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Maths-

General

Easy

Question

If A = $open square brackets table row 1 cell 1 divided by 2 end cell row 0 1 end table close square brackets$ then A⁶⁴ is :

$[\begin{matrix} 1 & 32 \\ 32 & 1 \end{matrix}]$
$[\begin{matrix} 1 & 0 \\ 32 & 1 \end{matrix}]$
$[\begin{matrix} 1 & 32 \\ 0 & 1 \end{matrix}]$
None

The correct answer is: $open square brackets table row 1 32 row 0 1 end table close square brackets$

A² = $open square brackets table row 1 cell 1 divided by 2 end cell row 0 1 end table close square brackets$ $open square brackets table row 1 cell 1 divided by 2 end cell row 0 1 end table close square brackets$ = $open square brackets table row 1 1 row 0 1 end table close square brackets$
A³ = $open square brackets table attributes columnalign left left columnspacing 1em end attributes row cell 1 1 end cell row cell 0 1 end cell end table close square brackets$ $open square brackets table row 1 cell 1 divided by 2 end cell row 0 1 end table close square brackets$ = $open square brackets table row 1 cell 3 divided by 2 end cell row 0 1 end table close square brackets$
$therefore$ A⁶⁴ = $open square brackets table row 1 32 row 0 1 end table close square brackets$

Related Questions to study

Which of the following is not a disproportionation reaction?
I) $NH subscript 4 NO subscript 3 not stretchy rightwards arrow with straight capital delta on top straight N subscript 2 straight O plus straight H subscript 2 straight O$
II) $straight P subscript 4 not stretchy rightwards arrow with straight capital delta on top PH subscript 3 plus HPO subscript 2 superscript ⊖$
III) $PCl subscript 5 not stretchy rightwards arrow with straight capital delta on top PCl subscript 3 plus Cl subscript 2$
IV) $IO subscript 3 superscript ⊖ plus straight I to the power of ⊖ not stretchy ⟶ straight I subscript 2$

Which of the following is not a disproportionation reaction?
I) $NH subscript 4 NO subscript 3 not stretchy rightwards arrow with straight capital delta on top straight N subscript 2 straight O plus straight H subscript 2 straight O$
II) $straight P subscript 4 not stretchy rightwards arrow with straight capital delta on top PH subscript 3 plus HPO subscript 2 superscript ⊖$
III) $PCl subscript 5 not stretchy rightwards arrow with straight capital delta on top PCl subscript 3 plus Cl subscript 2$
IV) $IO subscript 3 superscript ⊖ plus straight I to the power of ⊖ not stretchy ⟶ straight I subscript 2$

chemistry-General

Which is not true for ozone?

Which is not true for ozone?

chemistry-General

If A & B are two square matrices such that B = A^–1BA, then (A + B)² equals-

If A & B are two square matrices such that B = A^–1BA, then (A + B)² equals-

parallel

If A( $theta$ ) = $open square brackets table row 1 cell tan invisible function application theta end cell row cell negative tan invisible function application theta end cell 1 end table close square brackets$ and AB = I, then (sec² $theta$ ) B is equal to-

If A( $theta$ ) = $open square brackets table row 1 cell tan invisible function application theta end cell row cell negative tan invisible function application theta end cell 1 end table close square brackets$ and AB = I, then (sec² $theta$ ) B is equal to-

If A= $open square brackets table row i cell negative i end cell row cell negative i end cell i end table close square brackets$ , B = $open square brackets table row 1 cell negative 1 end cell row cell negative 1 end cell 1 end table close square brackets$ , then A⁸ equals

If A= $open square brackets table row i cell negative i end cell row cell negative i end cell i end table close square brackets$ , B = $open square brackets table row 1 cell negative 1 end cell row cell negative 1 end cell 1 end table close square brackets$ , then A⁸ equals

If $open square brackets table row alpha beta row gamma cell negative alpha end cell end table close square brackets$ is square root of I₂, then $alpha comma beta$ and $gamma$ will satisfy the relation-

If $open square brackets table row alpha beta row gamma cell negative alpha end cell end table close square brackets$ is square root of I₂, then $alpha comma beta$ and $gamma$ will satisfy the relation-

parallel

If A = $open square brackets table row 4 2 row cell negative 1 end cell 1 end table close square brackets$ , then (A –2I) (A –3I) =

If A = $open square brackets table row 4 2 row cell negative 1 end cell 1 end table close square brackets$ , then (A –2I) (A –3I) =

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 -

For the matrix A = $open square brackets table row 1 1 0 row 1 2 1 row 2 1 0 end table close square brackets$ , which is correct -

For the matrix A = $open square brackets table row 1 1 0 row 1 2 1 row 2 1 0 end table close square brackets$ , which is correct -

parallel

If X = $open square brackets table row 1 1 row 1 1 end table close square brackets$ , then Xⁿ, for n $element of$ N, is equal to -

If X = $open square brackets table row 1 1 row 1 1 end table close square brackets$ , then Xⁿ, for n $element of$ N, is equal to -

If A, B and C are three square matrices of the same size such that B = CAC^–1, then CA³C^–1 is equal to -

If A, B and C are three square matrices of the same size such that B = CAC^–1, then CA³C^–1 is equal to -

If A = $open square brackets table row i 0 row 0 i end table close square brackets$ , n $element of$ N, then A⁴ⁿ equals -

If A = $open square brackets table row i 0 row 0 i end table close square brackets$ , n $element of$ N, then A⁴ⁿ equals -

parallel

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. then P^T 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. then P^T Q²⁰⁰⁵ P is-

Let P be a non-singular matrix 1 + P + P² +….. + Pⁿ = O(O denotes the null matrix), then P^–1 is-

Let P be a non-singular matrix 1 + P + P² +….. + Pⁿ = O(O denotes the null matrix), then P^–1 is-

Let A = $open square brackets table row cell cos to the power of 2 end exponent invisible function application theta end cell cell sin invisible function application theta cos invisible function application theta end cell row cell cos invisible function application theta sin invisible function application theta end cell cell sin to the power of 2 end exponent invisible function application theta end cell end table close square brackets$ and B = $open square brackets table row cell cos to the power of 2 end exponent invisible function application phi end cell cell sin invisible function application phi cos invisible function application phi end cell row cell cos invisible function application phi sin invisible function application phi end cell cell sin to the power of 2 end exponent invisible function application phi end cell end table close square brackets$ then AB = 0, if-

Let A = $open square brackets table row cell cos to the power of 2 end exponent invisible function application theta end cell cell sin invisible function application theta cos invisible function application theta end cell row cell cos invisible function application theta sin invisible function application theta end cell cell sin to the power of 2 end exponent invisible function application theta end cell end table close square brackets$ and B = $open square brackets table row cell cos to the power of 2 end exponent invisible function application phi end cell cell sin invisible function application phi cos invisible function application phi end cell row cell cos invisible function application phi sin invisible function application phi end cell cell sin to the power of 2 end exponent invisible function application phi end cell end table close square brackets$ then AB = 0, if-

parallel

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