Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# For what value of x, is the matrix $A = \left( {\begin{array}{*{20}{c}} 0&1&{ - 1} \\ { - 1}&0&3 \\ x&{ - 3}&0 \end{array}} \right)$ a skew symmetric matrix?

Last updated date: 20th Jun 2024
Total views: 414.6k
Views today: 4.14k
Verified
414.6k+ views
Hint:For a matrix to be skew symmetric the condition which is to be satisfied is $A = - {A^T}$.First find $-{A^T}$ and then equate it to the given condition and compare the elements in matrix both in L.H.S and R.H.S and determine the value of x.

We are given that $A = \left( {\begin{array}{*{20}{c}} 0&1&{ - 1} \\ { - 1}&0&3 \\ x&{ - 3}&0 \end{array}} \right)$and we have to find the value of x for this matrix is skew symmetric.
For a matrix to be skew symmetric the condition which is to be satisfied is given as –
$A = - {A^T}$……. (1)
We will start by finding the value of ${A^T}$.
We find the transpose of a matrix is obtained by changing rows to columns and columns to rows. In other words, the transpose of a matrix is obtained by interchanging $A_{ij}$ to $A_{ji}$.
In simpler words Transpose of a Matrix is defined as “A Matrix which is formed by turning all the rows of a given matrix into columns and vice-versa.”
$A = \left( {\begin{array}{*{20}{c}} 0&1&{ - 1} \\ { - 1}&0&3 \\ x&{ - 3}&0 \end{array}} \right)$
Changing rows in columns and columns into row we get the transpose as shown –
${A^T} = \left( {\begin{array}{*{20}{c}} 0&{ - 1}&x \\ 1&0&{ - 3} \\ { - 1}&3&0 \end{array}} \right)$
Multiplying both sides by -1 we get,
$- {A^T} = ( - 1)\left( {\begin{array}{*{20}{c}} 0&{ - 1}&x \\ 1&0&{ - 3} \\ { - 1}&3&0 \end{array}} \right)$
To multiply -1 in a matrix we have to multiply -1 to every value inside the matrix by -1. Thus,
$- {A^T} = \left( {\begin{array}{*{20}{c}} 0&1&{ - x} \\ { - 1}&0&3 \\ 1&{ - 3}&0 \end{array}} \right)$…………. (2)
Substituting matrix $A$ and equation (2) in equation (1), we get
$A = - {A^T}$
$\left( {\begin{array}{*{20}{c}} 0&1&{ - 1} \\ { - 1}&0&3 \\ x&{ - 3}&0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0&1&{ - x} \\ { - 1}&0&3 \\ 1&{ - 3}&0 \end{array}} \right)$

Now we have to equate the matrix $A$ and $- {A^T}$. While equating the matrices we have to equate every element in a row of a column with corresponding value of the other matrix.
Equating ${a_{13}}$ we get,
$- 1 = - x$
Which implies, $x = 1$
Similarly, equating ${a_{31}}$ we get,
x = 1
Therefore, for x=1 the matrix $A$ is skew symmetric.

Note:While multiplying any constant to a matrix we have multiplied this constant to every element of the matrix.Students should remember for a matrix to be skew symmetric the condition which is to be satisfied is $A = - {A^T}$ and for a matrix to be symmetric the condition which is to be satisfied is$A = {A^T}$.