A symmetric matrix is always a square matrix. If A be any square matrix and if A = AT, then matrix A is called symmetric matrix.

Let A = [ aij ] be a square matrix and if A = AT, it means if we interchange rows and columns of A, then the matrix remains unchanged i.e. aij = aji, for all i and j. So, we say that, in a symmetric matrix, all elements are symmetric with respect to main diagonal.
A = $\begin{bmatrix} 2 &-1 &0 \\ -1 &-2 &3 \\ 0 &3 &4 \end{bmatrix}$ AT = $\begin{bmatrix} 2 &-1 &0 \\ -1 &-2 &3 \\ 0 &3 &4 \end{bmatrix}$

Here A =AT. So, A is a symmetric matrix. All diagonal matrices are symmetric because in diagonal matrix all non-diagonal elements are zero.

## Skew Symmetric Matrix

A square matrix say B = [ bij ] is said to skew-symmetric, if and only if B = -BT
i.e. [ bij ] = - [ bji ].

Let A = $\begin{bmatrix} 0&a &b \\ -a &0 &-c \\ -b &c &0 \end{bmatrix}$

Now AT = $\begin{bmatrix} 0&-a &-b \\ a &0 &c \\ b &-c &0 \end{bmatrix}$

so A = AT

For symmetric and skew-symmetric matrices, the necessary condition is that the matrices should be square matrices. And for skew-symmetric matrix, all the diagonal elements are zero.

## Inverse of Symmetric Matrix

To find the inverse of symmetric matrix, let A = $\begin{bmatrix} 1 &-3 &5 \\ -3 &2 &-1 \\ 5 &-1 &4 \end{bmatrix}$

We know that A-1 =$\frac{adjA}{\left | A \right |}$

First we have to find out I A I = ?
For this we use first row expansion method:

I A I = 1 $\begin{vmatrix} 2 &-1 \\ -1 &4 \end{vmatrix}$ - (-3) $\begin{vmatrix} -3 &-1 \\ 5 &4 \end{vmatrix}$ + 5 $\begin{vmatrix} -3&2 \\ 5&-1 \end{vmatrix}$

= 1(8 - 1) + 3(-12 + 5) + 5(3 - 10)
= 7 -21 - 35
= - 49 $\neq$ 0

Here I A I $\neq$ 0, hence inverse of matrix is exists.

For adj A, first calculate cofactor matrix of A say B and after that taking transpose of B,
which is said to be adj A.

Cofactor matrix of A = B = $\begin{bmatrix} 7 &-7 &-7 \\ 7 &-21 &-14 \\ -7 &-14 &-7 \end{bmatrix}$

Now adj A = BT = $\begin{bmatrix} 7 &7 &-7 \\ -7 &-21 &-14 \\ -7 &-14 &-7 \end{bmatrix}$

For A -1 = $\frac{1}{\left | A \right |}$ x adj A

= $\frac{1} {-49}$ x $\begin{bmatrix} 7 &7 &-7 \\ -7 &-21 &-14 \\ -7 &-14 &-7 \end{bmatrix}$

or A -1 = $\begin{bmatrix} 1/7 &1/7 &-1/7 \\ -1/7 &-3/7 &-2/7\\ -1/7 &-2/7 &-1/7 \end{bmatrix}$

## Determinant of Symmetric Matrix

Let C = $\begin{bmatrix} 1 &0 &2 \\ 0 &2 &-4 \\ 2 &-4 &3 \end{bmatrix}$be a square matrix. For det C, we choose first row for expansion.

I C I = 1 $\begin{vmatrix} 2 &-4 \\ -4&3 \end{vmatrix}$ - 0 $\begin{vmatrix} 0&-4 \\ 4 &3 \end{vmatrix}$ + 2 $\begin{vmatrix} 0 &2 \\ 2 &-4 \end{vmatrix}$

or I C I = 1( 6 - 16 ) + 0 + 2( 0 - 4)
or I C I = -10 - 8
or I C I = - 18

It is clear that finding the determinant of a symmetric matrix is the same as finding determinant of any square matrix.

## Properties of Symmetric Matrices

Listed below are some of the Symmetric Matrix Properties

Property 1
: If A and B are two symmetric matrix and (A + B)T = AT + BT = A + B
then A + B is a symmetric matrix.

Proof: Let A = $\begin{bmatrix} 3&2 &-1 \\ 2 &2 &5\\ -1 &5 &4 \end{bmatrix}$ and B = $\begin{bmatrix} 7 &0 &4 \\ 0&2 &-6 \\ 4 &-6 &1 \end{bmatrix}$ are two symmetric matrices of order 3x3.

Then AT = $\begin{bmatrix} 3&2 &-1 \\ 2 &2 &5\\ -1 &5 &4 \end{bmatrix}$ and BT = $\begin{bmatrix} 7 &0 &4 \\ 0&2 &-6 \\ 4 &-6 &1 \end{bmatrix}$

AT + BT = $\begin{bmatrix} 3&2 &-1 \\ 2 &2 &5\\ -1 &5 &4 \end{bmatrix}$ + $\begin{bmatrix} 7 &0 &4 \\ 0&2 &-6 \\ 4 &-6 &1 \end{bmatrix}$

= $\begin{bmatrix} 10 &2 &3\\ 2&4&-1 \\ 3 &-1 &5 \end{bmatrix}$ ...................................................(1)

( A + B ) = $\begin{bmatrix} 3&2 &-1 \\ 2 &2 &5\\ -1 &5 &4 \end{bmatrix}$ + $\begin{bmatrix} 7 &0 &4 \\ 0&2 &-6 \\ 4 &-6 &1 \end{bmatrix}$

= $\begin{bmatrix} 10 &2 &3\\ 2&4 &-1 \\ 3 &-1 &5 \end{bmatrix}$ ..................................................(2)

( A + B )T = $\begin{bmatrix} 10 &2 &3\\ 2&4 &-1 \\ 3 &-1 &5 \end{bmatrix}$ ................................................(3)

From (1), (2) and (3), we have ( A + B )T = AT + BT = A + B
Hence A + B is a symmetric matrix.

Property 2: If C and D are symmetric matrices, then (CD)T = DTCT = DC $\neq$ CD
Usually CD is not symmetric.

Proof: Let C = $\begin{bmatrix} 1 &2 &3\\ 2&-1 &4 \\ 3 &4&0 \end{bmatrix}$ and D = $\begin{bmatrix} -1 &3 &2\\ 3&0 &5 \\ 2 &5 &1 \end{bmatrix}$ are symmetric matrices of order 3x3.

Then CD = $\begin{bmatrix} 1 &2 &3\\ 2&-1 &4 \\ 3 &4&0 \end{bmatrix}$ x $\begin{bmatrix} -1 &3 &2\\ 3&0 &5 \\ 2 &5 &1 \end{bmatrix}$

= $\begin{bmatrix} 11 &18&15\\ 3&26 &3 \\ 9&9 &26 \end{bmatrix}$ ................................................(1)

( CD )T = $\begin{bmatrix} 11 &3 &9\\ 18&26 &9\\ 15&3 &26 \end{bmatrix}$ ..................................................(2)

Here CT = $\begin{bmatrix} 1 &2 &3\\ 2&-1 &4 \\ 3 &4&0 \end{bmatrix}$ , DT = $\begin{bmatrix} -1 &3 &2\\ 3&0 &5 \\ 2 &5 &1 \end{bmatrix}$

DTCT = $\begin{bmatrix} -1 &3 &2\\ 3&0 &5 \\ 2 &5 &1 \end{bmatrix}$ x $\begin{bmatrix} 1 &2 &3\\ 2&-1 &4 \\ 3 &4&0 \end{bmatrix}$

= $\begin{bmatrix} 11 &3 &9\\ 18&26 &9\\ 15&3 &26 \end{bmatrix}$ ................................................(3)

CD = $\begin{bmatrix} 1 &2 &3\\ 2&-1 &4 \\ 3 &4&0 \end{bmatrix}$ x $\begin{bmatrix} -1 &3 &2\\ 3&0 &5 \\ 2 &5 &1 \end{bmatrix}$

= $\begin{bmatrix} 11 &3 &9\\ 18&26 &9\\ 15&3 &26 \end{bmatrix}$ ...............................................(4)

From (1), it is clear that matrix CD is not symmetric and (CD)T = DTCT = DC $\neq$ CD

Property 3: If A is any symmetric matrix then B = ATA is symmetric and
( ATA)T = AT (AT)T = ATA

Proof: For this let A2x2 = $\begin{bmatrix} 1 &-2 \\ -2 &1 \end{bmatrix}$ be a matrix, then AT = $\begin{bmatrix} 1 &-2 \\ -2 &1 \end{bmatrix}$

Compute B = ATA = $\begin{bmatrix} 1 &-2 \\ -2 &1 \end{bmatrix}$ x $\begin{bmatrix} 1 &-2 \\ -2 &1 \end{bmatrix}$

= $\begin{bmatrix} 5 &-4\\ -4 &5 \end{bmatrix}$ ..........................................................(a)

So matrix B is a symmetric matrix of order 2x2.
Now ( ATA)T = $\begin{bmatrix} 5 &-4\\ -4 &5 \end{bmatrix}$ ...........................................................(b)

Again AT (AT)T = $\begin{bmatrix} 1 &-2 \\ -2 &1 \end{bmatrix}$ x $\begin{bmatrix} 1 &-2 \\ -2 &1 \end{bmatrix}$

= $\begin{bmatrix} 5 &-4\\ -4 &5 \end{bmatrix}$ .....................................................(c)

Using (a), (b) and (c), we get ( ATA)T = AT (AT)T = ATA