Question

Matrix A = \[\begin{bmatrix}0 & 2b & - 2 \\ 3 & 1 & 3 \\ 3a & 3 & - 1\end{bmatrix}\]  is given to be symmetric, find values of a and b.

 

Answer 1

We have

\[A = \begin{bmatrix}0 & 2b & - 2 \\ 3 & 1 & 3 \\ 3a & 3 & - 1\end{bmatrix}\] 

\[A' = \begin{bmatrix}0 & 3 & 3a \\ 2b & 1 & 3 \\ - 2 & 3 & - 1\end{bmatrix}\] 

We know that a matrix is symmetric if A = A'.

Thus ,

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

Now,

2b = 3 

`⇒ b = 3/2`

\[Also, \] 

\[3a = - 2\] 

\[ \Rightarrow a = \frac{- 2}{3}\]

\[Therefore, \] 

`a =(-2)/3  and b  = 3/2`