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

For a given matrix A = $open square brackets table row cell cos invisible function application theta end cell cell negative sin invisible function application theta end cell row cell sin invisible function application theta end cell cell cos invisible function application theta end cell end table close square brackets$ which of the following statement holds good:

A = A^–1 $\forall θ \in$ R
A is symmetric, for θ = (2n +1) $\frac{π}{2}$ , n $\in$ I
A is orthogonal matrix for θ $\in$ R
A is skew symmetric, for θ = n $π$ , n $\in$ I

The correct answer is: A is orthogonal matrix for θ $element of$ R

For θ = (2n +1) $fraction numerator pi over denominator 2 end fraction$
A= $open square brackets table row 0 cell negative-or-plus 1 end cell row cell plus-or-minus 1 end cell 0 end table close square brackets$ which is not symmetric for θ = n $pi$
A = $open square brackets table row cell plus-or-minus 1 end cell 0 row 0 cell plus-or-minus 1 end cell end table close square brackets$ which is not skew symmetric
A² = $open square brackets table row cell cos invisible function application 2 theta end cell cell negative sin invisible function application 2 theta end cell row cell sin invisible function application 2 theta end cell cell cos invisible function application 2 theta end cell end table close square brackets$ ≠ I so A ≠ A^–1

Related Questions to study

A and B are two given matrices such that the order of A is 3 × 4, If A'B and BA' are both defined then:

A and B are two given matrices such that the order of A is 3 × 4, If A'B and BA' are both defined then:

If A is an involuntary matrix given byA = $open square brackets table row 0 1 cell negative 1 end cell row 4 cell negative 3 end cell 4 row 3 cell negative 3 end cell 4 end table close square brackets$ then the inverse of $fraction numerator A over denominator 2 end fraction$ will be

If A is an involuntary matrix given byA = $open square brackets table row 0 1 cell negative 1 end cell row 4 cell negative 3 end cell 4 row 3 cell negative 3 end cell 4 end table close square brackets$ then the inverse of $fraction numerator A over denominator 2 end fraction$ will be

Matrix A = $open square brackets table row x 3 2 row 1 y 4 row 2 2 z end table close square brackets$ , If xyz = 60 and 8x + 4y + 3z = 20, then A(adj A) is equal to:

Matrix A = $open square brackets table row x 3 2 row 1 y 4 row 2 2 z end table close square brackets$ , If xyz = 60 and 8x + 4y + 3z = 20, then A(adj A) is equal to:

parallel

Let A = $open square brackets table row 1 0 0 row 2 1 0 row 3 2 1 end table close square brackets$ . If U₁, U₂, U₃ are column matrices satisfying AU₁ = $open square brackets table row 1 row 0 row 0 end table close square brackets$ , AU₂ = $open square brackets table row 2 row 3 row 0 end table close square brackets$ , AU₃ = $open square brackets table row 2 row 3 row 1 end table close square brackets$ and U is 3 × 3 matrix whose columns are U₁, U₂ and U₃. Then |U| =

Let A = $open square brackets table row 1 0 0 row 2 1 0 row 3 2 1 end table close square brackets$ . If U₁, U₂, U₃ are column matrices satisfying AU₁ = $open square brackets table row 1 row 0 row 0 end table close square brackets$ , AU₂ = $open square brackets table row 2 row 3 row 0 end table close square brackets$ , AU₃ = $open square brackets table row 2 row 3 row 1 end table close square brackets$ and U is 3 × 3 matrix whose columns are U₁, U₂ and U₃. Then |U| =

If the matrix A = $open square brackets table row 8 cell – 6 end cell 2 row cell – 6 end cell 7 cell – 4 end cell row 2 cell – 4 end cell lambda end table close square brackets$ is singular, then λ equal to -

If the matrix A = $open square brackets table row 8 cell – 6 end cell 2 row cell – 6 end cell 7 cell – 4 end cell row 2 cell – 4 end cell lambda end table close square brackets$ is singular, then λ equal to -

If AB = I and B = A' then :

If AB = I and B = A' then :

parallel

If A and B are square matrices of same size and |B| $not equal to$ 0, then (B^–1 AB)⁴ =

If A and B are square matrices of same size and |B| $not equal to$ 0, then (B^–1 AB)⁴ =

If B is non-singular matrix and A is a square matrix of same size, then det (B^–1AB) =

If B is non-singular matrix and A is a square matrix of same size, then det (B^–1AB) =

If A satisfies the equation x³ – 5x² + 4x + k = 0, then A^–1 exists if -

If A satisfies the equation x³ – 5x² + 4x + k = 0, then A^–1 exists if -

parallel

If 3A = $open square brackets table row 1 2 2 row 2 1 cell – 2 end cell row x 2 y end table close square brackets$ and A is orthogonal, then x + y =

If 3A = $open square brackets table row 1 2 2 row 2 1 cell – 2 end cell row x 2 y end table close square brackets$ and A is orthogonal, then x + y =

If A, B are symmetric matrices of the same order then (AB – BA) is -

If A, B are symmetric matrices of the same order then (AB – BA) is -

If A = $open square brackets table row 4 cell x plus 2 end cell row cell 2 x – 3 end cell cell x plus 1 end cell end table close square brackets$ is symmetric, then x =

If A = $open square brackets table row 4 cell x plus 2 end cell row cell 2 x – 3 end cell cell x plus 1 end cell end table close square brackets$ is symmetric, then x =

parallel

If the rank of the matrix $open square brackets table row 4 2 cell 1 minus x end cell row 5 k 1 row 6 3 cell 1 plus x end cell end table close square brackets$ is 2 then

If the rank of the matrix $open square brackets table row 4 2 cell 1 minus x end cell row 5 k 1 row 6 3 cell 1 plus x end cell end table close square brackets$ is 2 then

If A = $open square brackets table row cell cos invisible function application theta end cell cell – sin invisible function application theta end cell row cell sin invisible function application theta end cell cell cos invisible function application theta end cell end table close square brackets$ , then A^T + A = I₂, if –

If A = $open square brackets table row cell cos invisible function application theta end cell cell – sin invisible function application theta end cell row cell sin invisible function application theta end cell cell cos invisible function application theta end cell end table close square brackets$ , then A^T + A = I₂, if –

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ and B = $open square brackets table row 1 0 row 3 1 end table close square brackets$ , then value of $alpha$ for which A² = B is:

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ and B = $open square brackets table row 1 0 row 3 1 end table close square brackets$ , then value of $alpha$ for which A² = B is:

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.