If $a + b + c = 0$, then the solution of the equation $\left| {\begin{array}{*{20}{c}} {a - x}&c&b\\ c&{b - x}&a\\ b&a&{c - x} \end{array}} \right| = 0$ is

$ \pm \,\frac{3}{2}\left( {{a^2} + {b^2} + {c^2}} \right)$

$0, \pm \,\sqrt {\left( {{a^2} + {b^2} + {c^2}} \right)} $

If $a,b,c$ are non-zero real numbers, then $\left| {\begin{array}{*{20}{c}} {bc}&{ca}&{ab}\\ {ca}&{ab}&{bc}\\ {ab}&{bc}&{ca} \end{array}} \right|$ vanishes, when

$\frac{1}{b} + \frac{1}{c} - \frac{1}{a} = 0$

$\frac{1}{b} - \frac{1}{c} - \frac{1}{a} = 0$

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 · Medium

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Showing 31-60 of 408 questions

31 Medium

31 of 408

Analyzing +4 | -1

If the system of equations

2x + 3y + 5 = 0,x + ky + 5 = 0,kx - 12y - 14 = 0

be consistent, then value of

k

View answer and explanation

Correct option: C

- 6,\frac{{17}}{5}

Option C (

- 6,\frac{{17}}{5}

) is correct because these are the values of k for which the determinant of the coefficient matrix is non-zero, implying consistency.

32 Medium

32 of 408

Application +4 | -1

{a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0

such that

\left| {\begin{array}{cccccccccccccccccccc} {1 + a}&1&1\\ 1&{1 + b}&1\\ 1&1&{1 + c} \end{array}} \right| = \lambda

then value of

\lambda

View answer and explanation

Correct option: B

abc

Option B (

abc

) is correct because given the condition

{a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0

, the determinant simplifies to abc.

33 Medium

33 of 408

Understanding +4 | -1

Let

A

and

B

be two square matrices of the same dimension and let

[A, B]=A B-B A

. Then for three

2 \times 2

matrices

A, B, C

[[A, B], C]+[[B, C], A]+[[C, A], B]

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because the expression simplifies to zero due to the properties of the commutator and matrix algebra.

34 Medium

34 of 408

Analyzing +4 | -1

\left| {\begin{array}{cccccccccccccccccccc} {{a^2}}&{{b^2}}&{{c^2}}\\ {{{\left( {a + 1} \right)}^2}}&{{{\left( {b + 1} \right)}^2}}&{{{\left( {c + 1} \right)}^2}}\\ {{{\left( {a - 1} \right)}^2}}&{{{\left( {b - 1} \right)}^2}}&{{{\left( {c - 1} \right)}^2}} \end{array}} \right| = k\left| {\begin{array}{cccccccccccccccccccc} {{a^2}}&{{b^2}}&{{c^2}}\\ a&b&c\\ 1&1&1 \end{array}} \right|

, Then the value of

k

View answer and explanation

Correct option: D

Option D (4) is correct because after applying row operations, the determinant simplifies, and k is found to be 4.

35 Medium

35 of 408

Analyzing +4 | -1

a + b + c = 0

, then the solution of the equation

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

View answer and explanation

Correct option: C

0, \pm \,\sqrt {\frac{3}{2}\left( {{a^2} + {b^2} + {c^2}} \right)}

The correct answer is C (

0, \pm \sqrt {\frac{3}{2}\left( {{{{a^2} + {b^2} + {c^2}}}} \right)}

) because it represents the solutions to the determinant equation given the condition a + b + c = 0.

36 Medium

36 of 408

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.

37 Medium

37 of 408

Analyzing +4 | -1

a,b,c

are non-zero real numbers, then

\left| {\begin{array}{cccccccccccccccccccc} {bc}&{ca}&{ab}\\ {ca}&{ab}&{bc}\\ {ab}&{bc}&{ca} \end{array}} \right|

vanishes, when

View answer and explanation

Correct option: A

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0

Option A (

\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)\left( {x - \delta } \right)

) is correct because after applying column operations and expanding the determinant, we find that

f(x) = \left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)\left( {x - \delta } \right)

38 Medium

38 of 408

Analyzing +4 | -1

\omega

is an imaginary cube root of unity, then the value of

\left| {\begin{array}{cccccccccccccccccccc} a&{b{\omega ^2}}&{a\omega }\\ {b\omega }&c&{b{\omega ^2}}\\ {c{\omega ^2}}&{a\omega }&c \end{array}} \right|,

View answer and explanation

Correct option: C

Option C (0) is correct because applying the column operation C3 -> C3 - ωC1 results in a column of zeros, making the determinant zero.

39 Medium

39 of 408

Analyzing +4 | -1

a \ne b,b,c

satisfy

\left| {\begin{array}{cccccccccccccccccccc} a&{2b}&{2c}\\ 3&b&c\\ 4&a&b \end{array}} \right| = 0,

then

abc =

View answer and explanation

Correct option: C

{b^3}

Option C (

{b^3}

) is correct because from the given determinant equation, we derive

{b^2} = ac

, leading to

abc = {b^3}

40 Medium

40 of 408

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.

41 Medium

41 of 408

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.

42 Medium

42 of 408

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.

43 Medium

43 of 408

Application +4 | -1

\mathrm{A}=\left[\begin{array}{cc}1 & \cot \theta \\ -\cot \theta & 1\end{array}\right]

satisfies

\left(\mathrm{A}^{-1}\right)^2=f(\theta)\left[\begin{array}{cc}-\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & -\cos 2 \theta\end{array}\right]

, then

f(\theta)=

View answer and explanation

Correct option: A

\sin ^2 \theta

Option C is correct. By calculating A² and substituting it into the given equation, we can solve for k and find that it equals 5.

44 Medium

44 of 408

Analyzing +4 | -1

The determinant

\Delta=\left|\begin{array}{ccc}a & b & a \alpha+b \\ b & c & b \alpha+c \\ a \alpha+b & b \alpha+c & 0\end{array}\right|

is equal to zero if

View answer and explanation

Correct option: B

a, b, c$ are in G.P.

Option B, a, b, c are in G.P., is correct because the determinant is zero when

b^2 = ac

, which is the condition for a geometric progression.

45 Medium

45 of 408

Application +4 | -1

f(x)=\left|\begin{array}{ccc}\cos x & 1 & 0 \\ 1 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x\end{array}\right|

, then

\int\limits_{0}^{\pi / 2} f(x) d x

equals

View answer and explanation

Correct option: B

1 / 3

Option B (1/3) is correct because the integral of f(x) from 0 to π/2, after simplifying f(x) to cos(3x), results in 1/3.

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

46 Medium

46 of 408

Application +4 | -1

A=\left[\begin{array}{cc}i & -i \\ -i & i\end{array}\right]

and

B=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]

, then

A^8

equals

View answer and explanation

Correct option: A

128 B

Option A (

128B

) is correct because after calculating

A^2

A^4

, and

A^8

, we find that

A^8 = 128B

47 Medium

47 of 408

Analyzing +4 | -1

The value of

\left| {\begin{array}{cccccccccccccccccccc} 1&1&1\\ {bc}&{ca}&{ab}\\ {b + c}&{c + a}&{a + b} \end{array}} \right|

View answer and explanation

Correct option: C

\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)

The correct answer is C (

\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)

) because after applying column operations and simplifying, the determinant equals this expression.

48 Medium

48 of 408

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.

49 Medium

49 of 408

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

50 Medium

50 of 408

Application +4 | -1

If a non-zero solution of

x+3 y+z=0,2 x-y-z=0, k x+2 y+3 z=0

exist, then

k=

(x \neq 0, y \neq 0, z \neq 0)

View answer and explanation

Correct option: C

-\frac{15}{2}

Option C is correct. For a non-zero solution to exist, the determinant of the coefficient matrix must be zero, which leads to k = -15/2.

51 Medium

51 of 408

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

52 Medium

52 of 408

Application +4 | -1

A=\left(\begin{array}{cc}\frac{-1+i \sqrt{3}}{2 i} & \frac{-1-i \sqrt{3}}{2 i} \\ \frac{1+i \sqrt{3}}{2 i} & \frac{1-i \sqrt{3}}{2 i}\end{array}\right), i=\sqrt{-1}

and

f(x)=x^2+2

, then

f(A)

equals

View answer and explanation

Correct option: D

(2+i \sqrt{3})\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]

Option D (

(2+i \sqrt{{3}})\left[\begin{{array}}{{ll}}1 & 0 \\ 0 & 1\end{{array}}\right]

) is correct because after simplification and calculation, f(A) equals the given matrix.

53 Medium

53 of 408

Application +4 | -1

A=\left[\begin{array}{cc} a & b \\ c & 1-a \end{array}\right]

is idempotent matrix and

b c=1 / 4

then the value of

a

View answer and explanation

Correct option: C

1 / 2

Option C is correct. Since A is idempotent, A² = A. Using this property and the given condition bc = 1/4, we can solve for a and find that a = 1/2.

54 Medium

54 of 408

Application +4 | -1

Let the determinant of a

3 \times 3

matrix

A

be 6, then

B

is a matrix defined by

B = 5{A^2}

. Then, determinant of

B

View answer and explanation

Correct option: D

None of These

Option D (None of These) is correct because the determinant of B, calculated as det(5A²), equals 4500, which is not among the given options.

55 Medium

55 of 408

Application +4 | -1

The number of solutions of the matrix equation

A^{2}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]

is -

View answer and explanation

Correct option: D

Infinitely many

Option D (Infinitely many) is correct because there are infinitely many matrices that satisfy the given equation

A^2 = I

56 Medium

56 of 408

Analyzing +4 | -1

Let

A=\left[a_{i j}\right]

be a

3 \times 3

matrix, where

a_{i j}= \begin{cases}1, & \text { if } i=j \\ -x & \text { if }|i-j|=1 \\ 2 x+1, & \text { otherwise }\end{cases}

Let a function be defined as

f(x)=\operatorname{det}(A)

. Then the sum of maximum and minimum values of

f

{R}

is equal to

View answer and explanation

Correct option: C

-\frac{{88}}{{27}}

Option C (-88/27) is correct because it involves finding the determinant of the given matrix, finding its critical points using derivatives, and summing the maximum and minimum values.

57 Medium

57 of 408

Application +4 | -1

l, m

and

n

are positive real numbers and the matrix

A=\left[\begin{array}{rrr} 0 & 2 m & n \\ l & m & -n \\ l & -m & n \end{array}\right]

is such that

A A^{\mathrm{T}}=I

(unit matrix), then the ordered triple

(l, m, n)

may be

View answer and explanation

Correct option: D

\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{3}}\right)

Option C (

\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{6}}\right)

) is correct because it satisfies the condition

A A^{\mathrm{T}}=I

for the given matrix A.

58 Medium

58 of 408

Analyzing +4 | -1

The determinant

\Delta=\left|\begin{array}{ccc} 1 & \cos (\beta-\alpha) & \cos (\gamma-\alpha) \\ \cos (\alpha-\beta) & 1 & \cos (\gamma-\beta) \\ \cos (\alpha-\gamma) & \cos (\beta-\gamma) & 1 \end{array}\right| \text { is equal to }

View answer and explanation

Correct option: D

Option D (0) is correct because the given determinant can be expressed as the product of two determinants, each of which is zero.

59 Medium

59 of 408

Analyzing +4 | -1

The value of

\left| {\begin{array}{cccccccccccccccccccc} 1&{{{\log }_x}y}&{{{\log }_x}z}\\ {{{\log }_y}x}&1&{{{\log }_y}z}\\ {{{\log }_z}x}&{{{\log }_z}y}&1 \end{array}} \right|

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the determinant simplifies to 0 after applying logarithm properties.

60 Medium

60 of 408

Application +4 | -1

\left| {\begin{array}{cccccccccccccccccccc} {{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

\left| {\begin{array}{cccccccccccccccccccc} {{b_2}{c_3} - {b_3}{c_2}}&{{c_2}{a_3} - {c_3}{a_2}}&{{a_2}{b_3} - {c_3}{b_2}}\\ {{b_3}{c_1} - {b_1}{c_3}}&{{c_3}{a_1} - {c_1}{a_3}}&{{a_3}{b_1} - {a_1}{b_3}}\\ {{b_1}{c_2} - {b_2}{c_1}}&{{c_1}{a_2} - {c_2}{a_1}}&{{a_1}{b_2} - {a_2}{b_1}} \end{array}} \right|

View answer and explanation

Correct option: B

Option B (25) is correct because the given determinant is the adjoint of A, and |adj(A)| = |A|^(n-1) where n is the order of the matrix.

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