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

Assertion : If A is a skew symmetric matrix of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det $straight A to the power of straight prime$ = det (– $straight A to the power of straight prime$ )

If both (A) and (R) are true, and (R) is the correct explanation of (A).
If both (A) and (R) are true but (R) is not the correct explanation of (A).
If (A) is true but (R) is false.
If (A) is false but (R) is true.

The correct answer is: If (A) is true but (R) is false.

The reason R is false since
det $straight A to the power of straight prime$ = det (– $straight A to the power of straight prime$ ) is not true.
Indeed det (– $straight A to the power of straight prime$ ) = (–1)³ det $straight A to the power of straight prime$
Now as A = – $straight A to the power of straight prime$ (A is skew symmetric)
det A = det (– $straight A to the power of straight prime$ ) $not stretchy rightwards double arrow$ –det ( $straight A to the power of straight prime$ ) $not stretchy rightwards double arrow$ – det A
$not stretchy rightwards double arrow$ det A = 0
The assertion A is true.

Related Questions to study

Assertion : There are only finitely many 2 × 2 matrices which commute with the matrix
$open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets$
Reason : If A is non-singular then it commutes with I, adj A and A^–1

Assertion : There are only finitely many 2 × 2 matrices which commute with the matrix
$open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets$
Reason : If A is non-singular then it commutes with I, adj A and A^–1

Consider the system of equationsx – 2y + 3z = –1–x + y – 2z = kx – 3y + 4z = 1
Assertion : The system of equations has no solution for k $not equal to$ 3.and
Reason : The determinant $open vertical bar table row 1 3 cell negative 1 end cell row cell negative 1 end cell cell negative 2 end cell k row 1 4 1 end table close vertical bar$ $not equal to$ 0, for k $not equal to$ 3.

Consider the system of equationsx – 2y + 3z = –1–x + y – 2z = kx – 3y + 4z = 1
Assertion : The system of equations has no solution for k $not equal to$ 3.and
Reason : The determinant $open vertical bar table row 1 3 cell negative 1 end cell row cell negative 1 end cell cell negative 2 end cell k row 1 4 1 end table close vertical bar$ $not equal to$ 0, for k $not equal to$ 3.

Assertion : If a, b, c are distinct and x, y, z are not all zero given that ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 then a + b + c $not equal to$ 0
Reason : a² + b² + c² > ab + bc + ca if a, b, c are distinct

Assertion : If a, b, c are distinct and x, y, z are not all zero given that ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 then a + b + c $not equal to$ 0
Reason : a² + b² + c² > ab + bc + ca if a, b, c are distinct

parallel

Assertion : If A is a skew symmetric of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det $A to the power of straight prime$ = det (– $A to the power of straight prime$ )

Assertion : If A is a skew symmetric of order 3 then its determinant should be zero.
Reason : If A is square matrix then det A = det $A to the power of straight prime$ = det (– $A to the power of straight prime$ )

Assertion : There are only finitely many 2 ×2 matrices which commute with the matrix $open square brackets table row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets$
Reason : If A is non-singular then it commutes with I, Adj A and A^–1.

Assertion : There are only finitely many 2 ×2 matrices which commute with the matrix $open square brackets table row 1 2 row cell negative 1 end cell cell negative 1 end cell end table close square brackets$
Reason : If A is non-singular then it commutes with I, Adj A and A^–1.

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. Find PT 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. Find PT Q2005 P is

parallel

Let A = $open square brackets table row 1 0 0 row 0 1 1 row 0 cell negative 2 end cell 4 end table close square brackets$ & I = $open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets$ and A–1 = $fraction numerator 1 over denominator 6 end fraction$ [A²+ cA + dI], find ordered pair (c, d) ?]

Let A = $open square brackets table row 1 0 0 row 0 1 1 row 0 cell negative 2 end cell 4 end table close square brackets$ & I = $open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets$ and A–1 = $fraction numerator 1 over denominator 6 end fraction$ [A²+ cA + dI], find ordered pair (c, d) ?]

$open square brackets table row alpha 2 row 2 alpha end table close square brackets$ = A & | A³ | = 125, then $alpha$ is -

$open square brackets table row alpha 2 row 2 alpha end table close square brackets$ = A & | A³ | = 125, then $alpha$ is -

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ , B = $open square brackets table row 1 0 row 5 1 end table close square brackets$ and A² = B, then

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ , B = $open square brackets table row 1 0 row 5 1 end table close square brackets$ and A² = B, then

parallel

Let A be a square matrix all of whose entries are integers. Then which one of the following is true ?

Let A be a square matrix all of whose entries are integers. Then which one of the following is true ?

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

Let A = $open parentheses table row 1 2 row 3 4 end table close parentheses$ and B = $open parentheses table row a 0 row 0 b end table close parentheses$ , a, b $element of$ N. Then

Let A = $open parentheses table row 1 2 row 3 4 end table close parentheses$ and B = $open parentheses table row a 0 row 0 b end table close parentheses$ , a, b $element of$ N. Then

parallel

If A and B are square matrices of size n × n such that A² – B² = (A – B) (A + B), then which of the following will be always true –

If A and B are square matrices of size n × n such that A² – B² = (A – B) (A + B), then which of the following will be always true –

If A = $open square brackets table row 1 0 row 1 1 end table close square brackets$ and I = $open square brackets table row 1 0 row 0 1 end table close square brackets$ , then which one of the following holds for all n $greater or equal than$ 1, by the principle of mathematical induction -

If A = $open square brackets table row 1 0 row 1 1 end table close square brackets$ and I = $open square brackets table row 1 0 row 0 1 end table close square brackets$ , then which one of the following holds for all n $greater or equal than$ 1, by the principle of mathematical induction -

If A² – A + I = 0, then the inverse of A is

If A² – A + I = 0, then the inverse of A 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.