1
If A=[0    10    0],IA = \left[ {\begin{array}{cccccccccccccccccccc} {0\;\;1}\\ {0\;\;0} \end{array}} \right],I is the unit matrix of order 2 and a,ba,b are arbitrary constants, then (aI+bA)2{\left( {aI + bA} \right)^2} is equal to
View answer and explanation
Correct option: B
a2I+2abA{a^2}I + 2abA
Option B (a2I+2abA{a^2}I + 2abA) is correct because when expanding the square and using the property that A² = 0, we arrive at this result.
813 Medium
813 of 995
Analyzing +4 | -1
If a>0a>0 and discriminant of ax2+2bx+ca x^2+2 b x+c is negative, then
Δ=∣abax+bbcbx+cax+bbx+c0∣,\Delta=\left|\begin{array}{ccc}a & b & a x+b \\ b & c & b x+c \\ a x+b & b x+c & 0\end{array}\right|, is
View answer and explanation
Correct option: C
Negative
Option C (Negative) is correct because the determinant is equal to −(ac−b2)(ax2+2bx+c)-(ac-b^2)(ax^2+2bx+c), and since ac−b2>0ac-b^2 > 0 and ax2+2bx+c>0ax^2+2bx+c > 0, the determinant is negative.
814 Easy
814 of 995
Application +4 | -1
If [1−tan⁡θtan⁡θ1][1tan⁡θ−tan⁡θ1]−1=[a−bba],\left[ {\begin{array}{cccccccccccccccccccc} 1&{ - \tan \theta }\\ {\tan \theta }&1 \end{array}} \right]{\left[ {\begin{array}{cccccccccccccccccccc} 1&{\tan \theta }\\ { - \tan \theta }&1 \end{array}} \right]^{ - 1}} = \left[ {\begin{array}{cccccccccccccccccccc} a&{ - b}\\ b&a \end{array}} \right], then
View answer and explanation
Correct option: B
a=cos⁡2  θ,  b=sin⁡2  θa = \cos 2\;\theta ,\;b = \sin 2\;\theta
Option B (a=cos⁡2;θ,;b=sin⁡2;θa = \cos 2\\;\theta ,\\;b = \sin 2\\;\theta) is correct because after simplifying the matrix expression using trigonometric identities, the resulting matrix elements match these values.
815 Hard
815 of 995
Analyzing +4 | -1
If A,BA, B and CC are angles of a triangle, the value of ∣e2iAe−iCe−iBe−iCe2iBe−iAe−iBe−iBAe2iC∣\left|\begin{array}{lll}e^{2 i A} & e^{-i C} & e^{-i B} \\ e^{-i C} & e^{2 i B} & e^{-i A} \\ e^{-i B} & e^{-i B A} & e^{2 i C}\end{array}\right| is (where i=−1)\left.i=\sqrt{-1}\right)
View answer and explanation
Correct option: D
-4
Option D (-4) is correct because the determinant simplifies to -4 after applying properties of determinants and using the fact that A+B+C = π\pi.
816 Easy
816 of 995
Understanding +4 | -1
Which one of the following is true always for any two non-singular matrices AA and BB of same order?
View answer and explanation
Correct option: D
(AB)−1=B−1A−1{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}
Option D ((AB)−1=B−1A−1{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}) is correct because this is a fundamental property of the inverse of a product of matrices.
817 Easy
817 of 995
Application +4 | -1
If A=[ab],B=[−b−a]\mathrm{A}=\left[\begin{array}{ll}a & b\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}-b & -a\end{array}\right] and C=[a−a]\mathrm{C}=\left[\begin{array}{c}a \\ -a\end{array}\right], then the correct statement is
View answer and explanation
Correct option: C
AC=BC\mathrm{AC}=\mathrm{BC}
Option C (AC = BC) is correct because when the matrices are multiplied, the resulting matrices are equal.
818 Easy
818 of 995
Application +4 | -1
If A=[α22α]A=\left[\begin{array}{ll}\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 by using the property ∣A3∣=∣A∣3|A^3| = |A|^3 and solving the resulting equation, we find α=±3\alpha = \pm 3.
819 Medium
819 of 995
Application +4 | -1
Matrix AA is such that A2=2A−I{A^2} = 2A - I where II is the identity matrix, then for n≥2,  Ann \ge 2,\;{A^n} is equal to
View answer and explanation
Correct option: A
n  A−(n−1)In\;A - \left( {n - 1} \right)I
Option A (n  A−(n−1)In\;A - \left( {n - 1} \right)I) is correct because the pattern An=nA−(n−1)I{A^n} = nA - (n-1)I holds true based on the given relation A2=2A−I{A^2} = 2A - I.
820 Easy
820 of 995
Application +4 | -1
cos⁡θ[cos⁡θ        sin⁡θ−sin⁡θ          cos⁡θ]+sin⁡θ[sin⁡θ−cos⁡θcos⁡θ        sin⁡θ]\cos \theta \left[ {\begin{array}{cccccccccccccccccccc} {\cos \theta \;\;\;\;\sin \theta }\\ { - \sin \theta \;\;\;\;\;\cos \theta } \end{array}} \right] + \sin \theta \left[ {\begin{array}{cccccccccccccccccccc} {\sin \theta - \cos \theta }\\ {\cos \theta \;\;\;\;\sin \theta } \end{array}} \right] is equal to
View answer and explanation
Correct option: D
[1001]\left[ {\begin{array}{cccccccccccccccccccc} 1&0\\ 0&1 \end{array}} \right]
Option D \left[ {\begin{{array}}{{*{20}{c}}}1&0\\0&1\end{{array}}} \right]\left[ {\begin{{array}}{{*{20}{c}}}1&0\\0&1\end{{array}}} \right] is correct because after performing the matrix multiplication and addition, and using the trigonometric identity cos⁡2θ+sin⁡2θ=1{{\cos }^2}\theta + {{\sin }^2}\theta = 1, the result is the identity matrix.
821 Easy
821 of 995
Application +4 | -1
If ∣a1b1c1a2b2c2a3b3c3∣=5,\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=5, then the value of Δ=∣b2c3−b3c2a3c2−a2c3a2b3−a3b2b3c1−b1c3a1c3−a3c1a3b1−a1b3b1c2−b2c1a2c1−a1c2a1b2−a2b1∣,\Delta=\left|\begin{array}{lll} b_2 c_3-b_3 c_2 & a_3 c_2-a_2 c_3 & a_2 b_3-a_3 b_2 \\ b_3 c_1-b_1 c_3 & a_1 c_3 - a_3 c_1 & a_3 b_1-a_1 b_3 \\ b_1 c_2-b_2 c_1 & a_2 c_1 - a_1 c_2 & a_1 b_2-a_2 b_1 \end{array}\right|, is
View answer and explanation
Correct option: B
25
Option B (25) is correct because the determinant of the cofactor matrix is the square of the original determinant.
822 Easy
822 of 995
Application +4 | -1
If A=[abba]A=\left[\begin{array}{ll}a & b \\ b & a\end{array}\right] and A2=[αββα]A^{2}=\left[\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right], then
View answer and explanation
Correct option: C
α=a2+b2,β=2ab\alpha=a^{2}+b^{2}, \beta=2 a b
Option C (α=a2+b2,β=2ab\alpha=a^{2}+b^{2}, \beta=2 a b) is correct because when A2A^2 is calculated, the resulting matrix elements match these expressions.
823 Easy
823 of 995
Application +4 | -1
If A=12[13−31]A=\frac{1}{2}\left[\begin{array}{cc}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{array}\right], then:
View answer and explanation
Correct option: B
A30+A25−A=IA^{30}+A^{25}-A=I
Option B (A30+A25−A=IA^{30}+A^{25}-A=I) is correct because it is the result of simplifying the matrix powers using trigonometric identities.
824 Medium
824 of 995
Application +4 | -1
If A=[1235],A = \left[ {\begin{array}{cccccccccccccccccccc} 1&2\\ 3&5 \end{array}} \right], then the value of the determinant ∣A2009−5A2008∣\left| {{A^{2009}} - 5{A^{2008}}} \right|
View answer and explanation
Correct option: A
−6- 6
Option A (-6) is correct because simplifying the expression using determinant properties leads to |A2008| * |A - 5I| = 1 * (-6) = -6.
825 Easy
825 of 995
Application +4 | -1
In a 4×44 \times 4 matrix the sum of each row, column and both the main diagonals is α\alpha. Then the sum of the four corner elements
View answer and explanation
Correct option: A
is also α\alpha
Option A (is also α\alpha) is correct because the sum of the four corner elements is equal to α\alpha, derived by using the given conditions about the sum of each row, column, and main diagonals.

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
826 Medium
826 of 995
Application +4 | -1
None of these ∣x−1111x−1111x−1∣=0,\left| {\begin{array}{cccccccccccccccccccc} {x - 1}&1&1\\ 1&{x - 1}&1\\ 1&1&{x - 1} \end{array}} \right| = 0, are
View answer and explanation
Correct option: B
−1,2- 1,2
Option B ( - 1,2) is correct because these are the values of x that satisfy the determinant equation after simplification.
827 Easy
827 of 995
Application +4 | -1
If A=[cos⁡xsin⁡x−sin⁡xcos⁡x]A = \left[ {\begin{array}{cccccccccccccccccccc} {\cos x}&{\sin x}\\ { - \sin x}&{\cos x} \end{array}} \right] and A(adj  A)=k[1001],A\left( {\mathrm{adj\;}A} \right) = k\left[ {\begin{array}{cccccccccccccccccccc} 1&0\\ 0&1 \end{array}} \right], then the value of kk is
View answer and explanation
Correct option: B
1
Option B (1) is correct because A(adjA)=∣A∣IA(adj A) = |A|I, and in this case, ∣A∣=1|A| = 1.
828 Easy
828 of 995
Application +4 | -1
Matrix AA is such that A2=2A−I{A^2} = 2A - I where II is the identity matrix, then for n≥2,  Ann \ge 2,\;{A^n} is equal to
View answer and explanation
Correct option: A
n  A−(n−1)In\;A - \left( {n - 1} \right)I
Option A is correct. The recurrence relation can be solved to find a general formula for A^n in terms of A and I.
829 Easy
829 of 995
Application +4 | -1
If AA and BB are square matrices of order 3 such that ∣A∣=−1,∣B∣=3\left| A \right| = - 1,\left| B \right| = 3 then ∣3AB∣\left| {3AB} \right| is equal to
View answer and explanation
Correct option: B
−81- 81
Option B (-81) is correct because using the properties of determinants, |3AB| = 33 * |A| * |B| = 27 * (-1) * 3 = -81.
830 Easy
830 of 995
Application +4 | -1
If ∣x2+x3x−1−x+32x+12+x2x3−3x−3x2+43x∣\left| {\begin{array}{cccccccccccccccccccc} {{x^2} + x}&{3x - 1}&{ - x + 3}\\ {2x + 1}&{2 + {x^2}}&{{x^3} - 3}\\ {x - 3}&{{x^2} + 4}&{3x} \end{array}} \right| =a0+a1x+a2x2+.…+a7x7,= {a_0} + {a_1}x + {a_2}{x^2}{ + _.} \ldots + {a_7}{x^7}, The value of a0{a_0} is
View answer and explanation
Correct option: D
21
Option D (21) is correct because substituting x=0 into the determinant and evaluating gives the constant term a0{a_0}.
831 Easy
831 of 995
Application +4 | -1
If A2−3A+2I=0A^2-3 A+2 I=0, then A−1A^{-1} equals
View answer and explanation
Correct option: B
12(3I−A)\frac{1}{2}(3 I-A)
Option B (\begin{{bmatrix}}0 & 1 \\\\ 0 & 2+i \end{{bmatrix}}\begin{{bmatrix}}0 & 1 \\\\ 0 & 2+i \end{{bmatrix}}) is correct because it results from solving the matrix equation for X.
832 Easy
832 of 995
Application +4 | -1
If A=[2−3−41]A=\left[\begin{array}{cc}2 & -3 \\ -4 & 1\end{array}\right], then adj (3A2−12A)\left(3 A^2-12 A\right) is equal to
View answer and explanation
Correct option: B
[27−9−1224]\left[\begin{array}{cc}27 & -9 \\ -12 & 24\end{array}\right]
The correct answer is \left[\begin{{array}}{{cc}}27 & -9 \\ -12 & 24\end{{array}}\right]\left[\begin{{array}}{{cc}}27 & -9 \\ -12 & 24\end{{array}}\right] because it is the adjugate of the matrix 3A2−12A3A^2 - 12A after correct matrix multiplication and subtraction.
833 Easy
833 of 995
Analyzing +4 | -1
Let ω\omega be a complex cube root of unity with ω≠1\omega \neq 1 and P=[pij]P=\left[p_{i j}\right] bean ×\times nmatrix with pij=ωi+jp_{i j}=\omega^{i+j}. If P2=OP^2=O, then nn can take value
View answer and explanation
Correct option: A
57
Option A (57) is correct. The condition P2 = O implies that n must be a multiple of 3, and 57 is the only option that satisfies this condition.
834 Easy
834 of 995
Understanding +4 | -1
Let AA be a square matrix all of whose entries are integers. Then, which one of the following is true?
View answer and explanation
Correct option: D
If det (A)=±1\left( A \right) = \pm 1, then A−1{A^{ - 1}} exists and all its entries are integers
Option D is correct because if det(A) = ±1, then A⁻¹ exists and all its entries are integers, as A⁻¹ = (1/det(A)) * adj(A), and both det(A) and adj(A) have integer entries.
835 Easy
835 of 995
Application +4 | -1
If pλ4+qλ3+rλ2+sλ+t=∣λ2+3λλ−1λ+3λ+12−λλ−4λ−3λ+43λ∣p \lambda^4+q \lambda^3+r \lambda^2+s \lambda+t=\left|\begin{array}{ccc}\lambda^2+3 \lambda & \lambda-1 & \lambda+3 \\ \lambda+1 & 2-\lambda & \lambda-4 \\ \lambda-3 & \lambda+4 & 3 \lambda\end{array}\right|, the value of tt is
View answer and explanation
Correct option: B
18
Option B (18) is correct because by substituting λ=0\lambda = 0 into the given equation, we can directly solve for the value of tt.
836 Easy
836 of 995
Application +4 | -1
[2x+y4x5x−74x]=[77y−13yx+6]\left[\begin{array}{cc}2 x+y & 4 x \\ 5 x-7 & 4 x\end{array}\right]=\left[\begin{array}{cc}7 & 7 y-13 \\ y & x+6\end{array}\right], then the value of x+yx+y is
View answer and explanation
Correct option: B
x=2,y=3x=2, y=3
Option B (x=2,y=3x=2, y=3) is correct because by comparing the corresponding elements of the matrices, we can solve for x and y, resulting in x=2 and y=3.
837 Easy
837 of 995
Analyzing +4 | -1
The number of real values of λ\lambda, such that the system of linear equations
\begin{aligned}
&2 x-3 y+5 z=9 \\
&x+3 y-z=-18 \\
&3 x-y+\left(λ^2-|λ|\right) z=16
\end{aligned}
has no solutions, is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the system has no solutions when the determinant of the coefficient matrix is zero, leading to two possible values for lambda.
838 Medium
838 of 995
Application +4 | -1
If A=[1221]A = \left[ {\begin{array}{cccccccccccccccccccc} 1&2\\ 2&1 \end{array}} \right] and f(x)=1+x1−xf\left( x \right) = \frac{{1 + x}}{{1 - x}}, then f(A)f\left( A \right) is
View answer and explanation
Correct option: A
[−1−1−1−1]\left[ {\begin{array}{cccccccccccccccccccc} { - 1}&{ - 1}\\ { - 1}&{ - 1} \end{array}} \right]
Option A is correct because substituting the matrix A into the function f(x) and simplifying leads to the matrix with all elements equal to -1.
839 Easy
839 of 995
Understanding +4 | -1
AA is a square matrix of order 4 and II is a unit matrix, then it is true that
View answer and explanation
Correct option: B
det⁡(2 A)=16det⁡(A)\operatorname{det}(2 \mathrm{~A})=16 \operatorname{det}(\mathrm{A})
Option B (det(2A) = 16det(A)) is correct because for a square matrix A of order n, det(kA) = k^n det(A). In this case, n=4 and k=2.
840 Easy
840 of 995
Application +4 | -1
The matrix A2+4A−5IA^{2}+4 A-5 I, where II is identity matrix and A=[124−3]A=\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right], equals
View answer and explanation
Correct option: A
4[2120]4\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]
Option A (4\left[\begin{{array}}{{ll}}2 & 1 \\ 2 & 0\end{{array}}\right]4\left[\begin{{array}}{{ll}}2 & 1 \\ 2 & 0\end{{array}}\right]) is correct because the matrix A2+4A−5IA^2 + 4A - 5I simplifies to 4\left[\begin{{array}}{{ll}}2 & 1 \\ 2 & 0\end{{array}}\right]4\left[\begin{{array}}{{ll}}2 & 1 \\ 2 & 0\end{{array}}\right] after matrix multiplication and addition.
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