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 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| =

3
3/2
–3
2

The correct answer is: 3

U₁ = $open square brackets table row cell a subscript 1 end subscript end cell row cell b subscript 1 end subscript end cell row cell c subscript 1 end subscript end cell end table close square brackets$ , U₂ = $open square brackets table row cell a subscript 2 end subscript end cell row cell b subscript 2 end subscript end cell row cell c subscript 2 end subscript end cell end table close square brackets$ , U₃ = $open square brackets table row cell a subscript 3 end subscript end cell row cell b subscript 3 end subscript end cell row cell c subscript 3 end subscript end cell end table close square brackets$
A. U₁ = $open square brackets table attributes columnalign left left left columnspacing 1em end attributes row cell 1 0 0 end cell row cell 2 1 0 end cell row cell 3 2 1 end cell end table close square brackets$ $open square brackets table row cell a subscript 1 end subscript end cell row cell b subscript 1 end subscript end cell row cell c subscript 1 end subscript end cell end table close square brackets$ = $open square brackets table row 1 row 0 row 0 end table close square brackets$
= $open square brackets table row cell a subscript 1 end subscript end cell row cell 2 a subscript 1 end subscript plus b subscript 1 end subscript end cell row cell 3 a subscript 1 end subscript plus 2 b subscript 1 end subscript plus c subscript 1 end subscript end cell end table close square brackets$ = $open square brackets table row 1 row 0 row 0 end table close square brackets$
$rightwards double arrow$ a₁ = 1, b₁ = –2, c₁ = 1
$rightwards double arrow$ U₁ = $open square brackets table row 1 row cell – 2 end cell row 1 end table close square brackets$
Similarly U₂ = $open square brackets table row 2 row cell – 1 end cell row cell – 4 end cell end table close square brackets$ & U₃ = $open square brackets table row 2 row cell – 1 end cell row cell – 3 end cell end table close square brackets$
U = $open square brackets table row 1 2 2 row cell – 2 end cell cell – 1 end cell cell – 1 end cell row 1 cell – 4 end cell cell – 3 end cell end table close square brackets$ $rightwards double arrow$ |U| = 3

Related Questions to study

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 :

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 =

parallel

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 -

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 =

parallel

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 =

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

parallel

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:

Which of the following has no action with starch solution?

Which of the following has no action with starch solution?

chemistry-General

parallel

Let three matrices A = $open square brackets table row 2 1 row 4 1 end table close square brackets$ ; B = $open square brackets table row 3 4 row 2 3 end table close square brackets$ and C = $open square brackets table row 3 cell – 4 end cell row cell – 2 end cell 3 end table close square brackets$ then t_r(A) + t_r $open parentheses fraction numerator A B C over denominator 2 end fraction close parentheses$ + t_r $open parentheses fraction numerator A left parenthesis B C right parenthesis to the power of 2 end exponent over denominator 4 end fraction close parentheses$ + t_r $open parentheses fraction numerator A left parenthesis B C right parenthesis to the power of 3 end exponent over denominator 8 end fraction close parentheses$ + ....... + $straight infinity$ =

Let three matrices A = $open square brackets table row 2 1 row 4 1 end table close square brackets$ ; B = $open square brackets table row 3 4 row 2 3 end table close square brackets$ and C = $open square brackets table row 3 cell – 4 end cell row cell – 2 end cell 3 end table close square brackets$ then t_r(A) + t_r $open parentheses fraction numerator A B C over denominator 2 end fraction close parentheses$ + t_r $open parentheses fraction numerator A left parenthesis B C right parenthesis to the power of 2 end exponent over denominator 4 end fraction close parentheses$ + t_r $open parentheses fraction numerator A left parenthesis B C right parenthesis to the power of 3 end exponent over denominator 8 end fraction close parentheses$ + ....... + $straight infinity$ =

A = $open square brackets table row 1 cell tan invisible function application x end cell row cell – tan invisible function application x end cell 1 end table close square brackets$ then let us define a function f(x) = dt.(A^TA^–1) then which of the following can not be the value of $fraction numerator f open parentheses f open parentheses f open parentheses f............ f left parenthesis x right parenthesis close parentheses close parentheses close parentheses over denominator n t i m e s end fraction$ is (n ≥ 2)

A = $open square brackets table row 1 cell tan invisible function application x end cell row cell – tan invisible function application x end cell 1 end table close square brackets$ then let us define a function f(x) = dt.(A^TA^–1) then which of the following can not be the value of $fraction numerator f open parentheses f open parentheses f open parentheses f............ f left parenthesis x right parenthesis close parentheses close parentheses close parentheses over denominator n t i m e s end fraction$ is (n ≥ 2)

Let A, B, C, D be (not necessarily square) real matrices such that A^T = BCD; B^T = CDA; C^T = DAB and D^T = ABC for the matrix S = ABCD, consider the two statements.
I. S³ = S
II. S² = S⁴

Let A, B, C, D be (not necessarily square) real matrices such that A^T = BCD; B^T = CDA; C^T = DAB and D^T = ABC for the matrix S = ABCD, consider the two statements.
I. S³ = S
II. S² = S⁴

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.