
For what value of \[x\] , is the matrix \[A\] = \[\left[ {\begin{array}{*{20}{c}}
0&1&{ - 2} \\
{ - 1}&0&3 \\
x&{ - 3}&0
\end{array}} \right]\] a skew symmetric matrix ?
Answer
481.8k+ views
Hint: For a matrix to be skew symmetric the condition which is to be satisfied is \[A = - {A^T}\]. Here \[{A^T}\] denotes the transpose of the matrix \[A\] . Here we know the matrix \[A\] . From this we can calculate \[{A^T}\] and \[ - {A^T}\]. Then putting the value of \[A\] and \[ - {A^T}\] in the above condition we can obtain the value of \[x\] .
FORMULA USED :
For a skew symmetric matrix \[A = - {A^T}\].
Complete answer:We are given that \[A\] = \[\left[ {\begin{array}{*{20}{c}}
0&1&{ - 2} \\
{ - 1}&0&3 \\
x&{ - 3}&0
\end{array}} \right]\] we have to find the value of \[x\] such that the matrix satisfies the condition for the skew symmetric matrix .
We know the condition for the skew symmetric matrix \[A = - {A^T}\]. First we will obtain the value of \[{A^T}\] .
Transpose of a matrix is obtained by interchanging rows and columns of matrix, in other words we find the transpose of a matrix by changing rows to columns and columns to rows .
Every element of the matrix \[A\] can be denoted by \[{a_{ij}}\] where \[i\] denotes row number \[j\] denotes column number . \[{a_{ij}}\] = element of \[{i^{th}}\]row and \[{j^{th}}\] column .
Every element of the matrix \[{A^T}\] can be denoted by \[{a_{ji}}\] .
\[A\] = \[\left[ {\begin{array}{*{20}{c}}
0&1&{ - 2} \\
{ - 1}&0&3 \\
x&{ - 3}&0
\end{array}} \right]\]
By changing rows into column and vice versa we get
\[{A^T}\]= \[\left[ {\begin{array}{*{20}{c}}
0&{ - 1}&x \\
1&0&{ - 3} \\
{ - 2}&3&0
\end{array}} \right]\]
So \[ - {A^T}\]= \[\left[ {\begin{array}{*{20}{c}}
0&1&{ - x} \\
{ - 1}&0&3 \\
2&{ - 3}&0
\end{array}} \right]\]
Putting the values of \[A\] and \[ - {A^T}\] in condition we get
\[\left[ {\begin{array}{*{20}{c}}
0&1&{ - 2} \\
{ - 1}&0&3 \\
x&{ - 3}&0
\end{array}} \right]\]=\[\left[ {\begin{array}{*{20}{c}}
0&1&{ - x} \\
{ - 1}&0&3 \\
2&{ - 3}&0
\end{array}} \right]\]
Now we have to equate the matrix. While equating the matrices we must equate every element in a row of a column with the corresponding value of the other matrix.
\[{a_{13}}\] = element of \[{1^{st}}\]row and \[{3^{rd}}\] column .
\[{a_{31}}\] = element of \[{3^{rd}}\]row and \[{1^{st}}\]column .
Equating \[{a_{13}}\] we get
\[ - x = - 2\]
So \[x = 2\]
Equating \[{a_{31}}\] we get
\[x = 2\]
This is the desired answer .
Therefore for \[x = 2\] matrix \[A\] is a skew symmetric matrix .
Note:
Students should remember while multiplying any constant to a matrix we have multiplied this constant to every element of the matrix . Transpose of the matrix should be obtained carefully . Negative sign in the condition is very important ; without the negative sign the condition will become the condition for the symmetric matrix .
FORMULA USED :
For a skew symmetric matrix \[A = - {A^T}\].
Complete answer:We are given that \[A\] = \[\left[ {\begin{array}{*{20}{c}}
0&1&{ - 2} \\
{ - 1}&0&3 \\
x&{ - 3}&0
\end{array}} \right]\] we have to find the value of \[x\] such that the matrix satisfies the condition for the skew symmetric matrix .
We know the condition for the skew symmetric matrix \[A = - {A^T}\]. First we will obtain the value of \[{A^T}\] .
Transpose of a matrix is obtained by interchanging rows and columns of matrix, in other words we find the transpose of a matrix by changing rows to columns and columns to rows .
Every element of the matrix \[A\] can be denoted by \[{a_{ij}}\] where \[i\] denotes row number \[j\] denotes column number . \[{a_{ij}}\] = element of \[{i^{th}}\]row and \[{j^{th}}\] column .
Every element of the matrix \[{A^T}\] can be denoted by \[{a_{ji}}\] .
\[A\] = \[\left[ {\begin{array}{*{20}{c}}
0&1&{ - 2} \\
{ - 1}&0&3 \\
x&{ - 3}&0
\end{array}} \right]\]
By changing rows into column and vice versa we get
\[{A^T}\]= \[\left[ {\begin{array}{*{20}{c}}
0&{ - 1}&x \\
1&0&{ - 3} \\
{ - 2}&3&0
\end{array}} \right]\]
So \[ - {A^T}\]= \[\left[ {\begin{array}{*{20}{c}}
0&1&{ - x} \\
{ - 1}&0&3 \\
2&{ - 3}&0
\end{array}} \right]\]
Putting the values of \[A\] and \[ - {A^T}\] in condition we get
\[\left[ {\begin{array}{*{20}{c}}
0&1&{ - 2} \\
{ - 1}&0&3 \\
x&{ - 3}&0
\end{array}} \right]\]=\[\left[ {\begin{array}{*{20}{c}}
0&1&{ - x} \\
{ - 1}&0&3 \\
2&{ - 3}&0
\end{array}} \right]\]
Now we have to equate the matrix. While equating the matrices we must equate every element in a row of a column with the corresponding value of the other matrix.
\[{a_{13}}\] = element of \[{1^{st}}\]row and \[{3^{rd}}\] column .
\[{a_{31}}\] = element of \[{3^{rd}}\]row and \[{1^{st}}\]column .
Equating \[{a_{13}}\] we get
\[ - x = - 2\]
So \[x = 2\]
Equating \[{a_{31}}\] we get
\[x = 2\]
This is the desired answer .
Therefore for \[x = 2\] matrix \[A\] is a skew symmetric matrix .
Note:
Students should remember while multiplying any constant to a matrix we have multiplied this constant to every element of the matrix . Transpose of the matrix should be obtained carefully . Negative sign in the condition is very important ; without the negative sign the condition will become the condition for the symmetric matrix .
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