If in a square matrix A- a ij - we find that a ij - a ji - i-j - then A is

The correct option is A Symmetric #160;Given square matrix #160; A={ a _ ij } and #160; a _ ij = a _ ji We know that in a square matrix if #160 ...

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Standard X

Mathematics

Triangular Matrix

If in a squar...

Question

If in a square matrix $A = [a_{i j}]$ , we find that $a_{i j} = a_{j i} \forall i, j$ , then $A$ is

Symmetric

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Skew Symmetric

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Idempotent

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none of these

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Solution

The correct option is A Symmetric
Given square matrix $A = {a_{i j}}$ and $a_{i j} = a_{j i}$
We know that in a square matrix if $a_{i j} = a_{j i}$ for all i and j, then the matrix is a symmetric matrix
So A is a symmetric matrix

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Similar questions

Q. If in a square matrix

A = [a_{i j}]

, we find that

a_{i j} = a_{j i} \forall i, j

then

A

is a:

Q. A square matrix A such that

A^{θ} = A

is called Hermitian matrix i.e.

a_{i j} =_{j i}

for all values of i and j and a square matrix A such that

A^{θ} = - A

is called skew -Hermitain matrix i.e.

a_{i j} = -_{j i}

for all values of i and j. where is conjugate transpose matrix.
Let

f : M \to {1, - 1}, M

is set of all hermitian or Skew–hermitian matrixes, be a function defined as

f (A) = {\begin{matrix} 1 & A i s h e r m i t i a n - 1 & A i s s k e w h e r m i t i a n \end{matrix}

Let

Y = A^{n}

Q. A square matrix

A = [a_{i j}]_{n \times n}

, if

a_{i j} = 0

for

i > j,

then that matrix is known as

Q. Assertion :The determinants of a matrix

A = [a_{i j}]_{5 \times 5}

where

a_{i j} + a_{j i} = 0

for each i and j is zero.
Because Reason: The determinant of a skew symmetric matrix of odd order is zero.

Q. A square matrix A such that

A^{θ} = A

is called Hermitian matrix i.e.

a_{i j} =_{j i}

for all values of i and j and a square matrix A such that

A^{θ} = - A

is called skew –Hermitian matrix i.e.

a_{i j} = -_{j i}

for all values of i and j. where

A^{θ}

is conjugate transpose matrix.
Let f : M

\to

{1,-1}, M is set of all hermitian or Skew–hermitian matrices, be a function defined as

f (A) = {\begin{matrix} 1 A i s h e r m i t i a n - 1 A i s s k e w h e r m i t i a n \end{matrix}

f (A - A^{θ}) =

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If in a square matrix A=[aij], we find that aij=aji∀i,j , then A is

The correct option is A Symmetric Given square matrix A={aij} and aij=ajiWe know that in a square matrix if aij=aji for all i and j, then the matrix is a symmetric matrixSo A is a symmetric matrix

If in a square matrix $A = [a_{i j}]$ , we find that $a_{i j} = a_{j i} \forall i, j$ , then $A$ is

The correct option is A Symmetric
Given square matrix $A = {a_{i j}}$ and $a_{i j} = a_{j i}$
We know that in a square matrix if $a_{i j} = a_{j i}$ for all i and j, then the matrix is a symmetric matrix
So A is a symmetric matrix