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Test: Matrices & Determinants(May 7) - JEE MCQ


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15 Questions MCQ Test - Test: Matrices & Determinants(May 7)

Test: Matrices & Determinants(May 7) for JEE 2024 is part of JEE preparation. The Test: Matrices & Determinants(May 7) questions and answers have been prepared according to the JEE exam syllabus.The Test: Matrices & Determinants(May 7) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Matrices & Determinants(May 7) below.
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Test: Matrices & Determinants(May 7) - Question 1

If A and B are any two square matrices of the same order, then

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 1

By the property of transpose , (AB)’ = B’A’.

Test: Matrices & Determinants(May 7) - Question 2

The transformation ‘orthogonal projection on X-axis’ is given by the matrix

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 2

The orthogonal projection on x- axis is given by : 

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Test: Matrices & Determinants(May 7) - Question 3

The equations x + 2y + 2z = 1 and 2x + 4 y + 4z = 9 have

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 3

Test: Matrices & Determinants(May 7) - Question 4

The number of all the possible matrices of order 2 × 2 with each entry 0, 1 or 2 is

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 4

32x2 = 34 = 81

Test: Matrices & Determinants(May 7) - Question 5

A square matrix  A = [aij]n×n is called an upper triangular if aij = 0 for

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 5

Upper Triangular matrix is given by  :
.
 Here, aij=0 , if i is greater than j.and aij ≠ 0, if I is less than j.

Test: Matrices & Determinants(May 7) - Question 6

If A is any square matrix, then

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 6

For every square matrix (A + A’) is always symmetric.

Test: Matrices & Determinants(May 7) - Question 7

The equations, x + 4 y – 2 z = 3, 3 x + y + 5 z = 7, 2 x + 3y +z = 5 have

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 7

The given system of equations does not have solution if : 

- 0 ⇒ 1(-14) - 4(-7) -2(7) = 0

Test: Matrices & Determinants(May 7) - Question 8

If the system of equationsx + 4 ay + az = 0, x + 3by + bz = 0 andx + 2 cy +cz = 0 have a non-zero solution,then a, b, c are in

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 8

For a non trivial solution : 



⇒ bc + ab - 2ac = 0 ⇒  ∴ there, a , b ,c, are in H.P

Test: Matrices & Determinants(May 7) - Question 9

If I3 is the identity matrix of order 3 , then 13−1 is

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 9

Because , the inverse of an identity matrix is an identity matrix.

Test: Matrices & Determinants(May 7) - Question 10

If A and B are square matrices of same order and A’ denotes the transpose of A , then

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 10

By the property of transpose of a matrix ,(AB)’ = B’A’.

Test: Matrices & Determinants(May 7) - Question 11

A square matrix A is invertible iff det A is equal to

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 11

Only non-singular matrices possess inverse.

Test: Matrices & Determinants(May 7) - Question 12

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 12



Apply, C1→ C1+ C2+C3+C4,

Apply, R1 →R1 - R2,


Apply, R1 R- R2, R→ R- R3

=(x+3a) (a -x)3 (1) = (x+3a)(a-x)3

Test: Matrices & Determinants(May 7) - Question 13

If the entries in a 3 x 3 determinant are either 0 or 1 , then the greatest value of this determinant is :

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 13

Greatest value = 2 

Test: Matrices & Determinants(May 7) - Question 14

The roots of the equation  are

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 14

Operate,



Apply R3→R3- R1, R2 R2 -R1,

 

⇒ -6(5x2 - 20) +15(2x-4) = 0

⇒ (x- 2)(x+1) = 0⇒x=2, -1

⇒-6(5x2 - 20) + 15(2x-4) = 0

⇒(x-2)(x+1) = 0 ⇒ x=2, -1

 

Test: Matrices & Determinants(May 7) - Question 15

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 n determinants, where n has value

Detailed Solution for Test: Matrices & Determinants(May 7) - Question 15

N = 2 ×3 × 4 = 24.

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