If $P = \left[ {\begin{array}{*{20}{c}} i&0&{ - i}\\ 0&{ - i}&i\\ { - i}&i&0 \end{array}} \right]$ and $Q = \left[ {\begin{array}{*{20}{c}} { - i}&i\\ 0&0\\ i&{ - i} \end{array}} \right]$, then $PQ$ is equal to

$\left[ {\begin{array}{*{20}{c}} 2&{ - 2}\\ { - 1}&1 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]$

If $a,b,c$ are in AP, then the value of $\left| {\begin{array}{*{20}{c}} {x + 2}&{x + 3}&{x + a}\\ {x + 4}&{x + 5}&{x + b}\\ {x + 6}&{x + 7}&{x + c} \end{array}} \right|$ is

$x - \left( {a + b + c} \right)$

The matrix $\left[ {\begin{array}{*{20}{c}} \lambda &1&0\\ 0&2&3\\ 0&0&\lambda \end{array}} \right]$ is non- singular

For all real values of $\lambda $

Only when $\lambda = \pm \frac{1}{{\sqrt 2 }}$

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Matrices practice · Mixed difficulty

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Showing 91-120 of 995 questions

91 Easy

91 of 995

Understanding +4 | -1

A=\left(\begin{array}{ccc}2 & 3-i & -i \\ 3+i & \pi & 7+i \\ i & 7-i & e\end{array}\right)

, then

A

View answer and explanation

Correct option: B

Hermitian

Option B (Hermitian) is correct because

A^* = A

, meaning the conjugate transpose of A is equal to A.

92 Easy

92 of 995

Application +4 | -1

If for

a

matrix

A,|A|=6

and

\operatorname{adj} A=\left[\begin{array}{rrr}1 & -2 & 4 \\ 4 & 1 & 1 \\ -1 & k & 0\end{array}\right]

, then

k

is equal to

View answer and explanation

Correct option: B

Option D (2) is correct because by calculating the determinant of adj A and using the property |adj A| = |A|^(n-1), we can solve for k.

93 Medium

93 of 995

Analyzing +4 | -1

a, b, c

are three complex numbers such that

a^2+b^2+c^2=0

and

\Delta=\left|\begin{array}{ccc} b^2+c^2 & a b & a c \\ a b & c^2+a^2 & b c \\ a c & b c & a^2+b^2 \end{array}\right|=k a^2 b^2 c^2,

then the value of

k

View answer and explanation

Correct option: D

Option D (4) is correct because by substituting

a^2+b^2+c^2=0

into the determinant and simplifying, the value of k is found to be 4.

94 Easy

94 of 995

Application +4 | -1

The number of symmetric matrices of order 3, with all the entries from the set

\{0,1,2,3,4,5,6,7,8,9\}

View answer and explanation

Correct option: B

10^{6}

Option B (

10^6

) is correct because a symmetric matrix of order 3 has 6 independent entries, each of which can be any of the 10 digits.

95 Easy

95 of 995

Application +4 | -1

If the adjoint of a

3 \times 3

matrix

\mathrm{P}

\left[\begin{array}{ccc}1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3\end{array}\right]

, then the possible values of the determinant of

\mathrm{P}

are

View answer and explanation

Correct option: A

\pm 2

Option A (

\pm 2

) is correct because the determinant of the adjoint of a matrix P is equal to the square of the determinant of P, which leads to the possible values of

\pm 2

96 Easy

96 of 995

Analyzing +4 | -1

The values of

\lambda

and

\mu

such that the system of equations

x + y + z = 6,\,\,3x + 5y + 5z = 26,\,\,x + 2y + \lambda z = \mu

has no solution, are

View answer and explanation

Correct option: D

\lambda = 2,\,\mu \ne 10

Option D (

\lambda = 2,\,\mu \ne 10

) is correct because for the system to have no solution, the determinant of the coefficients must be zero, and the equations must be inconsistent.

97 Easy

97 of 995

Application +4 | -1

A^5=O

such that

A^n \neq I

for

1 \leq n \leq 4

, then

(I-A)^{-1}

equals

View answer and explanation

Correct option: D

none of these

Option D (none of these) is correct because none of the provided options satisfy the condition for being the inverse of (I-A) given

A^5 = O

and

A^n \neq I

for

1 \leq n \leq 4

98 Easy

98 of 995

Application +4 | -1

P = \left[ {\begin{array}{cccccccccccccccccccc} i&0&{ - i}\\ 0&{ - i}&i\\ { - i}&i&0 \end{array}} \right]

and

Q = \left[ {\begin{array}{cccccccccccccccccccc} { - i}&i\\ 0&0\\ i&{ - i} \end{array}} \right]

, then

PQ

is equal to

View answer and explanation

Correct option: B

\left[ {\begin{array}{cccccccccccccccccccc} 2&{ - 2}\\ { - 1}&1\\ { - 1}&1 \end{array}} \right]

Option B (

\left[ {\begin{{array}}{{{*{20}{c}}}} 2&{ - 2}\\ { - 1}&1\\ { - 1}&1 \end{{array}}} \right]

) is correct because it is the result of the matrix multiplication of P and Q.

99 Easy

99 of 995

Analyzing +4 | -1

\Delta=\left|\begin{array}{ccc}a+x & b & c \\ b & x+c & a \\ c & a & x+b\end{array}\right|.

Which of the following is a factor for the above determinant?

View answer and explanation

Correct option: B

x+(a+b+c)

Option B (

x+(a+b+c)

) is correct because applying a column operation allows factoring out this expression, indicating it's a factor of the determinant.

100 Easy

100 of 995

Application +4 | -1

Let

A

be a matrix of order 3 and let

\Delta

denotes the value of determinant

A

. Then determinant

(-2 A)=

View answer and explanation

Correct option: A

-8 \Delta

Option A (-8

\Delta

) is correct because when a matrix A of order 3 is multiplied by -2, its determinant is multiplied by

(-2)^3 = -8

101 Easy

101 of 995

Application +4 | -1

Let

A=\left[\begin{array}{cc}0 & -\tan (\alpha / 2) \\ \tan (\alpha / 2) & 0\end{array}\right]\alpha \neq(2 n+1) \pi, n \in {I},(I+A)(I-A)^{-1}

equals

View answer and explanation

Correct option: A

\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]

Option A (

\begin{{bmatrix}}\cos \alpha & -\sin \alpha \\\\ \sin \alpha & \cos \alpha \end{{bmatrix}}

) is correct because it results from simplifying the matrix expression

(I+A)(I-A)^{-1}

using trigonometric identities.

102 Easy

102 of 995

Analyzing +4 | -1

Let

A=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right)

. Then the sum of the diagonal elements of the matrix

(A+I)^{11}

is equal to

View answer and explanation

Correct option: D

4097

Option D (4097) is correct because the sum of the diagonal elements (trace) of the matrix

(A+I)^{11}

is calculated using the property

A^2 = A

103 Medium

103 of 995

Application +4 | -1

x \ne 0,\left| {\begin{array}{cccccccccccccccccccc} {x + 1}&{2x + 1}&{3x + 1}\\ {2x}&{4x + 3}&{6x + 3}\\ {4x + 4}&{6x + 4}&{8x + 4} \end{array}} \right| = 0

, then

x + 1

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because x+1 = 0 is the solution obtained after simplifying the determinant equation.

104 Medium

104 of 995

Application +4 | -1

\left| {\begin{array}{cccccccccccccccccccc} {a + x}&{a - x}&{a - x}\\ {a - x}&{a + x}&{a - x}\\ {a - x}&{a - x}&{a + x} \end{array}} \right| = 0,

then

x

is equal to

View answer and explanation

Correct option: C

0,3a

Option C (

0,3a

) is the correct answer because the determinant equation simplifies to

x = 0

x = 3a

after applying column operations and solving.

105 Easy

105 of 995

Application +4 | -1

If matrix

A=\left[a_y\right]_{2 \times 2}

, where

a_y=1

, if

i \neq j

and 0 if

i=j

, then

A^2

is equal to

View answer and explanation

Correct option: A

l

Option A (

l

) is correct because the given matrix A, when squared, results in the identity matrix, denoted as l.

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106 Medium

106 of 995

Application +4 | -1

a,b,c

are in AP, then the value of

\left| {\begin{array}{cccccccccccccccccccc} {x + 2}&{x + 3}&{x + a}\\ {x + 4}&{x + 5}&{x + b}\\ {x + 6}&{x + 7}&{x + c} \end{array}} \right|

View answer and explanation

Correct option: D

Option D (0) is correct because the determinant's value becomes zero when a, b, and c are in arithmetic progression.

107 Easy

107 of 995

Understanding +4 | -1

Let

A

be a

3 \times 3

matrix and

S=\left\{\left(\begin{array}{l} x \\ y \\ z \end{array}\right) x, y, z \in {R}\right\}

Define

f: S \rightarrow S

f\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=A\left(\begin{array}{l} x \\ y \\ z \end{array}\right)

Suppose

f\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right) \Rightarrow x=y=z=0

. Then

View answer and explanation

Correct option: A

f

is one-to-one

Option A (

f

is one-to-one) is correct because the condition

f(\begin{array}{l}x \\ y \\ z\end{array})=\left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right) \Rightarrow x=y=z=0

implies that the null space of A contains only the zero vector, meaning that the linear transformation is injective (one-to-one).

108 Easy

108 of 995

Application +4 | -1

The roots of the equation

\left| {\begin{array}{cccccccccccccccccccc} 1&4&{20}\\ 1&{ - 2}&5\\ 1&{2x}&{5{x^2}} \end{array}} \right| = 0

View answer and explanation

Correct option: B

- 1,2

Option B (

- 1,2

) is correct because these are the roots of the quadratic equation obtained after simplifying the determinant.

109 Easy

109 of 995

Application +4 | -1

The number of distinct values of a

2 \times 2

determinant whose entries are from the set

\{-1,0,1\}

, is

View answer and explanation

Correct option: C

Option C (5) is correct because there are 5 distinct possible values for the 2x2 determinant.

110 Easy

110 of 995

Application +4 | -1

A = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;\;\;\;2}\\ { - 4 - 1} \end{array}} \right],\;\mathrm{then}\;{A^{ - 1}}

View answer and explanation

Correct option: A

\frac{1}{7}\left[ {\begin{array}{cccccccccccccccccccc} { - 1\,\,\,\, - 2}\\ {4\;\;\;\;\;\;1} \end{array}} \right]

Option A (

\frac{1}{7}\left[ {\begin{{array}}{*{20}{c}} { - 1\,\,\,\, - 2}\\ {4\,\,\,\,\,\,1} \end{{array}} \right]

) is correct because it is the inverse matrix calculated using the formula A^-1 = (1/det(A)) * adj(A).

111 Easy

111 of 995

Application +4 | -1

Let

P(x)=\left|\begin{array}{ccc} x+1 & 2 & 3 \\ 1 & x+2 & 3 \\ 1 & 2 & x+3 \end{array}\right|,

the product of zeros of

P(x)

View answer and explanation

Correct option: A

The correct answer is A (0) because the product of the zeros of the polynomial P(x) is equal to -P(0), which is 0 in this case.

112 Medium

112 of 995

Analyzing +4 | -1

a^2+b^2+c^2=-2

and

f(x)=\left|\begin{array}{ccc}1+a^2 x & \left(1+b^2\right) x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & 1+b^2 x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & \left(1+b^2\right) x & 1+c^2 x\end{array}\right|

, then

f(x)

is a polynomial of degree

View answer and explanation

Correct option: B

Option B (2) is correct because after simplification, the function f(x) reduces to a polynomial of degree 2.

113 Easy

113 of 995

Application +4 | -1

A = \left[ {\begin{array}{cccccccccccccccccccc} 6&8&5\\ 4&2&3\\ 9&7&1 \end{array}} \right]

is the sum of a symmetric matrix

B

and skew-symmetric matrix,

C

, then

B

View answer and explanation

Correct option: A

\left[ {\begin{array}{cccccccccccccccccccc} 6&6&7\\ 6&2&5\\ 7&5&1 \end{array}} \right]

Option A is correct. The symmetric matrix B is calculated as (A + A')/2, where A' is the transpose of A.

114 Easy

114 of 995

Application +4 | -1

A = \left[ {\begin{array}{cccccccccccccccccccc} {x\;\;\; - 2}\\ {3\;\;\;\;\;\;\;7} \end{array}} \right]\mathrm{and}\;{A^{ - 1}} = \left[ {\begin{array}{cccccccccccccccccccc} {\frac{7}{{34}}\;\;\;\frac{1}{{17}}}\\ {\frac{{ - 3}}{{34}}\;\;\;\frac{2}{{17}}} \end{array}} \right],\;

then the value of

x

View answer and explanation

Correct option: D

Option D (4) is correct because by finding the inverse of matrix A and comparing it to the given inverse, we can solve for x.

115 Easy

115 of 995

Analyzing +4 | -1

The number of

q \in(0,4 \pi)

for which the system of linear equations

3(\sin 3 \theta) x-y+z=2

3(\cos 2 \theta) x+4 y+3 z=3

6 x+7 y+7 z=9

has no solution, is

View answer and explanation

Correct option: B

Option B (7) is correct because there are 7 values of

\theta

in the interval

(0, 4\pi)

for which the system of linear equations has no solution.

116 Medium

116 of 995

Application +4 | -1

Let

{P}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]

and

I

be the identity matrix of order 3. If

Q=\left[q_{i j}\right]

is a matrix such that

P^{50}-Q=I

, then

\frac{{q}_{31}+{q}_{32}}{103 {q}_{21}}

equals

View answer and explanation

Correct option: A

Option A (1) is correct because by calculating

P^{50}

using the binomial theorem and the given matrix P, we find the values of

q_{31}

q_{32}

, and

q_{21}

, and then

\frac{q_{31}+q_{32}}{103q_{21}} = 1

117 Easy

117 of 995

Application +4 | -1

A

and

B

are two non-singular matrices of the same order such that

B^r=I

for some positive integer

r>1.

Then

A^{-1} B^{r-1} A-A^{-1} B^{-1} A

is equal to

View answer and explanation

Correct option: C

Option C (0) is correct because given

B^r = I

, it implies

B^{r-1} = B^{-1}

, hence

A^{-1}B^{r-1}A - A^{-1}B^{-1}A = 0

118 Medium

118 of 995

Understanding +4 | -1

The matrix

\left[ {\begin{array}{cccccccccccccccccccc} \lambda &1&0\\ 0&2&3\\ 0&0&\lambda \end{array}} \right]

is non- singular

View answer and explanation

Correct option: C

Only when

\lambda \ne 0

Option C (Only when

\lambda \ne 0

) is correct because the determinant of the matrix must be non-zero for the matrix to be non-singular, which implies

\lambda \ne 0

119 Easy

119 of 995

Analyzing +4 | -1

A_1, A_2, A_3, \ldots, A_{2 n-1}

are n skew-symmetric matrices of same order, then

B = \sum\limits_{r = 1}^n {\left( {2r - 1} \right){{\left( {{A_{2r - 1}}} \right)}^{2r - 1}}}

will be

View answer and explanation

Correct option: B

Skew-symmetric

Option B (Skew-symmetric) is correct because when taking the transpose of the given expression, it results in the negative of the original expression, indicating a skew-symmetric matrix.

120 Easy

120 of 995

Understanding +4 | -1

A

is a skew Hermitian matrix, then the main diagonal elements of

A

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.

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