Maths-
General
Easy

Question

Assertion (A): open square brackets table row 5 0 0 row 0 3 0 row 0 0 9 end table close square brackets is a diagonal matrix
Reason (R) : A square matrix A = (aij) is a diagonal matrix if aij = 0 for all i not equal to j.

  1. If both (A) and (R) are true, and (R) is the correct explanation of (A).    
  2. If both (A) and (R) are true but (R) is not the correct explanation of (A).    
  3. If (A) is true but (R) is false.    
  4. If (A) is false but (R) is true.    

The correct answer is: If both (A) and (R) are true, and (R) is the correct explanation of (A).

Related Questions to study

General
maths-

Consider open vertical bar table row cell a subscript 1 end subscript end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript end cell end table close vertical bar= – 1, where ai. aj + bi. bj + ci.cj = open square brackets table row cell 0 semicolon end cell cell i not equal to j end cell row cell 1 semicolon end cell cell i equals j end cell end table close and i, j = 1,2,3
Assertion(A) : The value of open vertical bar table row cell a subscript 1 end subscript plus 1 end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript plus 1 end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript plus 1 end cell end table close vertical bar is equal to zero
Reason(R) : If A be square matrix of odd order such that AAT = I, then | A + I | = 0

Consider open vertical bar table row cell a subscript 1 end subscript end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript end cell end table close vertical bar= – 1, where ai. aj + bi. bj + ci.cj = open square brackets table row cell 0 semicolon end cell cell i not equal to j end cell row cell 1 semicolon end cell cell i equals j end cell end table close and i, j = 1,2,3
Assertion(A) : The value of open vertical bar table row cell a subscript 1 end subscript plus 1 end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript plus 1 end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript plus 1 end cell end table close vertical bar is equal to zero
Reason(R) : If A be square matrix of odd order such that AAT = I, then | A + I | = 0

maths-General
General
maths-

Assertion(A) : The inverse of the matrix A = [Aij]n × n where aij = 0, i greater or equal than j is B = [aij–1]n× n 
Reason(R): The inverse of singular matrix does not exist

Assertion(A) : The inverse of the matrix A = [Aij]n × n where aij = 0, i greater or equal than j is B = [aij–1]n× n 
Reason(R): The inverse of singular matrix does not exist

maths-General
General
maths-

Assertion : The product of two diagonal matrices of order 3 × 3 is also a diagonal matrix
Reason : matrix multiplicationis non commutative

Assertion : The product of two diagonal matrices of order 3 × 3 is also a diagonal matrix
Reason : matrix multiplicationis non commutative

maths-General
parallel
General
maths-

Assertion : If A is a skew symmetric of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det straight A to the power of straight prime = det (–straight A to the power of straight prime)

Assertion : If A is a skew symmetric of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det straight A to the power of straight prime = det (–straight A to the power of straight prime)

maths-General
General
maths-

Assertion : There are only finitely many 2 × 2 matrices which commute with the matrix open square brackets table row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets
Reason : If A is non-singular then it commutes with I, adj A and A–1

Assertion : There are only finitely many 2 × 2 matrices which commute with the matrix open square brackets table row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets
Reason : If A is non-singular then it commutes with I, adj A and A–1

maths-General
General
Maths-

Statement-I The equation square root of 3 cos space x minus sin space x equals 2 has exactly one solution in [0, 2straight pi].

Statement-II For equations of type a cos space theta plus b sin space theta equals c to have real solutions in left square bracket 0 comma 2 pi right square bracket comma vertical line c vertical line less or equal than square root of a squared plus b squared end root  should hold true.

In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not , is same like assertion and reason. Here, Start solving first Statement and try to prove it . Then solve the Statement-II . Remember cos a cosb -sin a sinb = cos ( a + b ) and sin a cosb + cosa sinb = sin( a+ b) .

Statement-I The equation square root of 3 cos space x minus sin space x equals 2 has exactly one solution in [0, 2straight pi].

Statement-II For equations of type a cos space theta plus b sin space theta equals c to have real solutions in left square bracket 0 comma 2 pi right square bracket comma vertical line c vertical line less or equal than square root of a squared plus b squared end root  should hold true.

Maths-General

In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not , is same like assertion and reason. Here, Start solving first Statement and try to prove it . Then solve the Statement-II . Remember cos a cosb -sin a sinb = cos ( a + b ) and sin a cosb + cosa sinb = sin( a+ b) .

parallel
General
maths-

Assertion : If A is a skew symmetric matrix of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det straight A to the power of straight prime  = det (–straight A to the power of straight prime)

Assertion : If A is a skew symmetric matrix of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det straight A to the power of straight prime  = det (–straight A to the power of straight prime)

maths-General
General
maths-

Assertion : There are only finitely many 2 × 2 matrices which commute with the matrix
open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets
Reason : If A is non-singular then it commutes with I, adj A and A–1

Assertion : There are only finitely many 2 × 2 matrices which commute with the matrix
open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets
Reason : If A is non-singular then it commutes with I, adj A and A–1

maths-General
General
maths-

Consider the system of equationsx – 2y + 3z = –1–x + y – 2z = kx – 3y + 4z = 1
Assertion : The system of equations has no solution for k not equal to3.and
Reason : The determinant open vertical bar table row 1 3 cell negative 1 end cell row cell negative 1 end cell cell negative 2 end cell k row 1 4 1 end table close vertical barnot equal to 0, for k not equal to 3.

Consider the system of equationsx – 2y + 3z = –1–x + y – 2z = kx – 3y + 4z = 1
Assertion : The system of equations has no solution for k not equal to3.and
Reason : The determinant open vertical bar table row 1 3 cell negative 1 end cell row cell negative 1 end cell cell negative 2 end cell k row 1 4 1 end table close vertical barnot equal to 0, for k not equal to 3.

maths-General
parallel
General
maths-

Assertion : If a, b, c are distinct and x, y, z are not all zero given that ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 then a + b + c not equal to 0
Reason : a2 + b2 + c2 > ab + bc + ca if a, b, c are distinct

Assertion : If a, b, c are distinct and x, y, z are not all zero given that ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 then a + b + c not equal to 0
Reason : a2 + b2 + c2 > ab + bc + ca if a, b, c are distinct

maths-General
General
maths-

Assertion : If A is a skew symmetric of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det A to the power of straight prime = det (–A to the power of straight prime)

Assertion : If A is a skew symmetric of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det A to the power of straight prime = det (–A to the power of straight prime)

maths-General
General
maths-

Assertion : There are only finitely many 2 ×2 matrices which commute with the matrix open square brackets table row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets
Reason : If A is non-singular then it commutes with I, Adj A and A–1.

Assertion : There are only finitely many 2 ×2 matrices which commute with the matrix open square brackets table row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets
Reason : If A is non-singular then it commutes with I, Adj A and A–1.

maths-General
parallel
General
maths-

Let a matrix A =open square brackets table row 1 1 row 0 1 end table close square brackets & P =open square brackets table row cell fraction numerator square root of 3 over denominator 2 end fraction end cell cell fraction numerator 1 over denominator 2 end fraction end cell row cell negative fraction numerator 1 over denominator 2 end fraction end cell cell fraction numerator square root of 3 over denominator 2 end fraction end cell end table close square brackets Q = PAPT where PT is transpose of matrix P. Find PT Q2005 P is

Let a matrix A =open square brackets table row 1 1 row 0 1 end table close square brackets & P =open square brackets table row cell fraction numerator square root of 3 over denominator 2 end fraction end cell cell fraction numerator 1 over denominator 2 end fraction end cell row cell negative fraction numerator 1 over denominator 2 end fraction end cell cell fraction numerator square root of 3 over denominator 2 end fraction end cell end table close square brackets Q = PAPT where PT is transpose of matrix P. Find PT Q2005 P is

maths-General
General
maths-

Let A = open square brackets table row 1 0 0 row 0 1 1 row 0 cell negative 2 end cell 4 end table close square brackets & I = open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets and A1 = fraction numerator 1 over denominator 6 end fraction [A2 + cA + dI], find ordered pair (c, d) ?]

Let A = open square brackets table row 1 0 0 row 0 1 1 row 0 cell negative 2 end cell 4 end table close square brackets & I = open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets and A1 = fraction numerator 1 over denominator 6 end fraction [A2 + cA + dI], find ordered pair (c, d) ?]

maths-General
General
maths-

open square brackets table row alpha 2 row 2 alpha end table close square brackets = A & | A3 | = 125, then alpha is -

open square brackets table row alpha 2 row 2 alpha end table close square brackets = A & | A3 | = 125, then alpha is -

maths-General
parallel

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