We are given that,

A = \(\begin{bmatrix} 0 & 2b & -2 \\[0.3em] 3& 1 &3 \\[0.3em] 3a & 3 &-1 \end{bmatrix}\)is symmetric matrix.

We need to find the values of a and b. 

We must understand what symmetric matrix is. 

A symmetric matrix is a square matrix that is equal to its transpose. 

A symmetric matrix ⇔ A = A^T 

This means, 

We need to find the transpose of matrix A. 

Let us take transpose of the matrix A. 

We know that, 

The transpose of a matrix is a new matrix whose rows are the columns of the original. 

We have, 

1^st row of matrix A = (0 ,2b, -2) 

2^nd row of matrix A = (3,1, 3) 

3^rd row of matrix A = (3a ,3 ,-1) 

For matrix A^T, it will become 

1^st column of A^T = 1^st row of A = (0 ,2b,-2)

2^nd column of A^T = 2^nd row of A = (3,1 ,3) 

3^rd column of A T = 3^rd row of A = (3a,3 ,-1)

∴ A^T = \(\begin{bmatrix} 0 & 3 & 3a \\[0.3em] 2b& 1 &3 \\[0.3em] -2 & 3 &-1 \end{bmatrix}\)

Now, 

as  A = A^T.

Substituting the matrices A and A^T, we get

\(\begin{bmatrix} 0 & 2b & -2 \\[0.3em] 3& 1 &3 \\[0.3em] 3a & 3 &-1 \end{bmatrix}\)= \(\begin{bmatrix} 0 & 3 & 3a \\[0.3em] 2b& 1 &3 \\[0.3em] -2 & 3 &-1 \end{bmatrix}\)

We know by the property of matrices,

 \(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\)

This implies, 

a₁₁ = b₁₁, 

a₁₂ = b₁₂, 

a₂₁ = b₂₁ and

 a₂₂ = b₂₂ 

Applying this property, we can write

2b = 3 …(i) 

-2 = 3a …(ii) 

3 = 2b 

3a = -2 

We can find a and b from equations (i) and (ii).

From equation (i),

2b = 3

⇒ b = \(\frac{ 3}{2}\)

From equation (ii), 

- 2 = 3a

 ⇒ a = \(-\frac{ 2}{3}\)

Thus, we get

 a = \(-\frac{ 2}{3}\) and  b = \(\frac{ 3}{2}\).