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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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121 Easy
121 of 922
Understanding +4 | -1
If AA is a skew Hermitian matrix, then the main diagonal elements of AA are all
View answer and explanation
Correct option: D
Purely imaginary
Option D (Purely imaginary) is correct because for a skew-Hermitian matrix, the diagonal elements are purely imaginary or zero.
122 Easy
122 of 922
Analyzing +4 | -1
The system of linear equations
λx+2y+2z=52λx+3y+5z=84x+λy+6z=10  has\begin{array}{l} \lambda x + 2y + 2z = 5\\ 2\lambda x + 3y + 5z = 8\\ 4x + \lambda y + 6z = 10\,\,\mathrm{has} \end{array}
View answer and explanation
Correct option: A
no solution when λ=2\lambda = 2
Option A (no solution when λ=2\lambda = 2) is correct because when λ=2\lambda = 2, the determinant of the coefficient matrix is zero, but at least one of the determinants formed by replacing a column with the constants is non-zero.
123 Easy
123 of 922
Analyzing +4 | -1
If f(x)=ax2+bx+c,a,b,c∈Rf(x)=a x^2+b x+c, a, b, c \in R and equation f(x)−x=0f(x)-x=0 has imaginary roots α,β,γ\alpha, \beta, \gamma and δ\delta be the roots of f(f(x))−x=0f(f(x))-x=0, then ∣2αδβ0αγβ1∣\left|\begin{array}{ccc}2 & \alpha & \delta \\ \beta & 0 & \alpha \\ \gamma & \beta & 1\end{array}\right| is
View answer and explanation
Correct option: B
purely real
The correct answer is B (purely real) because the determinant's conjugate is equal to itself, indicating it's a real number.
124 Easy
124 of 922
Application +4 | -1
If A=[a2abacabb2bcacbcc2]A=\left[\begin{array}{lll}a^2 & a b & a c \\ a b & b^2 & b c \\ a c & b c & c^2\end{array}\right] and B=[0c−b−c0ab−a0]B=\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right] then the product ABA B is equal to
View answer and explanation
Correct option: A
OO
Option A (O) is correct because the product of matrices A and B results in a zero matrix.
125 Easy
125 of 922
Analyzing +4 | -1
The number of solutions of the matrix equation X2=[1123]X^2=\left[\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right] is
View answer and explanation
Correct option: A
More than 2
Option A (More than 2) is correct because by solving the matrix equation X2=[1123]X^2 = \left[\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right], we find that there are more than two possible solutions for matrix XX.
126 Easy
126 of 922
Application +4 | -1
If AA and BB are square matrices such that A∗=AA^*=A and B∗=BB^*=B, where A∗A^* denotes the conjugate transpose of AA, then (AB−BA)∗=(A B-B A)^*=
View answer and explanation
Correct option: C
BA−ABB A-A B
Option C (BA−ABB A-A B) is correct because applying the conjugate transpose properties to (AB−BA)∗(A B-B A)^* and using the given conditions A∗=AA^*=A and B∗=BB^*=B results in BA−ABB A-A B.
127 Easy
127 of 922
Understanding +4 | -1
In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then, it can be decomposed into nn determinant, where nn has the value
View answer and explanation
Correct option: D
24
Option D is correct because each column's sum of terms multiplies to give the total number of possible determinants.
128 Easy
128 of 922
Application +4 | -1
If A=[−122−1]A = \left[ {\begin{array}{cccccccccccccccccccc} { - 1}&2\\ 2&{ - 1} \end{array}} \right] and B=[31],  Ax=BB = \left[ {\begin{array}{cccccccccccccccccccc} 3\\ 1 \end{array}} \right],\;Ax = B, then XX is equal to
View answer and explanation
Correct option: B
13[57]\frac{1}{3}\left[ {\begin{array}{cccccccccccccccccccc} 5\\ 7 \end{array}} \right]
Option B is correct. The solution involves finding the inverse of matrix A and then multiplying it by matrix B to solve for X.
129 Easy
129 of 922
Analyzing +4 | -1
A matrix of order 3×33 \times 3 are formed by using the elements of the set A={−3,−2,−1,0,1,2,3}A=\{-3,-2,-1,0,1,2,3\}, then find the probability that the matrix is either symmetric of skew symmetric is
View answer and explanation
Correct option: C
173+176−179\frac{1}{7^{3}}+\frac{1}{7^{6}}-\frac{1}{7^{9}}
Option C (173+176−179\frac{1}{7^{3}}+\frac{1}{7^{6}}-\frac{1}{7^{9}}) is correct because it calculates the probability of a matrix being symmetric or skew-symmetric, accounting for the overlap (the zero matrix).
130 Easy
130 of 922
Application +4 | -1
Let Sk=(1k01),k∈NS_k=\left(\begin{array}{ll}1 & k \\ 0 & 1\end{array}\right), k \in \mathbf{N}. Then (S2)n(Sk)−1\left(S_2\right)^n\left(S_k\right)^{-1} (where n∈Nn \in \mathbf{N} ) is equal to:
View answer and explanation
Correct option: B
S2n−kS_{2 n-k}
Option D is correct. By expressing S2S_2 as I+BI+B and using the binomial theorem, and then multiplying by the inverse of SkS_k, the result is S2n−kS_{2n-k}.
131 Easy
131 of 922
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 A is non-singular, it has an inverse, allowing us to multiply both sides of the equation AB=ACAB = AC by A−1A^{-1} to conclude that B=CB = C.
132 Easy
132 of 922
Analyzing +4 | -1
The value of Δ=∣11+ac1+bc11+ad1+bd11+ae1+be∣,\Delta=\left|\begin{array}{lll}1 & 1+a c & 1+b c \\ 1 & 1+a d & 1+b d \\ 1 & 1+ae & 1+b e\end{array}\right|, is
View answer and explanation
Correct option: B
0
Option B (0) is correct because after applying column operations, two columns of the determinant become identical, resulting in a determinant value of 0.
133 Easy
133 of 922
Analyzing +4 | -1
The value of aa and bb, for which the system of equations
2x+3y+6z=8,2x + 3y + 6z = 8, x+2y+az=5x + 2y + az = 5, 3x+5y+9z=b3x + 5y + 9z = b has no solution, are
View answer and explanation
Correct option: A
a=3,  b≠13a = 3,\,\,b \ne 13
Option A (a=3, b≠13a = 3,\,b \ne 13) is correct because for the system to have no solution, the determinant of the coefficient matrix must be zero, and at least one of the determinants formed by replacing a column with the constants must be non-zero.
134 Easy
134 of 922
Analyzing +4 | -1
If α,β≠0\alpha, \beta \neq 0 and f(n)=αn+βnf(n)=\alpha^n+\beta^n and ∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣=k(1−α)2(1−β)2(α−β)2\left|\begin{array}{ccc}3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4)\end{array}\right|=k(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2, then kk is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the determinant simplifies to (1−α)2(1−β)2(α−β)2(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2, making k=1k=1.
135 Easy
135 of 922
Application +4 | -1
If A=(2294)  and  I=(1001)A = \left( {\begin{array}{cccccccccccccccccccc} 2&2\\ 9&4 \end{array}} \right)\,\,\mathrm{and}\,\,I = \left( {\begin{array}{cccccccccccccccccccc} 1&0\\ 0&1 \end{array}} \right) the 10 A−110\,{A^{ - 1}} is equal to
View answer and explanation
Correct option: B
A−6IA - 6I
Option B (A−6IA - 6I) is correct because using the characteristic equation and Cayley-Hamilton theorem, we can find the expression for 10A−110A^{-1}.
136 Easy
136 of 922
Application +4 | -1
A square non-singular matrix AA satisfies A2−A+2I=0A^2-A+2 I=0, then A−1=A^{-1}=
View answer and explanation
Correct option: B
12(I−A)\frac{1}{2}(I-A)
Option B is correct because by manipulating the given equation and multiplying by the inverse, we can isolate A-1 and find it equals to 12\frac{1}{2}(I-A).
137 Easy
137 of 922
Understanding +4 | -1
If AA is an orthogonal matrix, then
View answer and explanation
Correct option: B
∣A∣=±1\left| A \right| = \pm 1
Option B (∣A∣=±1\left| A \right| = \pm 1) is correct because for an orthogonal matrix A, A*A^T = I, and therefore |A|*|A^T| = |I| = 1, which implies |A|^2 = 1, so |A| = ±1.
138 Easy
138 of 922
Analyzing +4 | -1
If A=[1     −32          k]  A = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\,\,\, - 3}\\ {2\;\;\;\;\;k} \end{array}} \right]\; and   A2−4A=10I=A  \;{A^2} - 4A = 10I = A\; then kk is equal to
View answer and explanation
Correct option: C
4 and not 1
Option C (4 and not 1) is correct because substituting the matrix A into the given equation and solving for k yields k = 4, but k = 1 does not satisfy the initial equation.
139 Easy
139 of 922
Application +4 | -1
If A=[42−11]A=\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right], then the value of (A−2I)(A−3I)(A-2 I)(A-3 I) is
View answer and explanation
Correct option: C
Null matrix
The correct answer is Null matrix because the product of (A-2I) and (A-3I) results in a matrix where all elements are zero.
140 Easy
140 of 922
Analyzing +4 | -1
If c<1c < 1 and the system of equations x+y−1=0,2x−y−c=0x + y - 1 = 0,2x - y - c = 0 and −bx+3by−c=- bx + 3by - c = is consistent, then the possible real values of bb are
View answer and explanation
Correct option: C
b∈(−34,3)b \in \left( { - \frac{3}{4},3} \right)
Option C (b∈(−34,3)b \in \left( { - \frac{3}{4},3} \right)) is correct because the condition for consistency leads to an inequality involving b, which defines this interval.
141 Easy
141 of 922
Analyzing +4 | -1
the system of simultaneous equations
kx+2y−z=1(k−1)y−2z=2(k+2)z=3\begin{array}{l} kx + 2y - z = 1\\ \left( {k - 1} \right)y - 2z = 2\\ \left( {k + 2} \right)z = 3 \end{array}
Have a unique solution if kk equals
View answer and explanation
Correct option: B
−1- 1
Option B (−1-1) is correct because for a unique solution, the determinant of the coefficient matrix must be non-zero, which leads to k≠0,1,−2k \ne 0, 1, -2.
142 Easy
142 of 922
Application +4 | -1
Let Sk=(1k01),k∈NS_k=\left(\begin{array}{ll}1 & k \\ 0 & 1\end{array}\right), k \in {N}. Then (S2)n(Sk)−1\left(S_2\right)^n\left(S_k\right)^{-1} (where n∈Nn \in {N} ) is equal to:
View answer and explanation
Correct option: B
S2n−kS_{2 n-k}
Option B (8∣A∣8|A|) is correct because it is derived from the property that ∣A2∣=∣A∣2|A^2| = |A|^2 and ∣kA∣=kn∣A∣|kA| = k^n|A| for a matrix of order n.
143 Easy
143 of 922
Application +4 | -1
If A=[a+ibc+id−c+ida−ib]A=\left[\begin{array}{cc}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right] and a2+b2+c2+d2=1a^2+b^2+c^2+d^2=1, then A−1A^{-1} is equal to
View answer and explanation
Correct option: C
[a−ib−c−idc−ida+ib]\left[\begin{array}{cc}a-i b & -c-i d \\ c-i d & a+i b\end{array}\right]
Option C ([a−ib−c−id−¸ida+ib]\left[\begin{array}{cc}a-i b & -c-i d \c-i d & a+i b\end{array}\right]) is correct because it is the inverse of the given matrix A, considering the given condition.
144 Easy
144 of 922
Analyzing +4 | -1
Let AA be a 3×33 \times 3 real matrix such that A(110)=(110);A(101)=(−101)  and  A(001)=(112).A\left( {\begin{array}{cccccccccccccccccccc} 1 \\ 1 \\ 0 \end{array}} \right) = \left( {\begin{array}{cccccccccccccccccccc} 1 \\ 1 \\ 0 \end{array}} \right);A\left( {\begin{array}{cccccccccccccccccccc} 1 \\ 0 \\ 1 \end{array}} \right) = \left( {\begin{array}{cccccccccccccccccccc} { - 1} \\ 0 \\ 1 \end{array}} \right)\,\,{\text{and}}\,\,A\left( {\begin{array}{cccccccccccccccccccc} 0 \\ 0 \\ 1 \end{array}} \right) = \left( {\begin{array}{cccccccccccccccccccc} 1 \\ 1 \\ 2 \end{array}} \right). If X=(x1,x2,x3)TX = {({x_1},{x_2},{x_3})^T} and II is an identity matrix of order 3, then the system (A−2I)X=(411)(A - 2I)X = \left( {\begin{array}{cccccccccccccccccccc} 4 \\ 1 \\ 1 \end{array}} \right) has:
View answer and explanation
Correct option: B
Infinite many solutions
Option B (Infinite many solutions) is correct because the system of equations derived from (A−2I)X=(411)(A - 2I)X = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix} has dependent equations, leading to infinite solutions.
145 Easy
145 of 922
Analyzing +4 | -1
For two 3×33 \times 3 matrices AA and BB, let A+B=2BTA+B=2 B^{\mathrm{T}} and 3A+2B=3 \mathrm{A}+2 \mathrm{B}= I3\mathrm{I}_{3}, where BT\mathrm{B}^{\mathrm{T}} is the transpose of B\mathrm{B} and I3\mathrm{I}_{3} is 3×33 \times 3 identity matrix. Then :
View answer and explanation
Correct option: B
10A+5B=3I310 \mathrm{A}+5 \mathrm{B}=3 \mathrm{I}_{3}
Option B (10A+5B=3I310 \mathrm{A}+5 \mathrm{B}=3 \mathrm{I}_{3}) is correct because it is derived from the given matrix equations and their transposes after solving for A and B.
146 Easy
146 of 922
Analyzing +4 | -1
Consider the following system of equations

αx+2y+z=12αx+3y+z=13x+ay+2z=b\begin{array}{l} \alpha x+2 y+z=1 \\ 2 \alpha x+3 y+z=1 \\ 3 x+a y+2 z=b \end{array}

For some α,β∈R\alpha, \beta \in \mathbb{R}. then which of the following is NOT correct?
View answer and explanation
Correct option: C
It has no solution for α=3\alpha=3 and for all β≠2\beta \neq 2
Option C (It has no solution for α=3\alpha=3 and for all β≠2\beta \neq 2) is NOT correct because when α=3\alpha=3 and β=2\beta=2, the system has infinitely many solutions, not no solution.
147 Easy
147 of 922
Application +4 | -1
If A=[log⁡x−1−log⁡x2]A=\left[\begin{array}{cc}\log x & -1 \\ -\log x & 2\end{array}\right] and if det⁡(A)=2\operatorname{det}(A)=2, then the value of xx is equal to
View answer and explanation
Correct option: B
e2e^2
Option B (e2e^2) is correct because by calculating the determinant of A and solving the resulting equation, we find x=e2x = e^2.
148 Easy
148 of 922
Application +4 | -1
If AA is a skew symmetric matrix of order nn and CC is a column matrix of order n×1,  then  CTAC  n \times 1,\;\mathrm{then}\;{C^T}AC\; is
View answer and explanation
Correct option: C
A zero matrix of order 1
Option C is correct. For a skew-symmetric matrix A, A^T = -A. Therefore, the expression simplifies to zero.
149 Easy
149 of 922
Remembering +4 | -1
If A=[abcd]A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] such that ad−bc≠0a d-b c \neq 0, then A−1A^{-1}, is
View answer and explanation
Correct option: A
1ad−bc[d−b−ca]\frac{1}{a d-b c}\left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right]
Option A (\frac{{1}}{{a d-b c}}\left[\begin{{array}}{{rr}}d & -b \\ -c & a\end{{array}}\right]\frac{{1}}{{a d-b c}}\left[\begin{{array}}{{rr}}d & -b \\ -c & a\end{{array}}\right]) is correct because the inverse of a 2x2 matrix is given by 1/determinant times the adjugate of the matrix.
150 Easy
150 of 922
Understanding +4 | -1
If AA is an idempotent matrix i.e., A2=AA^2=A and B=I−AB=I-A, then which of the following is not true?
View answer and explanation
Correct option: B
B2=IB^2=I
Option B (B2=IB^2=I) is correct because given A2=AA^2 = A and B=I−AB = I - A, it follows that B2=(I−A)(I−A)=I−2A+A2=I−2A+A=I−A=BB^2 = (I-A)(I-A) = I - 2A + A^2 = I - 2A + A = I - A = B, not II.
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