Question

Let \[A = {\left[ {{a_{ij}}} \right]_{n \times n}}\] ​ where \[{a_{ij}} = {i^2} - {j^2}\]. Show that A is a skew symmetric matrix.

Answer 1

Hint: Here, given matrix is a square matrix of order n × n, and\[{a_{ii}} = 0\], so for all ${a_{ij}}$ there exist${a_{ji}}$. Find ${a_{ji}}$ and compare it with ${a_{ij}}$, we can see that if I is equal to j then value of ${a_{ij}}$ is equal to 0. If \[{A^T} = - A\], then matrix is a symmetric matrix and if \[{A^T} = A\], then the matrix is a symmetric matrix.

Complete step-by-step answer:
A real square matrix A = ${a_{ij}}$ is said to be symmetric, if A′ = A. A real square matrix A = ${a_{ij}}$ is said to be skew-symmetric if A′ = –A. In a skew-symmetric matrix A = ${a_{ij}}$ all its diagonal elements are zero.
Alternatively, in this question all diagonal elements are zero and for non-diagonal elements if we interchange i by j and j by i only the result differs by negative sign. These types of matrices are always skew-symmetric. And if elements remain the same when we interchange i by j and j by I, then matrices are known as symmetric matrices.
Given, \[A = {\left[ {{a_{ij}}} \right]_{n \times n}}\] ​ where \[{a_{ij}} = {i^2} - {j^2}\]
Now, \[{a_{ji}} = {j^2} - {i^2} = - \left( {{i^2} - {j^2}} \right) = - {a_{ij}}\] ∀ i, j.
Also \[{a_{ii}} = 0\], ∀ i.
So, \[{A^T} = - A\] . [According to the definition of symmetric matrix]

Note: In these types of questions, use the given rule of elements and check whether the diagonal elements are 0 or not. In symmetric and skew-symmetric matrix diagonal elements remain the same.
Matrix is an ordered rectangular array of numbers or functions. The number of functions is called the elements or the entries of the matrix. It is denoted by the symbol [ ] or ( ).