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Question: If \(A\) is any square matrix, then \(\left( A+{{A}^{T}} \right)\) is a ............... matrix. Fill...

If AA is any square matrix, then (A+AT)\left( A+{{A}^{T}} \right) is a ............... matrix. Fill in the blank:
A. symmetric
B. skew symmetric
C. scalar
D. identity

Explanation

Solution

For this question we will first see what a square matrix is then we will assume a square matrix of any order (preferably 3) then we will write its transpose by changing rows of the original into the columns of the new matrix and vice-a-versa. Finally we will put both the original and transposed matrix into the given condition and find out the type of the resultant matrix.

Complete step by step answer:
First, we know that a square matrix is a matrix with the same number of rows and columns. That means a n×nn\times n matrix is known as a square matrix of ordernn.
Given that, AA is a square matrix.
Now, let matrix A=[abc dfg ehi ]A=\left[ \begin{matrix} a & b & c \\\ d & f & g \\\ e & h & i \\\ \end{matrix} \right]
Now AT{{A}^{T}} , stands for transpose of AA . Transpose of a matrix means a new matrix whose rows are the columns of the original matrix which means the columns of the new matrix are the rows of the original.
Therefore:
AT=[abc dfg ehi ]T=[ade bfh cgi ] AT=[ade bfh cgi ] \begin{aligned} & {{A}^{T}}={{\left[ \begin{matrix} a & b & c \\\ d & f & g \\\ e & h & i \\\ \end{matrix} \right]}^{T}}=\left[ \begin{matrix} a & d & e \\\ b & f & h \\\ c & g & i \\\ \end{matrix} \right] \\\ & {{A}^{T}}=\left[ \begin{matrix} a & d & e \\\ b & f & h \\\ c & g & i \\\ \end{matrix} \right] \\\ \end{aligned}

Now, it is given in the question that we have to find out what type of matrix (A+AT)\left( A+{{A}^{T}} \right) is?
Now, putting the values of AA and AT{{A}^{T}} in (A+AT)\left( A+{{A}^{T}} \right):
A+AT=[abc dfg ehi ]+[ade bfh cgi ]=[2a(b+d)(c+e) (d+b)2f(g+h) (e+c)(h+g)2i ] A+AT=[2a(b+d)(c+e) (b+d)2f(g+h) (c+e)(g+h)2i ] \begin{aligned} & A+{{A}^{T}}=\left[ \begin{matrix} a & b & c \\\ d & f & g \\\ e & h & i \\\ \end{matrix} \right]+\left[ \begin{matrix} a & d & e \\\ b & f & h \\\ c & g & i \\\ \end{matrix} \right]=\left[ \begin{matrix} 2a & \left( b+d \right) & \left( c+e \right) \\\ \left( d+b \right) & 2f & \left( g+h \right) \\\ \left( e+c \right) & \left( h+g \right) & 2i \\\ \end{matrix} \right] \\\ & A+{{A}^{T}}=\left[ \begin{matrix} 2a & \left( b+d \right) & \left( c+e \right) \\\ \left( b+d \right) & 2f & \left( g+h \right) \\\ \left( c+e \right) & \left( g+h \right) & 2i \\\ \end{matrix} \right] \\\ \end{aligned}
We can see the symmetry of (A+AT)\left( A+{{A}^{T}} \right) about its diagonal. Therefore it is a symmetric matrix.

So, the correct answer is “Option A”.

Note: Note that the sum and difference of two symmetric matrices is again symmetric.The resultant matrix is also called as persymmetric matrix which is a square matrix which is symmetric in the northeast-to-southwest diagonal. If the diagonal elements were all zeroes then it would have qualified as a skew symmetric matrix, in skew symmetric matrix: A=ATA=-{{A}^{T}} .