Determinants

The determinant is a single number attached to a square matrix. It tells whether the matrix is invertible (non-zero determinant), how the linear map scales area or volume, and, through the characteristic polynomial, what the eigenvalues are. The properties of determinants (effect of row operations, scaling, products, transposes, adjugates) give a faster route than a long expansion: simplify with row operations or apply a property.

For $n \times n$ $A$, the determinant is

a sum over all $n!$ permutations, each term taking one entry from every row and every column, with sign $+1$ for even and $-1$ for odd permutations. Small cases:

The diagonal (Sarrus) rule works only for $3 \times 3$ matrices.

The minor $M_{ij}$ is the determinant of the $(n - 1) \times (n - 1)$ matrix left after deleting row $i$ and column $j$. The cofactor is $C_{ij} = (-1)^{i+j} M_{ij}$; the signs form a chessboard starting with $+$ at the top left. Laplace (cofactor) expansion along any row $i$ or any column $j$:

Expand along the row or column with the most zeros. Expanding along row $i$ with the cofactors of a different row $k$ gives 0: $\sum_j a_{ij} C_{kj} = 0$ for $i \ne k$ (alien cofactors).

For $n \times n$ matrices $A, B$ and scalar $c$:

The adjugate $\operatorname{adj}(A)$ is the transpose of the cofactor matrix: $\operatorname{adj}(A)_{ij} = C_{ji}$. Combining expansion with alien cofactors gives

So $A^{-1} = \operatorname{adj}(A)/\det A$ when $\det A \ne 0$. Taking determinants, for $n \ge 2$:

If $\det A \ne 0$, the system $Ax = b$ has the unique solution $x_i = \det A_i / \det A$, where $A_i$ is $A$ with column $i$ replaced by $b$. It is useful for $2 \times 2$ and $3 \times 3$ systems and for proofs; for large systems elimination is far cheaper.

$\lvert \det A \rvert$ is the factor by which the linear map $x \mapsto Ax$ scales areas ($n = 2$) or volumes ($n = 3$); the sign records whether orientation is preserved. The area of the parallelogram spanned by $(a, c)$ and $(b, d)$ is $\lvert ad - bc \rvert$.

Computing from the permutation definition costs about $n \cdot n!$ operations; naive recursive cofactor expansion has factorial growth too (typically $\Theta(n!)$, versus $\Theta(n\,n!)$ for direct permutation evaluation). Row reduction to triangular form costs $O(n^3)$ and is how determinants are computed in practice (see LU Decomposition).

and in general $\prod_{i < j}(x_j - x_i)$. It is non-zero exactly when the $x_i$ are distinct, which is why a polynomial of degree less than $n$ is fixed by its values at $n$ distinct points.

$A = \begin{pmatrix} 2 & 1 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{pmatrix}$. Along row 1: $2(24 - 0) - 1(0 - 5) + 3(0 - 4) = 48 + 5 - 12 = 41$. Along column 1, which has a zero: $2 \cdot 24 + 1 \cdot (5 - 12) = 48 - 7 = 41$.

$A$ is $3 \times 3$ with $\det A = 2$. Then $\det(3A) = 27 \cdot 2 = 54$, $\det(A^{-1}) = 1/2$, $\det(A^TA) = 4$, $\det(\operatorname{adj} A) = 2^2 = 4$, and $\det(2A^{-1}) = 8/2 = 4$.

$\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix} = 0$, since $R_1 + R_3 = 2R_2$. No expansion needed.

$\begin{vmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}$. Add rows 2 and 3 to row 1: every entry of row 1 becomes $x + 2$; factor it out to leave a row of 1s. Subtract column 1 from columns 2 and 3 to get a lower triangular matrix with diagonal $1, x - 1, x - 1$. Result: $(x + 2)(x - 1)^2$. It vanishes at $x = 1$ (all rows equal) and $x = -2$ (rows sum to zero).

$3x + 2y = 7$, $x + 4y = 9$. $\det A = 10$; $\det A_1 = 7 \cdot 4 - 2 \cdot 9 = 10$; $\det A_2 = 3 \cdot 9 - 7 \cdot 1 = 20$. So $x = 1$, $y = 2$.

For any $3 \times 3$ skew-symmetric $K$, $\det K = \det(K^T) = \det(-K) = -\det K$, so $\det K = 0$.

The $n \times n$ matrix with $a$ on the diagonal and $b$ everywhere else has

by the same steps as Example 4: add all rows to the first, factor out $a + (n - 1)b$, then subtract. For $n = 4$, $a = 3$, $b = 1$: $2^3 \cdot 6 = 48$. The $n \times n$ all-ones matrix ($a = b = 1$) has determinant 0 for $n \ge 2$, and $J - I$ ($a = 0$, $b = 1$) has $(-1)^{n-1}(n - 1)$.

Let $D_n$ be the determinant of the $n \times n$ tridiagonal matrix with diagonal entries $a$, superdiagonal entries $b$ and subdiagonal entries $c$. Expanding along the first row gives $D_n = a D_{n-1} - bc\,D_{n-2}$, with $D_1 = a$ and $D_0 = 1$. With $a = 2$, $b = c = -1$ this is $D_n = 2D_{n-1} - D_{n-2}$, so $D_n = n + 1$: $D_3 = 4$ and $D_{10} = 11$.

$A$ is $3 \times 2$ and $B$ is $2 \times 3$. Then $AB$ is $3 \times 3$ with rank at most 2, so $\det(AB) = 0$ always, while $BA$ is $2 \times 2$ and can be invertible. With $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$, $BA = I_2$ has determinant 1 and $\det(AB) = 0$. The rule $\det(AB) = \det A \det B$ is for square matrices only.