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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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751 Easy
751 of 922
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.
752 Easy
752 of 922
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.
753 Easy
753 of 922
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.
754 Easy
754 of 922
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.
755 Easy
755 of 922
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.
756 Easy
756 of 922
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.
757 Easy
757 of 922
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.
758 Easy
758 of 922
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}.
759 Easy
759 of 922
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.
760 Easy
760 of 922
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.
761 Easy
761 of 922
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.
762 Easy
762 of 922
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.
763 Easy
763 of 922
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.
764 Easy
764 of 922
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.
765 Easy
765 of 922
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.

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
766 Easy
766 of 922
Application +4 | -1
If A=[100010ab−1]A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{array}\right], then A2A^2 is equal to
View answer and explanation
Correct option: A
Unit matrix
Option A (Unit matrix) is correct because when the matrix AA is multiplied by itself (A2A^2), the result is the identity matrix.
767 Easy
767 of 922
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.
768 Easy
768 of 922
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.
769 Easy
769 of 922
Understanding +4 | -1
If AA is any m×nm \times n matrix such that ABAB and BABA are both defined., then BB is an
View answer and explanation
Correct option: B
n×mn \times m matrix
Option B (n×mn \times m matrix) is correct because for AB and BA to be defined, B must have n rows and m columns.
770 Easy
770 of 922
Application +4 | -1
Let AA be an orthogonal non-singular matrix of order nn, then the determinant of matrix 'A−In′    ie,  ∣A−In∣A - {I'_n}\;\,\,ie,\;\left| {A - {I_n}} \right| is equal to
View answer and explanation
Correct option: C
∣A∣∣In−A∣\left| A \right|\left| {{I_n} - A} \right|
Option C (∣A∣∣In−A∣\left| A \right|\left| {{I_n} - A} \right|) is correct because it follows from the properties of orthogonal matrices and determinants.
771 Easy
771 of 922
Analyzing +4 | -1
The existence of the unique solution of the system of equations
x+y+z=λx+y+z=\lambda

5x−y+μz=105 x-y+\mu z=10

2x+3y−z=62 x+3 y-z=6
depends on
View answer and explanation
Correct option: A
μ\mu only
Option A is correct because the existence of a unique solution depends on the determinant of the coefficient matrix being non-zero, which is determined by μ\mu only.
772 Easy
772 of 922
Application +4 | -1
Let A=(aij)A=\left(a_{i j}\right) be a 3×33 \times 3 matrix whose determinant is 5. The determinant of the matrix B=(2i−jaij)B=\left(2^{i-j} a_{i j}\right) is
View answer and explanation
Correct option: A
5
Option A (5) is correct because the determinant of matrix B is equal to the determinant of matrix A.
773 Easy
773 of 922
Analyzing +4 | -1
If a,b,ca,b,c are the sides of a ΔABC\Delta ABC and A,B,CA,B,C are respectively the angles opposite to them, then ∣a2bsin⁡Acsin⁡Absin⁡A1cos⁡(B−C)csin⁡Acos⁡(B−C)1∣\left| {\begin{array}{cccccccccccccccccccc} {{a^2}}&{b\sin A}&{c\sin A}\\ {b\sin A}&1&{\cos \left( {B - C} \right)}\\ {c\sin A}&{\cos \left( {B - C} \right)}&1 \end{array}} \right| equals
View answer and explanation
Correct option: D
0
Option D (0) is correct because the determinant simplifies to zero when the triangle is isosceles, which is a condition derived from the given determinant equation.
774 Easy
774 of 922
Analyzing +4 | -1
The system of equations −kx+3y−14z=25,    −15x+4y−kz=3,   −4x+y+3z=4- kx + 3y - 14z = 25,\,\,\,\, - 15x + 4y - kz = 3,\,\,\, - 4x + y + 3z = 4 is consistent for all k in the set
View answer and explanation
Correct option: D
R−{−11,11}R - \{ - 11,11\}
Option B (R−{−11,13}R - \{ - 11,13\}) is incorrect. Option D (R−{−11,11}R - \{ - 11,11\}) is correct because the system is consistent when the determinant of the coefficient matrix is non-zero, which occurs for all k except ±11.
775 Easy
775 of 922
Analyzing +4 | -1
For what value of λ\lambda, the system of equations
x+y+z=6x+y+z=6

x+2y+3z=10x+2 y+3 z=10

x+2y+λz=12x+2 y+\lambda z=12
is inconsistent?
View answer and explanation
Correct option: D
λ=3\lambda=3
Option D is correct because the system is inconsistent when λ=3\lambda = 3, leading to a contradiction in the equations.
776 Easy
776 of 922
Application +4 | -1
Let A=[0α00]A=\left[\begin{array}{ll}0 & \alpha \\ 0 & 0\end{array}\right] and (A+I)50−50A=[abcd](A+I)^{50}-50 A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right], find a+b+c+da+b+\mathrm{c}+d.
View answer and explanation
Correct option: A
2
Option A (2) is correct because since A2=0A^2 = 0, (A+I)50=I+50A(A+I)^{50} = I + 50A. Therefore, (A+I)50−50A=I(A+I)^{50} - 50A = I, which means a=1,b=0,c=0,d=1a=1, b=0, c=0, d=1, and a+b+c+d=2a+b+c+d = 2.
777 Easy
777 of 922
Understanding +4 | -1
If AA is any square matrix then which of the following is not symmetric?
View answer and explanation
Correct option: B
A−A′A-A^{\prime}
Option B (A−A′A-A^{\prime}) is correct because the transpose of (A−A′)(A-A^{\prime}) is (A′−A)(A^{\prime}-A), which is the negative of (A−A′)(A-A^{\prime}), making it skew-symmetric, not symmetric.
778 Easy
778 of 922
Analyzing +4 | -1
If AA is a real skew-symmetric matrix such that A2+I=OA^2+I=O, then
View answer and explanation
Correct option: D
A is a skew-symmetric matrix of even order with ∣A∣=1|A|=1
The correct answer is A is a skew-symmetric matrix of even order with ∣A∣=1|A|=1 because A2+I=0A^2 + I = 0 implies ∣A∣=1|A| = 1 and A must be of even order since it's a real skew-symmetric matrix.
779 Easy
779 of 922
Application +4 | -1
Elements of a matrix AA of order 10×1010 \times 10 are defined as aij=ωi+ja_{i j}=\omega^{i+j} (where ω\omega is cube root of unity), then trace (A)(A) of the matrix is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the trace of the matrix A simplifies to ω2\omega^2, which is not among the given options.
780 Easy
780 of 922
Understanding +4 | -1
Let A=(abcd)A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right), where a,b,c,d∈Ra, b, c, d \in {R}. If ∣a∣,∣b∣,∣c∣,∣d∣≤k|a|,|b|,|c|,|d| \leq k, where k>0k > 0, then
View answer and explanation
Correct option: C
det⁡(A)≤2k2\operatorname{det}(A) \leq 2 k^2
Option C (det⁡(A)≤2k2\operatorname{det}(A) \leq 2 k^2) is correct because the determinant of A, which is ad−bcad-bc, is bounded by ∣ad∣+∣bc∣|ad| + |bc|, and since ∣a∣,∣b∣,∣c∣,∣d∣≤k|a|, |b|, |c|, |d| \leq k, the determinant is at most 2k22k^2.
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