Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

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

A = $open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets$ , then A³ – 4A² – 6A is equal to -

0
A
– A
1

The correct answer is: – A

A² = $open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets$ $open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets$ = $open square brackets table row 9 8 8 row 8 9 8 row 8 8 9 end table close square brackets$
A³ = $open square brackets table row 9 8 8 row 8 9 8 row 8 8 9 end table close square brackets$ = $open square brackets table row 41 42 42 row 42 41 42 row 42 42 41 end table close square brackets$
A³ – 4A² – 6A = $open square brackets table row 41 42 42 row 42 41 42 row 42 42 41 end table close square brackets$ – $open square brackets table row 36 32 32 row 32 36 32 row 32 32 36 end table close square brackets$
$– open square brackets table row 6 12 12 row 12 6 12 row 12 12 6 end table close square brackets$ $rightwards double arrow$ $– open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets$ = – A

Related Questions to study

Inverse of the matrix $open square brackets table row 3 cell – 2 end cell cell – 1 end cell row cell – 4 end cell 1 cell – 1 end cell row 2 0 1 end table close square brackets$ is

Inverse of the matrix $open square brackets table row 3 cell – 2 end cell cell – 1 end cell row cell – 4 end cell 1 cell – 1 end cell row 2 0 1 end table close square brackets$ is

If A = $open square brackets table row 2 cell – 1 end cell row cell – 1 end cell 2 end table close square brackets$ and A² – 4A – n I = 0, then n is equal to

If A = $open square brackets table row 2 cell – 1 end cell row cell – 1 end cell 2 end table close square brackets$ and A² – 4A – n I = 0, then n is equal to

The value of x for which the matrix product $open square brackets table attributes columnalign center center center columnspacing 1em end attributes row 2 0 7 row 0 1 0 row 1 cell negative 2 end cell 1 end table close square brackets$ $open square brackets table row cell – x end cell cell 14 x end cell cell 7 x end cell row 0 1 0 row x cell – 4 x end cell cell – 2 x end cell end table close square brackets$ equal an identity matrix is :

The value of x for which the matrix product $open square brackets table attributes columnalign center center center columnspacing 1em end attributes row 2 0 7 row 0 1 0 row 1 cell negative 2 end cell 1 end table close square brackets$ $open square brackets table row cell – x end cell cell 14 x end cell cell 7 x end cell row 0 1 0 row x cell – 4 x end cell cell – 2 x end cell end table close square brackets$ equal an identity matrix is :

parallel

If A = $open square brackets table row 1 2 row 3 5 end table close square brackets$ , then A^–1 is equal to :

If A = $open square brackets table row 1 2 row 3 5 end table close square brackets$ , then A^–1 is equal to :

If A is a singular matrix, then adj A is :

If A is a singular matrix, then adj A is :

Statement - I The value of x for which (sin x + cos x)^{1 + sin 2x} = 2, when 0 ≤ x ≤ , is $straight pi divided by 4$ only.
Statement - II The maximum value of sin x + cos x occurs when x = $straight pi divided by 4$

Statement - I The value of x for which (sin x + cos x)^{1 + sin 2x} = 2, when 0 ≤ x ≤ , is $straight pi divided by 4$ only.
Statement - II The maximum value of sin x + cos x occurs when x = $straight pi divided by 4$

parallel

If [1 2 3 ]A = [4 5], the order of matrix A is :

If [1 2 3 ]A = [4 5], the order of matrix A is :

Let A and B be 3 × 3 matrices, then AB = O implies:

Let A and B be 3 × 3 matrices, then AB = O implies:

If A = $open square brackets table row 1 2 row 3 cell negative 5 end cell end table close square brackets$ , B = $open square brackets table row 1 0 row 0 2 end table close square brackets$ and X is a matrix such that A = B X. Then X equals

If A = $open square brackets table row 1 2 row 3 cell negative 5 end cell end table close square brackets$ , B = $open square brackets table row 1 0 row 0 2 end table close square brackets$ and X is a matrix such that A = B X. Then X equals

parallel

If k $open square brackets table row cell negative 1 end cell 2 2 row 2 cell negative 1 end cell 2 row 2 2 cell negative 1 end cell end table close square brackets$ is an orthogonal matrix then k is equal to

If k $open square brackets table row cell negative 1 end cell 2 2 row 2 cell negative 1 end cell 2 row 2 2 cell negative 1 end cell end table close square brackets$ is an orthogonal matrix then k is equal to

If f(α) = $open square brackets table row cell cos invisible function application alpha end cell cell sin invisible function application alpha end cell row cell negative sin invisible function application alpha end cell cell cos invisible function application alpha end cell end table close square brackets$ and $alpha comma beta comma gamma$ are angles of triangle then f( $alpha$ ). f( $beta$ ).f ( $gamma$ ) =

If f(α) = $open square brackets table row cell cos invisible function application alpha end cell cell sin invisible function application alpha end cell row cell negative sin invisible function application alpha end cell cell cos invisible function application alpha end cell end table close square brackets$ and $alpha comma beta comma gamma$ are angles of triangle then f( $alpha$ ). f( $beta$ ).f ( $gamma$ ) =

If A is idempotent and A + B = I, then which of the following true?

If A is idempotent and A + B = I, then which of the following true?

parallel

Which of the following is not a disporportionation reaction?

Which of the following is not a disporportionation reaction?

chemistry-General

If f( $alpha$ ) = $open square brackets table row cell cos invisible function application alpha end cell cell negative sin invisible function application alpha end cell 0 row cell sin invisible function application alpha end cell cell cos invisible function application alpha end cell 0 row 0 0 1 end table close square brackets$ , then $open square brackets f left parenthesis alpha right parenthesis close square brackets to the power of negative 1 end exponent$ =

If f( $alpha$ ) = $open square brackets table row cell cos invisible function application alpha end cell cell negative sin invisible function application alpha end cell 0 row cell sin invisible function application alpha end cell cell cos invisible function application alpha end cell 0 row 0 0 1 end table close square brackets$ , then $open square brackets f left parenthesis alpha right parenthesis close square brackets to the power of negative 1 end exponent$ =

If A = $open square brackets table row 0 1 2 row 1 2 3 row 3 a 1 end table close square brackets$ and A^–1 = $open square brackets table row cell 1 divided by 2 end cell cell negative 1 divided by 2 end cell cell 1 divided by 2 end cell row cell negative 4 end cell 3 c row cell 5 divided by 2 end cell cell negative 3 divided by 2 end cell cell 1 divided by 2 end cell end table close square brackets$ then value of a and c is -

If A = $open square brackets table row 0 1 2 row 1 2 3 row 3 a 1 end table close square brackets$ and A^–1 = $open square brackets table row cell 1 divided by 2 end cell cell negative 1 divided by 2 end cell cell 1 divided by 2 end cell row cell negative 4 end cell 3 c row cell 5 divided by 2 end cell cell negative 3 divided by 2 end cell cell 1 divided by 2 end cell end table close square brackets$ then value of a and c 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.