Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Maths-

General

Easy

Question

Let $omega$ $not equal to 1$ be a cube root of unity and $S$ be the set of all non‐singular matrices of the form $open square brackets table row 1 a b row omega 1 c row cell omega to the power of 2 end exponent end cell omega 1 end table close square brackets$ , where each of a, b and $c$ is either $omega text or end text omega to the power of 2 end exponent$ Then the number of distinct matrices in the set $S$ is‐

2
6
4
8

The correct answer is: 2

Given questionLet $omega$ $not equal to 1$ be a cube root of unity and $S$ be the set of all non‐singular matrices of the form of matrix
Given Matrix
$open square brackets table row 1 a b row omega 1 c row cell omega squared end cell omega 1 end table close square brackets$
Determinant D is not equal to 0
$1 left parenthesis 1 minus omega c right parenthesis minus a left parenthesis omega minus omega squared c right parenthesis plus b left parenthesis omega squared space space minus omega squared right parenthesis not equal to 0 1 minus omega left parenthesis a plus c right parenthesis plus a c omega squared space not equal to 0 a plus c not equal to negative 1 a c not equal to 1 S o comma space a equals omega space o r space omega squared space space space a n d space c equals omega space o r space omega squared space space space space I f space c equals omega squared space comma space t h e n space capital delta equals 0. space S o space c not equal to omega squared space space space S o comma space c equals omega space space space S o comma space b y space e q u a t i o n space left parenthesis 1 right parenthesis comma space a not equal to omega squared space space. space space H e n c e comma space space a equals omega. space space S i n c e comma space t h e space d e t e r m i n a n t space v a l u e space i s space i n d e p e n d e n t space o f space b space. space S o space b space c a n space b e space omega space o r space omega squared space space. space space H e n c e comma space c equals omega comma a equals omega comma b equals omega space o r space omega squared space space. space space S o comma space n u m b e r space o f space m a t r i c e s space f o r m e d space w i l l space b e space 2.$

Hence no of matrices formed = 2

Related Questions to study

Let A be a $2 cross times 2$ matrix with non‐zero entries and let $A to the power of 2 end exponent equals I$ , where I is $2 cross times 2$ identity matrix Defi ne Tr(A) $equals s u m$ of diagonal elements $o f A$ and $vertical line A vertical line equals$ determinant of matrix A
Statement 1: Tr(A) $equals 0.$
Statement 2: $vertical line A vertical line equals 1.$

Hence option C is suitable option

Let A be a $2 cross times 2$ matrix with non‐zero entries and let $A to the power of 2 end exponent equals I$ , where I is $2 cross times 2$ identity matrix Defi ne Tr(A) $equals s u m$ of diagonal elements $o f A$ and $vertical line A vertical line equals$ determinant of matrix A
Statement 1: Tr(A) $equals 0.$
Statement 2: $vertical line A vertical line equals 1.$

Hence option C is suitable option

If A $equals left square bracket subscript 1 end subscript superscript 1 end superscript$ $01 right square bracket$ and I $equals left square bracket subscript 0 end subscript superscript 1 end superscript$ $01 right square bracket$ , then which one of the following holds for all $n greater or equal than 1$ , (by the principal of mathematical induction)

If A $equals left square bracket subscript 1 end subscript superscript 1 end superscript$ $01 right square bracket$ and I $equals left square bracket subscript 0 end subscript superscript 1 end superscript$ $01 right square bracket$ , then which one of the following holds for all $n greater or equal than 1$ , (by the principal of mathematical induction)

The equation $open curly brackets table attributes columnalign left end attributes row 1 2 2 row 1 3 4 row 3 4 k end table close curly brackets open curly brackets table attributes columnalign left end attributes row x row y row z end table close curly brackets equals open curly brackets table attributes columnalign left end attributes row 0 row 0 row 0 end table close curly brackets$ has a solution for (x, y, z) besides (0,0,0) The value of k equals

The equation $open curly brackets table attributes columnalign left end attributes row 1 2 2 row 1 3 4 row 3 4 k end table close curly brackets open curly brackets table attributes columnalign left end attributes row x row y row z end table close curly brackets equals open curly brackets table attributes columnalign left end attributes row 0 row 0 row 0 end table close curly brackets$ has a solution for (x, y, z) besides (0,0,0) The value of k equals

parallel

If $F left parenthesis alpha right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell c o s alpha end cell cell negative s i n alpha end cell 0 row cell s i n alpha end cell cell c o s alpha end cell 0 row 0 0 1 end table close curly brackets a n d G left parenthesis beta right parenthesis equals open square brackets table row cell c o s invisible function application beta end cell 0 cell s i n invisible function application beta end cell row 0 1 0 row cell negative s i n invisible function application beta end cell 0 cell c o s invisible function application beta end cell end table close square brackets$ , then $left square bracket F left parenthesis alpha right parenthesis G left parenthesis beta right parenthesis right square bracket to the power of negative 1 end exponent equals$

If $F left parenthesis alpha right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell c o s alpha end cell cell negative s i n alpha end cell 0 row cell s i n alpha end cell cell c o s alpha end cell 0 row 0 0 1 end table close curly brackets a n d G left parenthesis beta right parenthesis equals open square brackets table row cell c o s invisible function application beta end cell 0 cell s i n invisible function application beta end cell row 0 1 0 row cell negative s i n invisible function application beta end cell 0 cell c o s invisible function application beta end cell end table close square brackets$ , then $left square bracket F left parenthesis alpha right parenthesis G left parenthesis beta right parenthesis right square bracket to the power of negative 1 end exponent equals$

Matrix A is such that $A to the power of 2 end exponent equals 2 A minus I$ , where I is the identity matrix The for $n greater or equal than 2 comma$ $A to the power of n end exponent equals$

Matrix A is such that $A to the power of 2 end exponent equals 2 A minus I$ , where I is the identity matrix The for $n greater or equal than 2 comma$ $A to the power of n end exponent equals$

Statement‐I:: If $f left parenthesis x right parenthesis$ is increasing function with concavity upwards, then concavity of $f to the power of negative 1 end exponent left parenthesis x right parenthesis$ is also upwards
Statement‐II:: If $f left parenthesis x right parenthesis$ is decreasing function with concavity upwards, then concavity of $f to the power of negative 1 end exponent left parenthesis x right parenthesis$ is also upwards.

Statement‐I:: If $f left parenthesis x right parenthesis$ is increasing function with concavity upwards, then concavity of $f to the power of negative 1 end exponent left parenthesis x right parenthesis$ is also upwards
Statement‐II:: If $f left parenthesis x right parenthesis$ is decreasing function with concavity upwards, then concavity of $f to the power of negative 1 end exponent left parenthesis x right parenthesis$ is also upwards.

parallel

Statement‐I:: $fraction numerator 2 e to the power of x subscript 1 end subscript end exponent plus e to the power of x subscript 2 end subscript end exponent over denominator 3 end fraction greater than e left parenthesis fraction numerator 2 x subscript 1 to the power of plus X end exponent 2 end subscript over denominator 3 end fraction right parenthesis$ , where $e$ is Napier $’$ sconstant.
Statement‐II:: If f $e subscript i left parenthesis x right parenthesis end subscript$ and $f C C left parenthesis x right parenthesis$ is positive $for all x element of R$ , then $f left parenthesis x right parenthesis$ increases with concavity up $for all x element of R$ and any chord lies above the curve.

Statement‐I:: $fraction numerator 2 e to the power of x subscript 1 end subscript end exponent plus e to the power of x subscript 2 end subscript end exponent over denominator 3 end fraction greater than e left parenthesis fraction numerator 2 x subscript 1 to the power of plus X end exponent 2 end subscript over denominator 3 end fraction right parenthesis$ , where $e$ is Napier $’$ sconstant.
Statement‐II:: If f $e subscript i left parenthesis x right parenthesis end subscript$ and $f C C left parenthesis x right parenthesis$ is positive $for all x element of R$ , then $f left parenthesis x right parenthesis$ increases with concavity up $for all x element of R$ and any chord lies above the curve.

The beds of two rivers (within a certain region) are a parabola $y equals x to the power of 2 end exponent$ and a straight line y=x-2 These rivers are to be connected by a straight canal The coordinates of the ends of the shortest canal can be:

The beds of two rivers (within a certain region) are a parabola $y equals x to the power of 2 end exponent$ and a straight line y=x-2 These rivers are to be connected by a straight canal The coordinates of the ends of the shortest canal can be:

If f(x) is twice differentiable function f((1)=1, f(2)=4, f(3)=9

If f(x) is twice differentiable function f((1)=1, f(2)=4, f(3)=9

parallel

If f(x) $equals x to the power of 3 end exponent plus b x to the power of 2 end exponent plus c x plus d comma$ $0 less than b to the power of 2 end exponent less than c$ , then f(x)

If f(x) $equals x to the power of 3 end exponent plus b x to the power of 2 end exponent plus c x plus d comma$ $0 less than b to the power of 2 end exponent less than c$ , then f(x)

The set of value (s) of $a$ ’for which the function $f left parenthesis x right parenthesis equals fraction numerator a x to the power of 3 end exponent over denominator 3 end fraction plus left parenthesis a plus 2 right parenthesis x to the power of 2 end exponent plus left parenthesis a minus 1 right parenthesis x plus 2$ possess anegative point of inflection‐

The set of value (s) of $a$ ’for which the function $f left parenthesis x right parenthesis equals fraction numerator a x to the power of 3 end exponent over denominator 3 end fraction plus left parenthesis a plus 2 right parenthesis x to the power of 2 end exponent plus left parenthesis a minus 1 right parenthesis x plus 2$ possess anegative point of inflection‐

The greatest, the least values of the function, $f left parenthesis x right parenthesis equals 2 minus square root of 1 plus 2 x plus x to the power of 2 end exponent end root comma$ $x element of left square bracket 21 right square bracket$ are respectively

Hence the final answer is (2,0)

The greatest, the least values of the function, $f left parenthesis x right parenthesis equals 2 minus square root of 1 plus 2 x plus x to the power of 2 end exponent end root comma$ $x element of left square bracket 21 right square bracket$ are respectively

Hence the final answer is (2,0)

parallel

Suppose that $f$ is differentiable for allx and that $f to the power of ’ end exponent left parenthesis x right parenthesis less or equal than 2$ for all x If $f left parenthesis 1 right parenthesis equals 2$ and f(4)=8 then f(2) has the value equal to

Hence f(2)=4 is the suitable option

Suppose that $f$ is differentiable for allx and that $f to the power of ’ end exponent left parenthesis x right parenthesis less or equal than 2$ for all x If $f left parenthesis 1 right parenthesis equals 2$ and f(4)=8 then f(2) has the value equal to

Hence f(2)=4 is the suitable option

$f left parenthesis x right parenthesis equals not stretchy integral subscript blank superscript left parenthesis end superscript left parenthesis 2 minus fraction numerator 1 over denominator 1 plus x to the power of 2 end exponent end fraction minus fraction numerator 1 over denominator square root of 1 plus x to the power of 2 end exponent end root end fraction right parenthesis right parenthesis$ dxthenfis‐

$f left parenthesis x right parenthesis equals not stretchy integral subscript blank superscript left parenthesis end superscript left parenthesis 2 minus fraction numerator 1 over denominator 1 plus x to the power of 2 end exponent end fraction minus fraction numerator 1 over denominator square root of 1 plus x to the power of 2 end exponent end root end fraction right parenthesis right parenthesis$ dxthenfis‐

The tangent to the curve $3 x y to the power of 2 end exponent minus 2 x to the power of 2 end exponent y equals 1$ at (1,1) meets the curve again at the point‐

The final answer is option B

The tangent to the curve $3 x y to the power of 2 end exponent minus 2 x to the power of 2 end exponent y equals 1$ at (1,1) meets the curve again at the point‐

The final answer is option B

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.