Q.If the matrix $A = \begin{pmatrix} 0 & a & -3 \\ 2 & 0 & -1 \\ b & 1 & 0 \end{pmatrix}$ is skew symmetric, find the values of '$a$' and '$b$'.

With $A=\begin{bmatrix}1&2&-1\\-1&0&2\\1&2&0\end{bmatrix}$, multiply $A\cdot A$ entry by entry:

For example, the $(1,1)$ entry is $1(1)+2(-1)+(-1)(1)=-2$, and the $(2,2)$ entry is $-1(2)+0(0)+2(2)=2$.

Now add $A$ and $B$: …

The key idea is that symmetry is a condition on the entries: for a matrix $A$, $A = A^T$ means $A_{ij} = A_{ji}$ for every pair of indices. Once we extract the values of $a$, $b$, and $c$ from that condition, the problem reduces to a straightforward determinant calculation.

Let the given matrix be

For $M$ to be symmetric, we must have $M_{ij} = M_{ji}$ for all $i, j$.

1. Compare the $(1,2)$ and $(2,1)$ entries. $M_{12} = 2$ and $M_{21} = a$. Symmetry demands $2 = a$, so

2. Compare the $(1,3)$ and $(3,1)$ entries. $M_{13} = b$ and $M_{31} = 3$. Hence

3. Compare the $(2,3)$ and $(3,2)$ entries. $M_{23} = 6$ and $M_{32} = c$. Thus

All other positions are already symmetric by construction (the diagonal entries are free, and the remaining off-diagonal pairs match automatically once these three equalities hold).

Now substitute $a = 2$, $b = 3$, $c = 6$ into the determinant we need:

4. Evaluate the determinant. A clean way is to use the formula for a $3 \times 3$ determinant directly: …