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Maths-

General

Easy

Question

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.

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 false but (R) is true.

Related Questions to study

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

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

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) ?]

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) ?]

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

parallel

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

Let A be a square matrix all of whose entries are integers. Then which one of the following is true ?

Let A be a square matrix all of whose entries are integers. Then which one of the following is true ?

Let A = $open square brackets table row 5 cell 5 alpha end cell alpha row 0 alpha cell 5 alpha end cell row 0 0 5 end table close square brackets$ If |A2| = 25, then | $alpha$ | equals

Let A = $open square brackets table row 5 cell 5 alpha end cell alpha row 0 alpha cell 5 alpha end cell row 0 0 5 end table close square brackets$ If |A2| = 25, then | $alpha$ | equals

parallel

Let A = $open parentheses table row 1 2 row 3 4 end table close parentheses$ and B = $open parentheses table row a 0 row 0 b end table close parentheses$ , a, b $element of$ N. Then

Let A = $open parentheses table row 1 2 row 3 4 end table close parentheses$ and B = $open parentheses table row a 0 row 0 b end table close parentheses$ , a, b $element of$ N. Then

If A and B are square matrices of size n × n such that A² – B² = (A – B) (A + B), then which of the following will be always true –

If A and B are square matrices of size n × n such that A² – B² = (A – B) (A + B), then which of the following will be always true –

If A = $open square brackets table row 1 0 row 1 1 end table close square brackets$ and I = $open square brackets table row 1 0 row 0 1 end table close square brackets$ , then which one of the following holds for all n $greater or equal than$ 1, by the principle of mathematical induction -

If A = $open square brackets table row 1 0 row 1 1 end table close square brackets$ and I = $open square brackets table row 1 0 row 0 1 end table close square brackets$ , then which one of the following holds for all n $greater or equal than$ 1, by the principle of mathematical induction -

parallel

If A² – A + I = 0, then the inverse of A is

If A² – A + I = 0, then the inverse of A is

Let A = $open parentheses table row 1 cell negative 1 end cell 1 row 2 1 cell negative 3 end cell row 1 1 1 end table close parentheses$ and (10) B = $open parentheses table row 4 2 2 row cell negative 5 end cell 0 alpha row 1 cell negative 2 end cell 3 end table close parentheses$ . If B is the inverse of matrix A, then $alpha$ is

Let A = $open parentheses table row 1 cell negative 1 end cell 1 row 2 1 cell negative 3 end cell row 1 1 1 end table close parentheses$ and (10) B = $open parentheses table row 4 2 2 row cell negative 5 end cell 0 alpha row 1 cell negative 2 end cell 3 end table close parentheses$ . If B is the inverse of matrix A, then $alpha$ is

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Let A = $open parentheses table row 0 0 cell negative 1 end cell row 0 cell negative 1 end cell 0 row cell negative 1 end cell 0 0 end table close parentheses$ . The only correct statement about the matrix A is

parallel

Bleaching powder has the molecular formula:

Bleaching powder has the molecular formula:

chemistry-General

If A = $open square brackets table row a b row b a end table close square brackets$ and A² = $open square brackets table row alpha beta row beta alpha end table close square brackets$ , then

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If A= $open square brackets table row 1 2 3 row 3 1 2 row 2 3 1 end table close square brackets$ , B = $open square brackets table row cell negative 5 end cell 7 1 row 1 cell negative 5 end cell 7 row 7 1 cell negative 5 end cell end table close square brackets$ then AB =

If A= $open square brackets table row 1 2 3 row 3 1 2 row 2 3 1 end table close square brackets$ , B = $open square brackets table row cell negative 5 end cell 7 1 row 1 cell negative 5 end cell 7 row 7 1 cell negative 5 end cell end table close square brackets$ then AB =

parallel

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