Determinants: Evaluation & Properties

$\begin{vmatrix}a&b\\c&d\end{vmatrix} = ad - bc$. For $3\times3$, expand along a row: $\sum_j a_{1j}(-1)^{1+j}M_{1j}$, or use Sarrus — add the three down-right diagonal products, subtract the three down-left ones. Expanding along the row/column with the most zeros is fastest.

For $n\times n$ matrices:

• Product: $\det(AB) = \det A\,\det B$
• Transpose: $\det(A^T) = \det A$
• Scalar: $\det(kA) = k^n \det A$ (all $n$ rows scale)
• Power: $\det(A^m) = (\det A)^m$
• Inverse: $\det(A^{-1}) = 1/\det A$
• Conjugation: $\det(B^{-1}AB) = \det A$
• Gram: $\det(AA^T) = (\det A)^2$
• Rank-1 update (matrix-determinant lemma): $\det(I + AA^T) = 1 + A^T A$; for a column vector $A$, $A^T A$ is just the sum of squares of its entries.

• Transpose: $\det A^T = \det A$ (rows and columns play identical roles).
• Swap: swapping two rows/columns multiplies the determinant by $-1$.
• Equal/proportional: two identical or proportional rows/columns $\Rightarrow \det = 0$.
• Common factor: a factor common to a row/column pulls outside.
• Row operation: $R_i \to R_i + \lambda R_j$ leaves the determinant unchanged (the key simplification move).

$A$ is singular $\iff \det A = 0 \iff A^{-1}$ does not exist. Determinant equations $\det(\cdot) = 0$ become polynomial equations once expanded; simplify first with row operations to lower the degree of work.

If substituting $x = c$ makes two rows/columns identical, $(x - c)$ divides the determinant (factor theorem). The Vandermonde determinant $\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix} = (a-b)(b-c)(c-a)$. Use known factors plus a degree/leading-coefficient check to pin the constant.

$\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix} = -(a^3+b^3+c^3-3abc) = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. It equals 0 iff $a+b+c = 0$ (real case) or $a=b=c$. Recognising the cyclic pattern saves a full expansion.

Evaluate $\det(A_k)$ symbolically in $k$; it is frequently a constant, linear, or telescoping expression. Then sum with $\sum_{k=1}^{n} k = \tfrac{n(n+1)}{2}$, $\sum k^2 = \tfrac{n(n+1)(2n+1)}{6}$, or telescoping cancellation.

• Rank-1 patterns: if $a_{ij}$ factors as $f(i)g(j)$ (e.g. $a_{ij} = 2(i+j)$ is a sum of two rank-1 pieces), the $3\times3$ determinant collapses to 0.
• Bounded entries: a third-order determinant with entries all $\pm1$ lies in a small range; the maximum magnitude is 4.
• Counting determinants from a fixed set of numbers uses permutations of the placements.

For $f(x)=\begin{vmatrix}R_1\\R_2\\R_3\end{vmatrix}$ (rows $R_i$ of functions):

• Row-by-row rule: $f'(x)=\begin{vmatrix}R_1'\\R_2\\R_3\end{vmatrix}+\begin{vmatrix}R_1\\R_2'\\R_3\end{vmatrix}+\begin{vmatrix}R_1\\R_2\\R_3'\end{vmatrix}$ — differentiate ONE row per term.
• The same works column-by-column.
• A constant row differentiates to a zero row → that term's determinant is 0; ignore constant rows.
• Watch for the shortcut: after differentiating, if two rows become equal or proportional, the determinant is 0.

Pascal's identity: $\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}$. So a column of $\binom{n+1}{r+1}$ entries equals the sum of the matching $\binom{n}{r}$ and $\binom{n}{r+1}$ columns. The move: spot the Pascal relation among the columns, apply $C_i \to C_i - C_j - C_k$, and the column collapses to zeros (or to a condition on the unknown).