The adjoint matrix of $\left[ {\begin{array}{*{20}{c}} 3&{ - 3}&4\\ 2&{ - 3}&4\\ 0&{ - 1}&1 \end{array}} \right]$ is

$\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0\\ { - 2}&3&{ - 4}\\ { - 2}&3&{ - 3} \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 4&8&3\\ 2&1&6\\ 0&2&1 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} {11}&9&3\\ 1&2&8\\ 6&9&1 \end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}} 1&{ - 2}&1\\ { - 1}&3&3\\ { - 2}&3&{ - 3} \end{array}} \right]$

If $A = {\left[ {{a_{ij}}} \right]_{2 \times 2}}$, where ${a_{ij}} = i + j,\;{\rm{then\;}}A{\rm{\;is\;equal\;to}}$

$\left[ {\begin{array}{*{20}{c}} {2\;\,\,\,\;3}\\ {3\;\;\,\,4} \end{array}} \right]$

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

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

$\left[ {\begin{array}{*{20}{c}} {1\;\;\,\,\,2}\\ {3\;\,\,\,\,\;4} \end{array}} \right]$

If $A = {\left[ {{a_{ij}}} \right]_{m \times n}}$ is a matrix of rank $r$, then

$r = \min \left( {m,n} \right)$

$r < \min \left( {m,n} \right)$

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JEE • MATHEMATICS • MATRICES

Matrices practice · Easy

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421 Easy

421 of 922

Application +4 | -1

\mathrm{A}, \mathrm{B}

are two square matrices such that

\mathrm{AB}=\mathrm{A}

and

\mathrm{BA}=\mathrm{B}

, then

View answer and explanation

Correct option: B

A, B are idempotent

Option B, A, B are idempotent, is correct because given AB=A and BA=B, it can be proven that A²=A and B²=B, which is the definition of idempotent matrices.

422 Easy

422 of 922

Application +4 | -1

If each of third order determinant of value

\Lambda

is multiplied by 4 , then value of the new determinant is:

View answer and explanation

Correct option: C

64 \Delta

Option C (3

64 \Delta

) is correct because multiplying each element of a third-order determinant by 4 results in the determinant being multiplied by

4^3 = 64

423 Easy

423 of 922

Application +4 | -1

The adjoint matrix of

\left[ {\begin{array}{cccccccccccccccccccc} 3&{ - 3}&4\\ 2&{ - 3}&4\\ 0&{ - 1}&1 \end{array}} \right]

View answer and explanation

Correct option: B

\left[ {\begin{array}{cccccccccccccccccccc} 1&{ - 1}&0\\ { - 2}&3&{ - 4}\\ { - 2}&3&{ - 3} \end{array}} \right]

Option B is correct. The adjoint matrix is found by taking the transpose of the cofactor matrix.

424 Easy

424 of 922

Application +4 | -1

- 9

is a root of the equation

\left| {\begin{array}{cccccccccccccccccccc} x&3&7\\ 2&x&2\\ 7&6&x \end{array}} \right| = 0

, then the other two roots are

View answer and explanation

Correct option: A

2,7

Option A (2,7) is correct because factoring the determinant equation yields (x+9)(x-2)(x-7) = 0, so the other two roots are 2 and 7.

425 Easy

425 of 922

Application +4 | -1

Let

\mathrm{A}

be the set of all

3 \times 3

symmetric matrix all of whose entries are either 0 or 1 five of these entries are 1 and four of them are zero. Find the number of matrix in

\mathrm{A}

are-

View answer and explanation

Correct option: D

Option D (12) is correct because the number of such symmetric matrices is the sum of the ways to place the 1s on the diagonal and off-diagonal, which equals 12.

426 Easy

426 of 922

Application +4 | -1

X, Y

and

Z

are positive numbers such that

Y

and

Z

have respectively 1 and 0 at their unit's place and

\Delta = \left| {\begin{array}{llllllllllllllllllll} X&4&1 \\ Y&0&1 \\ Z&1&0 \end{array}} \right|

(\Delta+1)

is divisible by 10 , then

X

has at its unit's place

View answer and explanation

Correct option: C

Option C (2) is correct because the unit's place of X is 2, derived from the divisibility condition.

427 Easy

427 of 922

Analyzing +4 | -1

Determinant

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

is equal to

View answer and explanation

Correct option: C

4{a^2}{b^2}{c^2}

Option C (

4{a^2}{b^2}{c^2}

) is correct because it is the simplified value of the determinant after applying row operations and expanding.

428 Easy

428 of 922

Understanding +4 | -1

Assertion :

\Delta={a}_{11} {~A}_{11}+{a}_{12} {~A}_{12}+{a}_{13} {~A}_{13}

where,

{A}_{{ij}}

is cofactor of

{a}_{{ij}}

.

Reason :

\Delta{=}

Sum of the products of elements of any row (or column) with their corresponding cofactors.

View answer and explanation

Correct option: A

Assertion is correct, reason is correct; reason is a correct explanation for assertion.

Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because the determinant can be expanded using cofactors of any row or column.

429 Easy

429 of 922

Application +4 | -1

A=\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right], 6 A^{-1}=A^2+c A+d I

, then

(c, d)=

View answer and explanation

Correct option: A

(-6,11)

Option A ((-6,11)) is correct because the values of c and d are found by comparing the given equation with the characteristic equation of matrix A.

430 Easy

430 of 922

Analyzing +4 | -1

If the value of

\Delta=\left|\begin{array}{lll}{ }^{10} C_4 & { }^{10} C_5 & { }^{11} C_m \\ { }^{11} C_6 & { }^{11} C_7 & { }^{12} C_{m+2} \\ { }^{12} C_8 & { }^{12} C_9 & { }^{13} C_{m+4}\end{array}\right|

is equal to zero, then

m

is equal to

View answer and explanation

Correct option: C

Option C (5) is correct because using Pascal's rule and determinant properties, we find that

m=5

satisfies the condition for the determinant to be zero.

431 Easy

431 of 922

Analyzing +4 | -1

Let

A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]

and

B=\left[\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right], a, b \in N

. Then,

View answer and explanation

Correct option: D

there exist infinitely many

B^{\prime}

s such that

A B=B A

Option D is correct because there exist infinitely many matrices B such that AB = BA when a = b, where a and b are elements of matrix B.

432 Easy

432 of 922

Application +4 | -1

Let

A=\left(\begin{array}{cc}1 & 2 \\ -2 & -5\end{array}\right)

. Let

\alpha, \beta, \in \mathbb{R}

be such that

\alpha A^2

+\beta A=2I

. Then

\alpha+\beta

is equal to

View answer and explanation

Correct option: D

10

Option D is correct because by calculating A² and substituting into the given equation, we can solve for alpha and beta, and then find alpha + beta.

433 Easy

433 of 922

Analyzing +4 | -1

The system of linear equations

x+\lambda y-z=0

\lambda x-y-z=0

x+y-\lambda z=0

has non - trivial solutions for

View answer and explanation

Correct option: D

exactly three values of

\lambda

Option D is correct because the system has non-trivial solutions for exactly three values of

\lambda

: 0, 1, and -1.

434 Easy

434 of 922

Analyzing +4 | -1

\mathrm{a}

is real and

\sqrt{2} \mathrm{ax}+\sin \mathrm{By}+\cos \mathrm{Bz}=0

x+\cos B y+\sin B z=0,-x+\sin B y-\cos B z=0

, then the set of all values of a for which the system of linear equations has a non-trivial solution, is

View answer and explanation

Correct option: B

[-1,1]

Option B ([-1,1]) is correct because the determinant of the coefficient matrix must be zero for a non-trivial solution, leading to the condition

a \in [-1, 1]

435 Easy

435 of 922

Analyzing +4 | -1

The value of

\left|\begin{array}{ccc}x+y & y+z & z+x \\ x & y & z \\ x-y & y-z & z-x\end{array}\right|=

View answer and explanation

Correct option: D

Option D (0) is correct because the determinant is zero, as the first row is the sum of the second and third rows.

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436 Easy

436 of 922

Analyzing +4 | -1

Let

A

a 2 \times 2

matrix with real entries. Let

I

be the

2 \times 2

identity matrix. Denote bytr

(A)

, the sum of diagonal entries of

a

. Assume that

A^2=I

.

Statement-1 : If

A \neq I

and

A \neq-I

, then

\operatorname{det}(A)=-1

Statement-2 : If

A \neq I

and

A \neq-I

, then

\operatorname{tr}(A) \neq 0

View answer and explanation

Correct option: D

Statement -1 is true, Statement-2 is false

Option D (Statement -1 is true, Statement-2 is false) is correct because if A² = I and A is not I or -I, then det(A) = -1, but tr(A) can be 0.

437 Easy

437 of 922

Analyzing +4 | -1

s=a+b+c,

then the value of

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

View answer and explanation

Correct option: B

2 s^3

Option B (

2s^3

) is correct because after applying column operations and simplifying, the determinant evaluates to

2s^3

438 Easy

438 of 922

Analyzing +4 | -1

The system of linear equations

\begin{array}{l} x + y + z = 2\\ 2\;x + y - z = 3 \end{array}

3\;x + 2\;y + kz = 4

has a unique solution if

View answer and explanation

Correct option: A

k \ne 0

Option A (

k \ne 0

) is correct because for a unique solution, the determinant of the coefficient matrix must be non-zero, which implies k is not equal to 0.

439 Easy

439 of 922

Application +4 | -1

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

, where

{a_{ij}} = i + j,\;\mathrm{then\;}A\mathrm{\;is\;equal\;to}

View answer and explanation

Correct option: D

\left[ {\begin{array}{cccccccccccccccccccc} {2\;\,\,\,\;3}\\ {3\;\;\,\,4} \end{array}} \right]

Option D is correct. The matrix elements are defined by a_ij = i + j, and substituting the row and column indices gives the correct matrix.

440 Easy

440 of 922

Understanding +4 | -1

A = {\left[ {{a_{ij}}} \right]_{m \times n}}

is a matrix of rank

r

, then

View answer and explanation

Correct option: C

r \le \min \left( {m,n} \right)

Option C (r ≤ min(m,n)) is correct because the rank of a matrix cannot exceed the minimum of its number of rows and columns.

441 Easy

441 of 922

Analyzing +4 | -1

A

is a non-singular matrix satisfying

A B-B A=A

, then which one of the following holds true -

View answer and explanation

Correct option: D

\operatorname{det} .(\mathrm{B}+\mathrm{I})=\operatorname{det} .(\mathrm{B}-\mathrm{I})

Option D is correct because, through matrix manipulations and determinant properties, it can be shown that

\operatorname{det}.(B+I)=\operatorname{det}.(B-I)

holds true.

442 Easy

442 of 922

Application +4 | -1

A=\left(\begin{array}{rr}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right)

and

A+A^{\prime}=I

, then

\alpha

equals

View answer and explanation

Correct option: B

2 n \pi \pm \frac{\pi}{3}, n \in {I}

Option B (

2 n \pi \pm \frac{\pi}{3}, n \in {I}

) is correct because it satisfies the condition

A + A' = I

when solving for alpha.

443 Easy

443 of 922

Understanding +4 | -1

Let A be an invertible matrix. Which of the following is not true?

View answer and explanation

Correct option: B

A^{-1}=|A|^{-1}

Option B (

A^{-1}=|A|^{-1}

) is not true because

A^{-1}

is a matrix, while

|A|^{-1}

is a scalar. The correct formula is

A^{-1} = \frac{1}{|A|} adj(A)

444 Easy

444 of 922

Application +4 | -1

\left[ {\begin{array}{cccccccccccccccccccc} {x + y}&{2x + z}\\ {x - y}&{2z + w} \end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc} 4&7\\ 0&{10} \end{array}} \right]

, then the value of

x,\;y,\;z,\;w

are

View answer and explanation

Correct option: A

2,2,3,4

Option A (

2,2,3,4

) is correct because solving the system of equations derived from the matrix equality yields these values for

x, y, z,

and

w

445 Easy

445 of 922

Application +4 | -1

\mathrm{A}

is order 3 square matrix such that

|\mathrm{A}|=2

, then

|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} \mathrm{A}))|

is -

View answer and explanation

Correct option: B

256

Option B (256) is correct because the determinant of adj(adj(adj A)) is calculated using the property |adj(adj A)| = |A|^((n-1)²) and |adj A| = |A|^(n-1).

446 Easy

446 of 922

Application +4 | -1

Matrix

M_r

is defiend as

M_r=\left[\begin{array}{cc}r & r-1 \\ r-1 & r\end{array}\right], r \in N

value of

\operatorname{det}\left(M_1\right)+\operatorname{det}\left(M_2\right)+\operatorname{det}\left(M_3\right)+\ldots+

\operatorname{det}\left(\mathrm{M}_{2007}\right)

View answer and explanation

Correct option: D

2007^2

Option D (

2007^2

) is correct because the sum of the determinants simplifies to a telescoping series resulting in

2007^2

447 Easy

447 of 922

Application +4 | -1

f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} 1&1&1\\ {2x}&{\left( {x - 1} \right)}&x\\ {3x\left( {x - 1} \right)}&{\left( {x - 1} \right)\left( {x - 2} \right)}&{x\left( {x - 1} \right)} \end{array}} \right|,

then

f\left( {50} \right)

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the function f(x) simplifies to 0, making f(50) also equal to 0.

448 Easy

448 of 922

Application +4 | -1

Total number of possible matrices of order

3 \times 3

with each entry 2 or 0 is

View answer and explanation

Correct option: D

512

Option D (512) is correct because each of the 9 entries in the 3x3 matrix can be either 2 or 0, leading to

2^9 = 512

possible matrices.

449 Easy

449 of 922

Application +4 | -1

Suppose point

(x, y, z)

in space satisfies the equation

\left|\begin{array}{ccc} x^2+1 & x y & x z \\ y x & y^2+1 & y z \\ z x & z y & z^2+1 \end{array}\right|=5

Then

(x, y, z)

lies on a

View answer and explanation

Correct option: C

Sphere

The correct answer is C (Sphere) because simplifying the determinant leads to the equation

x^2 + y^2 + z^2 = 4

, which represents a sphere.

450 Easy

450 of 922

Understanding +4 | -1

If the matrices

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

and

B=\left[b_{i j}\right]

and

C=\left[c_{i j}\right]

are of the same order, say

m \times n

, satisfy Associative law, then

View answer and explanation

Correct option: A

(\mathrm{A}+\mathrm{B})+\mathrm{C}=\mathrm{A}+(\mathrm{B}+\mathrm{C})

Option A,

(\mathrm{A}+\mathrm{B})+\mathrm{C}=\mathrm{A}+(\mathrm{B}+\mathrm{C})

, is correct because it represents the associative law for matrix addition.

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