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JEE • MATHEMATICS • MATRICES

Matrices practice · Mixed difficulty

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Showing 511-540 of 995 questions

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511 Easy
511 of 995
Analyzing +4 | -1
Let MM and NN be two 3×33 \times 3 non-singular skew-symmetric matrices such that MN=NMM N=N M. If PTP^T denotes the transpose of PP, then M2N2(MTN)−1(MN−1)TM^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T is equal to
View answer and explanation
Correct option: C
−M2-M^2
Option C (-M2) is correct. After simplifying the expression using properties of skew-symmetric matrices and matrix transpose, the result is -M2.
512 Easy
512 of 995
Remembering +4 | -1
If AA is an orthogonal matrix, then A−1A^{-1}, equals
View answer and explanation
Correct option: B
A′{A}^{\prime}
Option B ({A}^{\'}{A}^{\'}) is correct because the inverse of an orthogonal matrix is equal to its transpose.
513 Medium
513 of 995
Application +4 | -1
Let A=[aij],B=[bij]A=\left[a_{i j}\right], B=\left[b_{i j}\right] be two 3×33 \times 3 matrices such that bij=λi+j−2aijb_{i j}=\lambda^{i+j-2} a_{i j} and ∣B∣=81|B|=81, then the value of ∣A∣|A|, if λ=3\lambda=3, is
View answer and explanation
Correct option: A
19\frac{1}{9}
Option A (1/9) is correct because it is the value of |A| calculated using the given relationship between the elements of matrices A and B and their determinants.
514 Easy
514 of 995
Application +4 | -1
If  A=[123],  B=[234]  and  C=[1502],  then  which  of   theI{\text{f}}\,\,A = \left[ {123} \right],\,\,B = \left[ \begin{gathered} 2 \quad \\ 3 \quad \\ 4 \quad \\ \end{gathered} \right]\,\,{\text{and}}\,\,C = \left[ {\begin{array}{cccccccccccccccccccc} 1&5 \\ 0&2 \end{array}} \right],\,\,{\text{then}}\,\,{\text{which}}\,\,{\text{of}}\;\,{\text{the}} following is defined
View answer and explanation
Correct option: A
AB\mathrm{AB}
Option A (AB) is correct because the number of columns in A matches the number of rows in B, making matrix multiplication AB defined.
515 Medium
515 of 995
Understanding +4 | -1
If AA is a 3×33 \times 3 skew-symmetric matrix, then trace of AA is equal to
View answer and explanation
Correct option: D
∣A∣|A|
Option D (∣A∣|A|) is correct because the trace of a skew-symmetric matrix is always 0, and the determinant of a 3x3 skew-symmetric matrix is also 0.
516 Easy
516 of 995
Analyzing +4 | -1
If AA is any square matrix, then det⁡(A−AT)T\det {\left( {A - {A^T}} \right)^T} is equal to
View answer and explanation
Correct option: C
Can be 0 or a perfect square
Option C (Can be 0 or a perfect square) is correct because the determinant of a skew-symmetric matrix can be 0 (if the order is odd) or a perfect square (if the order is even).
517 Easy
517 of 995
Analyzing +4 | -1
If x2+y2+z2≠0,  x=cy+bz,  y=az+cx{x^2} + {y^2} + {z^2} \ne 0,\;x = cy + bz,\;y = az + cx and z=bx+ay,z = bx + ay, then a2+b2+c2+2abc={a^2} + {b^2} + {c^2} + 2abc =
View answer and explanation
Correct option: C
1
Option C (1) is correct because the determinant of the coefficient matrix must be zero for non-trivial solutions, leading to the given expression equaling 1.
518 Easy
518 of 995
Analyzing +4 | -1
If =[ete−tcos⁡te−tsin⁡tet−e−tcos⁡t−e−t sin⁡t−e−tsin⁡t+e−tcos⁡tet2e−tsin⁡t−2e−tcos⁡t]= \left. {\left[ {\begin{array}{cccccccccccccccccccc} {{e^t}}&{{e^{ - t}}\cos t}&{{e^{ - t}}\sin t}\\ {{e^t}}&{ - {e^{ - t}}\cos t - {e^{ - t}}\,\sin t}&{ - {e^{ - t}}\sin t + {e^{ - t}}\cos t}\\ {{e^t}}&{2{e^{ - t}}\sin t}&{ - 2{e^{ - t}}\cos t} \end{array}} \right.} \right] then AA is
View answer and explanation
Correct option: B
Invertible for every t∈Rt \in R
Option B (Invertible for every t∈Rt \in R) is correct because the determinant of A is non-zero for all t in R, implying that A is always invertible.
519 Easy
519 of 995
Application +4 | -1
If A=13[12221−2a2b]A=\frac{1}{3}\left[\begin{array}{rrr}1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b\end{array}\right] is an orthogonal matrix,
View answer and explanation
Correct option: B
a=−2,b=−1a=-2, b=-1
Option D (a=−2,b=1a=-2, b=1) is correct because for A to be orthogonal, AAT=IAA^T = I. Solving the resulting equations gives a=−2a = -2 and b=−1b = -1.
520 Medium
520 of 995
Application +4 | -1
Suppose α,β,γ,θ∈R\alpha, \beta, \gamma, \theta \in {R} and A(α,β,γ,θ)=∣cos⁡(α+θ)sin⁡(α+θ)1cos⁡(β+θ)sin⁡(β+θ)1cos⁡(γ+θ)sin⁡(γ+θ)1∣A(\alpha, \beta, \gamma, \theta)=\left|\begin{array}{lll} \cos (\alpha+\theta) & \sin (\alpha+\theta) & 1 \\ \cos (\beta+\theta) & \sin (\beta+\theta) & 1 \\ \cos (\gamma+\theta) & \sin (\gamma+\theta) & 1 \end{array}\right| Numerical value of A(−π2,0,π2,2π13)A\left(-\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{2 \pi}{13}\right) is
View answer and explanation
Correct option: C
2
Option D (None of these) is incorrect. The determinant A is independent of θ\theta and its value is 2.
521 Easy
521 of 995
Understanding +4 | -1
The rank of the matrix A=[12344321]A=\left[\begin{array}{llll}1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1\end{array}\right], is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the matrix has two linearly independent rows (or columns), making its rank 2.
522 Easy
522 of 995
Application +4 | -1
If f(θ)=[cos⁡2θcos⁡θsin⁡θ−sin⁡θcos⁡θsin⁡θsin⁡2θcos⁡θsin⁡θ−cos⁡θ0],f(\theta)=\left[\begin{array}{ccc}\cos ^2 \theta & \cos \theta \sin \theta & -\sin \theta \\ \cos \theta \sin \theta & \sin ^2 \theta & \cos \theta \\ \sin \theta & -\cos \theta & 0\end{array}\right], then f(π7)f\left(\frac{\pi}{7}\right) is
View answer and explanation
Correct option: D
Non-singular
Option D (Non-singular) is correct because the determinant of the matrix f(θ) is 1, which is non-zero, indicating that the matrix is non-singular.
523 Easy
523 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], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] and Q=PAPTQ=P A P^T, then PTQ2005PP^T Q^{2005} P, is
View answer and explanation
Correct option: A
[1200501]\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]
Option A ([1200501]\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]) is correct because PTQ2005P=A2005P^T Q^{2005} P = A^{2005}, and An=[1n01]A^n = \left[\begin{array}{cc}1 & n \\ 0 & 1\end{array}\right].
524 Easy
524 of 995
Application +4 | -1
The value of ∣log⁡3512log⁡43log⁡38log⁡49∣×∣log⁡23log⁡83log⁡34log⁡34∣\left| {\begin{array}{cccccccccccccccccccc} {{{\log }_3}512}&{{{\log }_4}3}\\ {{{\log }_3}8}&{{{\log }_4}9} \end{array}} \right| \times \left| {\begin{array}{cccccccccccccccccccc} {{{\log }_2}3}&{{{\log }_8}3}\\ {{{\log }_3}4}&{{{\log }_3}4} \end{array}} \right| is
View answer and explanation
Correct option: B
10
Option B (10) is correct because after simplifying the logarithmic expressions and evaluating the determinant, the result is 10.
525 Easy
525 of 995
Application +4 | -1
If the matrices A=[213410]A = \left[ {\begin{array}{cccccccccccccccccccc} 2&1&3\\ 4&1&0 \end{array}} \right] and B=[1−10250]B = \left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 1}\\ 0&2\\ 5&0 \end{array}} \right], then ABAB
View answer and explanation
Correct option: A
[1704−2]\left[ {\begin{array}{cccccccccccccccccccc} {17}&0\\ 4&{ - 2} \end{array}} \right]
Option A ([17 0], [4 -2]) is correct because it is the result of the matrix multiplication A*B as shown in the solution.

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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
526 Medium
526 of 995
Analyzing +4 | -1
If a1,b2,c3≠0a_1, b_2, c_3 \neq 0 Δ1=∣a12+b1+c1a1a2+b2+c2a1a3+b3+c3b1b2+c1b22+c2b2b3+c3c3c1c3c2c32∣\Delta_1=\left|\begin{array}{ccc}a_1^2+b_1+c_1 & a_1 a_2+b_2+c_2 & a_1 a_3+b_3+c_3 \\ b_1 b_2+c_1 & b_2^2+c_2 & b_2 b_3+c_3 \\ c_3 c_1 & c_3 c_2 & c_3^2\end{array}\right| and Δ2=∣a1b1c1a2b2c2a3b3c3∣\Delta_2=\left|\begin{array}{ccc}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|, then Δ1Δ2\frac{\Delta_1}{\Delta_2} equal to
View answer and explanation
Correct option: A
a1b2c3a_1 b_2 c_3
Option A (a1b2c3a_1 b_2 c_3) is correct because after determinant operations, the ratio simplifies to this product.
527 Easy
527 of 995
Analyzing +4 | -1
If AA is a square matrix of order 2 such that A[1−1]=[−12]A\left[\begin{array}{r}1 \\ -1\end{array}\right]=\left[\begin{array}{r}-1 \\ 2\end{array}\right] and A2[1−1]=[10].A^2\left[\begin{array}{r}1 \\ -1\end{array}\right]=\left[\begin{array}{l}1 \\ 0\end{array}\right]. The sum of elements and product of elements of AA are SS and PP, then S+PS+P is
View answer and explanation
Correct option: D
5
Option D (5) is correct because by solving the matrix equations, we find the elements of matrix A, and then calculate S+P = 5.
528 Medium
528 of 995
Application +4 | -1
If ω\omega is an imaginary cube root of unity, then the value of ∣1+ωω2−ω1+ω2ω−ω2ω2+ωω−ω2∣\left| {\begin{array}{cccccccccccccccccccc} {1 + \omega }&{{\omega ^2}}&{ - \omega } \\ {1 + {\omega ^2}}&\omega &{ - {\omega ^2}} \\ {{\omega ^2} + \omega }&\omega &{ - {\omega ^2}} \end{array}} \right| is equal to
View answer and explanation
Correct option: D
−3ω2-3 \omega^2
Option D (-3 w2) is correct because after applying column operations and using the property 1 + w + w2 = 0, the determinant simplifies to -3 w2.
529 Easy
529 of 995
Analyzing +4 | -1
If A=[aij]A=\left[a_{i j}\right] is a square matrix of even order such that aij=i2−j2a_{i j}=i^2-j^2, then
View answer and explanation
Correct option: D
none of these
The correct answer is none of these because A is a skew-symmetric matrix, and the determinant of a skew-symmetric matrix of even order is a perfect square, not necessarily zero.
530 Easy
530 of 995
Analyzing +4 | -1
Let the system of linear equations
4x+λy+2z=04x + \lambda y + 2z = 0
2x−y+z=02x - y + z = 0
μx+2y+3z=0,  λ,  μ∈R\mu x + 2y + 3z = 0,\,\,\lambda ,\,\,\mu \in R.
has a non-trivial solution. Then which of the following is true?
View answer and explanation
Correct option: A
μ=6, λ∈R\mu = 6,\,\lambda \in R
Option A (μ = 6, λ ∈ R) is correct because for a non-trivial solution, the determinant of the coefficient matrix must be zero, leading to this condition.
531 Easy
531 of 995
Understanding +4 | -1
If A\mathrm{A} and B\mathrm{B} are 2×22 \times 2 matrices, then which of the following is true ?
View answer and explanation
Correct option: C
(A−B)(A+B)=A2+AB−BA−B2(A-B)(A+B)=A^2+A B-B A-B^2
Option C ((A-B)(A+B)=A²+AB-BA-B²) is correct because matrix multiplication is not commutative, so AB is not necessarily equal to BA.
532 Medium
532 of 995
Analyzing +4 | -1
Let a,b,c∈Ra, b, c \in R such that abc=2a b c=2 if A=(3abcb3caca3b)A=\left(\begin{array}{ccc}3 a & b & c \\ b & 3 c & a \\ c & a & 3 b\end{array}\right) is such that AAT=16IA A^T=16 I and ∣det⁡A∣|\operatorname{det} A| is a positive real number, then the value (a3+b3+c3)2\left(a^3+b^3+c^3\right)^2 equals
View answer and explanation
Correct option: C
4
Option C (4) is correct because by using the given conditions AAT=16IA A^T = 16I and abc=2abc=2, we can find the value of (a3+b3+c3)2(a^3+b^3+c^3)^2.
533 Easy
533 of 995
Understanding +4 | -1
Let A,B{A}, {B} and C{C} are three matrices of same order. Now, consider the following statements

I. If A=B{A}={B}, then AC=BC{AC}={BC}

II. If AC=BC{AC}={BC}, then A=B{A}={B}

Choose the correct option
View answer and explanation
Correct option: A
Only II is true
Option A (Only I is true) is correct because if A=B, then AC=BC, but the converse is not always true.
534 Easy
534 of 995
Application +4 | -1
Let A=(4−2αβ)A=\left(\begin{array}{cc}4 & -2 \\ \alpha & \beta\end{array}\right)

If A2+γA+18l=0A^2+\gamma A+18 l=0, then det⁡(A)\operatorname{det}(A) is equal to _____.
View answer and explanation
Correct option: B
18
Option B (18) is correct because by finding the characteristic equation and comparing it with the given equation, the determinant of A can be determined.
535 Easy
535 of 995
Application +4 | -1
If  A=[1011]  and  I=[1001],\mathrm{If\;}A = \left[ {\begin{array}{cccccccccccccccccccc} 1&0\\ 1&1 \end{array}} \right]\;\mathrm{and\;}I = \left[ {\begin{array}{cccccccccccccccccccc} 1&0\\ 0&1 \end{array}} \right], then which one of the following holds for all n≥1n \ge 1, by the principle of mathematical induction?
View answer and explanation
Correct option: D
An=nA−(n−1)I{A^n} = nA - \left( {n - 1} \right)I
Option D is correct because by induction, An=nA−(n−1)I{A^n} = nA - (n - 1)I holds true for all n >= 1.
536 Medium
536 of 995
Application +4 | -1
If A=[02yzxy−zx−yz]A=\left[\begin{array}{ccc}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right] satisfies A′=A−1A^{\prime}=A^{-1}, then
View answer and explanation
Correct option: B
x=±1/2,y=±1/6,z=±1/3x= \pm 1 / \sqrt{2}, y= \pm 1 / \sqrt{6}, z= \pm 1 / \sqrt{3}
Option B (x=±1/2,y=±1/6,z=±1/3x= \pm 1 / \sqrt{2}, y= \pm 1 / \sqrt{6}, z= \pm 1 / \sqrt{3}) is correct because these values satisfy the condition A′=A−1A' = A^{-1}, which implies AA′=IAA' = I.
537 Medium
537 of 995
Application +4 | -1
∣a+b+2cabc2a+b+cbcaa+2b+c∣\left| {\begin{array}{cccccccccccccccccccc} {a + b + 2c}&a&b\\ c&{2a + b + c}&b\\ c&a&{a + 2b + c} \end{array}} \right| is equal to
View answer and explanation
Correct option: D
2(a+b+c)32{\left( {a + b + c} \right)^3}
Option D (2(a+b+c)32(a + b + c)^3) is correct because applying column and row operations simplifies the determinant to 2(a+b+c)32(a + b + c)^3.
538 Easy
538 of 995
Analyzing +4 | -1
If the system of linear equations x−2y+kz=1,2x+y+z=2,3x−y−kz=3,x - 2y + kz = 1,2x + y + z = 2,3x - y - kz = 3, has a solution (x,y,z),z≠0,\left( {x,y,z} \right),\mathrm{ }z \ne 0, then (x,y)(x,y) lies on the straight line whose equation is
View answer and explanation
Correct option: C
4x−3y−4=04x - 3y - 4 = 0
Option C (4x−3y−4=04x - 3y - 4 = 0) is correct because adding equations (i) and (iii) eliminates zz and directly yields the equation of the line on which (x,y)(x, y) lies.
539 Easy
539 of 995
Understanding +4 | -1
Let AA be a 3×33 \times 3 matrix such that A2−5A+7I=0{A^2} - 5A + 7I = 0
Statement- I A−1=17(5I−A)A^{-1}=\frac{1}{7}(5 I-A)
Statement - II The polynomial A3−2A2−3A+IA^3-2 A^2-3 A+I can be reduced to, 5(A−4I)5(A - 4I)
Then
View answer and explanation
Correct option: A
Statement - 1 is false, but statement - 2 is true
Option A (Statement - 1 is false, but statement - 2 is true) is correct because Statement I is actually true, and Statement II is also true.
540 Hard
540 of 995
Application +4 | -1
If f(x)=∣12(x−1)3(x−1)(x−2)x−1(x−1)(x−2)(x−1)(x−2)(x−3)xx(x−1)x(x−1)(x−2)∣f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} 1&{2\left( {x - 1} \right)}&{3\left( {x - 1} \right)\left( {x - 2} \right)}\\ {x - 1}&{\left( {x - 1} \right)\left( {x - 2} \right)}&{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}\\ x&{x\left( {x - 1} \right)}&{x\left( {x - 1} \right)\left( {x - 2} \right)} \end{array}} \right| Then, the value of f(49)f\left( {49} \right) is
View answer and explanation
Correct option: C
0
Option C (0) is correct because after applying column operations, the determinant becomes zero, making f(x) = 0 for all x.
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