Maths-

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Question

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 a_i. a_j + b_i. b_j + c_i.c_j = $open square brackets table row cell 0 semicolon end cell cell i not equal to j end cell row cell 1 semicolon end cell cell i equals j end cell end table close$ 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 AA^T = I, then | A + I | = 0

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

The correct answer is: If (A) is true but (R) is false.

AA^T = I
|A| |A^T| = 1 $not stretchy rightwards double arrow$ |A|² = 1 $not stretchy rightwards double arrow$ |A| = ± 1
|A + I| = |A + AA^T| = |A|³ |I + A^T|
= A³ | I + A |
when |A| = – 1 $not stretchy rightwards double arrow$ |A + I| = – |A + I|
$not stretchy rightwards double arrow$ 2|A + I| = 0 $not stretchy rightwards double arrow$ |A + I| = 0
when |A| = 1 $not stretchy rightwards double arrow$ |A + I| = |A + I|
$therefore$ Reason is false
Let A = $open square brackets table row cell a subscript 1 end subscript end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript end cell end table close square brackets$
A + I = $open square brackets table row cell a subscript 1 end subscript plus 1 end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript plus 1 end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript plus 1 end cell end table close square brackets$
AA^T = $open square brackets table row cell a subscript 1 end subscript end cell cell a subscript 2 end subscript end cell cell a subscript 3 end subscript end cell row cell b subscript 1 end subscript end cell cell b subscript 2 end subscript end cell cell b subscript 3 end subscript end cell row cell c subscript 1 end subscript end cell cell c subscript 2 end subscript end cell cell c subscript 3 end subscript end cell end table close square brackets equals open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets$ = 1 |AA^T| = 1 $not stretchy rightwards double arrow$ |A| = ± 1 but |A| = – 1
$therefore$ |A + I| = |A + AA^T| = |A|³ |I + A^T| = – |A + I|
$therefore$ |A + I| = 0
So assertion is true.

Assertion(A) : The inverse of the matrix A = [Aij]_{n × n} where a_ij = 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

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 $straight A to the power of straight prime$ = 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

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, 2 $straight pi$ ].
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, 2 $straight pi$ ].
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

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 (– $straight A to the power of straight prime$ )

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

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 $not equal to$ 3.and
Reason : The determinant $open vertical bar table row 1 3 cell negative 1 end cell row cell negative 1 end cell cell negative 2 end cell k row 1 4 1 end table close vertical bar$ $not equal to$ 0, for k $not equal to$ 3.

maths-General

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

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 (– $A to the power of straight prime$ )

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.

maths-General

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 = PAP^T where P^T 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 A–1 = $fraction numerator 1 over denominator 6 end fraction$ [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 -

maths-General

General

maths-

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ , B = $open square brackets table row 1 0 row 5 1 end table close square brackets$ and A² = B, then

maths-General

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

The correct answer is: If (A) is true but (R) is false.

Related Questions to study

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

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

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 (– $straight A to the power of straight prime$ )

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

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 (– $straight A to the power of straight prime$ )

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

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 (– $A to the power of straight prime$ )

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.

$open square brackets table row alpha 2 row 2 alpha end table close square brackets$ = 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 -

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ , B = $open square brackets table row 1 0 row 5 1 end table close square brackets$ and A² = B, then

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ , B = $open square brackets table row 1 0 row 5 1 end table close square brackets$ and A² = B, then

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Consider = – 1, where ai. aj + bi. bj + ci.cj = and i, j = 1,2,3Assertion(A) : The value of is equal to zeroReason(R) : If A be square matrix of odd order such that AAT = I, then | A + I | = 0

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

The correct answer is: If (A) is true but (R) is false.

Related Questions to study

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

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

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

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 = det (–)

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 = det (–)

Assertion : There are only finitely many 2 × 2 matrices which commute with the matrix 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 Reason : If A is non-singular then it commutes with I, adj A and A–1

Statement-I The equation has exactly one solution in [0, 2].Statement-II For equations of type to have real solutions in should hold true.

Statement-I The equation has exactly one solution in [0, 2].Statement-II For equations of type to have real solutions in should hold true.

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 = det (–)

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 = det (–)

Assertion : There are only finitely many 2 × 2 matrices which commute with the matrixReason : 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 matrixReason : If A is non-singular then it commutes with I, adj A and A–1

Consider the system of equationsx – 2y + 3z = –1–x + y – 2z = kx – 3y + 4z = 1Assertion : The system of equations has no solution for k 3.andReason : The determinant 0, for k 3.

Consider the system of equationsx – 2y + 3z = –1–x + y – 2z = kx – 3y + 4z = 1Assertion : The system of equations has no solution for k 3.andReason : The determinant 0, for k 3.

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 0Reason : 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 0Reason : a2 + b2 + c2 > ab + bc + ca if a, b, c are distinct

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 = det (–)

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 = det (–)

Assertion : There are only finitely many 2 ×2 matrices which commute with the matrix 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 Reason : If A is non-singular then it commutes with I, Adj A and A–1.

Let a matrix A = & P = Q = PAPT where PT is transpose of matrix P. Find PT Q2005 P is

Let a matrix A = & P = Q = PAPT where PT is transpose of matrix P. Find PT Q2005 P is

Let A = & I = and A–1 = [A2 + cA + dI], find ordered pair (c, d) ?]

Let A = & I = and A–1 = [A2 + cA + dI], find ordered pair (c, d) ?]

= A & | A3 | = 125, then is -

= A & | A3 | = 125, then is -

If A = , B = and A2 = B, then

If A = , B = and A2 = B, then

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

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

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

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 (– $straight A to the power of straight prime$ )

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

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 (– $straight A to the power of straight prime$ )

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

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 (– $A to the power of straight prime$ )

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.

$open square brackets table row alpha 2 row 2 alpha end table close square brackets$ = 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 -

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ , B = $open square brackets table row 1 0 row 5 1 end table close square brackets$ and A² = B, then

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ , B = $open square brackets table row 1 0 row 5 1 end table close square brackets$ and A² = B, then