Q.If A is a matrix of order 3, then $\det(kA)$ is

These properties let a determinant be simplified — often to $0$ — without full expansion.

1. Transpose invariance: $|A^T|=|A|$ (row-wise and column-wise expansion agree).
2. Row/column swap: interchanging any two rows (or columns) changes the sign of the determinant, leaving its absolute value unchanged. More generally, $n$ interchanges multiply the determinant by $(-1)^n$.
3. Identical rows/columns: if two rows (or columns) are identical, $|A|=0$. (Proof idea: swapping the identical rows leaves the matrix unchanged but must flip the sign by Property 2, forcing $|A|=-|A|$, so $|A|=0$.)
4. Proportional rows/columns: if one row (or column) is a scalar multiple of another, $|A|=0$; in particular, an all-zero row/column forces $|A|=0$.
5. Scalar factor: multiplying every entry of one row (or column) by a scalar $k$ multiplies the whole determinant by $k$. Consequently $|kA| = k^n|A|$ for an $n\times n$ matrix $A$ (every one of the $n$ rows is scaled by $k$).
6. Sum splitting: if every entry of one row (or column) is a sum of two terms, the determinant splits as the sum of two determinants (one with each term in that row/column, all other rows/columns unchanged).
7. Row/column operations: adding to any row (column) a scalar multiple of another row (column) — e.g. $R_i \to R_i + pR_j + qR_k$ — leaves the determinant unchanged. This is the workhorse trick used to create zeros before expanding. …