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JEE • MATHEMATICS • MATRICES

Matrices practice · Medium

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121 Medium
121 of 408
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 determinants simplify to −x3-x^3, and their sum is −2x3-2x^3.
122 Medium
122 of 408
Application +4 | -1
The value of Δ=∣aa+ba+2ba+2baa+ba+ba+2ba∣\Delta = \left| {\begin{array}{cccccccccccccccccccc} a&{a + b}&{a + 2b}\\ {a + 2b}&a&{a + b}\\ {a + b}&{a + 2b}&a \end{array}} \right| is equal to
View answer and explanation
Correct option: B
9b2(a+b)9{b^2}\left( {a + b} \right)
Option B (9b2(a+b)9{b^2}(a + b)) is correct because after applying column operations and simplifying the determinant, the result is 9b2(a+b)9{b^2}(a + b).
123 Medium
123 of 408
Application +4 | -1
Let a,b,c,a,b,c, be positive and not all equal, the value of the Determinant ∣abcbcacab∣\left| {\begin{array}{cccccccccccccccccccc} a&b&c\\ b&c&a\\ c&a&b \end{array}} \right| is
View answer and explanation
Correct option: B
Negative
Option B (Negative) is correct because the determinant simplifies to a negative expression given the conditions on a, b, and c.
124 Medium
124 of 408
Understanding +4 | -1
If l1,m1,n1;l2,m2,n2l_1, m_1, n_1 ; l_2, m_2, n_2 and l3,m3,n3l_3, m_3, n_3 are direction cosines of three mutually perpendicular lines then, the value of ∣l1m1n1l2m2n2l3m3n3∣,\left|\begin{array}{lll}l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3\end{array}\right|, is
View answer and explanation
Correct option: B
±1\pm 1
Option B (±1\pm 1) is correct because the determinant of the direction cosines of three mutually perpendicular lines is either 1 or -1.
125 Medium
125 of 408
Analyzing +4 | -1
The number of triplets satisfying the equation ∣x3+1x2yx2zxy2y3+1y2zxz2yz2z3+1∣=11\left| {\begin{array}{cccccccccccccccccccc} {{x^3} + 1}&{{x^2}y}&{{x^2}z} \\ {x{y^2}}&{{y^3} + 1}&{{y^2}z} \\ {x{z^2}}&{y{z^2}}&{{z^3} + 1} \end{array}} \right| = 11 where x, y, zx, \, y, \, z are positive integers
View answer and explanation
Correct option: B
3
Option B (3) is correct because there are three triplets (2,1,1), (1,2,1), and (1,1,2) that satisfy the equation x3+y3+z3=10.
126 Medium
126 of 408
Understanding +4 | -1
From the matrix equation AB=ACAB = AC we can conclude B=CB = C provided that
View answer and explanation
Correct option: B
AA is non-singular
Option B (AA is non-singular) is correct because if AA is non-singular, it has an inverse, allowing us to multiply both sides of the equation AB=ACAB = AC by A−1A^{-1} to get B=CB = C.
127 Medium
127 of 408
Analyzing +4 | -1
If A,BA,B and CC are the angles of a triangle and ∣1111+sin⁡A1+sin⁡Bsin⁡Csin⁡A+sin⁡2Asin⁡B+sin⁡2Bsin⁡C+sin⁡2C∣=0\left| {\begin{array}{cccccccccccccccccccc} 1&1&1\\ {1 + \sin A}&{1 + \sin B}&{\sin C}\\ {\sin A + {{\sin }^2}A}&{\sin B + {{\sin }^2}B}&{\sin C + {{\sin }^2}C} \end{array}} \right| = 0 then the triangle must be
View answer and explanation
Correct option: B
Isosceles
The triangle must be isosceles because the determinant simplifies to a product of differences of sines of angles, which equals zero only when at least two angles have equal sines.
128 Medium
128 of 408
Analyzing +4 | -1
If x,y,zx, y, z are unequal and ∣x3(x+a)3(x−a)3y3(y−a)3(y−a)3z3(z+a)3(z−a)3∣=0\left|\begin{array}{ccc} x^3 & (x+a)^3 & (x-a)^3 \\ y^3 & (y-a)^3 & (y-a)^3 \\ z^3 & (z+a)^3 & (z-a)^3 \end{array}\right|=0 the the value of a2(x+y+z)a^2(x+y+z)
View answer and explanation
Correct option: B
3xyz3 x y z
Option B is correct because after applying column operations and simplifying the determinant, the equation leads to a2(x+y+z)=3xyza^2(x+y+z) = 3xyz.
129 Medium
129 of 408
Analyzing +4 | -1
If A=[21−1352161]A=\left[\begin{array}{rrr}2 & 1 & -1 \\ 3 & 5 & 2 \\ 1 & 6 & 1\end{array}\right], then tr⁡(Aadj⁡(adj⁡A)\operatorname{tr}(\operatorname{Aadj}(\operatorname{adj} \mathrm{A})) is equal to (where tr⁡(P)\operatorname{tr}(\mathrm{P}) denotes the trace of the matrix P i.e the sum of all the diagonal elements of the matrix P and adj (P) denotes the adjoint of matrix P)
View answer and explanation
Correct option: D
−1624-1624
Option D (-1624) is correct because it is the trace of the matrix Aadj(adj A) which can be simplified using properties of adjoint and determinant.
130 Medium
130 of 408
Analyzing +4 | -1
If α,β,γ∈R\alpha ,\beta ,\gamma \in R, then the determinant Δ=∣(eiα+e−iα)2(eiα−e−iα)24(eiβ+e−iβ)2(eiβ−e−iβ)24(eiγ+e−iγ)2(eiγ−e−iγ)24∣\Delta = \left| {\begin{array}{cccccccccccccccccccc} {{{\left( {{e^{i\alpha }} + {e^{ - i\alpha }}} \right)}^2}}&{{{\left( {{e^{i\alpha }} - {e^{ - i\alpha }}} \right)}^2}}&4\\ {{{\left( {{e^{i\beta }} + {e^{ - i\beta }}} \right)}^2}}&{{{\left( {{e^{i\beta }} - {e^{ - i\beta }}} \right)}^2}}&4\\ {{{\left( {{e^{i\gamma }} + {e^{ - i\gamma }}} \right)}^2}}&{{{\left( {{e^{i\gamma }} - {e^{ - i\gamma }}} \right)}^2}}&4 \end{array}} \right| is
View answer and explanation
Correct option: A
Independent of α,β\alpha ,\beta and γ\gamma
The correct answer is A (Independent of α,β\alpha ,\beta and γ\gamma) because the determinant simplifies to 0, which is independent of the variables.
131 Medium
131 of 408
Analyzing +4 | -1
Let A,B,CA, B, C be 3×33 \times 3 matrices such that AA is symmetric and BB and CC are skew-symmetric.

Consider the statements

(S1) A13B26−B26A13A^{13} B^{26}-B^{26} A^{13} is symmetric

(S2) A26C13−C13A26A^{26} C^{13}-C^{13} A^{26} is symmetric
Then,
View answer and explanation
Correct option: B
Only S2 is true
Option B (Only S2 is true) is correct because only A26C13−C13A26A^{26} C^{13}-C^{13} A^{26} is symmetric, while A13B26−B26A13A^{13} B^{26}-B^{26} A^{13} is skew-symmetric.
132 Medium
132 of 408
Understanding +4 | -1
Consider the system of equations a1x+b1y+c1z=0a_1 x+b_1 y+c_1 z=0, a2x+b2y+c2z=0,a3x+b3y+c3z=0a_2 x+b_2 y+c_2 z=0, a_3 x+b_3 y+c_3 z=0. If ∣a1b1c1a2b2c2a3c3c3∣=0\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & c_3 & c_3\end{array}\right|=0, then the system has
View answer and explanation
Correct option: A
More than 2 solutions
Option A (More than 2 solutions) is correct because when the determinant of the coefficient matrix is zero, the system has non-trivial solutions, implying more than one solution.
133 Medium
133 of 408
Analyzing +4 | -1
Suppose p,q,r≠0p, q, r \neq 0 and system of equations
(p+a)x+by+cz=0ax+(q+b)y+cz=0ax+by+(r+c)z=0\begin{array}{l} (p+a) x+b y+c z=0 \\ a x+(q+b) y+c z=0 \\ a x+b y+(r+c) z=0 \end{array}

has a non-trivial solution, then value of ap+bq+cr\frac{a}{p}+\frac{b}{q}+\frac{c}{r} is
View answer and explanation
Correct option: A
-1
Option A (-1) is correct because the determinant of the coefficient matrix must be zero for a non-trivial solution, leading to the equation a/p + b/q + c/r = -1.
134 Medium
134 of 408
Analyzing +4 | -1
If a1,a2,a3a_1, a_2, a_3, ana_n, are in G. P., then the determinant
Δ=∣log⁡anlog⁡an+1log⁡an+2log⁡an+3log⁡an+4log⁡an+5log⁡an+6log⁡an+7log⁡an+8∣\Delta=\left|\begin{array}{ccc} \log a_n & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8} \end{array}\right|
is equal to
View answer and explanation
Correct option: B
0
The correct answer is 0 because the rows of the determinant become linearly dependent when the terms are in a G.P.
135 Medium
135 of 408
Application +4 | -1
Let 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}. If PTP^{T} Q2007P=[abcd]Q^{2007} P=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right], then 2a+b−3c−4d2 a+b-3 c-4 d equal to
View answer and explanation
Correct option: B
2005
The correct answer is 2005 because after simplifying the matrix operations and using the properties of matrix exponentiation, the final expression evaluates to 2005.

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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
136 Medium
136 of 408
Application +4 | -1
A root of the equation ∣3−x−63−63−x333−6−x∣=0,\left| {\begin{array}{cccccccccccccccccccc} {3 - x}&{ - 6}&3\\ { - 6}&{3 - x}&3\\ 3&3&{ - 6 - x} \end{array}} \right| = 0, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because x=0 is a root of the equation after applying column and row operations and simplifying the determinant.
137 Medium
137 of 408
Application +4 | -1
If A,BA, B and CC are the angles of a non-right angled triangle ABCA B C, then the value of ∣tan⁡A111tan⁡B111tan⁡C∣\left|\begin{array}{ccc}\tan \mathrm{A} & 1 & 1 \\ 1 & \tan \mathrm{B} & 1 \\ 1 & 1 & \tan \mathrm{C}\end{array}\right| is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because using the trigonometric identity tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C\tan A + \tan B + \tan C = \tan A \tan B \tan C for a non-right angled triangle, the determinant simplifies to 2.
138 Medium
138 of 408
Application +4 | -1
Let A=(12−2−5)A=\left(\begin{array}{cc}1 & 2 \\ -2 & -5\end{array}\right). Let α,β∈R\alpha, \beta \in \mathbb{R} be such that αA2+βA=2I\alpha A^2+\beta A=2 I. Then α+β\alpha+\beta is equal to
View answer and explanation
Correct option: D
10
Option D (10) is correct because the characteristic equation of the matrix A can be used to find the relationship between A2 and A, which leads to the value of α + β.
139 Medium
139 of 408
Application +4 | -1
If A=[1111]A=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right] and det⁡(An−I)=1−λn,n∈N,\operatorname{det}\left(A^n-I\right)=1-\lambda^n, n \in N, then the value of λ,\lambda, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because after calculating AnA^n and the determinant of An−IA^n - I, we find that λ\lambda equals 2.
140 Medium
140 of 408
Analyzing +4 | -1
If a1b1c1,a2b2c2a_1 b_1 c_1, a_2 b_2 c_2 and a3b3c3a_3 b_3 c_3 are three-digit numbers, each of which is divisible by a non zero integer k{k}, then Δ=∣a1b1c1a2b2c2a3b3c3∣\Delta=\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| is always
View answer and explanation
Correct option: A
Divisible by kk
Option A (Divisible by k) is correct because the determinant can be expressed as k times another determinant with integer elements.
141 Medium
141 of 408
Application +4 | -1
The value of a,b,ca, b, c when [02bcab−ca−bc]\left[\begin{array}{ccc}0 & 2 b & c \\ a & b & -c \\ a & -b & c\end{array}\right] orthogonal, are
View answer and explanation
Correct option: C
±12,±16,±13\pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{6}}, \pm \frac{1}{\sqrt{3}}
Option C (±12,±16,±13\pm \frac{{1}}{{\sqrt{{2}}}}, \pm \frac{{1}}{{\sqrt{{6}}}}, \pm \frac{{1}}{{\sqrt{{3}}}}) is correct because the given matrix is orthogonal, implying that A*A^T = I, which leads to the specified values for a, b, and c after solving the resulting equations.
142 Medium
142 of 408
Application +4 | -1
If A=[4325]{A}=\left[\begin{array}{ll}4 & 3 \\ 2 & 5\end{array}\right] and A2−mA+nI=0{A}^2-m {A}+n {I}=0, then
View answer and explanation
Correct option: C
m=9,n=14m=9, n=14
Option C (m=9,n=14m=9, n=14) is correct because substituting these values into the equation A2−mA+nI=0A^2 - mA + nI = 0 satisfies the equation.
143 Medium
143 of 408
Application +4 | -1
Let
A=13[12221−2x2y]A=\frac{1}{3}\left[\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & -2 \\ x & 2 & y \end{array}\right]

If AAT=I3A A^{\mathrm{T}}=I_3, then
View answer and explanation
Correct option: C
x=−2,y=−1x=-2, y=-1
Option C is correct. By using the property AA^T = I, we can set up equations and solve for x and y, which results in x = -2 and y = -1.
144 Medium
144 of 408
Analyzing +4 | -1
The number of 3×33 \times 3 matrices AA whose entries are either 0 or 1 and for which the system A[xyz]=[100]A\left[ {\begin{array}{cccccccccccccccccccc} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc} 1\\ 0\\ 0 \end{array}} \right] has exactly two distinct solutions, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because a system of linear equations cannot have exactly two distinct solutions; it can have either zero, one, or infinitely many solutions.
145 Medium
145 of 408
Analyzing +4 | -1
The number of values of kk for which the system of linear equations, (k+2)x+10y=k,kx+(k+3)y=k−1(k+2) x+10 y=k, k x+(k+3) y=k-1 has no solution, is
View answer and explanation
Correct option: C
2
Option D (1) is correct because only one value of k (k=3) results in the system having no solution after checking for consistency.
146 Medium
146 of 408
Analyzing +4 | -1
If a,b,ca, b, c are sides of a scalene triangle, then the value of ∣abcbcacab∣\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right| is :
View answer and explanation
Correct option: B
negative
The correct answer is negative because the determinant of the given form for a scalene triangle is always negative.
147 Medium
147 of 408
Analyzing +4 | -1
If the value of the determinant ∣a111b111c∣\left| {\begin{array}{cccccccccccccccccccc} a&1&1\\ 1&b&1\\ 1&1&c \end{array}} \right| is positive, then
View answer and explanation
Correct option: B
abc>−8abc > - 8
Option B (abc>−8abc > -8) is correct because by using AM > GM inequality and simplifying the determinant condition, we arrive at abc>−8abc > -8.
148 Medium
148 of 408
Application +4 | -1
For non-singular square matrices A,  BA,\;B and CC of same order, (AB−1C)−1{\left( {A{B^{ - 1}}C} \right)^{ - 1}} is equal to
View answer and explanation
Correct option: D
C−1BA−1{C^{ - 1}}B{A^{ - 1}}
Option D is correct because applying the inverse properties to the expression results in C⁻¹BA⁻¹.
149 Medium
149 of 408
Analyzing +4 | -1
If the system of equations
x+y+z=5x + y + z = 5
x+2y+3z=9x + 2y + 3z = 9
x+3y+αz=β  x + 3y + \alpha z = \beta \,\,
has infinitely many solutions, then β−α\beta - \alpha equals
View answer and explanation
Correct option: A
8
Option A (8) is correct because when the system has infinitely many solutions, the determinant of the coefficient matrix and related determinants must be zero, leading to β−α=8\beta - \alpha = 8.
150 Medium
150 of 408
Application +4 | -1
If A=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right], then the matrix A−50A^{-50} when θ=π12\theta=\frac{\pi}{12}, is equal to
View answer and explanation
Correct option: A
[3/21/2−1/23/2]\left[\begin{array}{cc}\sqrt{3} / 2 & 1 / 2 \\ -1 / 2 & \sqrt{3} / 2\end{array}\right]
Option A ([3/21/2−1/23/2]\left[\begin{array}{cc}\sqrt{3} / 2 & 1 / 2 \\ -1 / 2 & \sqrt{3} / 2\end{array}\right]) is correct because it represents the matrix A raised to the power of -50 when theta is pi/12, obtained by using the properties of rotation matrices.
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