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JEE • MATHEMATICS • MATRICES

Matrices practice · Medium

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331 Medium
331 of 408
Analyzing +4 | -1
In Δ  ABC,if∣1ab1ca1bc∣=0\Delta \;ABC,\mathrm{if}\left| {\begin{array}{cccccccccccccccccccc} 1&a&b\\ 1&c&a\\ 1&b&c \end{array}} \right| = 0, then

sin⁡2A+sin⁡2B+sin⁡2C{\sin ^2}A + {\sin ^2}B + {\sin ^2}C is equal to
View answer and explanation
Correct option: B
94\frac{9}{4}
Option B (9/4) is correct because the determinant being zero implies the triangle is equilateral, and substituting 60 degrees into the expression yields 9/4.
332 Medium
332 of 408
Analyzing +4 | -1
Let pp be an odd prime number and TpT_p be the following set of 2×22 \times 2 matrices Tp={A=[abca]:a,b,c∈{0,1,2,…,(p−1)}}T_p=\left\{A=\left[\begin{array}{ll}a & b \\ c & a\end{array}\right]: a, b, c \in\{0,1,2, \ldots,(p-1)\}\right\}
The number of AA in TpT_p such that AA is either symmetric or skew-symmetric or both, and det (A)(A) divisible by pp is
View answer and explanation
Correct option: B
2(p−1)2(p-1)
The correct answer is R−{2,−3}R-\{2,-3\} because the system has only the trivial solution when the determinant of the coefficient matrix is non-zero, which means k≠2k \neq 2 and k≠−3k \neq -3.
333 Medium
333 of 408
Application +4 | -1
∣1aa21bb21cc2∣=\left|\begin{array}{lll}1 & \mathrm{a} & \mathrm{a}^{2} \\ 1 & \mathrm{b} & \mathrm{b}^{2} \\ 1 & \mathrm{c} & \mathrm{c}^{2}\end{array}\right|=
View answer and explanation
Correct option: C
(a−b)(b−c)(c−a)(a-b)(b-c)(c-a)
Option C (1/3) is correct because for the given matrix to be orthogonal, the product of the matrix and its transpose must equal the identity matrix, which is satisfied when k = 1/3.
334 Medium
334 of 408
Understanding +4 | -1
Consider the following statements

I. ∣A∣|{A}| is also called modulus of square matrix A.A.

II. Every matrix has determinant.
View answer and explanation
Correct option: D
Neither I nor II is true
Option D (Neither I nor II is true) is correct because |A| represents the determinant, not the modulus, and only square matrices have determinants.
335 Medium
335 of 408
Analyzing +4 | -1
If α,β\alpha ,\beta are the roots of equation ax2+bx+c=0,a{x^2} + bx + c = 0, then the value of the determinant
∣1cos⁡  (β−α)cos⁡αcos⁡  (α−β)1cos⁡βcos⁡αcos⁡β1∣,\left| {\begin{array}{cccccccccccccccccccc} 1&{\cos \;\left( {\beta - \alpha } \right)}&{\cos \alpha }\\ {\cos \;\left( {\alpha - \beta } \right)}&1&{\cos \beta }\\ {\cos \alpha }&{\cos \beta }&1 \end{array}} \right|, is
View answer and explanation
Correct option: D
None of these
Option C (2) is correct because by analyzing the equation in different intervals of [x], we find two solutions: x=0 and x=2.
336 Medium
336 of 408
Analyzing +4 | -1
Let a,b,ca,b,c be positive real numbers. The following system of equations in x,yx,y and zz x2a2+y2b2−z2c2=1,x2a2−y2b2+z2c2=1,−x2a2+y2b2+z2c2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1, - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1 has
View answer and explanation
Correct option: B
Unique solution
Option B (Unique solution) is correct because the coefficient matrix has a non-zero determinant, indicating a unique solution.
337 Medium
337 of 408
Application +4 | -1
The value of f(π6)f\left(\frac{\pi}{6}\right) where f(q)=∣cos⁡2θcos⁡θsin⁡θ−sin⁡θcos⁡θsin⁡θsin⁡2θcos⁡θsin⁡θ−cos⁡θ0∣f(q)=\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| is
View answer and explanation
Correct option: B
1
Option B (1) is correct because after applying column operations and simplifying, the determinant evaluates to 1, independent of the value of θ\theta.
338 Medium
338 of 408
Analyzing +4 | -1
The value of the determinant ∣cos⁡α−sin⁡α1sin⁡αcos⁡α1cos⁡(α+β)−sin⁡(α+β)1∣\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ \cos (\alpha+\beta) & -\sin (\alpha+\beta) & 1\end{array}\right| is
View answer and explanation
Correct option: A
Independent of α\alpha
Option A (Δ∈[1−2,1+2]\Delta \in[1-\sqrt{2}, 1+\sqrt{2}]) is correct because the range of the expression 1+sin(beta)-cos(beta) falls within this interval.
339 Medium
339 of 408
Application +4 | -1
Consider the matrix f(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]
Given below are two statements:
Statement I: f(−x)f(-x) is the inverse of the matrix f(x).f(x).
Statement II : f(x)f(y)=f(x+y).f(x) f(y)=f(x+y).
In the light of the above statements, choose the correct answer from the options given below.
View answer and explanation
Correct option: D
Both Statement I and Statement II are true
Option D is correct because both statements are true. Statement I is true because f(-x) is the inverse of f(x), and Statement II is true because f(x)f(y) = f(x+y).
340 Medium
340 of 408
Application +4 | -1
The solutions of the equation ∣x2−125x−12x∣=0,\left| {\begin{array}{cccccccccccccccccccc} x&2&{ - 1}\\ 2&5&x\\ { - 1}&2&x \end{array}} \right| = 0, are
View answer and explanation
Correct option: A
3,−13, - 1
Option A (3, -1) is correct because these are the roots of the quadratic equation obtained by expanding the determinant.
341 Medium
341 of 408
Application +4 | -1
If A=[−1111−1111−1]A=\left[\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right], then the value of Det⁡(AB+2I)\operatorname{Det}(A B+2I) where B=adj⁡AB=\operatorname{adj} A and II is identify matrix of order 3
View answer and explanation
Correct option: C
216
Option C (216) is correct because by using the properties of adjoint and determinant, the determinant of the given expression can be simplified and calculated.
342 Medium
342 of 408
Understanding +4 | -1
If ∣a1b1c1a2b2c2a3b3c3∣≠0\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| \neq 0, then the number of solutions of the system of equations a1x+b1y+c1z=0,a2x+b2y+c2z=0a_1 x+b_1 y+c_1 z=0, a_2 x+b_2 y+c_2 z=0 and a3x+b3y+c3z=0a_3 x+b_3 y+c_3 z=0 is
View answer and explanation
Correct option: B
Only one unique solution
Option B (Only one unique solution) is correct because the determinant of the coefficient matrix is non-zero, implying a unique (trivial) solution for the homogeneous system.
343 Medium
343 of 408
Analyzing +4 | -1
If maximum and minimum values of the determinant ∣1+Sin⁡2xCos⁡2xSin⁡2xSin⁡2x1+Cos⁡2xSin⁡2xSin⁡2xCos⁡2x1+Sin⁡2x∣\left|\begin{array}{lll}1+\operatorname{Sin}^2 x & \operatorname{Cos}^2 x & \operatorname{Sin} 2 x \\ \operatorname{Sin}^2 x & 1+\operatorname{Cos}^2 x & \operatorname{Sin} 2 x \\ \operatorname{Sin}^2 x & \operatorname{Cos}^2 x & 1+\operatorname{Sin} 2 x\end{array}\right| are α\alpha and β\beta respectively, then which of the statements are FALSE ?
View answer and explanation
Correct option: D
A triangle can be constructed having it's sides as α,β\alpha, \beta and α−β\alpha-\beta
Option D (A triangle can be constructed having it's sides as α,β\alpha, \beta and α−β\alpha-\beta) is false because the triangle inequality is not satisfied with the given values of α\alpha and β\beta.
344 Medium
344 of 408
Analyzing +4 | -1
The value of ∣x+yy+zz+xxyzx−yy−zz−x∣\left| {\begin{array}{cccccccccccccccccccc} {x + y}&{y + z}&{z + x}\\ x&y&z\\ {x - y}&{y - z}&{z - x} \end{array}} \right| is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because after applying column and row operations, two rows become identical, resulting in a determinant of 0.
345 Medium
345 of 408
Application +4 | -1
If A\mathrm{A} is 3×33 \times 3 matrix and if ∣A∣=3|\mathrm{A}|=3, then ∣adj⁡A∣=|\operatorname{adj} \mathrm{A}|=
View answer and explanation
Correct option: B
9
Option B (9) is correct because for a 3x3 matrix A, ∣adjA∣=∣A∣2|adj A| = |A|^2, and given ∣A∣=3|A| = 3, then ∣adjA∣=32=9|adj A| = 3^2 = 9.
346 Medium
346 of 408
Analyzing +4 | -1
Let a,b,ca, b, c be such that b(a+c)≠0b(a+c) \neq 0 if
∣aa+1a−1−bb+1b−1cc−1c+1∣+∣a+1b+1c−1a−1b−1c+1(−1)n+2a(−1)n+1b(−1)nc∣=0, \left|\begin{array}{ccc} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{array}\right|+\left|\begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^n c \end{array}\right|=0 \text {, }
then the value of nn is :
View answer and explanation
Correct option: B
any odd integer
The correct answer is any odd integer because the equation holds true only when n is an odd integer.
347 Medium
347 of 408
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.
348 Medium
348 of 408
Analyzing +4 | -1
If Δa=∣a−1n6(a−1)22n24n−2(a−1)33n33n2−3n∣,{\Delta _a} = \left| {\begin{array}{cccccccccccccccccccc} {a - 1}&n&6\\ {{{\left( {a - 1} \right)}^2}}&{2{n^2}}&{4n - 2}\\ {{{\left( {a - 1} \right)}^3}}&{3{n^3}}&{3{n^2} - 3n} \end{array}} \right|, then ∑a=1nΔa\sum\nolimits_{a = 1}^n {{\Delta _a}} is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the determinant can be simplified to have two identical columns, resulting in a value of 0.
349 Medium
349 of 408
Analyzing +4 | -1
∣a−b+c−a−b+c1a+b+2c−a+b+2c23c3c3∣\left| {\begin{array}{cccccccccccccccccccc} {a - b + c}&{ - a - b + c}&1\\ {a + b + 2c}&{ - a + b + 2c}&2\\ {3c}&{3c}&3 \end{array}} \right| is
View answer and explanation
Correct option: C
12ab12ab
Option C (12ab12ab) is correct because the determinant simplifies to this expression after applying row operations.
350 Medium
350 of 408
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.
351 Medium
351 of 408
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.
352 Medium
352 of 408
Analyzing +4 | -1
Consider the system of linear equations;

x1+2x2+x3=3x_1+2 x_2+x_3=3

2x1+3x2+x3=32 x_1+3 x_2+x_3=3

3x1+5x2+2x3=13 x_1+5 x_2+2 x_3=1

The system has
View answer and explanation
Correct option: C
no solution
Option C (no solution) is correct because the determinant D is zero, but at least one of Dx, Dy, or Dz is non-zero, indicating inconsistency.
353 Medium
353 of 408
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.
354 Medium
354 of 408
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.
355 Medium
355 of 408
Analyzing +4 | -1
Determinant ∣1xy2sin⁡x+2xsin⁡y+3y3cos⁡x+3xcos⁡y+3y∣\left| {\begin{array}{cccccccccccccccccccc} 1&x&y\\ 2&{\sin x + 2x}&{\sin y + 3y}\\ 3&{\cos x + 3x}&{\cos y + 3y} \end{array}} \right| is equal to
View answer and explanation
Correct option: A
sin⁡(x−y)\sin \left( {x - y} \right)
Option A (sin⁡(x−y)\sin \left( {x - y} \right)) is correct because after applying row operations, the determinant simplifies to sin⁡(x−y)\sin \left( {x - y} \right).
356 Medium
356 of 408
Analyzing +4 | -1
The factors of ∣xabaxbabx∣\left| {\begin{array}{cccccccccccccccccccc} x&a&b\\ a&x&b\\ a&b&x \end{array}} \right| are
View answer and explanation
Correct option: A
x−a,x−b,x - a,x - b, and x+a+bx + a + b
Option A (x−a,x−b,x - a,x - b, and x+a+bx + a + b) is the correct factorization of the given determinant, obtained through row and column operations.
357 Medium
357 of 408
Analyzing +4 | -1
Let the matrix A=∣010001100∣A=\left|\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right| and the matrix B0B_0 =A49+2A98=A^{49}+2 A^{98}. If Bn=Adj⁡(Bn−1)B_n=\operatorname{Adj}\left(B_{n-1}\right) for all n≥1n \geq 1, then det⁡(B4)\operatorname{det}\left(B_4\right) is equal to :
View answer and explanation
Correct option: C
3323^{32}
Option C (3323^{{32}}) is correct because after simplifying the matrix operations and using the properties of adjoint and determinants, the final determinant can be calculated.
358 Medium
358 of 408
Analyzing +4 | -1
If the system of equations
x+ay+az=0bx+y+bz=0cx+cy+z=0\begin{array}{l} x + a\mathrm{ }y + a\mathrm{ }z = 0\\ b\mathrm{ }x + y + b\mathrm{ }z = 0\\ c\mathrm{ }x + c\mathrm{ }y + z = 0 \end{array}
Where a,ba,b and cc are non-zero non-unity, has a non-trivial solution, then the value of a1−a+b1−b+c1−c\frac{a}{{1 - a}} + \frac{b}{{1 - b}} + \frac{c}{{1 - c}} is
View answer and explanation
Correct option: C
−1- 1
The correct answer is C (-1). For a non-trivial solution, the determinant of the coefficient matrix must be zero, which leads to the given expression equaling -1.
359 Medium
359 of 408
Application +4 | -1
If the product of nn matrices [1101][1201][1301]…[1n01]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] \ldots\left[\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right] is equal to the matrix [137801]\left[\begin{array}{cc}1 & 378 \\ 0 & 1\end{array}\right], then the value of nn is equal to
View answer and explanation
Correct option: A
27
Option A (27) is correct because the product of the matrices results in a matrix of the form [1sum01]\begin{bmatrix} 1 & sum \\ 0 & 1 \end{bmatrix}, where the sum is 1+2+3+...+n=n(n+1)21+2+3+...+n = \frac{n(n+1)}{2}. Setting this equal to 378 and solving for n gives n=27.
360 Medium
360 of 408
Analyzing +4 | -1
Let A{A} and B{B} be two matrices such that A=[aij]2×3=[x123x−114x2+1]A=\left[a_{i j}\right]_{2 \times 3}=\left[\begin{array}{ccc}x & 1 & 2 \\ 3 x & -1 & \frac{1}{4 x^2+1}\end{array}\right] and B=[bij]3×2=[1x21x2x23x]B=\left[b_{i j}\right]_{3 \times 2}=\left[\begin{array}{cc}\frac{1}{x^2} & \frac{1}{x} \\ 2 x & 2 \\ 3 & x\end{array}\right] where x>0x > 0. If matrix CC is defined as C=[cij]2×2C=\left[c_{i j}\right]_{2 \times 2}, where cij=∑r=13airbrj,∀1≤i,j≤2c_{i j}=\sum\limits_{{r=1}}^3 a_{i r} b_{r j}, \forall 1 \leq i, j \leq 2, then the minimum value of Δ(x)=∑1≤i≤j≤2cij\Delta(x)=\sum\limits_{{1 \leq i \leq j \leq 2}} c_{i j} is
View answer and explanation
Correct option: A
574\frac{57}{4}
Option A (574\frac{57}{4}) is correct because after calculating the matrix C and summing its elements, the minimum value of the resulting expression is 574\frac{57}{4}.
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